Generalized fractional-order Bernoulli–Legendre functions: an effective tool for solving two-dimensional fractional optimal control problems

Generalized fractional-order Bernoulli–Legendre functions: an effective tool for solving... Abstract In this article, we present a new method to solve a class of two-dimensional fractional optimal control problems. The fractional derivative is defined in the Caputo sense. Our approach is based upon to approximate the state and control functions by the elements of generalized fractional-order Bernoulli functions in space and generalized fractional-order Legendre functions in time with unknown coefficients. First the properties of these basis functions are presented. Second, the Riemann–Liouville fractional integral operators of generalized fractional-order Bernoulli–Legendre functions are proposed. Then we apply two-dimensional Legendre–Gauss quadrature rule to approximate double integral of the performance index functional. Next, the problem is converted into an equivalent non-linear unconstrained optimization problem. This problem is solved via the Newton’s iterative method. Finally, convergence of the proposed method is extensively investigated and two examples are included to demonstrate the validity and applicability of the presented new technique. 1. Introduction Fractional differential equations (FDEs) are generalized from integer order ones, which are obtained by replacing integer order derivatives by fractional order ones. In comparison with integer order differential equations, the fractional differential equations show many advantages over the simulation of problems in system biology (Yuste & Lindenberg, 2001), physics (Barkai et al., 2000), hydrology (Benson et al., 2000), chemistry and biochemistry (Yuste et al., 2004) and finance (Gorenflo et al., 2001). These new fractional-order models are more adequate than the previously used integer-order models, because fractional-order derivatives and integrals enable the description of the memory and hereditary properties of different substances (Podlubny, 1999). This is the most significant advantage of the fractional-order models in comparison with integer-order models, in which such effects are neglected. In the area of physics, the left and right space fractional derivatives allow the modelling of flow regime impacts from either side of the domain (Zaslavsky, 2002). Space fractional partial differential equations are used to model anomalous diffusive and super-diffusive systems, where a particle plume spread faster than the classical Brownian motion model predicts (Metzler & Klafter, 2000). The fractional advectiondispersion equation is used in groundwater hydrology to model the transport of passive tracers carried by fluid flow in a porous medium (Liu et al., 2004). It is known that some dynamical processes in gas absorption, water steam heating and air drying can be described by the Darboux equation (Marszalek, 1984). Therefore, constructing new analytical and numerical approaches for solving different types of fractional differential equations has become a strong topic to be considered, these approaches include the finite element method (Ma et al., 2014), the wavelet method (Rahimkhani et al., 2017a), the spectral tau method (Bhrawy et al., 2015a), the Gegenbauer spectral method (Izadkhah & Saber-Nadjafi, 2015), the iterative method (Daftardar-Gejji Bhalekar, 2010), the fractional-order wavelet method (Rahimkhani et al., 2016b, 2017b). The general definition of an optimal control problem refers to the minimization of a functional on a set of control and state variables (the performance index) subject to dynamic constraints on the states and controls. In case of there are fractional differential equations that used as the dynamic constraints, this leads to the fractional optimal control problem (FOCP). However very little work has been done in the area of fractional optimal control, FOCPs have gained much attention for their many applications in engineering and physics (Zamani et al., 2007). For that reason, finding accurate numerical methods for solving various types of FOCPs has become an active research undertaking. A general formulation and a solution scheme for FOCPs were first introduced in (Agrawal, 2004) where fractional derivatives were introduced in the Riemann–Liouville sense and FOCP formulation was expressed using the fractional variational principle and the Lagrange multiplier technique. In Agrawal (2008a,b), the FOCPs are formulated using the definition of fractional derivatives in the sense of Caputo, the FDEs are substituted into Volterra-type integral equations and a direct linear solver helps calculating the solution of the obtained algebraic equations. In Agrawal & Baleanu (2007) and Baleanu & Trujillo (2010), general necessary conditions of optimality have been derived for FOCPs via the Hamiltonian formulas with respect to the Riemann–Liouville fractional derivative and the Caputo fractional derivative. In Rabiei et al. (2016), the authors used Boubaker polynomials for solving FOCPs. Doha et al. (2015) used the Rayleigh–Ritz method with the operational matrix of fractional integrals based on the shifted Jacobi orthonormal polynomials for solving a FOCP in a general form. In Bhrawy et al. (2015b), the authors used the shifted Legendre orthonormal polynomials as basis functions of the Lagrange multiplier method for approximating the solution of different types of FOCPs. Rahimkhani et al. (2016a) introduced an efficient method based on the Bernoulli wavelets for solving delay FOCPs. Frederico & Torres (2006, 2008) formulated a Noether-type theorem in the general context of the fractional optimal control in the sense of Caputo and studied fractional conservation laws in FOCPs. Bhrawy & Ezz-Eldien (2016) derived new operational matrices for the shifted Legendre orthonormal polynomial for solving delay FOCPs. Nemati & Yousefi (2016) presented a numerical scheme for solving two-dimensional FOCPs by the Ritz method combined with fractional operational matrix. In this article, we intend to obtain a new numerical approach for approximating the solution of the two-dimensional FOCP \begin{equation} min\hspace{.3cm} J[\xi , u] = \int _{0}^{h_{2}}\int _{0}^{h_{1}}G \left(x, t, \xi (x, t), u(x, t), \frac{\partial \xi}{\partial x}(x, t), \frac{\partial \xi}{\partial t}(x, t)\right) dx dt, \label{Z25} \end{equation} (1.1) subject to \begin{equation} \frac{\partial ^{2}\xi}{\partial x \partial t}(x, t)= a \frac{\partial ^{\nu}\xi}{\partial x^{\nu}}(x, t) + b \frac{\partial ^{\gamma}\xi}{\partial t^{\gamma}}(x, t) + c \xi(x, t) + d u(x, t),\label{Z26} \end{equation} (1.2) with the following initial and boundary conditions \begin{equation} \xi (x, 0) = \zeta (x), \hspace{1cm} \xi (0, t) = \eta (t),\label{Z27} \end{equation} (1.3) where $$ 0 < \nu , \gamma \leq 1 $$ and the functions $$ G $$ and $$ \xi $$ are smooth and $$ d $$ is non-zero. In recent years a special attention has been given to applications of the operational matrices Walsh functions (Ordokhani, 2010), B-spline functions (Lakestani et al., 2012), Legendre polynomials (Saadatmandi & Dehghan, 2010; Nemati et al., 2013), Chebyshev polynomials (Doha et al., 2011), Bernstein polynomials (Saadatmandi, 2011), Legendre wavelet (Heydari et al., 2014) and Bernoulli wavelet (Rahimkhani et al., 2017a). Moreover, there are recent works on the operational matrices of fractional-order Legendre functions (Kazem et al., 2013; Chen et al., 2014), fractional-order Bernstein functions (Yuzbasi, 2013), fractional-order generalized Laguerre functions (Bhrawy et al., 2014) and fractional-order Bernoulli wavelet (Rahimkhani et al., 2016b, 2017b). In this article, we introduce a different approach for solving two-dimensional FOCP in (1.1)–(1.3). First, we transform the FOCP into an equivalent variational problem. Then we expand function $$ \frac{\partial ^{2}\xi}{\partial x \partial t}(x, t)$$ with generalized fractional-order Bernoulli–Legendre functions and the unknown coefficients. These generalized fractional-order Bernoulli–Legendre functions which consist of generalized fractional-order Bernoulli (GFBFs) and generalized fractional-order Legendre functions (GFLFs) are given. The Riemann–Liouville fractional integral operators for these functions are derived. By estimating double integral of the cost function via two-dimensional Legendre–Gauss (LG) quadrature rule, a common unconstrained optimization problem is obtained. After taking the necessary conditions of optimality into account, the Newton’s iterative method is used to find the solution of this problem. The proposed method is very convenient for solving such problems, since the initial and boundary conditions are taken into account automatically. Numerical results demonstrate the efficiency of the proposed method in solving FOCPs. The article is organized as follows. In Section 2, some necessary definitions of fractional calculus, generalized fractional-order Bernoulli, generalized fractional-order Legendre and their properties are defined. Riemann–Liouville fractional integral operators for GFBFs and GFLFs are introduced in Section 3. In Section 4, the numerical method for solving the under studying problems is presented. The convergence of the proposed method is discussed in Section 5. In Section 6, we report our numerical findings and demonstrate the accuracy of the proposed numerical scheme by considering two numerical examples. Finally, a conclusion is given in Section 7. 2. Preliminaries and notation In this section, we present some basic definitions needed in the remaining part of the article. We briefly recall fractional derivative and integral and some properties of GFBFs and GFLFs. 2.1. The fractional derivative and integral There are various definitions of fractional derivative and integration. The widely used definition of a fractional derivative is the Caputo definition, and a fractional integration is the Riemann–Liouville definition. Definition 2.1 The integral of order $$ \nu \geq 0 $$ (fractional) according to Riemann–Liouville is given by Rahimkhani et al. (2016b) \begin{equation} I^{\nu}f(x)=\left\{ \begin{array}{ll} \frac{1}{{\it{\Gamma}} (\nu)}\int _{0}^x \frac{f(s)}{(x-s)^{1-\nu}}ds=\frac{1}{{\it{\Gamma}} (\nu)}x^{\nu -1}\ast f(x), & \nu >0, x>0,\\ f(x) , \ \ \ \ \ \ \ \ \ \ \ \ & \nu =0,\label{E1} \end{array} \right. \end{equation} (2.1) where $$x^{\nu -1}\ast f(x)$$ is the convolution product of $$x^{\nu -1} $$ and $$f(x)$$. The operator $$ I^{\nu} $$ satisfies the following properties 1. $$I^{\nu}I^{\mu}f(x) = I^{\nu +\mu}f(x),$$ 2. $$I^{\nu }(\lambda f(x)+ \gamma g(x)) =\lambda I^{\nu }f(x)+ \gamma I^{\nu}g(x),$$ 3. $$I^{\nu }x^{\beta} = \frac{{\it{\Gamma}} (\beta +1)}{{\it{\Gamma}} (\beta +\nu +1)}x^{\nu +\beta}.$$ Definition 2.2 Caputo’s fractional derivative of order $$ \nu $$ is defined as (Rahimkhani et al., 2016b) \begin{equation} D^{\nu}f(x)=\frac{1}{{\it{\Gamma}} (n-\nu)}\int _{0}^{x}\frac{f^{(n)}(s)}{(x-s)^{\nu +1-n}}ds,\hspace{.5cm} n-1 <\nu \leq n, n\in N. \end{equation} (2.2) The operator $$ D^{\nu} $$ satisfies the following properties 1. $$ D^{\nu }I^{\nu}f(x)=f(x),$$ 2. $$ I^{\nu }D^{\nu}f(x)=f(x)-\sum _{i=0}^{n-1}f^{(i)}(0)\frac{x^{i}}{i!},$$ 3. $$ D^{\nu }(\lambda f(x)+\gamma g(x)) =\lambda D^{\nu}f(x)+ \gamma D^{\nu}g(x),$$ 4. $D^{\nu }x^{\beta} = \left\{\begin{array}{ll} 0, & \nu \in N_{0},\hspace{.1cm} \beta <\nu ,\\ \frac{{\it{\Gamma}} (\beta +1)}{{\it{\Gamma}} (\beta +1- \nu )}x^{\beta - \nu }, & \textrm{otherwise}, \end{array} \right. $ 5. $$D^{\nu } \lambda =0,$$ where $$ \lambda $$ is constant. Definition 2.3 The partial Caputo fractional derivative of order $$ \nu >0 $$ of the function $$ u(x, t) $$ with respect to its variables $$ x $$ and $$ t $$ are defined as follows, respectively (Nemati & Yousefi, 2016) \begin{equation} D^{\nu}_{x}u(x, t) = \frac{\partial ^{\nu}u(x, t)}{\partial x^{\nu}}= \frac{1}{{\it{\Gamma}} (n- \nu)}\int _{0}^{x}(x- \xi )^{n-\nu -1}\frac{\partial ^{n}u(\xi , t)}{\partial \xi ^{n}}d\xi , \hspace{.5cm}n-1 < \nu \leq n \end{equation} (2.3) and \begin{equation} D^{\nu}_{t}u(x, t) = \frac{\partial ^{\nu}u(x, t)}{\partial t^{\nu}}= \frac{1}{{\it{\Gamma}} (n- \nu)}\int _{0}^{t}(t- \theta)^{n-\nu -1}\frac{\partial ^{n}u(x , \theta)}{\partial \theta ^{n}}d\theta , \hspace{.5cm}n-1 < \nu \leq n. \end{equation} (2.4) Theorem 2.1 Let $$ \nu \in R, n-1 < \nu \leq n, n \in N $$ and $$ \lambda \in C. $$ Then the Caputo fractional derivative of the exponential and trigonometric functions are as follows (Nemati & Yousefi, 2016) \begin{gather*} D^{\nu}e^{\lambda x} = \lambda ^{n}x^{n- \nu} E_{1, n-\nu +1}(\lambda x),\\ D^{\nu}sin \lambda x = -\frac{1}{2}i(i \lambda) ^{n}x^{n-\nu} (E _{1, n- \nu +1}(i\lambda x) - (-1)^{n}E _{1, n- \nu +1}(- i\lambda x)),\\ D^{\nu}cos \lambda x = \frac{1}{2}(i \lambda) ^{n}x^{n-\nu} (E _{1, n- \nu +1}(i\lambda x) + (-1)^{n}E _{1, n- \nu +1}(- i\lambda x)), \end{gather*} where $$ E_{\nu , \gamma}(z) = \sum _{k=0}^{\infty} \frac{z^{k}}{{\it{\Gamma}} (\nu k + \gamma)} $$ is the two-parameter Mittag–Leffler function. 2.2. Generalized fractional-order Bernoulli functions The GFBFs are proposed by Rahimkhani et al. (2017b). These functions are constructed by change of variable $$ x $$ to $$(x/h)^{\alpha} $$, ($$ \alpha > 0 $$), on the Bernoulli polynomials. Let the GFBFs $$\beta_{m} ((x/h)^{\alpha}) $$ be denoted by $$\beta_{m} ^{h\alpha}(x)$$. The analytic form of $$ \beta_{m}^{h\alpha}(x) $$ of order $$ m\alpha $$, is given by \begin{equation} \beta_{m}^{h \alpha}(x)=\sum _{i=0}^{m} \left( {\begin{array}{*{5}c} m\\ i \\ \end{array}} \right)\frac{\beta_{m-i}^{\alpha}}{h^{i\alpha}} x^{i\alpha},\hspace{1cm}0 \leq x \leq h,\label{E11} \end{equation} (2.5) where $$\beta^{\alpha}_{i} :=\beta^{\alpha}_{i}(0) =\beta_{i},\hspace{.1cm} i= 0, 1, ..., m, $$ are Bernoulli numbers. Thus, the first four such functions are \begin{align*} \beta^{h\alpha}_{0}(x)&=1,\\ \beta^{h\alpha}_{1}(x)&= (x/h)^{\alpha}-\frac{1}{2},\\ \beta^{h\alpha}_{2}(x)&=(x/h)^{2\alpha}- (x/h)^{\alpha}+\frac{1}{6},\\ \beta^{h\alpha}_{3}(x)&=(x/h)^{3\alpha}-\frac{3}{2}(x/h)^{2\alpha}+\frac{1}{2}(x/h)^{\alpha}. \end{align*} These functions satisfy the following formula (Rahimkhani et al., 2017b) \begin{equation} \int _{0}^{h}\beta ^{h \alpha}_{n}(x)\beta ^{h \alpha}_{m}(x)x^{\alpha - 1}dx = \frac{h^{\alpha}}{\alpha} (-1)^{n-1}\frac{m!n!}{(m+n)!}\beta_{m+n}^{\alpha},\hspace{.5cm}m, n \geq 1.\label{M12} \end{equation} (2.6) Also, GFBFs form a complete basis over the interval [0, h] (Rahimkhani et al., 2017b). 2.3. Generalized fractional-order Legendre functions The GFLFs are proposed by Chen et al. (2014). These functions are defined by change of variable $$ t $$ to $$(t/h)^{\alpha} $$, ($$ \alpha > 0 $$), based on the shifted Legendre polynomials and are denoted by $$ L_{m}^{h\alpha}(t), \hspace{.2cm}m=0, 1, 2, \ldots $$. The GFLFs $$ L_{m}^{h\alpha}(t) $$ satisfy the following recursive formula \begin{align*} L_{m+1}^{h \alpha}(t) & = \frac{(2m+1)(2(t/h)^{\alpha}-1)}{m+1} L_{m}^{h \alpha}(t) - \frac{m}{m+1} L_{m-1}^{h \alpha}(t), \hspace{1cm}m=1, 2, \ldots,\\ L_{0}^{h \alpha}(t) & =1, \hspace{1cm} L_{1}^{h\alpha}(t) = 2(t/h)^{\alpha} - 1. \end{align*} The analytic form of $$ L_{m}^{h\alpha}(t) $$ of degree $$ m\alpha $$ given by \begin{equation} L_{m}^{h \alpha}(t) = \sum _{i=0}^{m}\frac{ b_{i,m}}{h^{i \alpha}}t^{i\alpha}, \hspace{1cm}m=0, 1, 2, \ldots,\label{W4} \end{equation} (2.7) where $$b_{i,m} = \frac{(-1)^{m+i}(m+i)!}{(m-i)!(i!)^{2}},$$ and $$ L_{m}^{h\alpha}(0)=(-1)^{m}, \hspace{.2cm} L_{m}^{h\alpha}(1)= 1.$$ The GFLFs are orthogonal with respect to the weight function $$ \omega (t) = t^{\alpha -1} $$ on the interval $$[0, h]$$, then the orthogonal condition is (Chen et al., 2014) \begin{equation} \int _{0}^{h}L_{n}^{h \alpha}(t) L_{m}^{h \alpha}(t) t^{\alpha -1}dt = \frac{h^{\alpha}}{(2m+1)\alpha}\delta _{nm},\hspace{1cm}m\geq n,\label{W5} \end{equation} (2.8) where $$ \delta _{nm} $$ is the Kronecker function. 2.4. Function approximation An arbitrary function of two variables $$ u(x, t) $$ defined over $$ L^{2}([0, h_{1}]\times [0, h_{2}]) $$, may be expanded into generalized fractional-order Bernoulli–Legendre functions as: \begin{equation} u(x, t) = \sum _{i=0}^{\infty}\sum _{j=0}^{\infty}u_{ij}\beta_{i}^{h_{1} \alpha}(x)L_{j}^{h_{2} \alpha}(t). \end{equation} (2.9) Therefore, one can consider the following truncated series for $$ u(x, t) $$ as: \begin{equation} u(x, t) \simeq \sum _{i=0}^{M}\sum _{j=0}^{M'}u_{ij}\beta_{i}^{h_{1} \alpha}(x)L_{j}^{h_{2} \alpha}(t). \label{Z22} \end{equation} (2.10) Now let $${\it{\Delta}} = [0, h_{1}] \times [0, h_{2}]$$ and $$ \lbrace \beta_{i}^{h_{1} \alpha}(x) L_{j}^{h_{2} \alpha}(t) \rbrace _{i,j=0}^{M, M'} $$ be the set of generalized fractional-order Bernoulli–Legendre functions in $$ L^{2}({\it{\Delta}})$$. We assume \begin{eqnarray*} {\it{\Omega}} _{M, M'}&=& span <\beta_{i}^{h_{1} \alpha}(x) L_{j}^{h_{2} \alpha}(t) > _{0 \leq i \leq M, 0 \leq j \leq M'}\\ &=& span <\beta_{0}^{h_{1} \alpha}(x) L_{0}^{h_{2} \alpha}(t), \ldots , \beta_{M}^{h_{1} \alpha}(x) L_{M'}^{h_{2} \alpha}(t)>\!. \end{eqnarray*} Let $$u(x, t)$$ be an arbitrary function in $$ L^{2}({\it{\Delta}}) $$. Since $${\it{\Omega}} _{M, M'}$$ is closed in the complete space $$ L^{2}({\it{\Delta}}) $$, so it is complete. Therefore, there is the best unique approximation out of $${\it{\Omega}} _{M, M'} $$ like $$ \tilde{u}(x, t) $$ that \begin{align*} \Vert u - \tilde{u} \Vert _{2} \leq \Vert u - f \Vert _{2},\hspace{1cm} \forall f \in {\it{\Omega}} _{M, M'}, \end{align*} where $$ \Vert u \Vert _{2} = \sqrt{< u, u>}$$. Since $$ \tilde{u}(x, t) \in {\it{\Omega}} _{M, M'} $$, so we can find coefficients $$ u_{ij}, i=0, 1, \ldots , M, j=0, 1, \ldots , M' $$ such that \begin{align*} \tilde{u}(x, t) = \sum _{i=0}^{M}\sum _{j=0}^{M'}u_{ij}\beta_{i}^{h_{1} \alpha}(x)L_{j}^{h_{2} \alpha}(t) = B^{h_{1}\alpha ,T}(x) U L^{h_{2}\alpha}(t), \end{align*} where $$T$$ indicates transposition, $$ B^{h_{1}\alpha }(x)$$ and $$ L^{h_{2} \alpha}(t) $$ are column vectors, given by \begin{equation} B^{h_{1}\alpha }(x) = [\beta_{0}^{h_{1} \alpha}(x), \beta_{1}^{h_{1} \alpha}(x), \ldots , \beta_{M}^{h_{1} \alpha}(x) ]^{T} \label{Z23} \end{equation} (2.11) and \begin{equation} L^{h_{2}\alpha }(t) = [L_{0}^{h_{2} \alpha}(t), L_{1}^{h_{2} \alpha}(t), \ldots ,L_{M'}^{h_{2} \alpha}(t) ]^{T}. \label{Z24} \end{equation} (2.12) For example, Fig. 1 shows graphs of $$ u(x, t)= \beta_{2}^{3 \alpha}(x) L_{3}^{3 \alpha}(t) $$ for various values of $$ \alpha $$. To compute double integral of a function, we state two-dimensional LG quadrature rule. Suppose that the LG nodes on the interval $$[a, b]$$ and $$[c, d]$$ are denoted by $$ \delta _{r} $$ and $$ \tau _{s} $$, $$r = 1, \ldots , l, s = 1, \ldots , l'$$, respectively. For the smooth function, $$ \xi (x, t) $$, the two-dimensional LG quadrature rule is constructed as follows (Nemati & Yousefi, 2016) \begin{align*} \int _{c}^{d}\int _{a}^{b} \xi (x, t) dx dt & \simeq \frac{ (b-a) (d-c)}{4}\sum _{r=1}^{l}\sum _{s=1}^{l'} \xi \left(a + (\delta _{r}+1)\frac{b-a}{2} , c+ (\tau _{s} +1)\frac{d-c}{2}\right) \omega _{r} \omega ' _{s}\\ & :=\kappa W^{T}{\it{\Omega}} W', \end{align*} where $$ \kappa =\frac{(b-a)(d-c)}{4} $$ and $$ W, W' $$ are column vectors of Christoffel numbers of size $$ l, l' (\omega _{r}= \frac{2}{(1-\delta _{r})^{2}(\dot{L}_{l}(\delta _{r}))^{2}}, \omega ' _{s}= \frac{2}{(1-\tau _{s})^{2}(\dot {L}_{l}(\tau _{s}))^{2}}) $$, and $$ {\it{\Omega}} $$ is $$ l \times l' $$ matrix in which its entries are defined as follows, \begin{align*} {\it{\Omega}} _{rs}= \xi \left(a + (\delta _{r}+1)\frac{b-a}{2}, c + (\tau_{s}+1)\frac{d-c}{2}\right)\!, \hspace{.2cm}1\leq r \leq l, 1 \leq s \leq l'. \end{align*} Fig. 1. View largeDownload slide Plots of $$ u(x, t)= \beta_{2}^{3 \alpha}(x) L_{3}^{3 \alpha}(t) $$ for (a) $$ \alpha = \frac{1}{4}, $$ (b) $$ \alpha = \frac{1}{2}, $$ (c) $$ \alpha = 1 $$ and (d) $$ \alpha = 2 $$. Fig. 1. View largeDownload slide Plots of $$ u(x, t)= \beta_{2}^{3 \alpha}(x) L_{3}^{3 \alpha}(t) $$ for (a) $$ \alpha = \frac{1}{4}, $$ (b) $$ \alpha = \frac{1}{2}, $$ (c) $$ \alpha = 1 $$ and (d) $$ \alpha = 2 $$. 3. Riemann–Liouville fractional integral operators In this section, we obtain Riemann–Liouville fractional integral operators for GFBFs and GFLFs. 3.1. Riemann–Liouville fractional integral operator for GFBFs We now derive the operator $$ I^{\nu} $$ for $$ B^{h\alpha}(x) $$ in Equation (2.11) given by \begin{equation} I^{\nu}B^{h\alpha}(x)= F^{h\alpha}( \nu , x), \label{M2} \end{equation} (3.1) where \begin{align*} F^{h\alpha}( \nu , x) = [I^{\nu}\beta^{h\alpha} _{0}(x), I^{\nu}\beta^{h\alpha} _{1}(x), I^{\nu}\beta^{h\alpha} _{2}(x), \ldots , I^{\nu}\beta^{h\alpha} _{M}(x) ]^{T}. \end{align*} To obtain $$ I^{\nu}\beta ^{h\alpha} _{m}(x), (m=0, 1, \ldots , M) $$ we use the Laplace transform. By taking the Laplace transform from Equation (2.5), we get \begin{eqnarray} L[\beta ^{h\alpha} _{m}(x)] &= & L\left[\sum _{i=0}^{m} \left( {\begin{array}{*{5}c} m\\ i \\ \end{array}} \right )\frac{ \beta_{m-i}^{\alpha}}{h^{i\alpha}} x^{i\alpha}\right]\nonumber\\ &=& \sum _{i=0}^{m} \left( {\begin{array}{*{5}c} m\\ i \\ \end{array}} \right )\frac{ \beta_{m-i}^{\alpha}}{h^{i\alpha}}L[x^{i\alpha}]\nonumber\\ &=&\sum _{i=0}^{m} \left( {\begin{array}{*{5}c} m\\ i \\ \end{array}} \right )\frac{ \beta_{m-i}^{\alpha}}{h^{i\alpha}} \frac{{\it{\Gamma}} (i \alpha +1)}{s^{i \alpha +1}}. \label{M4} \end{eqnarray} (3.2) From Equation (2.1), we get \begin{eqnarray} L[I^{\nu}\beta ^{h\alpha} _{m}(x)] &=& L \left[\frac{1}{{\it{\Gamma}} (\nu ) x^{1- \nu}}\ast \beta ^{h\alpha} _{m}(x)\right] \nonumber \\ &=& \frac{1}{{\it{\Gamma}} (\nu)} \frac{{\it{\Gamma}} (\nu)}{s^{\nu}} \sum _{i=0}^{m} \left( {\begin{array}{*{5}c}m\\ i \\ \end{array}} \right )\frac{ \beta_{m-i}^{\alpha}}{h^{i\alpha}} \frac{{\it{\Gamma}} (i \alpha +1)}{s^{i \alpha +1}} \nonumber \\ &=&\sum _{i=0}^{m} \left( {\begin{array}{*{5}c} m\\ i \\ \end{array}} \right )\frac{ \beta_{m-i}^{\alpha}}{h^{i\alpha}} \frac{{\it{\Gamma}} (i \alpha +1)}{s^{i \alpha + \nu +1}}.\label{M6} \end{eqnarray} (3.3) Taking the inverse Laplace transform of Eq. (3.3) yields \begin{equation} I^{\nu}\beta ^{h\alpha} _{m}(x) =\sum _{i=0}^{m} \left( {\begin{array}{*{5}c} m\\ i \\ \end{array}} \right )\frac{ \beta_{m-i}^{\alpha}}{h^{i\alpha}} \frac{{\it{\Gamma}} (i \alpha +1)}{{\it{\Gamma}} (i\alpha + \nu +1)}x^{i\alpha +\nu},\hspace{.5cm}m=0, 1, 2, \ldots ,M.\label{M11} \end{equation} (3.4) Therefore, we obtain the Riemann–Liouville fractional integral operator for the GFBFs. For example, for $$ M =5, h=3, \alpha = 0.5 , \nu = 0.5 $$ and $$ x=1.5 $$ the Riemann–Liouville fractional integral operator of the GFBFs can be expressed as \begin{align*} F^{3, 0.5}( 0.5, 1.5) = [1.38198, 0.0765067, -0.0765067, -0.0194301, 0.0232353, 0.0110308]^{T}. \end{align*} 3.2. Riemann–Liouville fractional integral operator for GFLFs We now derive the operator $$ I^{\nu} $$ for $$ L^{h\alpha}(t) $$ in Equation (2.12) given by \begin{equation} I^{\nu}L^{h\alpha}(t)= {\it{\Theta}} ^{h\alpha}( \nu , t), \label{Z2} \end{equation} (3.5) where \begin{equation} {\it{\Theta}}^{h\alpha}( \nu , t) = [I^{\nu}L^{h\alpha} _{0}(t), I^{\nu}L^{h\alpha} _{1}(t), I^{\nu}L^{h\alpha} _{2}(t), \ldots , I^{\nu}L^{h\alpha} _{M'}(t) ]^{T}. \end{equation} To obtain $$ I^{\nu}L^{h\alpha} _{m}(t) (m=0, 1, \ldots , M'), $$ we use the Laplace transform. By taking the Laplace transform from Equation (2.7), we get \begin{eqnarray} L[L ^{h\alpha} _{m}(t)] & = & L\left[\sum _{i=0}^{m} \frac{(-1)^{m+i}(m+i)!}{(m-i)!(i!)^{2}h^{i \alpha}} t^{i\alpha}\right]\nonumber\\ &=& \sum _{i=0}^{m} \frac{(-1)^{m+i}(m+i)!}{(m-i)!(i!)^{2}h^{i \alpha}} L[t^{i\alpha}]\nonumber\\ &= &\sum _{i=0}^{m} \frac{(-1)^{m+i}(m+i)!}{(m-i)!(i!)^{2}h^{i \alpha}} \frac{{\it{\Gamma}} (i \alpha +1)}{s^{i\alpha +1}}. \label{Z4} \end{eqnarray} (3.6) From Equation (2.1), we get \begin{eqnarray} L[I^{\nu}L^{h\alpha} _{m}(t)] &=& L \left[\frac{1}{{\it{\Gamma}} (\nu ) t^{1- \nu}}\ast L ^{h\alpha} _{m}(t)\right]\nonumber \\ &=& \frac{1}{{\it{\Gamma}} (\nu)} \frac{{\it{\Gamma}} (\nu)}{s^{\nu}} \sum _{i=0}^{m} \frac{(-1)^{m+i}(m+i)!}{(m-i)!(i!)^{2}h^{i \alpha}} \frac{{\it{\Gamma}} (i \alpha +1)}{s^{i\alpha +1}} \nonumber \\ &=&\sum _{i=0}^{m} \frac{(-1)^{m+i}(m+i)!}{(m-i)!(i!)^{2}h^{i \alpha}} \frac{{\it{\Gamma}} (i \alpha + 1)}{s^{i\alpha +\nu+1}}. \label{Z6} \end{eqnarray} (3.7) By taking the inverse Laplace transform of Equation (3.7) yields \begin{equation} I^{\nu}L_{m}^{h \alpha}(t) = \sum _{i=0}^{m} \frac{(-1)^{m+i}(m+i)!}{(m-i)!(i!)^{2}h^{i \alpha}} \frac{{\it{\Gamma}} (i \alpha + 1)}{{\it{\Gamma}} (i\alpha + \nu +1)}t^{i\alpha +\nu}, \hspace{ .5cm}m=0, 1, \ldots M'. \end{equation} (3.8) Therefore, we obtain the Riemann–Liouville fractional integral operator for the GFLFs. For example, for $$ M =5, h=3, \alpha = 1 , \nu = 1.5 $$ and $$ x=0.5 $$ the Riemann–Liouville fractional integral operator of the GFLFs can be expressed as \begin{align*} {\it{\Theta}} ^{3, 1}( 1.5, 0.5) = [0.265962, -0.2305, 0.169709, -0.100099, 0.0386484, 0.00282425 ]^{T}. \end{align*} 4. Problem statement and approximation method Consider the following two-dimensional FOCP be given: \begin{equation} min\hspace{.3cm} J[ \xi , u] = \int _{0}^{h_{2}}\int _{0}^{h_{1}}G \big ( x, t, \xi (x, t), u(x, t), \frac{\partial \xi}{\partial x}(x, t), \frac{\partial \xi}{\partial t}(x, t) \big) dx dt, \label{Z7} \end{equation} (4.1) subject to \begin{equation} \frac{\partial ^{2}\xi}{\partial x \partial t}(x, t)= a \frac{\partial ^{\nu}\xi}{\partial x^{\nu}}(x, t) + b \frac{\partial ^{\gamma}\xi}{\partial t^{\gamma}}(x, t) + c \xi(x, t) + d u(x, t), \label{Z8} \end{equation} (4.2) with the following initial and boundary conditions \begin{equation} \xi (x, 0) = \zeta (x), \hspace{1cm} \xi (0, t) = \eta (t), \label{Z9} \end{equation} (4.3) where $$ 0 < \nu , \gamma \leq 1. $$ The functions $$ G $$ and $$ \xi $$ are smooth and $$d $$ is non-zero. By computing the control function $$ u(x, t) $$ from Equation (4.2), and substituting into the index functional J, the following equivalent minimization problem is obtained: \begin{gather} min\hspace{.3cm} J[\xi] = \int _{0}^{h_{2}}\int _{0}^{h_{1}}G \big(x, t, \xi (x, t), \frac{1}{d} \Bigg[\frac{\partial ^{2}\xi}{\partial x \partial t}(x, t)- a \frac{\partial ^{\nu}\xi}{\partial x^{\nu}}(x, t) \nonumber\\ - b \frac{\partial ^{\gamma}\xi}{\partial t^{\gamma}}(x, t) -c \xi(x, t)\Bigg], \frac{\partial \xi}{\partial x}(x, t), \frac{\partial \xi}{\partial t}(x, t) \big) dx dt,\label{Z10}\\ \end{gather} (4.4) \begin{gather} \xi (x, 0) = \zeta (x), \hspace{1cm} \xi (0, t) = \eta (t).\label{Z11} \end{gather} (4.5) For solving this problem, we approximate: \begin{equation} \frac{\partial ^{2}\xi}{\partial x \partial t}(x, t) = B^{h _{1}\alpha ,T}(x) U L^{h_{2}\alpha}(t),\label{Z12} \end{equation} (4.6) where $$ U=[u_{i, j}]_{M\times M'} $$ is an unknown matrix which should be found. By integrating of (4.6) with respect to $$ t $$, we obtain: \begin{equation} \frac{\partial \xi}{\partial x}(x, t)=B^{h_{1}\alpha ,T}(x) U {\it{\Theta}} ^{h_{2}\alpha}(1, t) + \frac{\partial \xi }{\partial x}(x, t)\bigg \vert _{t=0}=B^{h_{1}\alpha , T}(x) U {\it{\Theta}} ^{h_{2}\alpha}(1, t) + \frac{\partial \zeta (x) }{\partial x}. \label{Z13} \end{equation} (4.7) Moreover, by integrating of (4.6) with respect to $$ x $$, we yield: \begin{equation} \frac{\partial \xi}{\partial t}(x, t)=F^{h_{1}\alpha ,T}(1, x) U L^{h_{2}\alpha}(t) + \frac{\partial \xi }{\partial t}(x, t)\bigg \vert _{x=0}=F^{h_{1}\alpha , T}(1, x) U L^{h_{2}\alpha}(t) + \frac{\partial \eta (t) }{\partial t}. \label{Z14} \end{equation} (4.8) Now by integrating of (4.7) with respect to $$ x $$, and considering (4.3), we get: \begin{equation} \xi (x, t) = F^{h_{1}\alpha ,T}(1, x) U {\it{\Theta}} ^{h_{2}\alpha} (1, t) + \zeta (x) +\eta (t)- \zeta (0). \label{Z15} \end{equation} (4.9) By fractional differentiation of order $$ \nu $$ of (4.9) with respect to $$ x $$, we have: \begin{equation} \frac{\partial ^{\nu}\xi}{\partial x^{\nu}}(x, t) = F^{h_{1}\alpha ,T}(1 - \nu, x) U {\it{\Theta}} ^{h_{2}\alpha} (1, t) + \frac{\partial ^{\nu}\zeta (x)}{\partial x^{\nu}}. \label{Z20} \end{equation} (4.10) Also by fractional differentiation of order $$ \gamma $$ of (4.9) with respect to $$ t $$, we obtain: \begin{equation} \frac{\partial ^{\gamma}\xi}{\partial t ^{\gamma}}(x, t) = F^{h_{1}\alpha ,T}(1, x) U {\it{\Theta}} ^{h_{2}\alpha} (1-\gamma , t) + \frac{\partial ^{\gamma}\eta (t)}{\partial t^{\gamma}}. \label{Z16} \end{equation} (4.11) From Equation (4.2) and approximations (4.6), (4.9)–(4.11), the control function is estimated as \begin{eqnarray} \tilde{u}(x, t)&=& \frac{1}{d} \Bigg[B^{h_{1}\alpha ,T}(x) U L^{h_{2}\alpha}(t) \nonumber\\ &&- a F^{h_{1}\alpha ,T}(1 - \nu, x) U {\it{\Theta}} ^{h_{2}\alpha} (1, t) - a\frac{\partial ^{\nu}\zeta (x)}{\partial x^{\nu}}\nonumber\\ &&-bF^{h_{1}\alpha ,T}(1, x) U {\it{\Theta}} ^{h_{2}\alpha} (1-\gamma , t) - b\frac{\partial ^{\gamma}\eta (t)}{\partial t^{\gamma}}\nonumber\\ &&-c F^{h_{1}\alpha ,T}(1, x) U {\it{\Theta}} ^{h_{2}\alpha} (1, t) - c\zeta (x) -c \eta (t) + c\zeta (0) \Bigg].\label{Z17} \end{eqnarray} (4.12) Now by substituting the approximate state function (4.9), control function (4.12), $$ \frac{\partial \xi}{\partial x}(x, t) $$ and $$ \frac{\partial \xi}{\partial t}(x, t) $$ from Equations (4.7) and (4.8) in the performance index function (4.4), our aim is to solve the resultant unconstrained minimization problem of $$ (M+1)\times(M'+1) $$ variables, $$ u_{ij}, i=0, 1, \ldots , M, j=0, 1, \ldots , M', $$ as follows \begin{eqnarray} min \hspace{.2cm}\bar{J}[U]&=& \int_{0}^{h_{2}}\int_{0}^{h_{1}}G\Bigg( x, t, F^{h_{1}\alpha ,T}(1, x) U {\it{\Theta}} ^{h_{2}\alpha} (1, t) + \zeta (x) \nonumber\\ &&+\eta (t)- \zeta (0), \frac{1}{d} \Bigg[ B^{h_{1}\alpha ,T}(x) U L^{h_{2}\alpha}(t) \nonumber\\ &&- a F^{h_{1}\alpha ,T}(1 - \nu, x) U {\it{\Theta}} ^{h_{2}\alpha} (1, t) - a \frac{\partial ^{\nu}\zeta (x)}{\partial x^{\nu}}\nonumber\\ &&- bF^{h_{1}\alpha ,T}(1, x) U {\it{\Theta}} ^{h_{2}\alpha} (1-\gamma , t) - b \frac{\partial ^{\gamma}\eta (t)}{\partial t^{\gamma}}\nonumber\\ &&- c F^{h_{1}\alpha ,T}(1, x) U {\it{\Theta}} ^{h_{2}\alpha} (1, t) - c \zeta (x) -c \eta (t) +c \zeta (0) \Bigg],\nonumber\\ && B^{h_{1}\alpha , T}(x) U {\it{\Theta}} ^{h_{2}\alpha}(1, t) + \frac{\partial \zeta (x) }{\partial x},\nonumber\\ & & F^{h_{1}\alpha , T}(1, x) U L^{h_{2}\alpha}(t) + \frac{\partial \eta (t) }{\partial t} \Bigg)dx dt. \label{Z18} \end{eqnarray} (4.13) By using the Riemann–Liouville fractional integral operators of GFBFs and GFLFs in Equation (4.13), we get \begin{equation} min \hspace{.2cm}\bar{J}[U]= \int_{0}^{h_{2}}\int_{0}^{h_{1}} Q(x, t, U) dxdt, \label{Z19} \end{equation} (4.14) where \begin{eqnarray} Q(x, t, U)&=&G\bigg( x, t, F^{h_{1}\alpha ,T}(1, x) U {\it{\Theta}} ^{h_{2}\alpha} (1, t) + \zeta (x) +\eta (t)- \zeta (0), \nonumber\\ && \frac{1}{d} \Bigg[ B^{h_{1}\alpha ,T}(x) U L^{h_{2}\alpha}(t) \nonumber\\ &&- a F^{h_{1}\alpha ,T}(1 - \nu, x) U {\it{\Theta}} ^{h_{2}\alpha} (1, t) - a \frac{\partial ^{\nu}\zeta (x)}{\partial x^{\nu}}\nonumber\\ &&- bF^{h_{1}\alpha ,T}(1, x) U {\it{\Theta}} ^{h_{2}\alpha} (1-\gamma , t) - b \frac{\partial ^{\gamma}\eta (t)}{\partial t^{\gamma}}\nonumber\\ &&- c F^{h_{1}\alpha ,T}(1, x) U {\it{\Theta}} ^{h_{2}\alpha} (1, t) - c \zeta (x) -c \eta (t) +c \zeta (0) \Bigg],\nonumber\\ && B^{h_{1}\alpha , T}(x) U {\it{\Theta}} ^{h_{2}\alpha}(1, t) + \frac{\partial \zeta (x) }{\partial x},\nonumber\\ && F^{h_{1}\alpha , T}(1, x) U L^{h_{2}\alpha}(t) + \frac{\partial \eta (t) }{\partial t} \bigg). \end{eqnarray} (4.15) To compute the integral (4.14), we use the two-dimensional LG quadrature rule (Nemati & Yousefi, 2016) as \begin{equation} \bar{J}[U] \simeq \tilde{J}[U]= \frac{h_{1}h_{2}}{4}\sum_{r=1}^{M}\sum_{s=1}^{M'}Q \left(\frac{h_{1}}{2}(\delta _{r}+1), \frac{h_{2}}{2}(\tau _{s}+1), U\right)\omega _{r}\omega _{s}:=\frac{h_{1} h_{2}}{4} W^{T}{\it{\Omega}} W', \end{equation} (4.16) where \begin{equation} ({\it{\Omega}}) _{r, s}= Q ((\delta _{r}+1)\frac{h_{1}}{2}, (\tau _{s}+1)\frac{h_{2}}{2}, U), \hspace{.5cm}1\leq r \leq M,\hspace{.1cm} 1 \leq s \leq M', \end{equation} (4.17) and $$ W $$ and $$ W' $$ are $$ M $$ and $$ M' $$-dimensional vectors of Christoffel weights of LG-nodes, respectively. Referring to differential calculus, we have the following necessary conditions for optimizing the functional $$ \tilde{J} $$, as \begin{equation} \frac{\partial \tilde{J}}{\partial u_{i, j}}[U]=0, \hspace{.5cm}0\leq i \leq M, \hspace{.2cm}0\leq j \leq M'. \label{Z21} \end{equation} (4.18) To find optimized matrix of $$ U $$, we shall solve the algebraic system (4.18). Therefore, the Newton’s iterative method is taken into account to find the solution. 5. Convergence of the new approach In this section, the convergence of the proposed method in the Sobolev space is investigated. We show that with an increase in the number of the generalized fractional-order Bernoulli–Legendre functions basis, the approximate optimum value approaches to the exact value. We show this fact in Theorem 5.3. Now we define our function space and provide some needed theorems, lemmas and corollaries. The Sobolev norm in the domain $${\it{\Delta}} = (a, b)^{d} $$ in $$R^{d} $$ with $$ d=2, 3 $$ for $$ \mu \geq 1 $$ is defined as (Canuto et al., 2006) \begin{equation} \Vert \xi \Vert _{H^{\mu}({\it{\Delta}})} = \left( \sum _{k=0}^{\mu} \sum _{i=1}^{d} \Vert D^{(k)}_{i} \xi \Vert ^{2}_{L^{2}({\it{\Delta}})}\right)^{\frac{1}{2}},\label{Z30} \end{equation} (5.1) where $$ D^{(k)}_{i} \xi $$ denotes the $$ k $$th derivative of $$ \xi $$ respect to variable of $$ i $$th. The symbol $$ \vert \xi \vert _{H^{\mu ; M}({\it{\Delta}})} $$ was defined by Canuto et al. (2006) \begin{align*} \vert \xi \vert _{H^{\mu ; M}({\it{\Delta}})} = \left( \sum _{k=min (\mu , M+1)}^{\mu} \sum _{i=1}^{d} \Vert D^{(k)}_{i} \xi \Vert ^{2}_{L^{2}({\it{\Delta}})}\right)^{\frac{1}{2}}. \end{align*} For simplicity of work, we assume $$ h_{1}=h_{2}=1 $$ and $$ M=M'. $$ Of course for $$ h_{1},h_{2}\neq 1 $$ and $$ M\neq M' $$ the procedure is similar. Theorem 5.1 Suppose that $$ \xi \in H^{\mu}({\it{\Delta}}) $$ with $$ \mu \geq 1 $$ and $$ {\it{\Delta}} = [0, 1]\times [0, 1]. $$ If $$ P_{M}^{\alpha}\xi = \sum _{i=0}^{M} \sum_{j=0}^{M} c_{ij}\beta _{i}^{\alpha}L_{j}^{\alpha},$$ is the best approximation of $$ \xi $$ then \begin{equation} \Vert \xi - P_{M}^{\alpha}\xi \Vert _{L^{2}({\it{\Delta}})} \leq c \alpha ^{1-\mu}M^{1-\mu} \vert \xi \vert _{H^{\mu ;M\alpha }({\it{\Delta}})}, \label{Z31} \end{equation} (5.2) and for $$ 1 \leq l \leq \mu $$, \begin{equation} \Vert \xi - P_{M}^{\alpha}\xi \Vert _{H^{l}({\it{\Delta}})}\leq c \alpha ^{ \varrho (l) - \mu } M^{\varrho (l) - \mu} \vert \xi \vert _{H^{\mu ;M\alpha }({\it{\Delta}})}, \label{Z35} \end{equation} (5.3) where $$ c $$ depends on $$ \mu $$ and \begin{align*} \varrho (l)=\left\{ \begin{array}{ll} 0, & l=0,\\ 2l-\frac{1}{2}, \ \ \ \ \ \ \ \ \ \ \ \ & l>0. \end{array} \right . \end{align*} Proof. Let $$ P_{N}=P_{N}({\it{\Pi}}) $$ be the space of all algebraic polynomials of degree up to $$ N $$ in each variable $$ x_{i} $$ for $$i=1, 2, \ldots , d $$. If $$ P'\xi $$ be the best approximation of $$ \xi $$ upon $$ P_{N}, $$ so for all $$ \xi \in H^{\mu}({\it{\Pi}}), \mu \geq 1,$$ we have (Canuto et al., 2006) \begin{equation} \Vert \xi - P'_{N}\xi \Vert _{L^{2}({\it{\Pi}})} \leq c N^{1-\mu} \vert \xi \vert _{H^{\mu ;N }({\it{\Pi}})}\label{Z33} \end{equation} (5.4) and for $$ 1 \leq l \leq \mu$$, \begin{equation} \Vert \xi - P'_{N}\xi \Vert _{H^{l}({\it{\Pi}})}\leq c N^{\varrho (l) - \mu} \vert \xi \vert _{H^{\mu ;N }({\it{\Pi}})}. \label{Z34} \end{equation} (5.5) Since the best approximation is unique (Kreyszig, 1978) and approximation with the fractional-order functions is from most degree $$ M\alpha $$ in each variable, therefore we have $$ N=M\alpha $$ and $$ {\it{\Pi}} ={\it{\Delta}} $$ in Equations (5.4) and (5.5) as \begin{align*} \Vert \xi - P_{M}^{\alpha}\xi \Vert _{L^{2}({\it{\Delta}})} = \Vert \xi - P'_{M \alpha}\xi \Vert _{L^{2}({\it{\Delta}})} \leq c \alpha ^{1-\mu} M^{1-\mu} \vert \xi \vert _{H^{\mu ;M \alpha }({\it{\Delta}})}, \end{align*} and for $$ 1 \leq l \leq \mu $$, \begin{align*} \Vert \xi - P_{M}^{\alpha}\xi \Vert _{H^{l}({\it{\Delta}})} = \Vert \xi - P'_{M\alpha }\xi \Vert _{H^{l}({\it{\Delta}})}\leq c \alpha ^{\varrho (l) - \mu} M^{\varrho (l) - \mu} \vert \xi \vert _{H^{\mu ;M\alpha }({\it{\Delta}})}, \end{align*} where \begin{align*} \varrho (l)=\left\{ \begin{array}{ll} 0, & l=0,\\ 2l-\frac{1}{2}, \ \ \ \ \ \ \ \ \ \ \ \ & l>0. \end{array} \right .\tag*{$\quad \square$} \end{align*} This result shows that in the case $$ \xi $$ is infinitely smooth, the rate of convergence of $$ P_{M}^{\alpha}\xi $$ to $$ \xi $$ is faster than of any power of $$ \frac{1}{M} $$, which is superior to that for the classical spectral methods (Canuto et al., 2006). Now, we obtain the error bound for the fractional and integer order derivative. Theorem 5.2 Suppose $$ \xi \in H^{\mu}({\it{\Delta}}) $$ with $$ \mu \geq 1 $$ and $$ 0 < \nu \leq 1 $$, then \begin{equation} \Vert D^{\nu}_{x} \xi (x, t) - D^{\nu}_{x}(P_{M}^{\alpha} \xi (x, t)) \Vert _{L^{2}({\it{\Delta}})} \leq \frac{c \alpha ^{\varrho (l)-\mu} M^{\varrho (l) - \mu}}{{\it{\Gamma}} (2 - \nu)} \vert \xi \vert _{H^{\mu ;M\alpha }({\it{\Delta}})}, \label{Z36} \end{equation} (5.6) where $$ 1 \leq l \leq \mu $$. Proof. By Equation (2.1) and Bass (2013) \begin{align*} \Vert f \ast g \Vert _{p} \leq \Vert f \Vert _{p} \Vert g \Vert _{1}, \end{align*} for $$ 0 < \nu \leq 1, $$ we obtain \begin{eqnarray*} && \Vert D^{\nu}_{x} \xi (x, t) - D^{\nu}_{x}(P_{M}^{\alpha}\xi (x, t)) \Vert _{L^{2}({\it{\Delta}})} ^{2}\\ &&\quad{}= \Vert I^{1-\nu }_{x}( D^{(1)}_{x} \xi (x, t) - D^{(1)}_{x}(P_{M}^{\alpha} \xi (x, t)))\Vert _{L^{2}({\it{\Delta}})} ^{2}\nonumber\\ &&\quad{}= \Vert \frac{1}{x^{\nu}{\it{\Gamma}} (1- \nu)} \ast ( D^{(1)}_{x} \xi (x, t) - D^{(1)}_{x}(P_{M}^{\alpha}\xi (x, t))) \Vert _{L^{2}({\it{\Delta}})} ^{2}\nonumber\\ &&\quad{}\leq \left(\frac{1}{(1- \nu) {\it{\Gamma}} (1-\nu)}\right)^{2}\Vert D^{(1)}_{x} \xi (x, t) - D^{(1)}_{x}(P_{M}^{\alpha} \xi (x, t)) \Vert _{L^{2}({\it{\Delta}})} ^{2}\nonumber\\ &&\quad{}\leq \left(\frac{1}{ {\it{\Gamma}} (2-\nu)}\right)^{2}\Vert \xi (x, t) - P_{M}^{\alpha} \xi (x, t) \Vert ^{2} _{H^{l}({\it{\Delta}})}, \end{eqnarray*} from Equation (5.3) we get Equation (5.6). □ Corollary 5.1 From Equation (5.6) for $$ \xi \in H^{\mu}({\it{\Delta}}) $$ with $$ \mu \geq 1 $$ and $$ 0 < \gamma \leq 1 $$ we can write \begin{equation} \Vert D^{\gamma }_{t} \xi (x, t) - D^{\gamma}_{t}(P_{M}^{\alpha}\xi (x, t)) \Vert _{L^{2}({\it{\Delta}})} \leq \frac{c \alpha ^{\varrho (l)-\mu} M^{\varrho (l) - \mu}}{{\it{\Gamma}} (2 - \gamma)} \vert \xi \vert _{H^{\mu ;M\alpha }({\it{\Delta}})}. \label{Z37} \end{equation} (5.7) Proof. It is an immediate consequence of Theorem 5.2. □ Lemma 5.1 Consider $$ \xi \in H^{\mu}({\it{\Delta}}) $$ with $$ \mu \geq 1 $$ and $$ 1 <l \leq \mu $$, then we have \begin{align*} \bigg \Vert \frac{\partial }{\partial x} \xi (x,t) - \frac{\partial }{\partial x} (P_{M}^{\alpha} \xi (x, t)) \bigg \Vert _{L^{2}({\it{\Delta}})} \leq c \alpha ^{ \varrho (l) - \mu } M^{\varrho (l) - \mu} \vert \xi \vert _{H^{\mu ;M\alpha }({\it{\Delta}})}. \end{align*} Proof. From Equation (5.3), we have \begin{align*} \bigg \Vert \frac{\partial }{\partial x} \xi (x,t) - \frac{\partial }{\partial x} (P_{M}^{\alpha} \xi (x, t)) \bigg \Vert _{L^{2}({\it{\Delta}})} & \leq \Vert \xi (x, t) - P_{M}^{\alpha} \xi (x, t) \Vert _{H^{l}({\it{\Delta}})}\nonumber\\ & \leq c \alpha ^{ \varrho (l) - \mu } M^{\varrho (l) - \mu} \vert \xi \vert _{H^{\mu ;M\alpha }({\it{\Delta}})}.\tag*{$\quad \square$} \end{align*} Corollary 5.2 Consider $$ \xi \in H^{\mu}({\it{\Delta}}) $$ with $$ \mu \geq 1 $$ and $$ 1 <l \leq \mu $$, then we have \begin{align*} \bigg \Vert \frac{\partial }{\partial t} \xi (x,t) - \frac{\partial }{\partial t} ( P_{M}^{\alpha}\xi (x, t)) \bigg \Vert _{L^{2}({\it{\Delta}})} \leq c \alpha ^{ \varrho (l) - \mu } M^{\varrho (l) - \mu} \vert \xi \vert _{H^{\mu ;M\alpha }({\it{\Delta}})}. \end{align*} Proof. It is an immediate consequence of Lemma 5.1. □ Corollary 5.3 Consider $$ \xi \in H^{\mu}({\it{\Delta}}) $$ with $$ \mu \geq 1 $$ and $$ 1 <l \leq \mu $$, then we get \begin{align*} \bigg \Vert \frac{\partial ^{2}}{\partial x \partial t} \xi (x,t) - \frac{\partial ^{2}}{\partial x \partial t} ( P_{M}^{\alpha}\xi (x, t)) \bigg \Vert _{L^{2}({\it{\Delta}})} \leq c \alpha ^{ \varrho (l) - \mu } M^{\varrho (l) - \mu} \big \vert \frac{\partial \xi }{\partial t}\big \vert _{H^{\mu ;M\alpha }({\it{\Delta}})}. \end{align*} Proof. From Equation (5.3), we obtain \begin{align*} \bigg \Vert \frac{\partial ^{2}}{\partial x \partial t} \xi (x,t) - \frac{\partial ^{2}}{\partial x \partial t} ( P_{M}^{\alpha}\xi (x, t)) \bigg \Vert _{L^{2}({\it{\Delta}})} &\leq \bigg \Vert \frac{\partial}{\partial x} \left(\frac{\partial}{\partial t} \xi (x, t)\right) - \frac{\partial}{\partial x}\left(\frac{\partial}{\partial t}P_{M}^{\alpha} \xi (x, t)\right) \bigg \Vert _{L^{2}({\it{\Delta}})}\\ &\leq \bigg \Vert \frac{\partial}{\partial t} \xi (x, t) - \frac{\partial}{\partial t}P_{M}^{\alpha} \xi (x, t) \bigg \Vert _{H^{l}({\it{\Delta}})} \\ &\leq c \alpha ^{ \varrho (l) - \mu } M^{\varrho (l) - \mu} \big \vert \frac{\partial \xi }{\partial t}\big \vert _{H^{\mu ;M\alpha }({\it{\Delta}})}.\tag*{$\quad \square$} \end{align*} Now we can show the convergence of the approximating method. Theorem 5.3 Assume that $$ \lbrace P_{M}^{\alpha}\xi ^{\ast} (x_{i}, t_{j}) , P_{M}^{\alpha}u ^{\ast} (x_{i}, t_{j}) , 0\leq i \leq M, 0 \leq j \leq M\rbrace _{M=M_{k}}^{\infty}$$ be a sequence of approximation optimal solutions and $$ \lbrace \xi ^{\ast} (x_{i}, t_{j}) ,u ^{\ast} (x_{i}, t_{j}) , 0\leq i \leq M, 0 \leq j \leq M \rbrace $$ be the exact optimal solution to problem (4.1) and $$ G $$ satisfy a Lipschitz condition with Lipschitz constant $$ \ell $$. Then we have \begin{equation} \tilde{J}[\xi ^{\ast}] = \lim _{M \rightarrow \infty} G_{M}( P_{M}^{\alpha}\xi ^{\ast}(x, t)), \end{equation} (5.8) where \begin{eqnarray*} G_{M}( P_{M}^{\alpha}\xi ^{\ast}(x, t)) &=& \frac{1}{4}\sum _{r=1}^{M}\sum _{s=1}^{M}G \Bigg( x _{r}, t_{s}, P_{M}^{\alpha} \xi ^{\ast} (x_{r}, t_{s}), \frac{1}{d} \Bigg[\frac{\partial ^{2}}{\partial x \partial t} P_{M}^{\alpha}\xi ^{\ast} (x_{r} , t_{s})- a \frac{\partial ^{\nu}}{\partial x^{\nu}}P_{M}^{\alpha} \xi ^{\ast} (x_{r}, t_{s})\\ &-& b \frac{\partial ^{\gamma} }{\partial t^{\gamma}}P_{M}^{\alpha} \xi ^{\ast}(x_{r}, t_{s}) -c P_{M}^{\alpha}\xi ^{\ast}(x_{r}, t_{s})\Bigg], \frac{\partial}{\partial x}P_{M}^{\alpha} \xi ^{\ast} (x_{r}, t_{s}), \frac{\partial }{\partial t}P_{M}^{\alpha} \xi ^{\ast} (x_{r}, t_{s}) \Bigg) \omega _{r}\omega _{s} \end{eqnarray*} and \begin{eqnarray*} J[\xi ^{\ast}] &=& \int _{0}^{1}\int _{0}^{1}G \left(x, t, \xi ^{\ast}(x, t), \frac{1}{d} \Bigg[\frac{\partial ^{2}\xi ^{\ast}}{\partial x \partial t}(x, t)- a \frac{\partial ^{\nu}\xi ^{\ast}}{\partial x^{\nu}}(x, t)\right. \nonumber\\ &&\left.- b \frac{\partial ^{\gamma}\xi ^{\ast}}{\partial t^{\gamma}}(x, t) -c \xi ^{\ast}(x, t)\Bigg], \frac{\partial \xi ^{\ast}}{\partial x}(x, t), \frac{\partial \xi ^{\ast}}{\partial t}(x, t) \right) dx dt. \end{eqnarray*} Proof. Since $$( P_{M}^{\alpha} \xi ^{\ast}, D_{x}^{\nu}P_{M}^{\alpha} \xi ^{\ast}, D_{t}^{\gamma}P_{M}^{\alpha} \xi ^{\ast}, \frac{\partial}{\partial x}P_{M}^{\alpha} \xi ^{\ast}, \frac{\partial}{\partial t}P_{M}^{\alpha} \xi ^{\ast}, \frac{\partial ^{2}}{\partial x \partial t}P_{M}^{\alpha} \xi ^{\ast})$$ converges to $$ (\xi ^{\ast}, D_{x}^{\nu}\xi ^{\ast}, D_{t}^{\gamma}\xi ^{\ast}, \frac{\partial}{\partial x} \xi ^{\ast}$$, $$\frac{\partial}{\partial t} \xi ^{\ast}, \frac{\partial ^{2}}{\partial x \partial t} \xi ^{\ast})$$ uniformly, we have \begin{gather} \lim _{M\rightarrow \infty} \Vert \xi ^{\ast} (x_{r}, t_{s}) - P_{M}^{\alpha} \xi ^{\ast} (x_{r}, t_{s}) \Vert _{L^{2}({\it{\Delta}})}=0,\label{Z40}\\ \end{gather} (5.9) \begin{gather} \lim _{M\rightarrow \infty} \Vert D_{x}^{\nu} \xi ^{\ast} (x_{r}, t_{s}) - D_{x}^{\nu} P_{M}^{\alpha} \xi ^{\ast} (x_{r}, t_{s}) \Vert _{L^{2}({\it{\Delta}})}=0,\\ \end{gather} (5.10) \begin{gather} \lim _{M\rightarrow \infty} \Vert D_{t}^{\gamma} \xi ^{\ast} (x_{r}, t_{s}) - D_{t}^{\gamma} P_{M}^{\alpha} \xi ^{\ast} (x_{r}, t_{s}) \Vert _{L^{2}({\it{\Delta}})}=0,\\ \end{gather} (5.11) \begin{gather} \lim _{M\rightarrow \infty}\bigg \Vert \frac{\partial}{\partial x} \xi ^{\ast} (x_{r}, t_{s}) - \frac{\partial}{\partial x} P_{M}^{\alpha} \xi ^{\ast} (x_{r}, t_{s}) \bigg \Vert _{L^{2}({\it{\Delta}})}=0,\\ \end{gather} (5.12) \begin{gather} \lim _{M\rightarrow \infty}\bigg \Vert \frac{\partial}{\partial t} \xi ^{\ast} (x_{r}, t_{s}) - \frac{\partial}{\partial t} P_{M}^{\alpha} \xi ^{\ast} (x_{r}, t_{s})\bigg \Vert _{L^{2}({\it{\Delta}})}=0,\\ \end{gather} (5.13) \begin{gather} \lim _{M\rightarrow \infty} \bigg \Vert \frac{\partial ^{2}}{\partial x \partial t} \xi ^{\ast} (x_{r}, t_{s}) - \frac{\partial ^{2}}{\partial x \partial t} P_{M}^{\alpha} \xi ^{\ast} (x_{r}, t_{s}) \bigg \Vert _{L^{2}({\it{\Delta}})}=0.\label{Z41} \end{gather} (5.14) On the other hand, since $$ G $$ is a continuously differentiable function and satisfy a Lipschitz condition with Lipschitz constant $$ \ell $$, thus we have \begin{align} &\Bigg \Vert G \Bigg( x_{r}, t_{s}, \xi ^{\ast}(x_{r}, t_{s}), \frac{1}{d} \Bigg[\frac{\partial ^{2}\xi ^{\ast}}{\partial x \partial t}(x_{r}, t_{s})- a \frac{\partial ^{\nu}\xi ^{\ast}}{\partial x^{\nu}}(x_{r}, t_{s}) \nonumber\\ &\quad{} - b \frac{\partial ^{\gamma}\xi ^{\ast}}{\partial t^{\gamma}}(x_{r}, t_{s}) -c \xi ^{\ast}(x_{r}, t_{s})\Bigg], \frac{\partial \xi ^{\ast}}{\partial x}(x_{r}, t_{s}), \frac{\partial \xi ^{\ast}}{\partial t}(x_{r}, t_{s}) \Bigg)\nonumber\\ &\quad- G \Bigg(x _{r}, t_{s}, P_{M}^{\alpha} \xi ^{\ast} (x_{r}, t_{s}), \frac{1}{d} \Bigg[\frac{\partial ^{2}}{\partial x \partial t} P_{M}^{\alpha}\xi ^{\ast} (x_{r} , t_{s})- a \frac{\partial ^{\nu}}{\partial x^{\nu}}P_{M}^{\alpha} \xi ^{\ast} (x_{r}, t_{s})\nonumber\\ &\quad- b \frac{\partial ^{\gamma} }{\partial t^{\gamma}}P_{M}^{\alpha} \xi ^{\ast}(x_{r}, t_{s}) -c P_{M}^{\alpha}\xi ^{\ast}(x_{r}, t_{s})\Bigg] , \frac{\partial}{\partial x}P_{M}^{\alpha} \xi ^{\ast} (x_{r}, t_{s}), \frac{\partial }{\partial t}P_{M}^{\alpha} \xi ^{\ast} (x_{r}, t_{s}) \Bigg) \Bigg \Vert _{L^{2}({\it{\Delta}})} \nonumber\\ &\leq \ell \Vert \xi ^{\ast} (x_{r}, t_{s}) - P_{M}^{\alpha} \xi ^{\ast} (x_{r}, t_{s}) \Vert _{L^{2}({\it{\Delta}})}+(\ell / d)\Vert \frac{\partial ^{2}}{\partial x \partial t} \xi ^{\ast} (x_{r}, t_{s}) \nonumber\\ &\quad- \frac{\partial ^{2}}{\partial x \partial t} P_{M}^{\alpha} \xi ^{\ast} (x_{r}, t_{s}) \Vert _{L^{2}({\it{\Delta}})}+(\ell a/d) \Vert D_{x}^{\nu} \xi ^{\ast} (x_{r}, t_{s}) - D_{x}^{\nu} P_{M}^{\alpha} \xi ^{\ast} (x_{r}, t_{s}) \Vert _{L^{2}({\it{\Delta}})}\nonumber\\ &\quad+ (\ell b/ d) \Vert D_{t}^{\gamma} \xi ^{\ast} (x_{r}, t_{s}) - D_{t}^{\gamma} P_{M}^{\alpha} \xi ^{\ast} (x_{r}, t_{s})+ (\ell c/d) \Vert \xi ^{\ast} (x_{r}, t_{s}) \nonumber\\ &\quad- P_{M}^{\alpha} \xi ^{\ast} (x_{r}, t_{s}) \Vert _{L^{2}({\it{\Delta}})}\Vert _{L^{2}({\it{\Delta}})}+ \ell \Vert \frac{\partial}{\partial x} \xi ^{\ast} (x_{r}, t_{s}) - \frac{\partial}{\partial x} P_{M}^{\alpha} \xi ^{\ast} (x_{r}, t_{s}) \Vert _{L^{2}({\it{\Delta}})}\nonumber\\ &\quad + \ell \Vert \frac{\partial}{\partial t} \xi ^{\ast} (x_{r}, t_{s}) - \frac{\partial}{\partial t} P_{M}^{\alpha} \xi ^{\ast} (x_{r}, t_{s}) \Vert _{L^{2}({\it{\Delta}})}, \hspace{.8cm}1 \leq r, s \leq M. \label{Z42} \end{align} (5.15) For simplicity, we let \begin{align*} \tilde{G}(x, t, \xi (x, t)) = G \left(\!\! x, t, \xi (x, t), \frac{1}{d} \left[\!\!\frac{\partial ^{2}\xi }{\partial x \partial t}(x, t)- a \frac{\partial ^{\nu}\xi }{\partial x^{\nu}}(x, t) - b \frac{\partial ^{\gamma}\xi }{\partial t^{\gamma}}(x, t) -c \xi (x, t)\!\!\right]\!, \frac{\partial \xi }{\partial x}(x, t), \frac{\partial \xi }{\partial t}(x, t) \!\!\right)\!. \end{align*} Furthermore, since $$ \tilde{G} $$ is a continuous function, we have (Canuto et al., 1988) \begin{align*} \int _{0}^{1}\int _{0}^{1}\tilde{G} \big ( x, t, \xi ^{\ast}(x, t) \big) dx dt= \lim _{M\rightarrow \infty }\frac{1}{4}\sum _{r=1}^{M}\sum _{s=1}^{M}\tilde{G} \big ( x _{r}, t_{s}, \xi ^{\ast} (x_{r}, t_{s}) \big) \omega _{r}\omega _{s}. \end{align*} Therefore \begin{eqnarray*} \int _{0}^{1}\int _{0}^{1} \tilde{G}(x, t, \xi ^{\ast}(x, t)) dxdt &=&\frac{1}{4} \lim _{M\rightarrow \infty} \left(\sum _{r=1}^{M}\sum _{s=1}^{M}\tilde{G}(x_{r}, t_{s}, P_{M}^{\alpha}\xi ^{\ast}(x_{r}, t_{s}))\omega _{r}\omega _{s}\right.\\ &&\qquad{}\left. + \sum _{r=1}^{M}\sum _{s=1}^{M}\big[ \tilde{G}(x_{r}, t_{s},\xi ^{\ast}(x_{r}, t_{s}))- \tilde{G}(x_{r}, t_{s}, P_{M}^{\alpha}\xi ^{\ast}(x_{r}, t_{s}))\big]\omega _{r}\omega _{s} \right)\!. \end{eqnarray*} From the uniform convergence of (5.9)–(5.14) and Equation (5.15), we have \begin{align*} \lim _{M\rightarrow \infty}\bigg \Vert \sum _{r=1}^{M}\sum _{s=1}^{M}\big[ \tilde{G}(x_{r}, t_{s},\xi ^{\ast}(x_{r}, t_{s}))- \tilde{G}(x_{r}, t_{s}, P_{M}^{\alpha}\xi ^{\ast}(x_{r}, t_{s}))\big]\omega _{r}\omega _{s} \bigg\Vert \hspace{.2cm}\leq 0. \end{align*} Thus \begin{align*} \int _{0}^{1}\int _{0}^{1} \tilde{G}(x, t, \xi ^{\ast}(x, t)) dxdt =\frac{1}{4} \lim _{M\rightarrow \infty} \left(\sum _{r=1}^{M}\sum _{s=1}^{M}\tilde{G}(x_{r}, t_{s}, P_{M}^{\alpha}\xi ^{\ast}(x_{r}, t_{s}))\omega _{r}\omega _{s} \right)\!, \end{align*} which confirms \begin{align*} \tilde{J}[\xi ^{\ast}] = \lim _{M \rightarrow \infty} G_{M}( P_{M}^{\alpha}\xi ^{\ast}(x, t)). \end{align*} This complete the proof. □ 6. Numerical examples In this section, we present two test problems and apply the method presented in Section 4 to solve them. The computations were performed on a personal computer, and the codes were written in Mathematica 10. Example 6.1 Consider the following two-dimensional optimal control problem (Tsai et al., 2002; Mamehrashi & Yousefi, 2016; Nemati & Yousefi, 2016) \begin{equation} min \hspace{.2cm}J[ \xi , u] = \int _{0}^{3} \int _{0}^{3} \bigg [ \bigg( \frac{\partial \xi }{\partial t}(x, t) + \xi(x, t) \bigg)^{2} + \xi ^{2}(x, t) + u^{2}(x, t)\bigg ] dx dt, \end{equation} (6.1) subject to \begin{gather} \frac{\partial ^{2}\xi}{\partial x \partial t}(x, t)= - \frac{\partial ^{\nu}\xi}{\partial x^{\nu}}(x, t) -3 \frac{\partial ^{\gamma}\xi}{\partial t^{\gamma}}(x, t) + 0.2 \xi(x, t) + 0.3 u(x, t),\\ \end{gather} (6.2) \begin{gather} \xi (x, 0) = e^{-3x}\cos (2 \pi x), \hspace{1cm} \xi (0, t) = e^{-2t}. \end{gather} (6.3) Here we solve this problem by using the proposed method for $$ l=M $$ and $$ l'=M' $$. In Table 1, the values of optimal performance index functional $$\tilde{J}$$ obtained by the present method are compared with the values of $$\tilde{J}$$ reported in Tsai et al. (2002), Mamehrashi & Yousefi (2016) and Nemati & Yousefi (2016),$$ \nu = \gamma =1$$. According to the values of $$\tilde{J}$$, we can say that our numerical solutions are better than the other methods (Tsai et al., 2002; Mamehrashi & Yousefi, 2016; Nemati & Yousefi, 2016). Also, Fig. 2 illustrates the optimal state and control functions of the problem with $$ M=1, M'=6 $$ and $$ \alpha =\frac{1}{5} $$ for $$ \nu = \gamma =1. $$ For partial fractional orders, two cases are considered ($$ \nu = \gamma = 0.8 $$ and $$ \nu =0.75, \gamma =0.8$$). Table 2 shows the values of the optimal performance index functional $$\tilde{J}$$ by the present method and Ref. (Nemati & Yousefi, 2016) for $$\nu = \gamma = 0.8$$. Also, Fig. 3 displays the optimal state and control functions for $$\nu = \gamma = 0.8$$ with $$\alpha =1$$. At last, we take $$ \nu =0.75 $$ and $$ \gamma =0.8 $$. The outcomes of optimal functional are shown in Table 3. Figure 4 depicts the state and control functions for $$ \nu =0.75 $$ and $$ \gamma =0.8 $$ with $$ \alpha =1. $$ From these figures and tables, it can be concluded that the proposed method is convenience for solving this problem. Therefore to show efficient of the parameter $$ \alpha $$, we show the values of optimal functional $$\tilde{J}$$ for $$ \nu = \gamma =1 $$ with various choices of $$ \alpha $$ in Table 4. Also, the computational results with $$ M=1, M' =6 $$ and various values of $$ \alpha $$ for $$ \nu = \gamma =1 $$ are displayed in Fig. 5. Fig. 2. View largeDownload slide (a) The state function and (b) the control input for $$ \nu =\gamma =1, \alpha = \frac{1}{5}$$ with $$ M=1, M' =6 $$ for Example 1. Fig. 2. View largeDownload slide (a) The state function and (b) the control input for $$ \nu =\gamma =1, \alpha = \frac{1}{5}$$ with $$ M=1, M' =6 $$ for Example 1. Fig. 3. View largeDownload slide (a) The state function and (b) the control input for $$ \nu = \gamma =0.8, \alpha =1$$ with $$ M=1, M' =6 $$ for Example 1. Fig. 3. View largeDownload slide (a) The state function and (b) the control input for $$ \nu = \gamma =0.8, \alpha =1$$ with $$ M=1, M' =6 $$ for Example 1. Fig. 4. View largeDownload slide (a) The state function and (b) the control input for $$ \nu =0.75, \gamma =0.8, \alpha =1$$ with $$ M=1, M' =6 $$ for Example 1. Fig. 4. View largeDownload slide (a) The state function and (b) the control input for $$ \nu =0.75, \gamma =0.8, \alpha =1$$ with $$ M=1, M' =6 $$ for Example 1. Fig. 5. View largeDownload slide The control function for (a) $$ \alpha = \frac{1}{5}, $$ (b) $$ \alpha = \frac{1}{4}, $$ (c) $$ \alpha = \frac{1}{2} $$ and (d) $$ \alpha = 1, $$ with $$ M=1, M' =6 $$ for Example 1. Fig. 5. View largeDownload slide The control function for (a) $$ \alpha = \frac{1}{5}, $$ (b) $$ \alpha = \frac{1}{4}, $$ (c) $$ \alpha = \frac{1}{2} $$ and (d) $$ \alpha = 1, $$ with $$ M=1, M' =6 $$ for Example 1. Table 1 Comparison of the estimated value of $$ \tilde{J} $$ for integer orders $$ \nu = \gamma =1 $$ and $$ \alpha =\frac{1}{5} $$ with different values of $$ M $$ and $$ M' $$ for Example 1 Method of Tsai et al. (2002) $$\tilde{J} $$ $$X=0.3, T= 0.3$$ 1.4979 $$X=0.2, T=0.2 $$ $$ 1.0953 $$ $$X=0.1, T=0.1 $$ $$0.7348 $$ $$X = 0.05, T=0.05$$ $$ 0.5510 $$ $$X=0.03, T=0.03 $$ $$ 0.4760 $$ Method of Mamehrashi & Yousefi (2016) $$\tilde{J}$$ $$M=6, M'= 1 $$ $$ 4.0203 $$ $$M=6, M'= 8 $$ $$2.2905 $$ $$M=9, M'= 2 $$ $$ 0.8024 $$ $$M=7, M'= 8 $$ $$ 0.6202 $$ $$M=8, M'= 3 $$ $$ 0.2792 $$ $$M=8, M'= 8 $$ $$0.2026 $$ Method of Nemati & Yousefi (2016) $$\tilde{J}$$ $$M=1, M'= 1 $$ $$6.3883 $$ $$M=2, M'= 3 $$ $$ 5.0251 $$ $$M=6, M'= 4 $$ $$ 2.2775 $$ $$M=7, M'= 5$$ $$0.5997 $$ $$M=8, M'= 5 $$ $$ 0.1906$$ $$M=9, M'= 5$$ $$0.1770$$ $$M=10, M'=5$$ $$0.0951$$ $$M=10, M'=6$$ $$0.0947$$ $$Present \; method$$ $$\tilde{J}$$ $$M=1, M' =2$$ $$ 3.4977\times 10^{-3}$$ $$M=1, M' =4 $$ $$6.0548 \times 10^{-5}$$ $$M=1, M' =6 $$ $$5.0340\times 10^{-7}$$ $$M=1, M' =8 $$ $$4.2535\times 10^{-8}$$ $$M=1, M' =10 $$ $$2.9455\times 10^{-9}$$ $$M=1, M' =12 $$ $$ 1.1270\times 10^{-9}$$ Method of Tsai et al. (2002) $$\tilde{J} $$ $$X=0.3, T= 0.3$$ 1.4979 $$X=0.2, T=0.2 $$ $$ 1.0953 $$ $$X=0.1, T=0.1 $$ $$0.7348 $$ $$X = 0.05, T=0.05$$ $$ 0.5510 $$ $$X=0.03, T=0.03 $$ $$ 0.4760 $$ Method of Mamehrashi & Yousefi (2016) $$\tilde{J}$$ $$M=6, M'= 1 $$ $$ 4.0203 $$ $$M=6, M'= 8 $$ $$2.2905 $$ $$M=9, M'= 2 $$ $$ 0.8024 $$ $$M=7, M'= 8 $$ $$ 0.6202 $$ $$M=8, M'= 3 $$ $$ 0.2792 $$ $$M=8, M'= 8 $$ $$0.2026 $$ Method of Nemati & Yousefi (2016) $$\tilde{J}$$ $$M=1, M'= 1 $$ $$6.3883 $$ $$M=2, M'= 3 $$ $$ 5.0251 $$ $$M=6, M'= 4 $$ $$ 2.2775 $$ $$M=7, M'= 5$$ $$0.5997 $$ $$M=8, M'= 5 $$ $$ 0.1906$$ $$M=9, M'= 5$$ $$0.1770$$ $$M=10, M'=5$$ $$0.0951$$ $$M=10, M'=6$$ $$0.0947$$ $$Present \; method$$ $$\tilde{J}$$ $$M=1, M' =2$$ $$ 3.4977\times 10^{-3}$$ $$M=1, M' =4 $$ $$6.0548 \times 10^{-5}$$ $$M=1, M' =6 $$ $$5.0340\times 10^{-7}$$ $$M=1, M' =8 $$ $$4.2535\times 10^{-8}$$ $$M=1, M' =10 $$ $$2.9455\times 10^{-9}$$ $$M=1, M' =12 $$ $$ 1.1270\times 10^{-9}$$ Table 1 Comparison of the estimated value of $$ \tilde{J} $$ for integer orders $$ \nu = \gamma =1 $$ and $$ \alpha =\frac{1}{5} $$ with different values of $$ M $$ and $$ M' $$ for Example 1 Method of Tsai et al. (2002) $$\tilde{J} $$ $$X=0.3, T= 0.3$$ 1.4979 $$X=0.2, T=0.2 $$ $$ 1.0953 $$ $$X=0.1, T=0.1 $$ $$0.7348 $$ $$X = 0.05, T=0.05$$ $$ 0.5510 $$ $$X=0.03, T=0.03 $$ $$ 0.4760 $$ Method of Mamehrashi & Yousefi (2016) $$\tilde{J}$$ $$M=6, M'= 1 $$ $$ 4.0203 $$ $$M=6, M'= 8 $$ $$2.2905 $$ $$M=9, M'= 2 $$ $$ 0.8024 $$ $$M=7, M'= 8 $$ $$ 0.6202 $$ $$M=8, M'= 3 $$ $$ 0.2792 $$ $$M=8, M'= 8 $$ $$0.2026 $$ Method of Nemati & Yousefi (2016) $$\tilde{J}$$ $$M=1, M'= 1 $$ $$6.3883 $$ $$M=2, M'= 3 $$ $$ 5.0251 $$ $$M=6, M'= 4 $$ $$ 2.2775 $$ $$M=7, M'= 5$$ $$0.5997 $$ $$M=8, M'= 5 $$ $$ 0.1906$$ $$M=9, M'= 5$$ $$0.1770$$ $$M=10, M'=5$$ $$0.0951$$ $$M=10, M'=6$$ $$0.0947$$ $$Present \; method$$ $$\tilde{J}$$ $$M=1, M' =2$$ $$ 3.4977\times 10^{-3}$$ $$M=1, M' =4 $$ $$6.0548 \times 10^{-5}$$ $$M=1, M' =6 $$ $$5.0340\times 10^{-7}$$ $$M=1, M' =8 $$ $$4.2535\times 10^{-8}$$ $$M=1, M' =10 $$ $$2.9455\times 10^{-9}$$ $$M=1, M' =12 $$ $$ 1.1270\times 10^{-9}$$ Method of Tsai et al. (2002) $$\tilde{J} $$ $$X=0.3, T= 0.3$$ 1.4979 $$X=0.2, T=0.2 $$ $$ 1.0953 $$ $$X=0.1, T=0.1 $$ $$0.7348 $$ $$X = 0.05, T=0.05$$ $$ 0.5510 $$ $$X=0.03, T=0.03 $$ $$ 0.4760 $$ Method of Mamehrashi & Yousefi (2016) $$\tilde{J}$$ $$M=6, M'= 1 $$ $$ 4.0203 $$ $$M=6, M'= 8 $$ $$2.2905 $$ $$M=9, M'= 2 $$ $$ 0.8024 $$ $$M=7, M'= 8 $$ $$ 0.6202 $$ $$M=8, M'= 3 $$ $$ 0.2792 $$ $$M=8, M'= 8 $$ $$0.2026 $$ Method of Nemati & Yousefi (2016) $$\tilde{J}$$ $$M=1, M'= 1 $$ $$6.3883 $$ $$M=2, M'= 3 $$ $$ 5.0251 $$ $$M=6, M'= 4 $$ $$ 2.2775 $$ $$M=7, M'= 5$$ $$0.5997 $$ $$M=8, M'= 5 $$ $$ 0.1906$$ $$M=9, M'= 5$$ $$0.1770$$ $$M=10, M'=5$$ $$0.0951$$ $$M=10, M'=6$$ $$0.0947$$ $$Present \; method$$ $$\tilde{J}$$ $$M=1, M' =2$$ $$ 3.4977\times 10^{-3}$$ $$M=1, M' =4 $$ $$6.0548 \times 10^{-5}$$ $$M=1, M' =6 $$ $$5.0340\times 10^{-7}$$ $$M=1, M' =8 $$ $$4.2535\times 10^{-8}$$ $$M=1, M' =10 $$ $$2.9455\times 10^{-9}$$ $$M=1, M' =12 $$ $$ 1.1270\times 10^{-9}$$ Table 2 Comparison of the estimated value of $$ \tilde{J} $$ for orders $$ \nu = \gamma =0.8 $$ and $$ \alpha =\frac{1}{5} $$ with different values of $$ M $$ and $$ M' $$ for Example $$1$$ Method of Nemati & Yousefi (2016) $$\tilde{J}$$ $$M=4, M'= 1 $$ $$ 7.5622 $$ $$M=5, M'= 2 $$ $$5.0050 $$ $$M=5, M'= 3 $$ $$ 3.9465 $$ $$M=7, M'= 3$$ $$1.6332$$ $$M=8, M'= 4 $$ $$ 0.3464$$ $$M=9, M'= 4$$ $$0.2383 $$ $$M=9, M'=5$$ $$0.1579 $$ $$M=9, M'=6$$ $$0.1331 $$ $$Present \; method$$ $$\tilde{J}$$ $$M=1, M' =3 $$ $$ 1.0799\times 10^{-3} -4.2697\times 10^{-29} i $$ $$M=1, M' =5 $$ $$4.7624 \times 10^{-6} -4.5606\times 10^{-33} i $$ $$M=1, M' =7 $$ $$1.5109\times 10^{-8}-1.7117 \times 10^{-31} i $$ $$M=1, M' =9 $$ $$ 3.5017\times 10^{-10}+ 7.7166\times 10^{-29} i $$ $$M=1, M' =11 $$ $$7.3526\times 10^{-11}-1.5310\times 10^{-22} i $$ Method of Nemati & Yousefi (2016) $$\tilde{J}$$ $$M=4, M'= 1 $$ $$ 7.5622 $$ $$M=5, M'= 2 $$ $$5.0050 $$ $$M=5, M'= 3 $$ $$ 3.9465 $$ $$M=7, M'= 3$$ $$1.6332$$ $$M=8, M'= 4 $$ $$ 0.3464$$ $$M=9, M'= 4$$ $$0.2383 $$ $$M=9, M'=5$$ $$0.1579 $$ $$M=9, M'=6$$ $$0.1331 $$ $$Present \; method$$ $$\tilde{J}$$ $$M=1, M' =3 $$ $$ 1.0799\times 10^{-3} -4.2697\times 10^{-29} i $$ $$M=1, M' =5 $$ $$4.7624 \times 10^{-6} -4.5606\times 10^{-33} i $$ $$M=1, M' =7 $$ $$1.5109\times 10^{-8}-1.7117 \times 10^{-31} i $$ $$M=1, M' =9 $$ $$ 3.5017\times 10^{-10}+ 7.7166\times 10^{-29} i $$ $$M=1, M' =11 $$ $$7.3526\times 10^{-11}-1.5310\times 10^{-22} i $$ Table 2 Comparison of the estimated value of $$ \tilde{J} $$ for orders $$ \nu = \gamma =0.8 $$ and $$ \alpha =\frac{1}{5} $$ with different values of $$ M $$ and $$ M' $$ for Example $$1$$ Method of Nemati & Yousefi (2016) $$\tilde{J}$$ $$M=4, M'= 1 $$ $$ 7.5622 $$ $$M=5, M'= 2 $$ $$5.0050 $$ $$M=5, M'= 3 $$ $$ 3.9465 $$ $$M=7, M'= 3$$ $$1.6332$$ $$M=8, M'= 4 $$ $$ 0.3464$$ $$M=9, M'= 4$$ $$0.2383 $$ $$M=9, M'=5$$ $$0.1579 $$ $$M=9, M'=6$$ $$0.1331 $$ $$Present \; method$$ $$\tilde{J}$$ $$M=1, M' =3 $$ $$ 1.0799\times 10^{-3} -4.2697\times 10^{-29} i $$ $$M=1, M' =5 $$ $$4.7624 \times 10^{-6} -4.5606\times 10^{-33} i $$ $$M=1, M' =7 $$ $$1.5109\times 10^{-8}-1.7117 \times 10^{-31} i $$ $$M=1, M' =9 $$ $$ 3.5017\times 10^{-10}+ 7.7166\times 10^{-29} i $$ $$M=1, M' =11 $$ $$7.3526\times 10^{-11}-1.5310\times 10^{-22} i $$ Method of Nemati & Yousefi (2016) $$\tilde{J}$$ $$M=4, M'= 1 $$ $$ 7.5622 $$ $$M=5, M'= 2 $$ $$5.0050 $$ $$M=5, M'= 3 $$ $$ 3.9465 $$ $$M=7, M'= 3$$ $$1.6332$$ $$M=8, M'= 4 $$ $$ 0.3464$$ $$M=9, M'= 4$$ $$0.2383 $$ $$M=9, M'=5$$ $$0.1579 $$ $$M=9, M'=6$$ $$0.1331 $$ $$Present \; method$$ $$\tilde{J}$$ $$M=1, M' =3 $$ $$ 1.0799\times 10^{-3} -4.2697\times 10^{-29} i $$ $$M=1, M' =5 $$ $$4.7624 \times 10^{-6} -4.5606\times 10^{-33} i $$ $$M=1, M' =7 $$ $$1.5109\times 10^{-8}-1.7117 \times 10^{-31} i $$ $$M=1, M' =9 $$ $$ 3.5017\times 10^{-10}+ 7.7166\times 10^{-29} i $$ $$M=1, M' =11 $$ $$7.3526\times 10^{-11}-1.5310\times 10^{-22} i $$ Table 3 Comparison of the estimated value of $$ \tilde{J} $$ for orders $$ \nu = 0.75, \gamma =0.8 $$ and $$ \alpha =\frac{1}{5} $$ with different values of $$ M $$ and $$ M' $$ for Example 1 Method of Nemati & Yousefi (2016) $$\tilde{J}$$ $$M=2, M'= 1 $$ $$10.3484 $$ $$M=3, M'= 2 $$ $$7.3672 $$ $$M=4, M'= 3 $$ $$4.5423 $$ $$M=4, M'= 4$$ $$4.2553$$ $$M=6, M'= 5 $$ $$ 3.1165$$ $$M=8, M'= 5 $$ $$ 0.3131 $$ $$M=9, M'=6 $$ $$ 0.1368 $$ $$Present \; method$$ $$\tilde{J}$$ $$M=1, M' =3 $$ $$ 1.0799\times 10^{-3} -9.9194\times 10^{-34} i$$ $$M=1, M' =5 $$ $$4.7624\times 10^{-6} -7.1948 \times 10^{-33}i$$ $$M=1, M' =7 $$ $$1.5109\times 10^{-8}- 5.6134\times 10^{-34}i$$ $$M=1, M' =9 $$ $$ 3.5017\times 10^{-10}-1.6718\times 10^{-32}i$$ $$M=1, M' =11 $$ $$1.5838\times 10^{-11}+9.1743\times 10^{-23}i$$ Method of Nemati & Yousefi (2016) $$\tilde{J}$$ $$M=2, M'= 1 $$ $$10.3484 $$ $$M=3, M'= 2 $$ $$7.3672 $$ $$M=4, M'= 3 $$ $$4.5423 $$ $$M=4, M'= 4$$ $$4.2553$$ $$M=6, M'= 5 $$ $$ 3.1165$$ $$M=8, M'= 5 $$ $$ 0.3131 $$ $$M=9, M'=6 $$ $$ 0.1368 $$ $$Present \; method$$ $$\tilde{J}$$ $$M=1, M' =3 $$ $$ 1.0799\times 10^{-3} -9.9194\times 10^{-34} i$$ $$M=1, M' =5 $$ $$4.7624\times 10^{-6} -7.1948 \times 10^{-33}i$$ $$M=1, M' =7 $$ $$1.5109\times 10^{-8}- 5.6134\times 10^{-34}i$$ $$M=1, M' =9 $$ $$ 3.5017\times 10^{-10}-1.6718\times 10^{-32}i$$ $$M=1, M' =11 $$ $$1.5838\times 10^{-11}+9.1743\times 10^{-23}i$$ Table 3 Comparison of the estimated value of $$ \tilde{J} $$ for orders $$ \nu = 0.75, \gamma =0.8 $$ and $$ \alpha =\frac{1}{5} $$ with different values of $$ M $$ and $$ M' $$ for Example 1 Method of Nemati & Yousefi (2016) $$\tilde{J}$$ $$M=2, M'= 1 $$ $$10.3484 $$ $$M=3, M'= 2 $$ $$7.3672 $$ $$M=4, M'= 3 $$ $$4.5423 $$ $$M=4, M'= 4$$ $$4.2553$$ $$M=6, M'= 5 $$ $$ 3.1165$$ $$M=8, M'= 5 $$ $$ 0.3131 $$ $$M=9, M'=6 $$ $$ 0.1368 $$ $$Present \; method$$ $$\tilde{J}$$ $$M=1, M' =3 $$ $$ 1.0799\times 10^{-3} -9.9194\times 10^{-34} i$$ $$M=1, M' =5 $$ $$4.7624\times 10^{-6} -7.1948 \times 10^{-33}i$$ $$M=1, M' =7 $$ $$1.5109\times 10^{-8}- 5.6134\times 10^{-34}i$$ $$M=1, M' =9 $$ $$ 3.5017\times 10^{-10}-1.6718\times 10^{-32}i$$ $$M=1, M' =11 $$ $$1.5838\times 10^{-11}+9.1743\times 10^{-23}i$$ Method of Nemati & Yousefi (2016) $$\tilde{J}$$ $$M=2, M'= 1 $$ $$10.3484 $$ $$M=3, M'= 2 $$ $$7.3672 $$ $$M=4, M'= 3 $$ $$4.5423 $$ $$M=4, M'= 4$$ $$4.2553$$ $$M=6, M'= 5 $$ $$ 3.1165$$ $$M=8, M'= 5 $$ $$ 0.3131 $$ $$M=9, M'=6 $$ $$ 0.1368 $$ $$Present \; method$$ $$\tilde{J}$$ $$M=1, M' =3 $$ $$ 1.0799\times 10^{-3} -9.9194\times 10^{-34} i$$ $$M=1, M' =5 $$ $$4.7624\times 10^{-6} -7.1948 \times 10^{-33}i$$ $$M=1, M' =7 $$ $$1.5109\times 10^{-8}- 5.6134\times 10^{-34}i$$ $$M=1, M' =9 $$ $$ 3.5017\times 10^{-10}-1.6718\times 10^{-32}i$$ $$M=1, M' =11 $$ $$1.5838\times 10^{-11}+9.1743\times 10^{-23}i$$ Table 4 Comparison of the estimated value of $$ \tilde{J} $$ for integer orders $$ \nu =\gamma =1 $$ with different values of $$ \alpha $$ for Example $$1$$ $$M, M'$$ $$\alpha = \frac{1}{5} $$ $$ \alpha = \frac{1}{4}$$ $$ \alpha = \frac{1}{2}$$ $$\alpha = 1$$ $$M=1, M' =2$$ $$3.4977\times 10^{-3}$$ $$2.1816\times 10^{-3} $$ $$ 2.0015\times 10^{-4} $$ $$ 1.0650\times 10^{-2}$$ $$M=1, M' =4 $$ $$ 6.0548\times 10^{-5} $$ $$3.7983\times 10^{-5}$$ $$3.7836\times 10^{-5}$$ $$8.7676\times 10^{-5}$$ $$M=1, M' =6$$ $$5.0340\times 10^{-7}$$ $$9.5888\times 10^{-8} $$ $$ 5.2497\times 10^{-6}$$ $$1.7175\times 10^{-5}$$ $$M=1, M' =8$$ $$4.2535\times 10^{-8}$$ $$1.8969\times 10^{-8}$$ $$3.0392 \times 10^{-7}$$ $$1.5806\times 10^{-5}$$ $$M=1, M' =10 $$ $$ 2.9455\times 10^{-9}$$ $$1.1887\times 10^{-11}$$ $$6.8830 \times 10^{-8}$$ $$1.4988\times 10^{-5}$$ $$M=1, M' =12$$ $$1.1270\times 10^{-9}$$ $$1.0570\times 10^{-10} $$ $$2.0730 \times 10^{-8} $$ $$1.4379\times 10^{-5}$$ $$M, M'$$ $$\alpha = \frac{1}{5} $$ $$ \alpha = \frac{1}{4}$$ $$ \alpha = \frac{1}{2}$$ $$\alpha = 1$$ $$M=1, M' =2$$ $$3.4977\times 10^{-3}$$ $$2.1816\times 10^{-3} $$ $$ 2.0015\times 10^{-4} $$ $$ 1.0650\times 10^{-2}$$ $$M=1, M' =4 $$ $$ 6.0548\times 10^{-5} $$ $$3.7983\times 10^{-5}$$ $$3.7836\times 10^{-5}$$ $$8.7676\times 10^{-5}$$ $$M=1, M' =6$$ $$5.0340\times 10^{-7}$$ $$9.5888\times 10^{-8} $$ $$ 5.2497\times 10^{-6}$$ $$1.7175\times 10^{-5}$$ $$M=1, M' =8$$ $$4.2535\times 10^{-8}$$ $$1.8969\times 10^{-8}$$ $$3.0392 \times 10^{-7}$$ $$1.5806\times 10^{-5}$$ $$M=1, M' =10 $$ $$ 2.9455\times 10^{-9}$$ $$1.1887\times 10^{-11}$$ $$6.8830 \times 10^{-8}$$ $$1.4988\times 10^{-5}$$ $$M=1, M' =12$$ $$1.1270\times 10^{-9}$$ $$1.0570\times 10^{-10} $$ $$2.0730 \times 10^{-8} $$ $$1.4379\times 10^{-5}$$ Table 4 Comparison of the estimated value of $$ \tilde{J} $$ for integer orders $$ \nu =\gamma =1 $$ with different values of $$ \alpha $$ for Example $$1$$ $$M, M'$$ $$\alpha = \frac{1}{5} $$ $$ \alpha = \frac{1}{4}$$ $$ \alpha = \frac{1}{2}$$ $$\alpha = 1$$ $$M=1, M' =2$$ $$3.4977\times 10^{-3}$$ $$2.1816\times 10^{-3} $$ $$ 2.0015\times 10^{-4} $$ $$ 1.0650\times 10^{-2}$$ $$M=1, M' =4 $$ $$ 6.0548\times 10^{-5} $$ $$3.7983\times 10^{-5}$$ $$3.7836\times 10^{-5}$$ $$8.7676\times 10^{-5}$$ $$M=1, M' =6$$ $$5.0340\times 10^{-7}$$ $$9.5888\times 10^{-8} $$ $$ 5.2497\times 10^{-6}$$ $$1.7175\times 10^{-5}$$ $$M=1, M' =8$$ $$4.2535\times 10^{-8}$$ $$1.8969\times 10^{-8}$$ $$3.0392 \times 10^{-7}$$ $$1.5806\times 10^{-5}$$ $$M=1, M' =10 $$ $$ 2.9455\times 10^{-9}$$ $$1.1887\times 10^{-11}$$ $$6.8830 \times 10^{-8}$$ $$1.4988\times 10^{-5}$$ $$M=1, M' =12$$ $$1.1270\times 10^{-9}$$ $$1.0570\times 10^{-10} $$ $$2.0730 \times 10^{-8} $$ $$1.4379\times 10^{-5}$$ $$M, M'$$ $$\alpha = \frac{1}{5} $$ $$ \alpha = \frac{1}{4}$$ $$ \alpha = \frac{1}{2}$$ $$\alpha = 1$$ $$M=1, M' =2$$ $$3.4977\times 10^{-3}$$ $$2.1816\times 10^{-3} $$ $$ 2.0015\times 10^{-4} $$ $$ 1.0650\times 10^{-2}$$ $$M=1, M' =4 $$ $$ 6.0548\times 10^{-5} $$ $$3.7983\times 10^{-5}$$ $$3.7836\times 10^{-5}$$ $$8.7676\times 10^{-5}$$ $$M=1, M' =6$$ $$5.0340\times 10^{-7}$$ $$9.5888\times 10^{-8} $$ $$ 5.2497\times 10^{-6}$$ $$1.7175\times 10^{-5}$$ $$M=1, M' =8$$ $$4.2535\times 10^{-8}$$ $$1.8969\times 10^{-8}$$ $$3.0392 \times 10^{-7}$$ $$1.5806\times 10^{-5}$$ $$M=1, M' =10 $$ $$ 2.9455\times 10^{-9}$$ $$1.1887\times 10^{-11}$$ $$6.8830 \times 10^{-8}$$ $$1.4988\times 10^{-5}$$ $$M=1, M' =12$$ $$1.1270\times 10^{-9}$$ $$1.0570\times 10^{-10} $$ $$2.0730 \times 10^{-8} $$ $$1.4379\times 10^{-5}$$ Example 6.2 Consider the following two-dimensional optimal control problem (Mamehrashi & Yousefi, 2016) \begin{equation} min \hspace{.2cm}J[\xi , u] = \frac{1}{2} \int _{0}^{5} \int _{0}^{5} \bigg [ 10^{7} \big( \xi (x, t)- \sin (x+t) \big )^{2} + u^{2}(x, t)\bigg] dx dt, \end{equation} (6.4) subject to \begin{gather} \frac{\partial ^{2}\xi}{\partial x \partial t}(x, t)= - \frac{\partial ^{\nu}\xi}{\partial x^{\nu}}(x, t) -3 \frac{\partial ^{\gamma}\xi}{\partial t^{\gamma}}(x, t) + 0.2 \xi(x, t) + 0.3 u(x, t),\\ \end{gather} (6.5) \begin{gather} \xi (x, 0) = e^{-3x}\cos (2 \pi x), \hspace{1cm} \xi (0, t) = e^{-2t}. \end{gather} (6.6) Here we solve this problem by using the proposed method for $$ l=M $$ and $$ l'=M' $$. In Table 5, a comparison is made between the values of $$ \tilde{J} $$ obtained by the present method with the values of $$\tilde{J}$$ reported in Mamehrashi & Yousefi (2016) with different values of $$ M, M' $$ for $$ \nu = \gamma = 1 $$ and $$ \alpha =1. $$Figure 6(a) and (b) shows the state and control functions for $$ \nu = \gamma =1 $$ with $$ M=1, M'=5 $$ and $$ \alpha =1, $$ respectively. Table 6 presents the values of optimal functional $$\tilde{J}$$ for $$ \nu = \gamma =1 $$ with different choices of $$ \alpha $$. By varying the value of $$ \alpha $$ the obtained state and control functions for $$ \nu = \gamma =1 $$ with $$ M=1, M'=5 $$ are shown, in Figs 7 and 8, respectively. For partial fractional orders, two cases are considered. Initially we take $$ \nu =\gamma =0.8 $$. Table 7 shows the values of optimal performance index functional $$ \tilde{J} $$ for various values of $$M $$ and $$ M' $$ with $$ \alpha =\frac{1}{2}, 1 $$. In addition, Fig. 9(a) and (b) displays state and control functions for $$ \nu = \gamma =0.8 $$ with $$ M=1, M'=5 $$ and $$ \alpha =1, $$ respectively. At last, we take $$ \nu =0.75 $$ and $$ \gamma =0.8 $$. The outcomes of optimal functional are shown in Table 8. Fig. 6. View largeDownload slide (a) The state function and (b) the control input for $$ \nu =\gamma =1, \alpha = 1$$ with $$ M=1, M' =5 $$ for Example 2. Fig. 6. View largeDownload slide (a) The state function and (b) the control input for $$ \nu =\gamma =1, \alpha = 1$$ with $$ M=1, M' =5 $$ for Example 2. Fig. 7. View largeDownload slide The state function for (a) $$ \alpha = \frac{1}{5}, $$ (b) $$ \alpha = \frac{1}{2}, $$ (c) $$ \alpha = 1 $$ and (d) $$ \alpha = 2, $$ with $$ \nu = \gamma =1 $$ and $$ M=1, M' =5 $$ for Example 2. Fig. 7. View largeDownload slide The state function for (a) $$ \alpha = \frac{1}{5}, $$ (b) $$ \alpha = \frac{1}{2}, $$ (c) $$ \alpha = 1 $$ and (d) $$ \alpha = 2, $$ with $$ \nu = \gamma =1 $$ and $$ M=1, M' =5 $$ for Example 2. Fig. 8. View largeDownload slide The control function for (a) $$ \alpha = \frac{1}{5}, $$ (b) $$ \alpha = \frac{1}{2}, $$ (c) $$ \alpha = 1 $$ and (d) $$ \alpha = 2, $$ with $$ \nu = \gamma =1 $$ and $$ M=1, M' =5 $$ for Example 2. Fig. 8. View largeDownload slide The control function for (a) $$ \alpha = \frac{1}{5}, $$ (b) $$ \alpha = \frac{1}{2}, $$ (c) $$ \alpha = 1 $$ and (d) $$ \alpha = 2, $$ with $$ \nu = \gamma =1 $$ and $$ M=1, M' =5 $$ for Example 2. Fig. 9. View largeDownload slide (a) The state function and (b) the control input for $$ \nu =\gamma =0.8, \alpha = 1$$ with $$ M=1, M' =5 $$ for Example 2. Fig. 9. View largeDownload slide (a) The state function and (b) the control input for $$ \nu =\gamma =0.8, \alpha = 1$$ with $$ M=1, M' =5 $$ for Example 2. Table 5 Comparison of the estimated values of $$ \tilde{J} $$ for integer orders $$ \nu = \gamma =1 $$ and $$ \alpha =1 $$ with different values of $$ M $$ and $$ M' $$ for Example 2 Method of Mamehrashi & Yousefi, 2016 $$\tilde{J}$$ $$M=6, M'=6 $$ $$ 2.03080\times 10^{6} $$ $$M=7, M'= 6 $$ $$ 1.82721\times 10^{6}$$ $$M=6, M'= 8 $$ $$ 1.65595\times 10^{6} $$ $$M=7, M'= 7 $$ $$ 1.60773\times 10^{6} $$ $$M=6, M'=9 $$ $$ 1.54056\times 10^{6} $$ $$M=7, M'=8 $$ $$ 1.45170\times 10^{6} $$ $$M=7, M'=9 $$ $$ 1.33534\times 10^{6} $$ $$M=8, M'= 8 $$ $$ 1.30907\times 10^{6} $$ $$Present \; method $$ $$\tilde{J} $$ $$M=1, M' =1 $$ $$1.5407\times 10^{-24} $$ $$M=1, M' =2 $$ $$ 1.1707\times 10^{-23}$$ $$M=1, M' =4 $$ $$ 8.9944\times 10^{-23}$$ $$M=1, M' =5 $$ $$1.2681\times 10^{-22}$$ $$M=1, M' =8 $$ $$1.3831\times 10^{-21}$$ $$M=1, M' =10 $$ $$1.0888\times 10^{-21}$$ $$M=1, M' =12 $$ $$1.0542\times 10^{-21}$$ Method of Mamehrashi & Yousefi, 2016 $$\tilde{J}$$ $$M=6, M'=6 $$ $$ 2.03080\times 10^{6} $$ $$M=7, M'= 6 $$ $$ 1.82721\times 10^{6}$$ $$M=6, M'= 8 $$ $$ 1.65595\times 10^{6} $$ $$M=7, M'= 7 $$ $$ 1.60773\times 10^{6} $$ $$M=6, M'=9 $$ $$ 1.54056\times 10^{6} $$ $$M=7, M'=8 $$ $$ 1.45170\times 10^{6} $$ $$M=7, M'=9 $$ $$ 1.33534\times 10^{6} $$ $$M=8, M'= 8 $$ $$ 1.30907\times 10^{6} $$ $$Present \; method $$ $$\tilde{J} $$ $$M=1, M' =1 $$ $$1.5407\times 10^{-24} $$ $$M=1, M' =2 $$ $$ 1.1707\times 10^{-23}$$ $$M=1, M' =4 $$ $$ 8.9944\times 10^{-23}$$ $$M=1, M' =5 $$ $$1.2681\times 10^{-22}$$ $$M=1, M' =8 $$ $$1.3831\times 10^{-21}$$ $$M=1, M' =10 $$ $$1.0888\times 10^{-21}$$ $$M=1, M' =12 $$ $$1.0542\times 10^{-21}$$ Table 5 Comparison of the estimated values of $$ \tilde{J} $$ for integer orders $$ \nu = \gamma =1 $$ and $$ \alpha =1 $$ with different values of $$ M $$ and $$ M' $$ for Example 2 Method of Mamehrashi & Yousefi, 2016 $$\tilde{J}$$ $$M=6, M'=6 $$ $$ 2.03080\times 10^{6} $$ $$M=7, M'= 6 $$ $$ 1.82721\times 10^{6}$$ $$M=6, M'= 8 $$ $$ 1.65595\times 10^{6} $$ $$M=7, M'= 7 $$ $$ 1.60773\times 10^{6} $$ $$M=6, M'=9 $$ $$ 1.54056\times 10^{6} $$ $$M=7, M'=8 $$ $$ 1.45170\times 10^{6} $$ $$M=7, M'=9 $$ $$ 1.33534\times 10^{6} $$ $$M=8, M'= 8 $$ $$ 1.30907\times 10^{6} $$ $$Present \; method $$ $$\tilde{J} $$ $$M=1, M' =1 $$ $$1.5407\times 10^{-24} $$ $$M=1, M' =2 $$ $$ 1.1707\times 10^{-23}$$ $$M=1, M' =4 $$ $$ 8.9944\times 10^{-23}$$ $$M=1, M' =5 $$ $$1.2681\times 10^{-22}$$ $$M=1, M' =8 $$ $$1.3831\times 10^{-21}$$ $$M=1, M' =10 $$ $$1.0888\times 10^{-21}$$ $$M=1, M' =12 $$ $$1.0542\times 10^{-21}$$ Method of Mamehrashi & Yousefi, 2016 $$\tilde{J}$$ $$M=6, M'=6 $$ $$ 2.03080\times 10^{6} $$ $$M=7, M'= 6 $$ $$ 1.82721\times 10^{6}$$ $$M=6, M'= 8 $$ $$ 1.65595\times 10^{6} $$ $$M=7, M'= 7 $$ $$ 1.60773\times 10^{6} $$ $$M=6, M'=9 $$ $$ 1.54056\times 10^{6} $$ $$M=7, M'=8 $$ $$ 1.45170\times 10^{6} $$ $$M=7, M'=9 $$ $$ 1.33534\times 10^{6} $$ $$M=8, M'= 8 $$ $$ 1.30907\times 10^{6} $$ $$Present \; method $$ $$\tilde{J} $$ $$M=1, M' =1 $$ $$1.5407\times 10^{-24} $$ $$M=1, M' =2 $$ $$ 1.1707\times 10^{-23}$$ $$M=1, M' =4 $$ $$ 8.9944\times 10^{-23}$$ $$M=1, M' =5 $$ $$1.2681\times 10^{-22}$$ $$M=1, M' =8 $$ $$1.3831\times 10^{-21}$$ $$M=1, M' =10 $$ $$1.0888\times 10^{-21}$$ $$M=1, M' =12 $$ $$1.0542\times 10^{-21}$$ Table 6 Comparison of the estimated values of $$ \tilde{J} $$ for integer orders $$ \nu =\gamma =1 $$ with different values of $$ \alpha $$ for Example 2 $$M, M'$$ $$\alpha = \frac{1}{2}$$ $$\alpha = \frac{1}{4}$$ $$\alpha = 1$$ $$\alpha = 2$$ $$M=1, M' =2 $$ $$1.2907 \times 10^{-29}$$ $$1.1500\times 10^{-23}$$ $$1.1707\times 10^{-23}$$ $$4.6001 \times 10^{-23}$$ $$M=1, M' =4$$ $$2.3121\times 10^{-22}$$ $$1.4715\times 10^{-20}$$ $$8.9944\times 10^{-23}$$ $$9.3637 \times 10^{8}$$ $$M=1, M' =5$$ $$1.5279\times 10^{-23}$$ $$6.6134\times 10^{-20}$$ $$1.2681\times 10^{-22}$$ $$1.1247 \times 10^{9}$$ $$M=1, M' =8$$ $$3.9518\times 10^{-22}$$ $$1.5329 \times 10^{9}$$ $$1.3831 \times 10^{-21}$$ $$1.5329 \times 10^{9}$$ $$M=1, M' =10$$ $$1.7069 \times 10^{-21}$$ $$1.8313\times 10^{-4}$$ $$1.0888 \times 10^{-21}$$ $$1.7312\times 10^{9}$$ $$M=1, M' =12$$ $$5.4848\times 10^{-21}$$ $$1.0738\times 10^{-5}$$ $$1.0542 \times 10^{-21}$$ $$1.8949 \times 10^{9}$$ $$M, M'$$ $$\alpha = \frac{1}{2}$$ $$\alpha = \frac{1}{4}$$ $$\alpha = 1$$ $$\alpha = 2$$ $$M=1, M' =2 $$ $$1.2907 \times 10^{-29}$$ $$1.1500\times 10^{-23}$$ $$1.1707\times 10^{-23}$$ $$4.6001 \times 10^{-23}$$ $$M=1, M' =4$$ $$2.3121\times 10^{-22}$$ $$1.4715\times 10^{-20}$$ $$8.9944\times 10^{-23}$$ $$9.3637 \times 10^{8}$$ $$M=1, M' =5$$ $$1.5279\times 10^{-23}$$ $$6.6134\times 10^{-20}$$ $$1.2681\times 10^{-22}$$ $$1.1247 \times 10^{9}$$ $$M=1, M' =8$$ $$3.9518\times 10^{-22}$$ $$1.5329 \times 10^{9}$$ $$1.3831 \times 10^{-21}$$ $$1.5329 \times 10^{9}$$ $$M=1, M' =10$$ $$1.7069 \times 10^{-21}$$ $$1.8313\times 10^{-4}$$ $$1.0888 \times 10^{-21}$$ $$1.7312\times 10^{9}$$ $$M=1, M' =12$$ $$5.4848\times 10^{-21}$$ $$1.0738\times 10^{-5}$$ $$1.0542 \times 10^{-21}$$ $$1.8949 \times 10^{9}$$ Table 6 Comparison of the estimated values of $$ \tilde{J} $$ for integer orders $$ \nu =\gamma =1 $$ with different values of $$ \alpha $$ for Example 2 $$M, M'$$ $$\alpha = \frac{1}{2}$$ $$\alpha = \frac{1}{4}$$ $$\alpha = 1$$ $$\alpha = 2$$ $$M=1, M' =2 $$ $$1.2907 \times 10^{-29}$$ $$1.1500\times 10^{-23}$$ $$1.1707\times 10^{-23}$$ $$4.6001 \times 10^{-23}$$ $$M=1, M' =4$$ $$2.3121\times 10^{-22}$$ $$1.4715\times 10^{-20}$$ $$8.9944\times 10^{-23}$$ $$9.3637 \times 10^{8}$$ $$M=1, M' =5$$ $$1.5279\times 10^{-23}$$ $$6.6134\times 10^{-20}$$ $$1.2681\times 10^{-22}$$ $$1.1247 \times 10^{9}$$ $$M=1, M' =8$$ $$3.9518\times 10^{-22}$$ $$1.5329 \times 10^{9}$$ $$1.3831 \times 10^{-21}$$ $$1.5329 \times 10^{9}$$ $$M=1, M' =10$$ $$1.7069 \times 10^{-21}$$ $$1.8313\times 10^{-4}$$ $$1.0888 \times 10^{-21}$$ $$1.7312\times 10^{9}$$ $$M=1, M' =12$$ $$5.4848\times 10^{-21}$$ $$1.0738\times 10^{-5}$$ $$1.0542 \times 10^{-21}$$ $$1.8949 \times 10^{9}$$ $$M, M'$$ $$\alpha = \frac{1}{2}$$ $$\alpha = \frac{1}{4}$$ $$\alpha = 1$$ $$\alpha = 2$$ $$M=1, M' =2 $$ $$1.2907 \times 10^{-29}$$ $$1.1500\times 10^{-23}$$ $$1.1707\times 10^{-23}$$ $$4.6001 \times 10^{-23}$$ $$M=1, M' =4$$ $$2.3121\times 10^{-22}$$ $$1.4715\times 10^{-20}$$ $$8.9944\times 10^{-23}$$ $$9.3637 \times 10^{8}$$ $$M=1, M' =5$$ $$1.5279\times 10^{-23}$$ $$6.6134\times 10^{-20}$$ $$1.2681\times 10^{-22}$$ $$1.1247 \times 10^{9}$$ $$M=1, M' =8$$ $$3.9518\times 10^{-22}$$ $$1.5329 \times 10^{9}$$ $$1.3831 \times 10^{-21}$$ $$1.5329 \times 10^{9}$$ $$M=1, M' =10$$ $$1.7069 \times 10^{-21}$$ $$1.8313\times 10^{-4}$$ $$1.0888 \times 10^{-21}$$ $$1.7312\times 10^{9}$$ $$M=1, M' =12$$ $$5.4848\times 10^{-21}$$ $$1.0738\times 10^{-5}$$ $$1.0542 \times 10^{-21}$$ $$1.8949 \times 10^{9}$$ Table 7 Comparison of the estimated values of $$ \tilde{J} $$ for orders $$ \nu =\gamma =0.8 $$ with $$ \alpha = \frac{1}{2}, 1 $$ for Example 2 $$M, M' $$ $$\alpha = \frac{1}{2} $$ $$ \alpha = 1$$ $$M=1, M' =1 $$ $$ 0 $$ $$ 2.95\times 10^{-30} $$ $$M=1, M' =3 $$ $$ 2.17\times 10^{-25}-1.11\times 10^{-30}i $$ $$ 6.12\times 10^{-24}+6.71\times 10^{-29}i $$ $$M=1, M' =5 $$ $$ 8.59\times 10^{-16}-2.63\times 10^{-24}i $$ $$ 6.71\times 10^{-23}-4.97\times 10^{-31}i $$ $$M=1, M' =7 $$ $$ 2.95\times 10^{-22}+2.55\times 10^{-30}i $$ $$ 5.90\times 10^{-23}-6.50\times 10^{-31}i $$ $$M=1, M' =9 $$ $$ 2.49\times 10^{-21}-5.08\times 10^{-29}i $$ $$ 5.06\times 10^{-22}+2.29\times 10^{-29}i $$ $$M=1, M' =11$$ $$5.84\times 10^{-21}+1.11\times 10^{-29}i $$ $$2.47\times 10^{-23}+1.70\times 10^{-29}i$$ $$M, M' $$ $$\alpha = \frac{1}{2} $$ $$ \alpha = 1$$ $$M=1, M' =1 $$ $$ 0 $$ $$ 2.95\times 10^{-30} $$ $$M=1, M' =3 $$ $$ 2.17\times 10^{-25}-1.11\times 10^{-30}i $$ $$ 6.12\times 10^{-24}+6.71\times 10^{-29}i $$ $$M=1, M' =5 $$ $$ 8.59\times 10^{-16}-2.63\times 10^{-24}i $$ $$ 6.71\times 10^{-23}-4.97\times 10^{-31}i $$ $$M=1, M' =7 $$ $$ 2.95\times 10^{-22}+2.55\times 10^{-30}i $$ $$ 5.90\times 10^{-23}-6.50\times 10^{-31}i $$ $$M=1, M' =9 $$ $$ 2.49\times 10^{-21}-5.08\times 10^{-29}i $$ $$ 5.06\times 10^{-22}+2.29\times 10^{-29}i $$ $$M=1, M' =11$$ $$5.84\times 10^{-21}+1.11\times 10^{-29}i $$ $$2.47\times 10^{-23}+1.70\times 10^{-29}i$$ Table 7 Comparison of the estimated values of $$ \tilde{J} $$ for orders $$ \nu =\gamma =0.8 $$ with $$ \alpha = \frac{1}{2}, 1 $$ for Example 2 $$M, M' $$ $$\alpha = \frac{1}{2} $$ $$ \alpha = 1$$ $$M=1, M' =1 $$ $$ 0 $$ $$ 2.95\times 10^{-30} $$ $$M=1, M' =3 $$ $$ 2.17\times 10^{-25}-1.11\times 10^{-30}i $$ $$ 6.12\times 10^{-24}+6.71\times 10^{-29}i $$ $$M=1, M' =5 $$ $$ 8.59\times 10^{-16}-2.63\times 10^{-24}i $$ $$ 6.71\times 10^{-23}-4.97\times 10^{-31}i $$ $$M=1, M' =7 $$ $$ 2.95\times 10^{-22}+2.55\times 10^{-30}i $$ $$ 5.90\times 10^{-23}-6.50\times 10^{-31}i $$ $$M=1, M' =9 $$ $$ 2.49\times 10^{-21}-5.08\times 10^{-29}i $$ $$ 5.06\times 10^{-22}+2.29\times 10^{-29}i $$ $$M=1, M' =11$$ $$5.84\times 10^{-21}+1.11\times 10^{-29}i $$ $$2.47\times 10^{-23}+1.70\times 10^{-29}i$$ $$M, M' $$ $$\alpha = \frac{1}{2} $$ $$ \alpha = 1$$ $$M=1, M' =1 $$ $$ 0 $$ $$ 2.95\times 10^{-30} $$ $$M=1, M' =3 $$ $$ 2.17\times 10^{-25}-1.11\times 10^{-30}i $$ $$ 6.12\times 10^{-24}+6.71\times 10^{-29}i $$ $$M=1, M' =5 $$ $$ 8.59\times 10^{-16}-2.63\times 10^{-24}i $$ $$ 6.71\times 10^{-23}-4.97\times 10^{-31}i $$ $$M=1, M' =7 $$ $$ 2.95\times 10^{-22}+2.55\times 10^{-30}i $$ $$ 5.90\times 10^{-23}-6.50\times 10^{-31}i $$ $$M=1, M' =9 $$ $$ 2.49\times 10^{-21}-5.08\times 10^{-29}i $$ $$ 5.06\times 10^{-22}+2.29\times 10^{-29}i $$ $$M=1, M' =11$$ $$5.84\times 10^{-21}+1.11\times 10^{-29}i $$ $$2.47\times 10^{-23}+1.70\times 10^{-29}i$$ Table 8 Comparison of the estimated values of $$ \tilde{J} $$ for orders $$ \nu =0.75, \gamma =0.8 $$ with $$ \alpha =\frac{1}{2}, 1 $$ for Example 2 $$M, M'$$ $$ \alpha = \frac{1}{2} $$ $$ \alpha = 1$$ $$M=1, M' =1 $$ $$ 3.70\times 10^{-22} $$ $$ 2.17 \times 10^{-21} $$ $$M=1, M' =3 $$ $$ 9.02\times 10^{-18}+3.28\times 10^{-25}i $$ $$ 3.31\times 10^{-23}-7.25\times 10^{-31}i $$ $$M=1, M' =5 $$ $$ 4.78\times 10^{-22}+2.88\times 10^{-29}i $$ $$ 1.69\times 10^{-23}+1.92\times 10^{-29}i $$ $$M=1, M' =7 $$ $$ 5.76\times 10^{-22}+1.18\times 10^{-29}i $$ $$ 2.00\times 10^{-22}-3.66\times 10^{-38}i $$ $$M=1, M' =9 $$ $$ 3.66\times 10^{-22}-7.02\times 10^{-29}i $$ $$ 2.91\times 10^{-22}+4.23\times 10^{-29}i $$ $$M=1, M' =11$$ $$3.90\times 10^{-21}-3.25\times 10^{-31}i$$ $$1.68 \times 10^{-22}-3.84\times 10^{-30}i$$ $$M, M'$$ $$ \alpha = \frac{1}{2} $$ $$ \alpha = 1$$ $$M=1, M' =1 $$ $$ 3.70\times 10^{-22} $$ $$ 2.17 \times 10^{-21} $$ $$M=1, M' =3 $$ $$ 9.02\times 10^{-18}+3.28\times 10^{-25}i $$ $$ 3.31\times 10^{-23}-7.25\times 10^{-31}i $$ $$M=1, M' =5 $$ $$ 4.78\times 10^{-22}+2.88\times 10^{-29}i $$ $$ 1.69\times 10^{-23}+1.92\times 10^{-29}i $$ $$M=1, M' =7 $$ $$ 5.76\times 10^{-22}+1.18\times 10^{-29}i $$ $$ 2.00\times 10^{-22}-3.66\times 10^{-38}i $$ $$M=1, M' =9 $$ $$ 3.66\times 10^{-22}-7.02\times 10^{-29}i $$ $$ 2.91\times 10^{-22}+4.23\times 10^{-29}i $$ $$M=1, M' =11$$ $$3.90\times 10^{-21}-3.25\times 10^{-31}i$$ $$1.68 \times 10^{-22}-3.84\times 10^{-30}i$$ Table 8 Comparison of the estimated values of $$ \tilde{J} $$ for orders $$ \nu =0.75, \gamma =0.8 $$ with $$ \alpha =\frac{1}{2}, 1 $$ for Example 2 $$M, M'$$ $$ \alpha = \frac{1}{2} $$ $$ \alpha = 1$$ $$M=1, M' =1 $$ $$ 3.70\times 10^{-22} $$ $$ 2.17 \times 10^{-21} $$ $$M=1, M' =3 $$ $$ 9.02\times 10^{-18}+3.28\times 10^{-25}i $$ $$ 3.31\times 10^{-23}-7.25\times 10^{-31}i $$ $$M=1, M' =5 $$ $$ 4.78\times 10^{-22}+2.88\times 10^{-29}i $$ $$ 1.69\times 10^{-23}+1.92\times 10^{-29}i $$ $$M=1, M' =7 $$ $$ 5.76\times 10^{-22}+1.18\times 10^{-29}i $$ $$ 2.00\times 10^{-22}-3.66\times 10^{-38}i $$ $$M=1, M' =9 $$ $$ 3.66\times 10^{-22}-7.02\times 10^{-29}i $$ $$ 2.91\times 10^{-22}+4.23\times 10^{-29}i $$ $$M=1, M' =11$$ $$3.90\times 10^{-21}-3.25\times 10^{-31}i$$ $$1.68 \times 10^{-22}-3.84\times 10^{-30}i$$ $$M, M'$$ $$ \alpha = \frac{1}{2} $$ $$ \alpha = 1$$ $$M=1, M' =1 $$ $$ 3.70\times 10^{-22} $$ $$ 2.17 \times 10^{-21} $$ $$M=1, M' =3 $$ $$ 9.02\times 10^{-18}+3.28\times 10^{-25}i $$ $$ 3.31\times 10^{-23}-7.25\times 10^{-31}i $$ $$M=1, M' =5 $$ $$ 4.78\times 10^{-22}+2.88\times 10^{-29}i $$ $$ 1.69\times 10^{-23}+1.92\times 10^{-29}i $$ $$M=1, M' =7 $$ $$ 5.76\times 10^{-22}+1.18\times 10^{-29}i $$ $$ 2.00\times 10^{-22}-3.66\times 10^{-38}i $$ $$M=1, M' =9 $$ $$ 3.66\times 10^{-22}-7.02\times 10^{-29}i $$ $$ 2.91\times 10^{-22}+4.23\times 10^{-29}i $$ $$M=1, M' =11$$ $$3.90\times 10^{-21}-3.25\times 10^{-31}i$$ $$1.68 \times 10^{-22}-3.84\times 10^{-30}i$$ 7. Conclusion In the present work, a new scheme for two-dimensional FOCPs is introduced. In this method, general formulations for the Riemann–Liouville fractional integral operators of generalized fractional-order Bernoulli and generalized fractional-order Legendre have been derived. First the given problem is transformed into an equivalent variational problem, then the variational problem is solved approximately by utilizing the generalized fractional-order Bernoulli–Legendre functions, Riemann-Liouville fractional integral operators, two-dimensional LG quadrature rule and Newton’s iterative formula for solving the non-linear system of equations. Illustrative examples are given to demonstrate the validity and applicability of the proposed method. Moreover, only a small number of generalized fractional-order Bernoulli–Legendre functions is needed to obtain a satisfactory result. 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Google Scholar CrossRef Search ADS © The authors 2017. Published by Oxford University Press on behalf of the Institute of Mathematics and its Applications. All rights reserved. http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png IMA Journal of Mathematical Control and Information Oxford University Press

Generalized fractional-order Bernoulli–Legendre functions: an effective tool for solving two-dimensional fractional optimal control problems

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Abstract

Abstract In this article, we present a new method to solve a class of two-dimensional fractional optimal control problems. The fractional derivative is defined in the Caputo sense. Our approach is based upon to approximate the state and control functions by the elements of generalized fractional-order Bernoulli functions in space and generalized fractional-order Legendre functions in time with unknown coefficients. First the properties of these basis functions are presented. Second, the Riemann–Liouville fractional integral operators of generalized fractional-order Bernoulli–Legendre functions are proposed. Then we apply two-dimensional Legendre–Gauss quadrature rule to approximate double integral of the performance index functional. Next, the problem is converted into an equivalent non-linear unconstrained optimization problem. This problem is solved via the Newton’s iterative method. Finally, convergence of the proposed method is extensively investigated and two examples are included to demonstrate the validity and applicability of the presented new technique. 1. Introduction Fractional differential equations (FDEs) are generalized from integer order ones, which are obtained by replacing integer order derivatives by fractional order ones. In comparison with integer order differential equations, the fractional differential equations show many advantages over the simulation of problems in system biology (Yuste & Lindenberg, 2001), physics (Barkai et al., 2000), hydrology (Benson et al., 2000), chemistry and biochemistry (Yuste et al., 2004) and finance (Gorenflo et al., 2001). These new fractional-order models are more adequate than the previously used integer-order models, because fractional-order derivatives and integrals enable the description of the memory and hereditary properties of different substances (Podlubny, 1999). This is the most significant advantage of the fractional-order models in comparison with integer-order models, in which such effects are neglected. In the area of physics, the left and right space fractional derivatives allow the modelling of flow regime impacts from either side of the domain (Zaslavsky, 2002). Space fractional partial differential equations are used to model anomalous diffusive and super-diffusive systems, where a particle plume spread faster than the classical Brownian motion model predicts (Metzler & Klafter, 2000). The fractional advectiondispersion equation is used in groundwater hydrology to model the transport of passive tracers carried by fluid flow in a porous medium (Liu et al., 2004). It is known that some dynamical processes in gas absorption, water steam heating and air drying can be described by the Darboux equation (Marszalek, 1984). Therefore, constructing new analytical and numerical approaches for solving different types of fractional differential equations has become a strong topic to be considered, these approaches include the finite element method (Ma et al., 2014), the wavelet method (Rahimkhani et al., 2017a), the spectral tau method (Bhrawy et al., 2015a), the Gegenbauer spectral method (Izadkhah & Saber-Nadjafi, 2015), the iterative method (Daftardar-Gejji Bhalekar, 2010), the fractional-order wavelet method (Rahimkhani et al., 2016b, 2017b). The general definition of an optimal control problem refers to the minimization of a functional on a set of control and state variables (the performance index) subject to dynamic constraints on the states and controls. In case of there are fractional differential equations that used as the dynamic constraints, this leads to the fractional optimal control problem (FOCP). However very little work has been done in the area of fractional optimal control, FOCPs have gained much attention for their many applications in engineering and physics (Zamani et al., 2007). For that reason, finding accurate numerical methods for solving various types of FOCPs has become an active research undertaking. A general formulation and a solution scheme for FOCPs were first introduced in (Agrawal, 2004) where fractional derivatives were introduced in the Riemann–Liouville sense and FOCP formulation was expressed using the fractional variational principle and the Lagrange multiplier technique. In Agrawal (2008a,b), the FOCPs are formulated using the definition of fractional derivatives in the sense of Caputo, the FDEs are substituted into Volterra-type integral equations and a direct linear solver helps calculating the solution of the obtained algebraic equations. In Agrawal & Baleanu (2007) and Baleanu & Trujillo (2010), general necessary conditions of optimality have been derived for FOCPs via the Hamiltonian formulas with respect to the Riemann–Liouville fractional derivative and the Caputo fractional derivative. In Rabiei et al. (2016), the authors used Boubaker polynomials for solving FOCPs. Doha et al. (2015) used the Rayleigh–Ritz method with the operational matrix of fractional integrals based on the shifted Jacobi orthonormal polynomials for solving a FOCP in a general form. In Bhrawy et al. (2015b), the authors used the shifted Legendre orthonormal polynomials as basis functions of the Lagrange multiplier method for approximating the solution of different types of FOCPs. Rahimkhani et al. (2016a) introduced an efficient method based on the Bernoulli wavelets for solving delay FOCPs. Frederico & Torres (2006, 2008) formulated a Noether-type theorem in the general context of the fractional optimal control in the sense of Caputo and studied fractional conservation laws in FOCPs. Bhrawy & Ezz-Eldien (2016) derived new operational matrices for the shifted Legendre orthonormal polynomial for solving delay FOCPs. Nemati & Yousefi (2016) presented a numerical scheme for solving two-dimensional FOCPs by the Ritz method combined with fractional operational matrix. In this article, we intend to obtain a new numerical approach for approximating the solution of the two-dimensional FOCP \begin{equation} min\hspace{.3cm} J[\xi , u] = \int _{0}^{h_{2}}\int _{0}^{h_{1}}G \left(x, t, \xi (x, t), u(x, t), \frac{\partial \xi}{\partial x}(x, t), \frac{\partial \xi}{\partial t}(x, t)\right) dx dt, \label{Z25} \end{equation} (1.1) subject to \begin{equation} \frac{\partial ^{2}\xi}{\partial x \partial t}(x, t)= a \frac{\partial ^{\nu}\xi}{\partial x^{\nu}}(x, t) + b \frac{\partial ^{\gamma}\xi}{\partial t^{\gamma}}(x, t) + c \xi(x, t) + d u(x, t),\label{Z26} \end{equation} (1.2) with the following initial and boundary conditions \begin{equation} \xi (x, 0) = \zeta (x), \hspace{1cm} \xi (0, t) = \eta (t),\label{Z27} \end{equation} (1.3) where $$ 0 < \nu , \gamma \leq 1 $$ and the functions $$ G $$ and $$ \xi $$ are smooth and $$ d $$ is non-zero. In recent years a special attention has been given to applications of the operational matrices Walsh functions (Ordokhani, 2010), B-spline functions (Lakestani et al., 2012), Legendre polynomials (Saadatmandi & Dehghan, 2010; Nemati et al., 2013), Chebyshev polynomials (Doha et al., 2011), Bernstein polynomials (Saadatmandi, 2011), Legendre wavelet (Heydari et al., 2014) and Bernoulli wavelet (Rahimkhani et al., 2017a). Moreover, there are recent works on the operational matrices of fractional-order Legendre functions (Kazem et al., 2013; Chen et al., 2014), fractional-order Bernstein functions (Yuzbasi, 2013), fractional-order generalized Laguerre functions (Bhrawy et al., 2014) and fractional-order Bernoulli wavelet (Rahimkhani et al., 2016b, 2017b). In this article, we introduce a different approach for solving two-dimensional FOCP in (1.1)–(1.3). First, we transform the FOCP into an equivalent variational problem. Then we expand function $$ \frac{\partial ^{2}\xi}{\partial x \partial t}(x, t)$$ with generalized fractional-order Bernoulli–Legendre functions and the unknown coefficients. These generalized fractional-order Bernoulli–Legendre functions which consist of generalized fractional-order Bernoulli (GFBFs) and generalized fractional-order Legendre functions (GFLFs) are given. The Riemann–Liouville fractional integral operators for these functions are derived. By estimating double integral of the cost function via two-dimensional Legendre–Gauss (LG) quadrature rule, a common unconstrained optimization problem is obtained. After taking the necessary conditions of optimality into account, the Newton’s iterative method is used to find the solution of this problem. The proposed method is very convenient for solving such problems, since the initial and boundary conditions are taken into account automatically. Numerical results demonstrate the efficiency of the proposed method in solving FOCPs. The article is organized as follows. In Section 2, some necessary definitions of fractional calculus, generalized fractional-order Bernoulli, generalized fractional-order Legendre and their properties are defined. Riemann–Liouville fractional integral operators for GFBFs and GFLFs are introduced in Section 3. In Section 4, the numerical method for solving the under studying problems is presented. The convergence of the proposed method is discussed in Section 5. In Section 6, we report our numerical findings and demonstrate the accuracy of the proposed numerical scheme by considering two numerical examples. Finally, a conclusion is given in Section 7. 2. Preliminaries and notation In this section, we present some basic definitions needed in the remaining part of the article. We briefly recall fractional derivative and integral and some properties of GFBFs and GFLFs. 2.1. The fractional derivative and integral There are various definitions of fractional derivative and integration. The widely used definition of a fractional derivative is the Caputo definition, and a fractional integration is the Riemann–Liouville definition. Definition 2.1 The integral of order $$ \nu \geq 0 $$ (fractional) according to Riemann–Liouville is given by Rahimkhani et al. (2016b) \begin{equation} I^{\nu}f(x)=\left\{ \begin{array}{ll} \frac{1}{{\it{\Gamma}} (\nu)}\int _{0}^x \frac{f(s)}{(x-s)^{1-\nu}}ds=\frac{1}{{\it{\Gamma}} (\nu)}x^{\nu -1}\ast f(x), & \nu >0, x>0,\\ f(x) , \ \ \ \ \ \ \ \ \ \ \ \ & \nu =0,\label{E1} \end{array} \right. \end{equation} (2.1) where $$x^{\nu -1}\ast f(x)$$ is the convolution product of $$x^{\nu -1} $$ and $$f(x)$$. The operator $$ I^{\nu} $$ satisfies the following properties 1. $$I^{\nu}I^{\mu}f(x) = I^{\nu +\mu}f(x),$$ 2. $$I^{\nu }(\lambda f(x)+ \gamma g(x)) =\lambda I^{\nu }f(x)+ \gamma I^{\nu}g(x),$$ 3. $$I^{\nu }x^{\beta} = \frac{{\it{\Gamma}} (\beta +1)}{{\it{\Gamma}} (\beta +\nu +1)}x^{\nu +\beta}.$$ Definition 2.2 Caputo’s fractional derivative of order $$ \nu $$ is defined as (Rahimkhani et al., 2016b) \begin{equation} D^{\nu}f(x)=\frac{1}{{\it{\Gamma}} (n-\nu)}\int _{0}^{x}\frac{f^{(n)}(s)}{(x-s)^{\nu +1-n}}ds,\hspace{.5cm} n-1 <\nu \leq n, n\in N. \end{equation} (2.2) The operator $$ D^{\nu} $$ satisfies the following properties 1. $$ D^{\nu }I^{\nu}f(x)=f(x),$$ 2. $$ I^{\nu }D^{\nu}f(x)=f(x)-\sum _{i=0}^{n-1}f^{(i)}(0)\frac{x^{i}}{i!},$$ 3. $$ D^{\nu }(\lambda f(x)+\gamma g(x)) =\lambda D^{\nu}f(x)+ \gamma D^{\nu}g(x),$$ 4. $D^{\nu }x^{\beta} = \left\{\begin{array}{ll} 0, & \nu \in N_{0},\hspace{.1cm} \beta <\nu ,\\ \frac{{\it{\Gamma}} (\beta +1)}{{\it{\Gamma}} (\beta +1- \nu )}x^{\beta - \nu }, & \textrm{otherwise}, \end{array} \right. $ 5. $$D^{\nu } \lambda =0,$$ where $$ \lambda $$ is constant. Definition 2.3 The partial Caputo fractional derivative of order $$ \nu >0 $$ of the function $$ u(x, t) $$ with respect to its variables $$ x $$ and $$ t $$ are defined as follows, respectively (Nemati & Yousefi, 2016) \begin{equation} D^{\nu}_{x}u(x, t) = \frac{\partial ^{\nu}u(x, t)}{\partial x^{\nu}}= \frac{1}{{\it{\Gamma}} (n- \nu)}\int _{0}^{x}(x- \xi )^{n-\nu -1}\frac{\partial ^{n}u(\xi , t)}{\partial \xi ^{n}}d\xi , \hspace{.5cm}n-1 < \nu \leq n \end{equation} (2.3) and \begin{equation} D^{\nu}_{t}u(x, t) = \frac{\partial ^{\nu}u(x, t)}{\partial t^{\nu}}= \frac{1}{{\it{\Gamma}} (n- \nu)}\int _{0}^{t}(t- \theta)^{n-\nu -1}\frac{\partial ^{n}u(x , \theta)}{\partial \theta ^{n}}d\theta , \hspace{.5cm}n-1 < \nu \leq n. \end{equation} (2.4) Theorem 2.1 Let $$ \nu \in R, n-1 < \nu \leq n, n \in N $$ and $$ \lambda \in C. $$ Then the Caputo fractional derivative of the exponential and trigonometric functions are as follows (Nemati & Yousefi, 2016) \begin{gather*} D^{\nu}e^{\lambda x} = \lambda ^{n}x^{n- \nu} E_{1, n-\nu +1}(\lambda x),\\ D^{\nu}sin \lambda x = -\frac{1}{2}i(i \lambda) ^{n}x^{n-\nu} (E _{1, n- \nu +1}(i\lambda x) - (-1)^{n}E _{1, n- \nu +1}(- i\lambda x)),\\ D^{\nu}cos \lambda x = \frac{1}{2}(i \lambda) ^{n}x^{n-\nu} (E _{1, n- \nu +1}(i\lambda x) + (-1)^{n}E _{1, n- \nu +1}(- i\lambda x)), \end{gather*} where $$ E_{\nu , \gamma}(z) = \sum _{k=0}^{\infty} \frac{z^{k}}{{\it{\Gamma}} (\nu k + \gamma)} $$ is the two-parameter Mittag–Leffler function. 2.2. Generalized fractional-order Bernoulli functions The GFBFs are proposed by Rahimkhani et al. (2017b). These functions are constructed by change of variable $$ x $$ to $$(x/h)^{\alpha} $$, ($$ \alpha > 0 $$), on the Bernoulli polynomials. Let the GFBFs $$\beta_{m} ((x/h)^{\alpha}) $$ be denoted by $$\beta_{m} ^{h\alpha}(x)$$. The analytic form of $$ \beta_{m}^{h\alpha}(x) $$ of order $$ m\alpha $$, is given by \begin{equation} \beta_{m}^{h \alpha}(x)=\sum _{i=0}^{m} \left( {\begin{array}{*{5}c} m\\ i \\ \end{array}} \right)\frac{\beta_{m-i}^{\alpha}}{h^{i\alpha}} x^{i\alpha},\hspace{1cm}0 \leq x \leq h,\label{E11} \end{equation} (2.5) where $$\beta^{\alpha}_{i} :=\beta^{\alpha}_{i}(0) =\beta_{i},\hspace{.1cm} i= 0, 1, ..., m, $$ are Bernoulli numbers. Thus, the first four such functions are \begin{align*} \beta^{h\alpha}_{0}(x)&=1,\\ \beta^{h\alpha}_{1}(x)&= (x/h)^{\alpha}-\frac{1}{2},\\ \beta^{h\alpha}_{2}(x)&=(x/h)^{2\alpha}- (x/h)^{\alpha}+\frac{1}{6},\\ \beta^{h\alpha}_{3}(x)&=(x/h)^{3\alpha}-\frac{3}{2}(x/h)^{2\alpha}+\frac{1}{2}(x/h)^{\alpha}. \end{align*} These functions satisfy the following formula (Rahimkhani et al., 2017b) \begin{equation} \int _{0}^{h}\beta ^{h \alpha}_{n}(x)\beta ^{h \alpha}_{m}(x)x^{\alpha - 1}dx = \frac{h^{\alpha}}{\alpha} (-1)^{n-1}\frac{m!n!}{(m+n)!}\beta_{m+n}^{\alpha},\hspace{.5cm}m, n \geq 1.\label{M12} \end{equation} (2.6) Also, GFBFs form a complete basis over the interval [0, h] (Rahimkhani et al., 2017b). 2.3. Generalized fractional-order Legendre functions The GFLFs are proposed by Chen et al. (2014). These functions are defined by change of variable $$ t $$ to $$(t/h)^{\alpha} $$, ($$ \alpha > 0 $$), based on the shifted Legendre polynomials and are denoted by $$ L_{m}^{h\alpha}(t), \hspace{.2cm}m=0, 1, 2, \ldots $$. The GFLFs $$ L_{m}^{h\alpha}(t) $$ satisfy the following recursive formula \begin{align*} L_{m+1}^{h \alpha}(t) & = \frac{(2m+1)(2(t/h)^{\alpha}-1)}{m+1} L_{m}^{h \alpha}(t) - \frac{m}{m+1} L_{m-1}^{h \alpha}(t), \hspace{1cm}m=1, 2, \ldots,\\ L_{0}^{h \alpha}(t) & =1, \hspace{1cm} L_{1}^{h\alpha}(t) = 2(t/h)^{\alpha} - 1. \end{align*} The analytic form of $$ L_{m}^{h\alpha}(t) $$ of degree $$ m\alpha $$ given by \begin{equation} L_{m}^{h \alpha}(t) = \sum _{i=0}^{m}\frac{ b_{i,m}}{h^{i \alpha}}t^{i\alpha}, \hspace{1cm}m=0, 1, 2, \ldots,\label{W4} \end{equation} (2.7) where $$b_{i,m} = \frac{(-1)^{m+i}(m+i)!}{(m-i)!(i!)^{2}},$$ and $$ L_{m}^{h\alpha}(0)=(-1)^{m}, \hspace{.2cm} L_{m}^{h\alpha}(1)= 1.$$ The GFLFs are orthogonal with respect to the weight function $$ \omega (t) = t^{\alpha -1} $$ on the interval $$[0, h]$$, then the orthogonal condition is (Chen et al., 2014) \begin{equation} \int _{0}^{h}L_{n}^{h \alpha}(t) L_{m}^{h \alpha}(t) t^{\alpha -1}dt = \frac{h^{\alpha}}{(2m+1)\alpha}\delta _{nm},\hspace{1cm}m\geq n,\label{W5} \end{equation} (2.8) where $$ \delta _{nm} $$ is the Kronecker function. 2.4. Function approximation An arbitrary function of two variables $$ u(x, t) $$ defined over $$ L^{2}([0, h_{1}]\times [0, h_{2}]) $$, may be expanded into generalized fractional-order Bernoulli–Legendre functions as: \begin{equation} u(x, t) = \sum _{i=0}^{\infty}\sum _{j=0}^{\infty}u_{ij}\beta_{i}^{h_{1} \alpha}(x)L_{j}^{h_{2} \alpha}(t). \end{equation} (2.9) Therefore, one can consider the following truncated series for $$ u(x, t) $$ as: \begin{equation} u(x, t) \simeq \sum _{i=0}^{M}\sum _{j=0}^{M'}u_{ij}\beta_{i}^{h_{1} \alpha}(x)L_{j}^{h_{2} \alpha}(t). \label{Z22} \end{equation} (2.10) Now let $${\it{\Delta}} = [0, h_{1}] \times [0, h_{2}]$$ and $$ \lbrace \beta_{i}^{h_{1} \alpha}(x) L_{j}^{h_{2} \alpha}(t) \rbrace _{i,j=0}^{M, M'} $$ be the set of generalized fractional-order Bernoulli–Legendre functions in $$ L^{2}({\it{\Delta}})$$. We assume \begin{eqnarray*} {\it{\Omega}} _{M, M'}&=& span <\beta_{i}^{h_{1} \alpha}(x) L_{j}^{h_{2} \alpha}(t) > _{0 \leq i \leq M, 0 \leq j \leq M'}\\ &=& span <\beta_{0}^{h_{1} \alpha}(x) L_{0}^{h_{2} \alpha}(t), \ldots , \beta_{M}^{h_{1} \alpha}(x) L_{M'}^{h_{2} \alpha}(t)>\!. \end{eqnarray*} Let $$u(x, t)$$ be an arbitrary function in $$ L^{2}({\it{\Delta}}) $$. Since $${\it{\Omega}} _{M, M'}$$ is closed in the complete space $$ L^{2}({\it{\Delta}}) $$, so it is complete. Therefore, there is the best unique approximation out of $${\it{\Omega}} _{M, M'} $$ like $$ \tilde{u}(x, t) $$ that \begin{align*} \Vert u - \tilde{u} \Vert _{2} \leq \Vert u - f \Vert _{2},\hspace{1cm} \forall f \in {\it{\Omega}} _{M, M'}, \end{align*} where $$ \Vert u \Vert _{2} = \sqrt{< u, u>}$$. Since $$ \tilde{u}(x, t) \in {\it{\Omega}} _{M, M'} $$, so we can find coefficients $$ u_{ij}, i=0, 1, \ldots , M, j=0, 1, \ldots , M' $$ such that \begin{align*} \tilde{u}(x, t) = \sum _{i=0}^{M}\sum _{j=0}^{M'}u_{ij}\beta_{i}^{h_{1} \alpha}(x)L_{j}^{h_{2} \alpha}(t) = B^{h_{1}\alpha ,T}(x) U L^{h_{2}\alpha}(t), \end{align*} where $$T$$ indicates transposition, $$ B^{h_{1}\alpha }(x)$$ and $$ L^{h_{2} \alpha}(t) $$ are column vectors, given by \begin{equation} B^{h_{1}\alpha }(x) = [\beta_{0}^{h_{1} \alpha}(x), \beta_{1}^{h_{1} \alpha}(x), \ldots , \beta_{M}^{h_{1} \alpha}(x) ]^{T} \label{Z23} \end{equation} (2.11) and \begin{equation} L^{h_{2}\alpha }(t) = [L_{0}^{h_{2} \alpha}(t), L_{1}^{h_{2} \alpha}(t), \ldots ,L_{M'}^{h_{2} \alpha}(t) ]^{T}. \label{Z24} \end{equation} (2.12) For example, Fig. 1 shows graphs of $$ u(x, t)= \beta_{2}^{3 \alpha}(x) L_{3}^{3 \alpha}(t) $$ for various values of $$ \alpha $$. To compute double integral of a function, we state two-dimensional LG quadrature rule. Suppose that the LG nodes on the interval $$[a, b]$$ and $$[c, d]$$ are denoted by $$ \delta _{r} $$ and $$ \tau _{s} $$, $$r = 1, \ldots , l, s = 1, \ldots , l'$$, respectively. For the smooth function, $$ \xi (x, t) $$, the two-dimensional LG quadrature rule is constructed as follows (Nemati & Yousefi, 2016) \begin{align*} \int _{c}^{d}\int _{a}^{b} \xi (x, t) dx dt & \simeq \frac{ (b-a) (d-c)}{4}\sum _{r=1}^{l}\sum _{s=1}^{l'} \xi \left(a + (\delta _{r}+1)\frac{b-a}{2} , c+ (\tau _{s} +1)\frac{d-c}{2}\right) \omega _{r} \omega ' _{s}\\ & :=\kappa W^{T}{\it{\Omega}} W', \end{align*} where $$ \kappa =\frac{(b-a)(d-c)}{4} $$ and $$ W, W' $$ are column vectors of Christoffel numbers of size $$ l, l' (\omega _{r}= \frac{2}{(1-\delta _{r})^{2}(\dot{L}_{l}(\delta _{r}))^{2}}, \omega ' _{s}= \frac{2}{(1-\tau _{s})^{2}(\dot {L}_{l}(\tau _{s}))^{2}}) $$, and $$ {\it{\Omega}} $$ is $$ l \times l' $$ matrix in which its entries are defined as follows, \begin{align*} {\it{\Omega}} _{rs}= \xi \left(a + (\delta _{r}+1)\frac{b-a}{2}, c + (\tau_{s}+1)\frac{d-c}{2}\right)\!, \hspace{.2cm}1\leq r \leq l, 1 \leq s \leq l'. \end{align*} Fig. 1. View largeDownload slide Plots of $$ u(x, t)= \beta_{2}^{3 \alpha}(x) L_{3}^{3 \alpha}(t) $$ for (a) $$ \alpha = \frac{1}{4}, $$ (b) $$ \alpha = \frac{1}{2}, $$ (c) $$ \alpha = 1 $$ and (d) $$ \alpha = 2 $$. Fig. 1. View largeDownload slide Plots of $$ u(x, t)= \beta_{2}^{3 \alpha}(x) L_{3}^{3 \alpha}(t) $$ for (a) $$ \alpha = \frac{1}{4}, $$ (b) $$ \alpha = \frac{1}{2}, $$ (c) $$ \alpha = 1 $$ and (d) $$ \alpha = 2 $$. 3. Riemann–Liouville fractional integral operators In this section, we obtain Riemann–Liouville fractional integral operators for GFBFs and GFLFs. 3.1. Riemann–Liouville fractional integral operator for GFBFs We now derive the operator $$ I^{\nu} $$ for $$ B^{h\alpha}(x) $$ in Equation (2.11) given by \begin{equation} I^{\nu}B^{h\alpha}(x)= F^{h\alpha}( \nu , x), \label{M2} \end{equation} (3.1) where \begin{align*} F^{h\alpha}( \nu , x) = [I^{\nu}\beta^{h\alpha} _{0}(x), I^{\nu}\beta^{h\alpha} _{1}(x), I^{\nu}\beta^{h\alpha} _{2}(x), \ldots , I^{\nu}\beta^{h\alpha} _{M}(x) ]^{T}. \end{align*} To obtain $$ I^{\nu}\beta ^{h\alpha} _{m}(x), (m=0, 1, \ldots , M) $$ we use the Laplace transform. By taking the Laplace transform from Equation (2.5), we get \begin{eqnarray} L[\beta ^{h\alpha} _{m}(x)] &= & L\left[\sum _{i=0}^{m} \left( {\begin{array}{*{5}c} m\\ i \\ \end{array}} \right )\frac{ \beta_{m-i}^{\alpha}}{h^{i\alpha}} x^{i\alpha}\right]\nonumber\\ &=& \sum _{i=0}^{m} \left( {\begin{array}{*{5}c} m\\ i \\ \end{array}} \right )\frac{ \beta_{m-i}^{\alpha}}{h^{i\alpha}}L[x^{i\alpha}]\nonumber\\ &=&\sum _{i=0}^{m} \left( {\begin{array}{*{5}c} m\\ i \\ \end{array}} \right )\frac{ \beta_{m-i}^{\alpha}}{h^{i\alpha}} \frac{{\it{\Gamma}} (i \alpha +1)}{s^{i \alpha +1}}. \label{M4} \end{eqnarray} (3.2) From Equation (2.1), we get \begin{eqnarray} L[I^{\nu}\beta ^{h\alpha} _{m}(x)] &=& L \left[\frac{1}{{\it{\Gamma}} (\nu ) x^{1- \nu}}\ast \beta ^{h\alpha} _{m}(x)\right] \nonumber \\ &=& \frac{1}{{\it{\Gamma}} (\nu)} \frac{{\it{\Gamma}} (\nu)}{s^{\nu}} \sum _{i=0}^{m} \left( {\begin{array}{*{5}c}m\\ i \\ \end{array}} \right )\frac{ \beta_{m-i}^{\alpha}}{h^{i\alpha}} \frac{{\it{\Gamma}} (i \alpha +1)}{s^{i \alpha +1}} \nonumber \\ &=&\sum _{i=0}^{m} \left( {\begin{array}{*{5}c} m\\ i \\ \end{array}} \right )\frac{ \beta_{m-i}^{\alpha}}{h^{i\alpha}} \frac{{\it{\Gamma}} (i \alpha +1)}{s^{i \alpha + \nu +1}}.\label{M6} \end{eqnarray} (3.3) Taking the inverse Laplace transform of Eq. (3.3) yields \begin{equation} I^{\nu}\beta ^{h\alpha} _{m}(x) =\sum _{i=0}^{m} \left( {\begin{array}{*{5}c} m\\ i \\ \end{array}} \right )\frac{ \beta_{m-i}^{\alpha}}{h^{i\alpha}} \frac{{\it{\Gamma}} (i \alpha +1)}{{\it{\Gamma}} (i\alpha + \nu +1)}x^{i\alpha +\nu},\hspace{.5cm}m=0, 1, 2, \ldots ,M.\label{M11} \end{equation} (3.4) Therefore, we obtain the Riemann–Liouville fractional integral operator for the GFBFs. For example, for $$ M =5, h=3, \alpha = 0.5 , \nu = 0.5 $$ and $$ x=1.5 $$ the Riemann–Liouville fractional integral operator of the GFBFs can be expressed as \begin{align*} F^{3, 0.5}( 0.5, 1.5) = [1.38198, 0.0765067, -0.0765067, -0.0194301, 0.0232353, 0.0110308]^{T}. \end{align*} 3.2. Riemann–Liouville fractional integral operator for GFLFs We now derive the operator $$ I^{\nu} $$ for $$ L^{h\alpha}(t) $$ in Equation (2.12) given by \begin{equation} I^{\nu}L^{h\alpha}(t)= {\it{\Theta}} ^{h\alpha}( \nu , t), \label{Z2} \end{equation} (3.5) where \begin{equation} {\it{\Theta}}^{h\alpha}( \nu , t) = [I^{\nu}L^{h\alpha} _{0}(t), I^{\nu}L^{h\alpha} _{1}(t), I^{\nu}L^{h\alpha} _{2}(t), \ldots , I^{\nu}L^{h\alpha} _{M'}(t) ]^{T}. \end{equation} To obtain $$ I^{\nu}L^{h\alpha} _{m}(t) (m=0, 1, \ldots , M'), $$ we use the Laplace transform. By taking the Laplace transform from Equation (2.7), we get \begin{eqnarray} L[L ^{h\alpha} _{m}(t)] & = & L\left[\sum _{i=0}^{m} \frac{(-1)^{m+i}(m+i)!}{(m-i)!(i!)^{2}h^{i \alpha}} t^{i\alpha}\right]\nonumber\\ &=& \sum _{i=0}^{m} \frac{(-1)^{m+i}(m+i)!}{(m-i)!(i!)^{2}h^{i \alpha}} L[t^{i\alpha}]\nonumber\\ &= &\sum _{i=0}^{m} \frac{(-1)^{m+i}(m+i)!}{(m-i)!(i!)^{2}h^{i \alpha}} \frac{{\it{\Gamma}} (i \alpha +1)}{s^{i\alpha +1}}. \label{Z4} \end{eqnarray} (3.6) From Equation (2.1), we get \begin{eqnarray} L[I^{\nu}L^{h\alpha} _{m}(t)] &=& L \left[\frac{1}{{\it{\Gamma}} (\nu ) t^{1- \nu}}\ast L ^{h\alpha} _{m}(t)\right]\nonumber \\ &=& \frac{1}{{\it{\Gamma}} (\nu)} \frac{{\it{\Gamma}} (\nu)}{s^{\nu}} \sum _{i=0}^{m} \frac{(-1)^{m+i}(m+i)!}{(m-i)!(i!)^{2}h^{i \alpha}} \frac{{\it{\Gamma}} (i \alpha +1)}{s^{i\alpha +1}} \nonumber \\ &=&\sum _{i=0}^{m} \frac{(-1)^{m+i}(m+i)!}{(m-i)!(i!)^{2}h^{i \alpha}} \frac{{\it{\Gamma}} (i \alpha + 1)}{s^{i\alpha +\nu+1}}. \label{Z6} \end{eqnarray} (3.7) By taking the inverse Laplace transform of Equation (3.7) yields \begin{equation} I^{\nu}L_{m}^{h \alpha}(t) = \sum _{i=0}^{m} \frac{(-1)^{m+i}(m+i)!}{(m-i)!(i!)^{2}h^{i \alpha}} \frac{{\it{\Gamma}} (i \alpha + 1)}{{\it{\Gamma}} (i\alpha + \nu +1)}t^{i\alpha +\nu}, \hspace{ .5cm}m=0, 1, \ldots M'. \end{equation} (3.8) Therefore, we obtain the Riemann–Liouville fractional integral operator for the GFLFs. For example, for $$ M =5, h=3, \alpha = 1 , \nu = 1.5 $$ and $$ x=0.5 $$ the Riemann–Liouville fractional integral operator of the GFLFs can be expressed as \begin{align*} {\it{\Theta}} ^{3, 1}( 1.5, 0.5) = [0.265962, -0.2305, 0.169709, -0.100099, 0.0386484, 0.00282425 ]^{T}. \end{align*} 4. Problem statement and approximation method Consider the following two-dimensional FOCP be given: \begin{equation} min\hspace{.3cm} J[ \xi , u] = \int _{0}^{h_{2}}\int _{0}^{h_{1}}G \big ( x, t, \xi (x, t), u(x, t), \frac{\partial \xi}{\partial x}(x, t), \frac{\partial \xi}{\partial t}(x, t) \big) dx dt, \label{Z7} \end{equation} (4.1) subject to \begin{equation} \frac{\partial ^{2}\xi}{\partial x \partial t}(x, t)= a \frac{\partial ^{\nu}\xi}{\partial x^{\nu}}(x, t) + b \frac{\partial ^{\gamma}\xi}{\partial t^{\gamma}}(x, t) + c \xi(x, t) + d u(x, t), \label{Z8} \end{equation} (4.2) with the following initial and boundary conditions \begin{equation} \xi (x, 0) = \zeta (x), \hspace{1cm} \xi (0, t) = \eta (t), \label{Z9} \end{equation} (4.3) where $$ 0 < \nu , \gamma \leq 1. $$ The functions $$ G $$ and $$ \xi $$ are smooth and $$d $$ is non-zero. By computing the control function $$ u(x, t) $$ from Equation (4.2), and substituting into the index functional J, the following equivalent minimization problem is obtained: \begin{gather} min\hspace{.3cm} J[\xi] = \int _{0}^{h_{2}}\int _{0}^{h_{1}}G \big(x, t, \xi (x, t), \frac{1}{d} \Bigg[\frac{\partial ^{2}\xi}{\partial x \partial t}(x, t)- a \frac{\partial ^{\nu}\xi}{\partial x^{\nu}}(x, t) \nonumber\\ - b \frac{\partial ^{\gamma}\xi}{\partial t^{\gamma}}(x, t) -c \xi(x, t)\Bigg], \frac{\partial \xi}{\partial x}(x, t), \frac{\partial \xi}{\partial t}(x, t) \big) dx dt,\label{Z10}\\ \end{gather} (4.4) \begin{gather} \xi (x, 0) = \zeta (x), \hspace{1cm} \xi (0, t) = \eta (t).\label{Z11} \end{gather} (4.5) For solving this problem, we approximate: \begin{equation} \frac{\partial ^{2}\xi}{\partial x \partial t}(x, t) = B^{h _{1}\alpha ,T}(x) U L^{h_{2}\alpha}(t),\label{Z12} \end{equation} (4.6) where $$ U=[u_{i, j}]_{M\times M'} $$ is an unknown matrix which should be found. By integrating of (4.6) with respect to $$ t $$, we obtain: \begin{equation} \frac{\partial \xi}{\partial x}(x, t)=B^{h_{1}\alpha ,T}(x) U {\it{\Theta}} ^{h_{2}\alpha}(1, t) + \frac{\partial \xi }{\partial x}(x, t)\bigg \vert _{t=0}=B^{h_{1}\alpha , T}(x) U {\it{\Theta}} ^{h_{2}\alpha}(1, t) + \frac{\partial \zeta (x) }{\partial x}. \label{Z13} \end{equation} (4.7) Moreover, by integrating of (4.6) with respect to $$ x $$, we yield: \begin{equation} \frac{\partial \xi}{\partial t}(x, t)=F^{h_{1}\alpha ,T}(1, x) U L^{h_{2}\alpha}(t) + \frac{\partial \xi }{\partial t}(x, t)\bigg \vert _{x=0}=F^{h_{1}\alpha , T}(1, x) U L^{h_{2}\alpha}(t) + \frac{\partial \eta (t) }{\partial t}. \label{Z14} \end{equation} (4.8) Now by integrating of (4.7) with respect to $$ x $$, and considering (4.3), we get: \begin{equation} \xi (x, t) = F^{h_{1}\alpha ,T}(1, x) U {\it{\Theta}} ^{h_{2}\alpha} (1, t) + \zeta (x) +\eta (t)- \zeta (0). \label{Z15} \end{equation} (4.9) By fractional differentiation of order $$ \nu $$ of (4.9) with respect to $$ x $$, we have: \begin{equation} \frac{\partial ^{\nu}\xi}{\partial x^{\nu}}(x, t) = F^{h_{1}\alpha ,T}(1 - \nu, x) U {\it{\Theta}} ^{h_{2}\alpha} (1, t) + \frac{\partial ^{\nu}\zeta (x)}{\partial x^{\nu}}. \label{Z20} \end{equation} (4.10) Also by fractional differentiation of order $$ \gamma $$ of (4.9) with respect to $$ t $$, we obtain: \begin{equation} \frac{\partial ^{\gamma}\xi}{\partial t ^{\gamma}}(x, t) = F^{h_{1}\alpha ,T}(1, x) U {\it{\Theta}} ^{h_{2}\alpha} (1-\gamma , t) + \frac{\partial ^{\gamma}\eta (t)}{\partial t^{\gamma}}. \label{Z16} \end{equation} (4.11) From Equation (4.2) and approximations (4.6), (4.9)–(4.11), the control function is estimated as \begin{eqnarray} \tilde{u}(x, t)&=& \frac{1}{d} \Bigg[B^{h_{1}\alpha ,T}(x) U L^{h_{2}\alpha}(t) \nonumber\\ &&- a F^{h_{1}\alpha ,T}(1 - \nu, x) U {\it{\Theta}} ^{h_{2}\alpha} (1, t) - a\frac{\partial ^{\nu}\zeta (x)}{\partial x^{\nu}}\nonumber\\ &&-bF^{h_{1}\alpha ,T}(1, x) U {\it{\Theta}} ^{h_{2}\alpha} (1-\gamma , t) - b\frac{\partial ^{\gamma}\eta (t)}{\partial t^{\gamma}}\nonumber\\ &&-c F^{h_{1}\alpha ,T}(1, x) U {\it{\Theta}} ^{h_{2}\alpha} (1, t) - c\zeta (x) -c \eta (t) + c\zeta (0) \Bigg].\label{Z17} \end{eqnarray} (4.12) Now by substituting the approximate state function (4.9), control function (4.12), $$ \frac{\partial \xi}{\partial x}(x, t) $$ and $$ \frac{\partial \xi}{\partial t}(x, t) $$ from Equations (4.7) and (4.8) in the performance index function (4.4), our aim is to solve the resultant unconstrained minimization problem of $$ (M+1)\times(M'+1) $$ variables, $$ u_{ij}, i=0, 1, \ldots , M, j=0, 1, \ldots , M', $$ as follows \begin{eqnarray} min \hspace{.2cm}\bar{J}[U]&=& \int_{0}^{h_{2}}\int_{0}^{h_{1}}G\Bigg( x, t, F^{h_{1}\alpha ,T}(1, x) U {\it{\Theta}} ^{h_{2}\alpha} (1, t) + \zeta (x) \nonumber\\ &&+\eta (t)- \zeta (0), \frac{1}{d} \Bigg[ B^{h_{1}\alpha ,T}(x) U L^{h_{2}\alpha}(t) \nonumber\\ &&- a F^{h_{1}\alpha ,T}(1 - \nu, x) U {\it{\Theta}} ^{h_{2}\alpha} (1, t) - a \frac{\partial ^{\nu}\zeta (x)}{\partial x^{\nu}}\nonumber\\ &&- bF^{h_{1}\alpha ,T}(1, x) U {\it{\Theta}} ^{h_{2}\alpha} (1-\gamma , t) - b \frac{\partial ^{\gamma}\eta (t)}{\partial t^{\gamma}}\nonumber\\ &&- c F^{h_{1}\alpha ,T}(1, x) U {\it{\Theta}} ^{h_{2}\alpha} (1, t) - c \zeta (x) -c \eta (t) +c \zeta (0) \Bigg],\nonumber\\ && B^{h_{1}\alpha , T}(x) U {\it{\Theta}} ^{h_{2}\alpha}(1, t) + \frac{\partial \zeta (x) }{\partial x},\nonumber\\ & & F^{h_{1}\alpha , T}(1, x) U L^{h_{2}\alpha}(t) + \frac{\partial \eta (t) }{\partial t} \Bigg)dx dt. \label{Z18} \end{eqnarray} (4.13) By using the Riemann–Liouville fractional integral operators of GFBFs and GFLFs in Equation (4.13), we get \begin{equation} min \hspace{.2cm}\bar{J}[U]= \int_{0}^{h_{2}}\int_{0}^{h_{1}} Q(x, t, U) dxdt, \label{Z19} \end{equation} (4.14) where \begin{eqnarray} Q(x, t, U)&=&G\bigg( x, t, F^{h_{1}\alpha ,T}(1, x) U {\it{\Theta}} ^{h_{2}\alpha} (1, t) + \zeta (x) +\eta (t)- \zeta (0), \nonumber\\ && \frac{1}{d} \Bigg[ B^{h_{1}\alpha ,T}(x) U L^{h_{2}\alpha}(t) \nonumber\\ &&- a F^{h_{1}\alpha ,T}(1 - \nu, x) U {\it{\Theta}} ^{h_{2}\alpha} (1, t) - a \frac{\partial ^{\nu}\zeta (x)}{\partial x^{\nu}}\nonumber\\ &&- bF^{h_{1}\alpha ,T}(1, x) U {\it{\Theta}} ^{h_{2}\alpha} (1-\gamma , t) - b \frac{\partial ^{\gamma}\eta (t)}{\partial t^{\gamma}}\nonumber\\ &&- c F^{h_{1}\alpha ,T}(1, x) U {\it{\Theta}} ^{h_{2}\alpha} (1, t) - c \zeta (x) -c \eta (t) +c \zeta (0) \Bigg],\nonumber\\ && B^{h_{1}\alpha , T}(x) U {\it{\Theta}} ^{h_{2}\alpha}(1, t) + \frac{\partial \zeta (x) }{\partial x},\nonumber\\ && F^{h_{1}\alpha , T}(1, x) U L^{h_{2}\alpha}(t) + \frac{\partial \eta (t) }{\partial t} \bigg). \end{eqnarray} (4.15) To compute the integral (4.14), we use the two-dimensional LG quadrature rule (Nemati & Yousefi, 2016) as \begin{equation} \bar{J}[U] \simeq \tilde{J}[U]= \frac{h_{1}h_{2}}{4}\sum_{r=1}^{M}\sum_{s=1}^{M'}Q \left(\frac{h_{1}}{2}(\delta _{r}+1), \frac{h_{2}}{2}(\tau _{s}+1), U\right)\omega _{r}\omega _{s}:=\frac{h_{1} h_{2}}{4} W^{T}{\it{\Omega}} W', \end{equation} (4.16) where \begin{equation} ({\it{\Omega}}) _{r, s}= Q ((\delta _{r}+1)\frac{h_{1}}{2}, (\tau _{s}+1)\frac{h_{2}}{2}, U), \hspace{.5cm}1\leq r \leq M,\hspace{.1cm} 1 \leq s \leq M', \end{equation} (4.17) and $$ W $$ and $$ W' $$ are $$ M $$ and $$ M' $$-dimensional vectors of Christoffel weights of LG-nodes, respectively. Referring to differential calculus, we have the following necessary conditions for optimizing the functional $$ \tilde{J} $$, as \begin{equation} \frac{\partial \tilde{J}}{\partial u_{i, j}}[U]=0, \hspace{.5cm}0\leq i \leq M, \hspace{.2cm}0\leq j \leq M'. \label{Z21} \end{equation} (4.18) To find optimized matrix of $$ U $$, we shall solve the algebraic system (4.18). Therefore, the Newton’s iterative method is taken into account to find the solution. 5. Convergence of the new approach In this section, the convergence of the proposed method in the Sobolev space is investigated. We show that with an increase in the number of the generalized fractional-order Bernoulli–Legendre functions basis, the approximate optimum value approaches to the exact value. We show this fact in Theorem 5.3. Now we define our function space and provide some needed theorems, lemmas and corollaries. The Sobolev norm in the domain $${\it{\Delta}} = (a, b)^{d} $$ in $$R^{d} $$ with $$ d=2, 3 $$ for $$ \mu \geq 1 $$ is defined as (Canuto et al., 2006) \begin{equation} \Vert \xi \Vert _{H^{\mu}({\it{\Delta}})} = \left( \sum _{k=0}^{\mu} \sum _{i=1}^{d} \Vert D^{(k)}_{i} \xi \Vert ^{2}_{L^{2}({\it{\Delta}})}\right)^{\frac{1}{2}},\label{Z30} \end{equation} (5.1) where $$ D^{(k)}_{i} \xi $$ denotes the $$ k $$th derivative of $$ \xi $$ respect to variable of $$ i $$th. The symbol $$ \vert \xi \vert _{H^{\mu ; M}({\it{\Delta}})} $$ was defined by Canuto et al. (2006) \begin{align*} \vert \xi \vert _{H^{\mu ; M}({\it{\Delta}})} = \left( \sum _{k=min (\mu , M+1)}^{\mu} \sum _{i=1}^{d} \Vert D^{(k)}_{i} \xi \Vert ^{2}_{L^{2}({\it{\Delta}})}\right)^{\frac{1}{2}}. \end{align*} For simplicity of work, we assume $$ h_{1}=h_{2}=1 $$ and $$ M=M'. $$ Of course for $$ h_{1},h_{2}\neq 1 $$ and $$ M\neq M' $$ the procedure is similar. Theorem 5.1 Suppose that $$ \xi \in H^{\mu}({\it{\Delta}}) $$ with $$ \mu \geq 1 $$ and $$ {\it{\Delta}} = [0, 1]\times [0, 1]. $$ If $$ P_{M}^{\alpha}\xi = \sum _{i=0}^{M} \sum_{j=0}^{M} c_{ij}\beta _{i}^{\alpha}L_{j}^{\alpha},$$ is the best approximation of $$ \xi $$ then \begin{equation} \Vert \xi - P_{M}^{\alpha}\xi \Vert _{L^{2}({\it{\Delta}})} \leq c \alpha ^{1-\mu}M^{1-\mu} \vert \xi \vert _{H^{\mu ;M\alpha }({\it{\Delta}})}, \label{Z31} \end{equation} (5.2) and for $$ 1 \leq l \leq \mu $$, \begin{equation} \Vert \xi - P_{M}^{\alpha}\xi \Vert _{H^{l}({\it{\Delta}})}\leq c \alpha ^{ \varrho (l) - \mu } M^{\varrho (l) - \mu} \vert \xi \vert _{H^{\mu ;M\alpha }({\it{\Delta}})}, \label{Z35} \end{equation} (5.3) where $$ c $$ depends on $$ \mu $$ and \begin{align*} \varrho (l)=\left\{ \begin{array}{ll} 0, & l=0,\\ 2l-\frac{1}{2}, \ \ \ \ \ \ \ \ \ \ \ \ & l>0. \end{array} \right . \end{align*} Proof. Let $$ P_{N}=P_{N}({\it{\Pi}}) $$ be the space of all algebraic polynomials of degree up to $$ N $$ in each variable $$ x_{i} $$ for $$i=1, 2, \ldots , d $$. If $$ P'\xi $$ be the best approximation of $$ \xi $$ upon $$ P_{N}, $$ so for all $$ \xi \in H^{\mu}({\it{\Pi}}), \mu \geq 1,$$ we have (Canuto et al., 2006) \begin{equation} \Vert \xi - P'_{N}\xi \Vert _{L^{2}({\it{\Pi}})} \leq c N^{1-\mu} \vert \xi \vert _{H^{\mu ;N }({\it{\Pi}})}\label{Z33} \end{equation} (5.4) and for $$ 1 \leq l \leq \mu$$, \begin{equation} \Vert \xi - P'_{N}\xi \Vert _{H^{l}({\it{\Pi}})}\leq c N^{\varrho (l) - \mu} \vert \xi \vert _{H^{\mu ;N }({\it{\Pi}})}. \label{Z34} \end{equation} (5.5) Since the best approximation is unique (Kreyszig, 1978) and approximation with the fractional-order functions is from most degree $$ M\alpha $$ in each variable, therefore we have $$ N=M\alpha $$ and $$ {\it{\Pi}} ={\it{\Delta}} $$ in Equations (5.4) and (5.5) as \begin{align*} \Vert \xi - P_{M}^{\alpha}\xi \Vert _{L^{2}({\it{\Delta}})} = \Vert \xi - P'_{M \alpha}\xi \Vert _{L^{2}({\it{\Delta}})} \leq c \alpha ^{1-\mu} M^{1-\mu} \vert \xi \vert _{H^{\mu ;M \alpha }({\it{\Delta}})}, \end{align*} and for $$ 1 \leq l \leq \mu $$, \begin{align*} \Vert \xi - P_{M}^{\alpha}\xi \Vert _{H^{l}({\it{\Delta}})} = \Vert \xi - P'_{M\alpha }\xi \Vert _{H^{l}({\it{\Delta}})}\leq c \alpha ^{\varrho (l) - \mu} M^{\varrho (l) - \mu} \vert \xi \vert _{H^{\mu ;M\alpha }({\it{\Delta}})}, \end{align*} where \begin{align*} \varrho (l)=\left\{ \begin{array}{ll} 0, & l=0,\\ 2l-\frac{1}{2}, \ \ \ \ \ \ \ \ \ \ \ \ & l>0. \end{array} \right .\tag*{$\quad \square$} \end{align*} This result shows that in the case $$ \xi $$ is infinitely smooth, the rate of convergence of $$ P_{M}^{\alpha}\xi $$ to $$ \xi $$ is faster than of any power of $$ \frac{1}{M} $$, which is superior to that for the classical spectral methods (Canuto et al., 2006). Now, we obtain the error bound for the fractional and integer order derivative. Theorem 5.2 Suppose $$ \xi \in H^{\mu}({\it{\Delta}}) $$ with $$ \mu \geq 1 $$ and $$ 0 < \nu \leq 1 $$, then \begin{equation} \Vert D^{\nu}_{x} \xi (x, t) - D^{\nu}_{x}(P_{M}^{\alpha} \xi (x, t)) \Vert _{L^{2}({\it{\Delta}})} \leq \frac{c \alpha ^{\varrho (l)-\mu} M^{\varrho (l) - \mu}}{{\it{\Gamma}} (2 - \nu)} \vert \xi \vert _{H^{\mu ;M\alpha }({\it{\Delta}})}, \label{Z36} \end{equation} (5.6) where $$ 1 \leq l \leq \mu $$. Proof. By Equation (2.1) and Bass (2013) \begin{align*} \Vert f \ast g \Vert _{p} \leq \Vert f \Vert _{p} \Vert g \Vert _{1}, \end{align*} for $$ 0 < \nu \leq 1, $$ we obtain \begin{eqnarray*} && \Vert D^{\nu}_{x} \xi (x, t) - D^{\nu}_{x}(P_{M}^{\alpha}\xi (x, t)) \Vert _{L^{2}({\it{\Delta}})} ^{2}\\ &&\quad{}= \Vert I^{1-\nu }_{x}( D^{(1)}_{x} \xi (x, t) - D^{(1)}_{x}(P_{M}^{\alpha} \xi (x, t)))\Vert _{L^{2}({\it{\Delta}})} ^{2}\nonumber\\ &&\quad{}= \Vert \frac{1}{x^{\nu}{\it{\Gamma}} (1- \nu)} \ast ( D^{(1)}_{x} \xi (x, t) - D^{(1)}_{x}(P_{M}^{\alpha}\xi (x, t))) \Vert _{L^{2}({\it{\Delta}})} ^{2}\nonumber\\ &&\quad{}\leq \left(\frac{1}{(1- \nu) {\it{\Gamma}} (1-\nu)}\right)^{2}\Vert D^{(1)}_{x} \xi (x, t) - D^{(1)}_{x}(P_{M}^{\alpha} \xi (x, t)) \Vert _{L^{2}({\it{\Delta}})} ^{2}\nonumber\\ &&\quad{}\leq \left(\frac{1}{ {\it{\Gamma}} (2-\nu)}\right)^{2}\Vert \xi (x, t) - P_{M}^{\alpha} \xi (x, t) \Vert ^{2} _{H^{l}({\it{\Delta}})}, \end{eqnarray*} from Equation (5.3) we get Equation (5.6). □ Corollary 5.1 From Equation (5.6) for $$ \xi \in H^{\mu}({\it{\Delta}}) $$ with $$ \mu \geq 1 $$ and $$ 0 < \gamma \leq 1 $$ we can write \begin{equation} \Vert D^{\gamma }_{t} \xi (x, t) - D^{\gamma}_{t}(P_{M}^{\alpha}\xi (x, t)) \Vert _{L^{2}({\it{\Delta}})} \leq \frac{c \alpha ^{\varrho (l)-\mu} M^{\varrho (l) - \mu}}{{\it{\Gamma}} (2 - \gamma)} \vert \xi \vert _{H^{\mu ;M\alpha }({\it{\Delta}})}. \label{Z37} \end{equation} (5.7) Proof. It is an immediate consequence of Theorem 5.2. □ Lemma 5.1 Consider $$ \xi \in H^{\mu}({\it{\Delta}}) $$ with $$ \mu \geq 1 $$ and $$ 1 <l \leq \mu $$, then we have \begin{align*} \bigg \Vert \frac{\partial }{\partial x} \xi (x,t) - \frac{\partial }{\partial x} (P_{M}^{\alpha} \xi (x, t)) \bigg \Vert _{L^{2}({\it{\Delta}})} \leq c \alpha ^{ \varrho (l) - \mu } M^{\varrho (l) - \mu} \vert \xi \vert _{H^{\mu ;M\alpha }({\it{\Delta}})}. \end{align*} Proof. From Equation (5.3), we have \begin{align*} \bigg \Vert \frac{\partial }{\partial x} \xi (x,t) - \frac{\partial }{\partial x} (P_{M}^{\alpha} \xi (x, t)) \bigg \Vert _{L^{2}({\it{\Delta}})} & \leq \Vert \xi (x, t) - P_{M}^{\alpha} \xi (x, t) \Vert _{H^{l}({\it{\Delta}})}\nonumber\\ & \leq c \alpha ^{ \varrho (l) - \mu } M^{\varrho (l) - \mu} \vert \xi \vert _{H^{\mu ;M\alpha }({\it{\Delta}})}.\tag*{$\quad \square$} \end{align*} Corollary 5.2 Consider $$ \xi \in H^{\mu}({\it{\Delta}}) $$ with $$ \mu \geq 1 $$ and $$ 1 <l \leq \mu $$, then we have \begin{align*} \bigg \Vert \frac{\partial }{\partial t} \xi (x,t) - \frac{\partial }{\partial t} ( P_{M}^{\alpha}\xi (x, t)) \bigg \Vert _{L^{2}({\it{\Delta}})} \leq c \alpha ^{ \varrho (l) - \mu } M^{\varrho (l) - \mu} \vert \xi \vert _{H^{\mu ;M\alpha }({\it{\Delta}})}. \end{align*} Proof. It is an immediate consequence of Lemma 5.1. □ Corollary 5.3 Consider $$ \xi \in H^{\mu}({\it{\Delta}}) $$ with $$ \mu \geq 1 $$ and $$ 1 <l \leq \mu $$, then we get \begin{align*} \bigg \Vert \frac{\partial ^{2}}{\partial x \partial t} \xi (x,t) - \frac{\partial ^{2}}{\partial x \partial t} ( P_{M}^{\alpha}\xi (x, t)) \bigg \Vert _{L^{2}({\it{\Delta}})} \leq c \alpha ^{ \varrho (l) - \mu } M^{\varrho (l) - \mu} \big \vert \frac{\partial \xi }{\partial t}\big \vert _{H^{\mu ;M\alpha }({\it{\Delta}})}. \end{align*} Proof. From Equation (5.3), we obtain \begin{align*} \bigg \Vert \frac{\partial ^{2}}{\partial x \partial t} \xi (x,t) - \frac{\partial ^{2}}{\partial x \partial t} ( P_{M}^{\alpha}\xi (x, t)) \bigg \Vert _{L^{2}({\it{\Delta}})} &\leq \bigg \Vert \frac{\partial}{\partial x} \left(\frac{\partial}{\partial t} \xi (x, t)\right) - \frac{\partial}{\partial x}\left(\frac{\partial}{\partial t}P_{M}^{\alpha} \xi (x, t)\right) \bigg \Vert _{L^{2}({\it{\Delta}})}\\ &\leq \bigg \Vert \frac{\partial}{\partial t} \xi (x, t) - \frac{\partial}{\partial t}P_{M}^{\alpha} \xi (x, t) \bigg \Vert _{H^{l}({\it{\Delta}})} \\ &\leq c \alpha ^{ \varrho (l) - \mu } M^{\varrho (l) - \mu} \big \vert \frac{\partial \xi }{\partial t}\big \vert _{H^{\mu ;M\alpha }({\it{\Delta}})}.\tag*{$\quad \square$} \end{align*} Now we can show the convergence of the approximating method. Theorem 5.3 Assume that $$ \lbrace P_{M}^{\alpha}\xi ^{\ast} (x_{i}, t_{j}) , P_{M}^{\alpha}u ^{\ast} (x_{i}, t_{j}) , 0\leq i \leq M, 0 \leq j \leq M\rbrace _{M=M_{k}}^{\infty}$$ be a sequence of approximation optimal solutions and $$ \lbrace \xi ^{\ast} (x_{i}, t_{j}) ,u ^{\ast} (x_{i}, t_{j}) , 0\leq i \leq M, 0 \leq j \leq M \rbrace $$ be the exact optimal solution to problem (4.1) and $$ G $$ satisfy a Lipschitz condition with Lipschitz constant $$ \ell $$. Then we have \begin{equation} \tilde{J}[\xi ^{\ast}] = \lim _{M \rightarrow \infty} G_{M}( P_{M}^{\alpha}\xi ^{\ast}(x, t)), \end{equation} (5.8) where \begin{eqnarray*} G_{M}( P_{M}^{\alpha}\xi ^{\ast}(x, t)) &=& \frac{1}{4}\sum _{r=1}^{M}\sum _{s=1}^{M}G \Bigg( x _{r}, t_{s}, P_{M}^{\alpha} \xi ^{\ast} (x_{r}, t_{s}), \frac{1}{d} \Bigg[\frac{\partial ^{2}}{\partial x \partial t} P_{M}^{\alpha}\xi ^{\ast} (x_{r} , t_{s})- a \frac{\partial ^{\nu}}{\partial x^{\nu}}P_{M}^{\alpha} \xi ^{\ast} (x_{r}, t_{s})\\ &-& b \frac{\partial ^{\gamma} }{\partial t^{\gamma}}P_{M}^{\alpha} \xi ^{\ast}(x_{r}, t_{s}) -c P_{M}^{\alpha}\xi ^{\ast}(x_{r}, t_{s})\Bigg], \frac{\partial}{\partial x}P_{M}^{\alpha} \xi ^{\ast} (x_{r}, t_{s}), \frac{\partial }{\partial t}P_{M}^{\alpha} \xi ^{\ast} (x_{r}, t_{s}) \Bigg) \omega _{r}\omega _{s} \end{eqnarray*} and \begin{eqnarray*} J[\xi ^{\ast}] &=& \int _{0}^{1}\int _{0}^{1}G \left(x, t, \xi ^{\ast}(x, t), \frac{1}{d} \Bigg[\frac{\partial ^{2}\xi ^{\ast}}{\partial x \partial t}(x, t)- a \frac{\partial ^{\nu}\xi ^{\ast}}{\partial x^{\nu}}(x, t)\right. \nonumber\\ &&\left.- b \frac{\partial ^{\gamma}\xi ^{\ast}}{\partial t^{\gamma}}(x, t) -c \xi ^{\ast}(x, t)\Bigg], \frac{\partial \xi ^{\ast}}{\partial x}(x, t), \frac{\partial \xi ^{\ast}}{\partial t}(x, t) \right) dx dt. \end{eqnarray*} Proof. Since $$( P_{M}^{\alpha} \xi ^{\ast}, D_{x}^{\nu}P_{M}^{\alpha} \xi ^{\ast}, D_{t}^{\gamma}P_{M}^{\alpha} \xi ^{\ast}, \frac{\partial}{\partial x}P_{M}^{\alpha} \xi ^{\ast}, \frac{\partial}{\partial t}P_{M}^{\alpha} \xi ^{\ast}, \frac{\partial ^{2}}{\partial x \partial t}P_{M}^{\alpha} \xi ^{\ast})$$ converges to $$ (\xi ^{\ast}, D_{x}^{\nu}\xi ^{\ast}, D_{t}^{\gamma}\xi ^{\ast}, \frac{\partial}{\partial x} \xi ^{\ast}$$, $$\frac{\partial}{\partial t} \xi ^{\ast}, \frac{\partial ^{2}}{\partial x \partial t} \xi ^{\ast})$$ uniformly, we have \begin{gather} \lim _{M\rightarrow \infty} \Vert \xi ^{\ast} (x_{r}, t_{s}) - P_{M}^{\alpha} \xi ^{\ast} (x_{r}, t_{s}) \Vert _{L^{2}({\it{\Delta}})}=0,\label{Z40}\\ \end{gather} (5.9) \begin{gather} \lim _{M\rightarrow \infty} \Vert D_{x}^{\nu} \xi ^{\ast} (x_{r}, t_{s}) - D_{x}^{\nu} P_{M}^{\alpha} \xi ^{\ast} (x_{r}, t_{s}) \Vert _{L^{2}({\it{\Delta}})}=0,\\ \end{gather} (5.10) \begin{gather} \lim _{M\rightarrow \infty} \Vert D_{t}^{\gamma} \xi ^{\ast} (x_{r}, t_{s}) - D_{t}^{\gamma} P_{M}^{\alpha} \xi ^{\ast} (x_{r}, t_{s}) \Vert _{L^{2}({\it{\Delta}})}=0,\\ \end{gather} (5.11) \begin{gather} \lim _{M\rightarrow \infty}\bigg \Vert \frac{\partial}{\partial x} \xi ^{\ast} (x_{r}, t_{s}) - \frac{\partial}{\partial x} P_{M}^{\alpha} \xi ^{\ast} (x_{r}, t_{s}) \bigg \Vert _{L^{2}({\it{\Delta}})}=0,\\ \end{gather} (5.12) \begin{gather} \lim _{M\rightarrow \infty}\bigg \Vert \frac{\partial}{\partial t} \xi ^{\ast} (x_{r}, t_{s}) - \frac{\partial}{\partial t} P_{M}^{\alpha} \xi ^{\ast} (x_{r}, t_{s})\bigg \Vert _{L^{2}({\it{\Delta}})}=0,\\ \end{gather} (5.13) \begin{gather} \lim _{M\rightarrow \infty} \bigg \Vert \frac{\partial ^{2}}{\partial x \partial t} \xi ^{\ast} (x_{r}, t_{s}) - \frac{\partial ^{2}}{\partial x \partial t} P_{M}^{\alpha} \xi ^{\ast} (x_{r}, t_{s}) \bigg \Vert _{L^{2}({\it{\Delta}})}=0.\label{Z41} \end{gather} (5.14) On the other hand, since $$ G $$ is a continuously differentiable function and satisfy a Lipschitz condition with Lipschitz constant $$ \ell $$, thus we have \begin{align} &\Bigg \Vert G \Bigg( x_{r}, t_{s}, \xi ^{\ast}(x_{r}, t_{s}), \frac{1}{d} \Bigg[\frac{\partial ^{2}\xi ^{\ast}}{\partial x \partial t}(x_{r}, t_{s})- a \frac{\partial ^{\nu}\xi ^{\ast}}{\partial x^{\nu}}(x_{r}, t_{s}) \nonumber\\ &\quad{} - b \frac{\partial ^{\gamma}\xi ^{\ast}}{\partial t^{\gamma}}(x_{r}, t_{s}) -c \xi ^{\ast}(x_{r}, t_{s})\Bigg], \frac{\partial \xi ^{\ast}}{\partial x}(x_{r}, t_{s}), \frac{\partial \xi ^{\ast}}{\partial t}(x_{r}, t_{s}) \Bigg)\nonumber\\ &\quad- G \Bigg(x _{r}, t_{s}, P_{M}^{\alpha} \xi ^{\ast} (x_{r}, t_{s}), \frac{1}{d} \Bigg[\frac{\partial ^{2}}{\partial x \partial t} P_{M}^{\alpha}\xi ^{\ast} (x_{r} , t_{s})- a \frac{\partial ^{\nu}}{\partial x^{\nu}}P_{M}^{\alpha} \xi ^{\ast} (x_{r}, t_{s})\nonumber\\ &\quad- b \frac{\partial ^{\gamma} }{\partial t^{\gamma}}P_{M}^{\alpha} \xi ^{\ast}(x_{r}, t_{s}) -c P_{M}^{\alpha}\xi ^{\ast}(x_{r}, t_{s})\Bigg] , \frac{\partial}{\partial x}P_{M}^{\alpha} \xi ^{\ast} (x_{r}, t_{s}), \frac{\partial }{\partial t}P_{M}^{\alpha} \xi ^{\ast} (x_{r}, t_{s}) \Bigg) \Bigg \Vert _{L^{2}({\it{\Delta}})} \nonumber\\ &\leq \ell \Vert \xi ^{\ast} (x_{r}, t_{s}) - P_{M}^{\alpha} \xi ^{\ast} (x_{r}, t_{s}) \Vert _{L^{2}({\it{\Delta}})}+(\ell / d)\Vert \frac{\partial ^{2}}{\partial x \partial t} \xi ^{\ast} (x_{r}, t_{s}) \nonumber\\ &\quad- \frac{\partial ^{2}}{\partial x \partial t} P_{M}^{\alpha} \xi ^{\ast} (x_{r}, t_{s}) \Vert _{L^{2}({\it{\Delta}})}+(\ell a/d) \Vert D_{x}^{\nu} \xi ^{\ast} (x_{r}, t_{s}) - D_{x}^{\nu} P_{M}^{\alpha} \xi ^{\ast} (x_{r}, t_{s}) \Vert _{L^{2}({\it{\Delta}})}\nonumber\\ &\quad+ (\ell b/ d) \Vert D_{t}^{\gamma} \xi ^{\ast} (x_{r}, t_{s}) - D_{t}^{\gamma} P_{M}^{\alpha} \xi ^{\ast} (x_{r}, t_{s})+ (\ell c/d) \Vert \xi ^{\ast} (x_{r}, t_{s}) \nonumber\\ &\quad- P_{M}^{\alpha} \xi ^{\ast} (x_{r}, t_{s}) \Vert _{L^{2}({\it{\Delta}})}\Vert _{L^{2}({\it{\Delta}})}+ \ell \Vert \frac{\partial}{\partial x} \xi ^{\ast} (x_{r}, t_{s}) - \frac{\partial}{\partial x} P_{M}^{\alpha} \xi ^{\ast} (x_{r}, t_{s}) \Vert _{L^{2}({\it{\Delta}})}\nonumber\\ &\quad + \ell \Vert \frac{\partial}{\partial t} \xi ^{\ast} (x_{r}, t_{s}) - \frac{\partial}{\partial t} P_{M}^{\alpha} \xi ^{\ast} (x_{r}, t_{s}) \Vert _{L^{2}({\it{\Delta}})}, \hspace{.8cm}1 \leq r, s \leq M. \label{Z42} \end{align} (5.15) For simplicity, we let \begin{align*} \tilde{G}(x, t, \xi (x, t)) = G \left(\!\! x, t, \xi (x, t), \frac{1}{d} \left[\!\!\frac{\partial ^{2}\xi }{\partial x \partial t}(x, t)- a \frac{\partial ^{\nu}\xi }{\partial x^{\nu}}(x, t) - b \frac{\partial ^{\gamma}\xi }{\partial t^{\gamma}}(x, t) -c \xi (x, t)\!\!\right]\!, \frac{\partial \xi }{\partial x}(x, t), \frac{\partial \xi }{\partial t}(x, t) \!\!\right)\!. \end{align*} Furthermore, since $$ \tilde{G} $$ is a continuous function, we have (Canuto et al., 1988) \begin{align*} \int _{0}^{1}\int _{0}^{1}\tilde{G} \big ( x, t, \xi ^{\ast}(x, t) \big) dx dt= \lim _{M\rightarrow \infty }\frac{1}{4}\sum _{r=1}^{M}\sum _{s=1}^{M}\tilde{G} \big ( x _{r}, t_{s}, \xi ^{\ast} (x_{r}, t_{s}) \big) \omega _{r}\omega _{s}. \end{align*} Therefore \begin{eqnarray*} \int _{0}^{1}\int _{0}^{1} \tilde{G}(x, t, \xi ^{\ast}(x, t)) dxdt &=&\frac{1}{4} \lim _{M\rightarrow \infty} \left(\sum _{r=1}^{M}\sum _{s=1}^{M}\tilde{G}(x_{r}, t_{s}, P_{M}^{\alpha}\xi ^{\ast}(x_{r}, t_{s}))\omega _{r}\omega _{s}\right.\\ &&\qquad{}\left. + \sum _{r=1}^{M}\sum _{s=1}^{M}\big[ \tilde{G}(x_{r}, t_{s},\xi ^{\ast}(x_{r}, t_{s}))- \tilde{G}(x_{r}, t_{s}, P_{M}^{\alpha}\xi ^{\ast}(x_{r}, t_{s}))\big]\omega _{r}\omega _{s} \right)\!. \end{eqnarray*} From the uniform convergence of (5.9)–(5.14) and Equation (5.15), we have \begin{align*} \lim _{M\rightarrow \infty}\bigg \Vert \sum _{r=1}^{M}\sum _{s=1}^{M}\big[ \tilde{G}(x_{r}, t_{s},\xi ^{\ast}(x_{r}, t_{s}))- \tilde{G}(x_{r}, t_{s}, P_{M}^{\alpha}\xi ^{\ast}(x_{r}, t_{s}))\big]\omega _{r}\omega _{s} \bigg\Vert \hspace{.2cm}\leq 0. \end{align*} Thus \begin{align*} \int _{0}^{1}\int _{0}^{1} \tilde{G}(x, t, \xi ^{\ast}(x, t)) dxdt =\frac{1}{4} \lim _{M\rightarrow \infty} \left(\sum _{r=1}^{M}\sum _{s=1}^{M}\tilde{G}(x_{r}, t_{s}, P_{M}^{\alpha}\xi ^{\ast}(x_{r}, t_{s}))\omega _{r}\omega _{s} \right)\!, \end{align*} which confirms \begin{align*} \tilde{J}[\xi ^{\ast}] = \lim _{M \rightarrow \infty} G_{M}( P_{M}^{\alpha}\xi ^{\ast}(x, t)). \end{align*} This complete the proof. □ 6. Numerical examples In this section, we present two test problems and apply the method presented in Section 4 to solve them. The computations were performed on a personal computer, and the codes were written in Mathematica 10. Example 6.1 Consider the following two-dimensional optimal control problem (Tsai et al., 2002; Mamehrashi & Yousefi, 2016; Nemati & Yousefi, 2016) \begin{equation} min \hspace{.2cm}J[ \xi , u] = \int _{0}^{3} \int _{0}^{3} \bigg [ \bigg( \frac{\partial \xi }{\partial t}(x, t) + \xi(x, t) \bigg)^{2} + \xi ^{2}(x, t) + u^{2}(x, t)\bigg ] dx dt, \end{equation} (6.1) subject to \begin{gather} \frac{\partial ^{2}\xi}{\partial x \partial t}(x, t)= - \frac{\partial ^{\nu}\xi}{\partial x^{\nu}}(x, t) -3 \frac{\partial ^{\gamma}\xi}{\partial t^{\gamma}}(x, t) + 0.2 \xi(x, t) + 0.3 u(x, t),\\ \end{gather} (6.2) \begin{gather} \xi (x, 0) = e^{-3x}\cos (2 \pi x), \hspace{1cm} \xi (0, t) = e^{-2t}. \end{gather} (6.3) Here we solve this problem by using the proposed method for $$ l=M $$ and $$ l'=M' $$. In Table 1, the values of optimal performance index functional $$\tilde{J}$$ obtained by the present method are compared with the values of $$\tilde{J}$$ reported in Tsai et al. (2002), Mamehrashi & Yousefi (2016) and Nemati & Yousefi (2016),$$ \nu = \gamma =1$$. According to the values of $$\tilde{J}$$, we can say that our numerical solutions are better than the other methods (Tsai et al., 2002; Mamehrashi & Yousefi, 2016; Nemati & Yousefi, 2016). Also, Fig. 2 illustrates the optimal state and control functions of the problem with $$ M=1, M'=6 $$ and $$ \alpha =\frac{1}{5} $$ for $$ \nu = \gamma =1. $$ For partial fractional orders, two cases are considered ($$ \nu = \gamma = 0.8 $$ and $$ \nu =0.75, \gamma =0.8$$). Table 2 shows the values of the optimal performance index functional $$\tilde{J}$$ by the present method and Ref. (Nemati & Yousefi, 2016) for $$\nu = \gamma = 0.8$$. Also, Fig. 3 displays the optimal state and control functions for $$\nu = \gamma = 0.8$$ with $$\alpha =1$$. At last, we take $$ \nu =0.75 $$ and $$ \gamma =0.8 $$. The outcomes of optimal functional are shown in Table 3. Figure 4 depicts the state and control functions for $$ \nu =0.75 $$ and $$ \gamma =0.8 $$ with $$ \alpha =1. $$ From these figures and tables, it can be concluded that the proposed method is convenience for solving this problem. Therefore to show efficient of the parameter $$ \alpha $$, we show the values of optimal functional $$\tilde{J}$$ for $$ \nu = \gamma =1 $$ with various choices of $$ \alpha $$ in Table 4. Also, the computational results with $$ M=1, M' =6 $$ and various values of $$ \alpha $$ for $$ \nu = \gamma =1 $$ are displayed in Fig. 5. Fig. 2. View largeDownload slide (a) The state function and (b) the control input for $$ \nu =\gamma =1, \alpha = \frac{1}{5}$$ with $$ M=1, M' =6 $$ for Example 1. Fig. 2. View largeDownload slide (a) The state function and (b) the control input for $$ \nu =\gamma =1, \alpha = \frac{1}{5}$$ with $$ M=1, M' =6 $$ for Example 1. Fig. 3. View largeDownload slide (a) The state function and (b) the control input for $$ \nu = \gamma =0.8, \alpha =1$$ with $$ M=1, M' =6 $$ for Example 1. Fig. 3. View largeDownload slide (a) The state function and (b) the control input for $$ \nu = \gamma =0.8, \alpha =1$$ with $$ M=1, M' =6 $$ for Example 1. Fig. 4. View largeDownload slide (a) The state function and (b) the control input for $$ \nu =0.75, \gamma =0.8, \alpha =1$$ with $$ M=1, M' =6 $$ for Example 1. Fig. 4. View largeDownload slide (a) The state function and (b) the control input for $$ \nu =0.75, \gamma =0.8, \alpha =1$$ with $$ M=1, M' =6 $$ for Example 1. Fig. 5. View largeDownload slide The control function for (a) $$ \alpha = \frac{1}{5}, $$ (b) $$ \alpha = \frac{1}{4}, $$ (c) $$ \alpha = \frac{1}{2} $$ and (d) $$ \alpha = 1, $$ with $$ M=1, M' =6 $$ for Example 1. Fig. 5. View largeDownload slide The control function for (a) $$ \alpha = \frac{1}{5}, $$ (b) $$ \alpha = \frac{1}{4}, $$ (c) $$ \alpha = \frac{1}{2} $$ and (d) $$ \alpha = 1, $$ with $$ M=1, M' =6 $$ for Example 1. Table 1 Comparison of the estimated value of $$ \tilde{J} $$ for integer orders $$ \nu = \gamma =1 $$ and $$ \alpha =\frac{1}{5} $$ with different values of $$ M $$ and $$ M' $$ for Example 1 Method of Tsai et al. (2002) $$\tilde{J} $$ $$X=0.3, T= 0.3$$ 1.4979 $$X=0.2, T=0.2 $$ $$ 1.0953 $$ $$X=0.1, T=0.1 $$ $$0.7348 $$ $$X = 0.05, T=0.05$$ $$ 0.5510 $$ $$X=0.03, T=0.03 $$ $$ 0.4760 $$ Method of Mamehrashi & Yousefi (2016) $$\tilde{J}$$ $$M=6, M'= 1 $$ $$ 4.0203 $$ $$M=6, M'= 8 $$ $$2.2905 $$ $$M=9, M'= 2 $$ $$ 0.8024 $$ $$M=7, M'= 8 $$ $$ 0.6202 $$ $$M=8, M'= 3 $$ $$ 0.2792 $$ $$M=8, M'= 8 $$ $$0.2026 $$ Method of Nemati & Yousefi (2016) $$\tilde{J}$$ $$M=1, M'= 1 $$ $$6.3883 $$ $$M=2, M'= 3 $$ $$ 5.0251 $$ $$M=6, M'= 4 $$ $$ 2.2775 $$ $$M=7, M'= 5$$ $$0.5997 $$ $$M=8, M'= 5 $$ $$ 0.1906$$ $$M=9, M'= 5$$ $$0.1770$$ $$M=10, M'=5$$ $$0.0951$$ $$M=10, M'=6$$ $$0.0947$$ $$Present \; method$$ $$\tilde{J}$$ $$M=1, M' =2$$ $$ 3.4977\times 10^{-3}$$ $$M=1, M' =4 $$ $$6.0548 \times 10^{-5}$$ $$M=1, M' =6 $$ $$5.0340\times 10^{-7}$$ $$M=1, M' =8 $$ $$4.2535\times 10^{-8}$$ $$M=1, M' =10 $$ $$2.9455\times 10^{-9}$$ $$M=1, M' =12 $$ $$ 1.1270\times 10^{-9}$$ Method of Tsai et al. (2002) $$\tilde{J} $$ $$X=0.3, T= 0.3$$ 1.4979 $$X=0.2, T=0.2 $$ $$ 1.0953 $$ $$X=0.1, T=0.1 $$ $$0.7348 $$ $$X = 0.05, T=0.05$$ $$ 0.5510 $$ $$X=0.03, T=0.03 $$ $$ 0.4760 $$ Method of Mamehrashi & Yousefi (2016) $$\tilde{J}$$ $$M=6, M'= 1 $$ $$ 4.0203 $$ $$M=6, M'= 8 $$ $$2.2905 $$ $$M=9, M'= 2 $$ $$ 0.8024 $$ $$M=7, M'= 8 $$ $$ 0.6202 $$ $$M=8, M'= 3 $$ $$ 0.2792 $$ $$M=8, M'= 8 $$ $$0.2026 $$ Method of Nemati & Yousefi (2016) $$\tilde{J}$$ $$M=1, M'= 1 $$ $$6.3883 $$ $$M=2, M'= 3 $$ $$ 5.0251 $$ $$M=6, M'= 4 $$ $$ 2.2775 $$ $$M=7, M'= 5$$ $$0.5997 $$ $$M=8, M'= 5 $$ $$ 0.1906$$ $$M=9, M'= 5$$ $$0.1770$$ $$M=10, M'=5$$ $$0.0951$$ $$M=10, M'=6$$ $$0.0947$$ $$Present \; method$$ $$\tilde{J}$$ $$M=1, M' =2$$ $$ 3.4977\times 10^{-3}$$ $$M=1, M' =4 $$ $$6.0548 \times 10^{-5}$$ $$M=1, M' =6 $$ $$5.0340\times 10^{-7}$$ $$M=1, M' =8 $$ $$4.2535\times 10^{-8}$$ $$M=1, M' =10 $$ $$2.9455\times 10^{-9}$$ $$M=1, M' =12 $$ $$ 1.1270\times 10^{-9}$$ Table 1 Comparison of the estimated value of $$ \tilde{J} $$ for integer orders $$ \nu = \gamma =1 $$ and $$ \alpha =\frac{1}{5} $$ with different values of $$ M $$ and $$ M' $$ for Example 1 Method of Tsai et al. (2002) $$\tilde{J} $$ $$X=0.3, T= 0.3$$ 1.4979 $$X=0.2, T=0.2 $$ $$ 1.0953 $$ $$X=0.1, T=0.1 $$ $$0.7348 $$ $$X = 0.05, T=0.05$$ $$ 0.5510 $$ $$X=0.03, T=0.03 $$ $$ 0.4760 $$ Method of Mamehrashi & Yousefi (2016) $$\tilde{J}$$ $$M=6, M'= 1 $$ $$ 4.0203 $$ $$M=6, M'= 8 $$ $$2.2905 $$ $$M=9, M'= 2 $$ $$ 0.8024 $$ $$M=7, M'= 8 $$ $$ 0.6202 $$ $$M=8, M'= 3 $$ $$ 0.2792 $$ $$M=8, M'= 8 $$ $$0.2026 $$ Method of Nemati & Yousefi (2016) $$\tilde{J}$$ $$M=1, M'= 1 $$ $$6.3883 $$ $$M=2, M'= 3 $$ $$ 5.0251 $$ $$M=6, M'= 4 $$ $$ 2.2775 $$ $$M=7, M'= 5$$ $$0.5997 $$ $$M=8, M'= 5 $$ $$ 0.1906$$ $$M=9, M'= 5$$ $$0.1770$$ $$M=10, M'=5$$ $$0.0951$$ $$M=10, M'=6$$ $$0.0947$$ $$Present \; method$$ $$\tilde{J}$$ $$M=1, M' =2$$ $$ 3.4977\times 10^{-3}$$ $$M=1, M' =4 $$ $$6.0548 \times 10^{-5}$$ $$M=1, M' =6 $$ $$5.0340\times 10^{-7}$$ $$M=1, M' =8 $$ $$4.2535\times 10^{-8}$$ $$M=1, M' =10 $$ $$2.9455\times 10^{-9}$$ $$M=1, M' =12 $$ $$ 1.1270\times 10^{-9}$$ Method of Tsai et al. (2002) $$\tilde{J} $$ $$X=0.3, T= 0.3$$ 1.4979 $$X=0.2, T=0.2 $$ $$ 1.0953 $$ $$X=0.1, T=0.1 $$ $$0.7348 $$ $$X = 0.05, T=0.05$$ $$ 0.5510 $$ $$X=0.03, T=0.03 $$ $$ 0.4760 $$ Method of Mamehrashi & Yousefi (2016) $$\tilde{J}$$ $$M=6, M'= 1 $$ $$ 4.0203 $$ $$M=6, M'= 8 $$ $$2.2905 $$ $$M=9, M'= 2 $$ $$ 0.8024 $$ $$M=7, M'= 8 $$ $$ 0.6202 $$ $$M=8, M'= 3 $$ $$ 0.2792 $$ $$M=8, M'= 8 $$ $$0.2026 $$ Method of Nemati & Yousefi (2016) $$\tilde{J}$$ $$M=1, M'= 1 $$ $$6.3883 $$ $$M=2, M'= 3 $$ $$ 5.0251 $$ $$M=6, M'= 4 $$ $$ 2.2775 $$ $$M=7, M'= 5$$ $$0.5997 $$ $$M=8, M'= 5 $$ $$ 0.1906$$ $$M=9, M'= 5$$ $$0.1770$$ $$M=10, M'=5$$ $$0.0951$$ $$M=10, M'=6$$ $$0.0947$$ $$Present \; method$$ $$\tilde{J}$$ $$M=1, M' =2$$ $$ 3.4977\times 10^{-3}$$ $$M=1, M' =4 $$ $$6.0548 \times 10^{-5}$$ $$M=1, M' =6 $$ $$5.0340\times 10^{-7}$$ $$M=1, M' =8 $$ $$4.2535\times 10^{-8}$$ $$M=1, M' =10 $$ $$2.9455\times 10^{-9}$$ $$M=1, M' =12 $$ $$ 1.1270\times 10^{-9}$$ Table 2 Comparison of the estimated value of $$ \tilde{J} $$ for orders $$ \nu = \gamma =0.8 $$ and $$ \alpha =\frac{1}{5} $$ with different values of $$ M $$ and $$ M' $$ for Example $$1$$ Method of Nemati & Yousefi (2016) $$\tilde{J}$$ $$M=4, M'= 1 $$ $$ 7.5622 $$ $$M=5, M'= 2 $$ $$5.0050 $$ $$M=5, M'= 3 $$ $$ 3.9465 $$ $$M=7, M'= 3$$ $$1.6332$$ $$M=8, M'= 4 $$ $$ 0.3464$$ $$M=9, M'= 4$$ $$0.2383 $$ $$M=9, M'=5$$ $$0.1579 $$ $$M=9, M'=6$$ $$0.1331 $$ $$Present \; method$$ $$\tilde{J}$$ $$M=1, M' =3 $$ $$ 1.0799\times 10^{-3} -4.2697\times 10^{-29} i $$ $$M=1, M' =5 $$ $$4.7624 \times 10^{-6} -4.5606\times 10^{-33} i $$ $$M=1, M' =7 $$ $$1.5109\times 10^{-8}-1.7117 \times 10^{-31} i $$ $$M=1, M' =9 $$ $$ 3.5017\times 10^{-10}+ 7.7166\times 10^{-29} i $$ $$M=1, M' =11 $$ $$7.3526\times 10^{-11}-1.5310\times 10^{-22} i $$ Method of Nemati & Yousefi (2016) $$\tilde{J}$$ $$M=4, M'= 1 $$ $$ 7.5622 $$ $$M=5, M'= 2 $$ $$5.0050 $$ $$M=5, M'= 3 $$ $$ 3.9465 $$ $$M=7, M'= 3$$ $$1.6332$$ $$M=8, M'= 4 $$ $$ 0.3464$$ $$M=9, M'= 4$$ $$0.2383 $$ $$M=9, M'=5$$ $$0.1579 $$ $$M=9, M'=6$$ $$0.1331 $$ $$Present \; method$$ $$\tilde{J}$$ $$M=1, M' =3 $$ $$ 1.0799\times 10^{-3} -4.2697\times 10^{-29} i $$ $$M=1, M' =5 $$ $$4.7624 \times 10^{-6} -4.5606\times 10^{-33} i $$ $$M=1, M' =7 $$ $$1.5109\times 10^{-8}-1.7117 \times 10^{-31} i $$ $$M=1, M' =9 $$ $$ 3.5017\times 10^{-10}+ 7.7166\times 10^{-29} i $$ $$M=1, M' =11 $$ $$7.3526\times 10^{-11}-1.5310\times 10^{-22} i $$ Table 2 Comparison of the estimated value of $$ \tilde{J} $$ for orders $$ \nu = \gamma =0.8 $$ and $$ \alpha =\frac{1}{5} $$ with different values of $$ M $$ and $$ M' $$ for Example $$1$$ Method of Nemati & Yousefi (2016) $$\tilde{J}$$ $$M=4, M'= 1 $$ $$ 7.5622 $$ $$M=5, M'= 2 $$ $$5.0050 $$ $$M=5, M'= 3 $$ $$ 3.9465 $$ $$M=7, M'= 3$$ $$1.6332$$ $$M=8, M'= 4 $$ $$ 0.3464$$ $$M=9, M'= 4$$ $$0.2383 $$ $$M=9, M'=5$$ $$0.1579 $$ $$M=9, M'=6$$ $$0.1331 $$ $$Present \; method$$ $$\tilde{J}$$ $$M=1, M' =3 $$ $$ 1.0799\times 10^{-3} -4.2697\times 10^{-29} i $$ $$M=1, M' =5 $$ $$4.7624 \times 10^{-6} -4.5606\times 10^{-33} i $$ $$M=1, M' =7 $$ $$1.5109\times 10^{-8}-1.7117 \times 10^{-31} i $$ $$M=1, M' =9 $$ $$ 3.5017\times 10^{-10}+ 7.7166\times 10^{-29} i $$ $$M=1, M' =11 $$ $$7.3526\times 10^{-11}-1.5310\times 10^{-22} i $$ Method of Nemati & Yousefi (2016) $$\tilde{J}$$ $$M=4, M'= 1 $$ $$ 7.5622 $$ $$M=5, M'= 2 $$ $$5.0050 $$ $$M=5, M'= 3 $$ $$ 3.9465 $$ $$M=7, M'= 3$$ $$1.6332$$ $$M=8, M'= 4 $$ $$ 0.3464$$ $$M=9, M'= 4$$ $$0.2383 $$ $$M=9, M'=5$$ $$0.1579 $$ $$M=9, M'=6$$ $$0.1331 $$ $$Present \; method$$ $$\tilde{J}$$ $$M=1, M' =3 $$ $$ 1.0799\times 10^{-3} -4.2697\times 10^{-29} i $$ $$M=1, M' =5 $$ $$4.7624 \times 10^{-6} -4.5606\times 10^{-33} i $$ $$M=1, M' =7 $$ $$1.5109\times 10^{-8}-1.7117 \times 10^{-31} i $$ $$M=1, M' =9 $$ $$ 3.5017\times 10^{-10}+ 7.7166\times 10^{-29} i $$ $$M=1, M' =11 $$ $$7.3526\times 10^{-11}-1.5310\times 10^{-22} i $$ Table 3 Comparison of the estimated value of $$ \tilde{J} $$ for orders $$ \nu = 0.75, \gamma =0.8 $$ and $$ \alpha =\frac{1}{5} $$ with different values of $$ M $$ and $$ M' $$ for Example 1 Method of Nemati & Yousefi (2016) $$\tilde{J}$$ $$M=2, M'= 1 $$ $$10.3484 $$ $$M=3, M'= 2 $$ $$7.3672 $$ $$M=4, M'= 3 $$ $$4.5423 $$ $$M=4, M'= 4$$ $$4.2553$$ $$M=6, M'= 5 $$ $$ 3.1165$$ $$M=8, M'= 5 $$ $$ 0.3131 $$ $$M=9, M'=6 $$ $$ 0.1368 $$ $$Present \; method$$ $$\tilde{J}$$ $$M=1, M' =3 $$ $$ 1.0799\times 10^{-3} -9.9194\times 10^{-34} i$$ $$M=1, M' =5 $$ $$4.7624\times 10^{-6} -7.1948 \times 10^{-33}i$$ $$M=1, M' =7 $$ $$1.5109\times 10^{-8}- 5.6134\times 10^{-34}i$$ $$M=1, M' =9 $$ $$ 3.5017\times 10^{-10}-1.6718\times 10^{-32}i$$ $$M=1, M' =11 $$ $$1.5838\times 10^{-11}+9.1743\times 10^{-23}i$$ Method of Nemati & Yousefi (2016) $$\tilde{J}$$ $$M=2, M'= 1 $$ $$10.3484 $$ $$M=3, M'= 2 $$ $$7.3672 $$ $$M=4, M'= 3 $$ $$4.5423 $$ $$M=4, M'= 4$$ $$4.2553$$ $$M=6, M'= 5 $$ $$ 3.1165$$ $$M=8, M'= 5 $$ $$ 0.3131 $$ $$M=9, M'=6 $$ $$ 0.1368 $$ $$Present \; method$$ $$\tilde{J}$$ $$M=1, M' =3 $$ $$ 1.0799\times 10^{-3} -9.9194\times 10^{-34} i$$ $$M=1, M' =5 $$ $$4.7624\times 10^{-6} -7.1948 \times 10^{-33}i$$ $$M=1, M' =7 $$ $$1.5109\times 10^{-8}- 5.6134\times 10^{-34}i$$ $$M=1, M' =9 $$ $$ 3.5017\times 10^{-10}-1.6718\times 10^{-32}i$$ $$M=1, M' =11 $$ $$1.5838\times 10^{-11}+9.1743\times 10^{-23}i$$ Table 3 Comparison of the estimated value of $$ \tilde{J} $$ for orders $$ \nu = 0.75, \gamma =0.8 $$ and $$ \alpha =\frac{1}{5} $$ with different values of $$ M $$ and $$ M' $$ for Example 1 Method of Nemati & Yousefi (2016) $$\tilde{J}$$ $$M=2, M'= 1 $$ $$10.3484 $$ $$M=3, M'= 2 $$ $$7.3672 $$ $$M=4, M'= 3 $$ $$4.5423 $$ $$M=4, M'= 4$$ $$4.2553$$ $$M=6, M'= 5 $$ $$ 3.1165$$ $$M=8, M'= 5 $$ $$ 0.3131 $$ $$M=9, M'=6 $$ $$ 0.1368 $$ $$Present \; method$$ $$\tilde{J}$$ $$M=1, M' =3 $$ $$ 1.0799\times 10^{-3} -9.9194\times 10^{-34} i$$ $$M=1, M' =5 $$ $$4.7624\times 10^{-6} -7.1948 \times 10^{-33}i$$ $$M=1, M' =7 $$ $$1.5109\times 10^{-8}- 5.6134\times 10^{-34}i$$ $$M=1, M' =9 $$ $$ 3.5017\times 10^{-10}-1.6718\times 10^{-32}i$$ $$M=1, M' =11 $$ $$1.5838\times 10^{-11}+9.1743\times 10^{-23}i$$ Method of Nemati & Yousefi (2016) $$\tilde{J}$$ $$M=2, M'= 1 $$ $$10.3484 $$ $$M=3, M'= 2 $$ $$7.3672 $$ $$M=4, M'= 3 $$ $$4.5423 $$ $$M=4, M'= 4$$ $$4.2553$$ $$M=6, M'= 5 $$ $$ 3.1165$$ $$M=8, M'= 5 $$ $$ 0.3131 $$ $$M=9, M'=6 $$ $$ 0.1368 $$ $$Present \; method$$ $$\tilde{J}$$ $$M=1, M' =3 $$ $$ 1.0799\times 10^{-3} -9.9194\times 10^{-34} i$$ $$M=1, M' =5 $$ $$4.7624\times 10^{-6} -7.1948 \times 10^{-33}i$$ $$M=1, M' =7 $$ $$1.5109\times 10^{-8}- 5.6134\times 10^{-34}i$$ $$M=1, M' =9 $$ $$ 3.5017\times 10^{-10}-1.6718\times 10^{-32}i$$ $$M=1, M' =11 $$ $$1.5838\times 10^{-11}+9.1743\times 10^{-23}i$$ Table 4 Comparison of the estimated value of $$ \tilde{J} $$ for integer orders $$ \nu =\gamma =1 $$ with different values of $$ \alpha $$ for Example $$1$$ $$M, M'$$ $$\alpha = \frac{1}{5} $$ $$ \alpha = \frac{1}{4}$$ $$ \alpha = \frac{1}{2}$$ $$\alpha = 1$$ $$M=1, M' =2$$ $$3.4977\times 10^{-3}$$ $$2.1816\times 10^{-3} $$ $$ 2.0015\times 10^{-4} $$ $$ 1.0650\times 10^{-2}$$ $$M=1, M' =4 $$ $$ 6.0548\times 10^{-5} $$ $$3.7983\times 10^{-5}$$ $$3.7836\times 10^{-5}$$ $$8.7676\times 10^{-5}$$ $$M=1, M' =6$$ $$5.0340\times 10^{-7}$$ $$9.5888\times 10^{-8} $$ $$ 5.2497\times 10^{-6}$$ $$1.7175\times 10^{-5}$$ $$M=1, M' =8$$ $$4.2535\times 10^{-8}$$ $$1.8969\times 10^{-8}$$ $$3.0392 \times 10^{-7}$$ $$1.5806\times 10^{-5}$$ $$M=1, M' =10 $$ $$ 2.9455\times 10^{-9}$$ $$1.1887\times 10^{-11}$$ $$6.8830 \times 10^{-8}$$ $$1.4988\times 10^{-5}$$ $$M=1, M' =12$$ $$1.1270\times 10^{-9}$$ $$1.0570\times 10^{-10} $$ $$2.0730 \times 10^{-8} $$ $$1.4379\times 10^{-5}$$ $$M, M'$$ $$\alpha = \frac{1}{5} $$ $$ \alpha = \frac{1}{4}$$ $$ \alpha = \frac{1}{2}$$ $$\alpha = 1$$ $$M=1, M' =2$$ $$3.4977\times 10^{-3}$$ $$2.1816\times 10^{-3} $$ $$ 2.0015\times 10^{-4} $$ $$ 1.0650\times 10^{-2}$$ $$M=1, M' =4 $$ $$ 6.0548\times 10^{-5} $$ $$3.7983\times 10^{-5}$$ $$3.7836\times 10^{-5}$$ $$8.7676\times 10^{-5}$$ $$M=1, M' =6$$ $$5.0340\times 10^{-7}$$ $$9.5888\times 10^{-8} $$ $$ 5.2497\times 10^{-6}$$ $$1.7175\times 10^{-5}$$ $$M=1, M' =8$$ $$4.2535\times 10^{-8}$$ $$1.8969\times 10^{-8}$$ $$3.0392 \times 10^{-7}$$ $$1.5806\times 10^{-5}$$ $$M=1, M' =10 $$ $$ 2.9455\times 10^{-9}$$ $$1.1887\times 10^{-11}$$ $$6.8830 \times 10^{-8}$$ $$1.4988\times 10^{-5}$$ $$M=1, M' =12$$ $$1.1270\times 10^{-9}$$ $$1.0570\times 10^{-10} $$ $$2.0730 \times 10^{-8} $$ $$1.4379\times 10^{-5}$$ Table 4 Comparison of the estimated value of $$ \tilde{J} $$ for integer orders $$ \nu =\gamma =1 $$ with different values of $$ \alpha $$ for Example $$1$$ $$M, M'$$ $$\alpha = \frac{1}{5} $$ $$ \alpha = \frac{1}{4}$$ $$ \alpha = \frac{1}{2}$$ $$\alpha = 1$$ $$M=1, M' =2$$ $$3.4977\times 10^{-3}$$ $$2.1816\times 10^{-3} $$ $$ 2.0015\times 10^{-4} $$ $$ 1.0650\times 10^{-2}$$ $$M=1, M' =4 $$ $$ 6.0548\times 10^{-5} $$ $$3.7983\times 10^{-5}$$ $$3.7836\times 10^{-5}$$ $$8.7676\times 10^{-5}$$ $$M=1, M' =6$$ $$5.0340\times 10^{-7}$$ $$9.5888\times 10^{-8} $$ $$ 5.2497\times 10^{-6}$$ $$1.7175\times 10^{-5}$$ $$M=1, M' =8$$ $$4.2535\times 10^{-8}$$ $$1.8969\times 10^{-8}$$ $$3.0392 \times 10^{-7}$$ $$1.5806\times 10^{-5}$$ $$M=1, M' =10 $$ $$ 2.9455\times 10^{-9}$$ $$1.1887\times 10^{-11}$$ $$6.8830 \times 10^{-8}$$ $$1.4988\times 10^{-5}$$ $$M=1, M' =12$$ $$1.1270\times 10^{-9}$$ $$1.0570\times 10^{-10} $$ $$2.0730 \times 10^{-8} $$ $$1.4379\times 10^{-5}$$ $$M, M'$$ $$\alpha = \frac{1}{5} $$ $$ \alpha = \frac{1}{4}$$ $$ \alpha = \frac{1}{2}$$ $$\alpha = 1$$ $$M=1, M' =2$$ $$3.4977\times 10^{-3}$$ $$2.1816\times 10^{-3} $$ $$ 2.0015\times 10^{-4} $$ $$ 1.0650\times 10^{-2}$$ $$M=1, M' =4 $$ $$ 6.0548\times 10^{-5} $$ $$3.7983\times 10^{-5}$$ $$3.7836\times 10^{-5}$$ $$8.7676\times 10^{-5}$$ $$M=1, M' =6$$ $$5.0340\times 10^{-7}$$ $$9.5888\times 10^{-8} $$ $$ 5.2497\times 10^{-6}$$ $$1.7175\times 10^{-5}$$ $$M=1, M' =8$$ $$4.2535\times 10^{-8}$$ $$1.8969\times 10^{-8}$$ $$3.0392 \times 10^{-7}$$ $$1.5806\times 10^{-5}$$ $$M=1, M' =10 $$ $$ 2.9455\times 10^{-9}$$ $$1.1887\times 10^{-11}$$ $$6.8830 \times 10^{-8}$$ $$1.4988\times 10^{-5}$$ $$M=1, M' =12$$ $$1.1270\times 10^{-9}$$ $$1.0570\times 10^{-10} $$ $$2.0730 \times 10^{-8} $$ $$1.4379\times 10^{-5}$$ Example 6.2 Consider the following two-dimensional optimal control problem (Mamehrashi & Yousefi, 2016) \begin{equation} min \hspace{.2cm}J[\xi , u] = \frac{1}{2} \int _{0}^{5} \int _{0}^{5} \bigg [ 10^{7} \big( \xi (x, t)- \sin (x+t) \big )^{2} + u^{2}(x, t)\bigg] dx dt, \end{equation} (6.4) subject to \begin{gather} \frac{\partial ^{2}\xi}{\partial x \partial t}(x, t)= - \frac{\partial ^{\nu}\xi}{\partial x^{\nu}}(x, t) -3 \frac{\partial ^{\gamma}\xi}{\partial t^{\gamma}}(x, t) + 0.2 \xi(x, t) + 0.3 u(x, t),\\ \end{gather} (6.5) \begin{gather} \xi (x, 0) = e^{-3x}\cos (2 \pi x), \hspace{1cm} \xi (0, t) = e^{-2t}. \end{gather} (6.6) Here we solve this problem by using the proposed method for $$ l=M $$ and $$ l'=M' $$. In Table 5, a comparison is made between the values of $$ \tilde{J} $$ obtained by the present method with the values of $$\tilde{J}$$ reported in Mamehrashi & Yousefi (2016) with different values of $$ M, M' $$ for $$ \nu = \gamma = 1 $$ and $$ \alpha =1. $$Figure 6(a) and (b) shows the state and control functions for $$ \nu = \gamma =1 $$ with $$ M=1, M'=5 $$ and $$ \alpha =1, $$ respectively. Table 6 presents the values of optimal functional $$\tilde{J}$$ for $$ \nu = \gamma =1 $$ with different choices of $$ \alpha $$. By varying the value of $$ \alpha $$ the obtained state and control functions for $$ \nu = \gamma =1 $$ with $$ M=1, M'=5 $$ are shown, in Figs 7 and 8, respectively. For partial fractional orders, two cases are considered. Initially we take $$ \nu =\gamma =0.8 $$. Table 7 shows the values of optimal performance index functional $$ \tilde{J} $$ for various values of $$M $$ and $$ M' $$ with $$ \alpha =\frac{1}{2}, 1 $$. In addition, Fig. 9(a) and (b) displays state and control functions for $$ \nu = \gamma =0.8 $$ with $$ M=1, M'=5 $$ and $$ \alpha =1, $$ respectively. At last, we take $$ \nu =0.75 $$ and $$ \gamma =0.8 $$. The outcomes of optimal functional are shown in Table 8. Fig. 6. View largeDownload slide (a) The state function and (b) the control input for $$ \nu =\gamma =1, \alpha = 1$$ with $$ M=1, M' =5 $$ for Example 2. Fig. 6. View largeDownload slide (a) The state function and (b) the control input for $$ \nu =\gamma =1, \alpha = 1$$ with $$ M=1, M' =5 $$ for Example 2. Fig. 7. View largeDownload slide The state function for (a) $$ \alpha = \frac{1}{5}, $$ (b) $$ \alpha = \frac{1}{2}, $$ (c) $$ \alpha = 1 $$ and (d) $$ \alpha = 2, $$ with $$ \nu = \gamma =1 $$ and $$ M=1, M' =5 $$ for Example 2. Fig. 7. View largeDownload slide The state function for (a) $$ \alpha = \frac{1}{5}, $$ (b) $$ \alpha = \frac{1}{2}, $$ (c) $$ \alpha = 1 $$ and (d) $$ \alpha = 2, $$ with $$ \nu = \gamma =1 $$ and $$ M=1, M' =5 $$ for Example 2. Fig. 8. View largeDownload slide The control function for (a) $$ \alpha = \frac{1}{5}, $$ (b) $$ \alpha = \frac{1}{2}, $$ (c) $$ \alpha = 1 $$ and (d) $$ \alpha = 2, $$ with $$ \nu = \gamma =1 $$ and $$ M=1, M' =5 $$ for Example 2. Fig. 8. View largeDownload slide The control function for (a) $$ \alpha = \frac{1}{5}, $$ (b) $$ \alpha = \frac{1}{2}, $$ (c) $$ \alpha = 1 $$ and (d) $$ \alpha = 2, $$ with $$ \nu = \gamma =1 $$ and $$ M=1, M' =5 $$ for Example 2. Fig. 9. View largeDownload slide (a) The state function and (b) the control input for $$ \nu =\gamma =0.8, \alpha = 1$$ with $$ M=1, M' =5 $$ for Example 2. Fig. 9. View largeDownload slide (a) The state function and (b) the control input for $$ \nu =\gamma =0.8, \alpha = 1$$ with $$ M=1, M' =5 $$ for Example 2. Table 5 Comparison of the estimated values of $$ \tilde{J} $$ for integer orders $$ \nu = \gamma =1 $$ and $$ \alpha =1 $$ with different values of $$ M $$ and $$ M' $$ for Example 2 Method of Mamehrashi & Yousefi, 2016 $$\tilde{J}$$ $$M=6, M'=6 $$ $$ 2.03080\times 10^{6} $$ $$M=7, M'= 6 $$ $$ 1.82721\times 10^{6}$$ $$M=6, M'= 8 $$ $$ 1.65595\times 10^{6} $$ $$M=7, M'= 7 $$ $$ 1.60773\times 10^{6} $$ $$M=6, M'=9 $$ $$ 1.54056\times 10^{6} $$ $$M=7, M'=8 $$ $$ 1.45170\times 10^{6} $$ $$M=7, M'=9 $$ $$ 1.33534\times 10^{6} $$ $$M=8, M'= 8 $$ $$ 1.30907\times 10^{6} $$ $$Present \; method $$ $$\tilde{J} $$ $$M=1, M' =1 $$ $$1.5407\times 10^{-24} $$ $$M=1, M' =2 $$ $$ 1.1707\times 10^{-23}$$ $$M=1, M' =4 $$ $$ 8.9944\times 10^{-23}$$ $$M=1, M' =5 $$ $$1.2681\times 10^{-22}$$ $$M=1, M' =8 $$ $$1.3831\times 10^{-21}$$ $$M=1, M' =10 $$ $$1.0888\times 10^{-21}$$ $$M=1, M' =12 $$ $$1.0542\times 10^{-21}$$ Method of Mamehrashi & Yousefi, 2016 $$\tilde{J}$$ $$M=6, M'=6 $$ $$ 2.03080\times 10^{6} $$ $$M=7, M'= 6 $$ $$ 1.82721\times 10^{6}$$ $$M=6, M'= 8 $$ $$ 1.65595\times 10^{6} $$ $$M=7, M'= 7 $$ $$ 1.60773\times 10^{6} $$ $$M=6, M'=9 $$ $$ 1.54056\times 10^{6} $$ $$M=7, M'=8 $$ $$ 1.45170\times 10^{6} $$ $$M=7, M'=9 $$ $$ 1.33534\times 10^{6} $$ $$M=8, M'= 8 $$ $$ 1.30907\times 10^{6} $$ $$Present \; method $$ $$\tilde{J} $$ $$M=1, M' =1 $$ $$1.5407\times 10^{-24} $$ $$M=1, M' =2 $$ $$ 1.1707\times 10^{-23}$$ $$M=1, M' =4 $$ $$ 8.9944\times 10^{-23}$$ $$M=1, M' =5 $$ $$1.2681\times 10^{-22}$$ $$M=1, M' =8 $$ $$1.3831\times 10^{-21}$$ $$M=1, M' =10 $$ $$1.0888\times 10^{-21}$$ $$M=1, M' =12 $$ $$1.0542\times 10^{-21}$$ Table 5 Comparison of the estimated values of $$ \tilde{J} $$ for integer orders $$ \nu = \gamma =1 $$ and $$ \alpha =1 $$ with different values of $$ M $$ and $$ M' $$ for Example 2 Method of Mamehrashi & Yousefi, 2016 $$\tilde{J}$$ $$M=6, M'=6 $$ $$ 2.03080\times 10^{6} $$ $$M=7, M'= 6 $$ $$ 1.82721\times 10^{6}$$ $$M=6, M'= 8 $$ $$ 1.65595\times 10^{6} $$ $$M=7, M'= 7 $$ $$ 1.60773\times 10^{6} $$ $$M=6, M'=9 $$ $$ 1.54056\times 10^{6} $$ $$M=7, M'=8 $$ $$ 1.45170\times 10^{6} $$ $$M=7, M'=9 $$ $$ 1.33534\times 10^{6} $$ $$M=8, M'= 8 $$ $$ 1.30907\times 10^{6} $$ $$Present \; method $$ $$\tilde{J} $$ $$M=1, M' =1 $$ $$1.5407\times 10^{-24} $$ $$M=1, M' =2 $$ $$ 1.1707\times 10^{-23}$$ $$M=1, M' =4 $$ $$ 8.9944\times 10^{-23}$$ $$M=1, M' =5 $$ $$1.2681\times 10^{-22}$$ $$M=1, M' =8 $$ $$1.3831\times 10^{-21}$$ $$M=1, M' =10 $$ $$1.0888\times 10^{-21}$$ $$M=1, M' =12 $$ $$1.0542\times 10^{-21}$$ Method of Mamehrashi & Yousefi, 2016 $$\tilde{J}$$ $$M=6, M'=6 $$ $$ 2.03080\times 10^{6} $$ $$M=7, M'= 6 $$ $$ 1.82721\times 10^{6}$$ $$M=6, M'= 8 $$ $$ 1.65595\times 10^{6} $$ $$M=7, M'= 7 $$ $$ 1.60773\times 10^{6} $$ $$M=6, M'=9 $$ $$ 1.54056\times 10^{6} $$ $$M=7, M'=8 $$ $$ 1.45170\times 10^{6} $$ $$M=7, M'=9 $$ $$ 1.33534\times 10^{6} $$ $$M=8, M'= 8 $$ $$ 1.30907\times 10^{6} $$ $$Present \; method $$ $$\tilde{J} $$ $$M=1, M' =1 $$ $$1.5407\times 10^{-24} $$ $$M=1, M' =2 $$ $$ 1.1707\times 10^{-23}$$ $$M=1, M' =4 $$ $$ 8.9944\times 10^{-23}$$ $$M=1, M' =5 $$ $$1.2681\times 10^{-22}$$ $$M=1, M' =8 $$ $$1.3831\times 10^{-21}$$ $$M=1, M' =10 $$ $$1.0888\times 10^{-21}$$ $$M=1, M' =12 $$ $$1.0542\times 10^{-21}$$ Table 6 Comparison of the estimated values of $$ \tilde{J} $$ for integer orders $$ \nu =\gamma =1 $$ with different values of $$ \alpha $$ for Example 2 $$M, M'$$ $$\alpha = \frac{1}{2}$$ $$\alpha = \frac{1}{4}$$ $$\alpha = 1$$ $$\alpha = 2$$ $$M=1, M' =2 $$ $$1.2907 \times 10^{-29}$$ $$1.1500\times 10^{-23}$$ $$1.1707\times 10^{-23}$$ $$4.6001 \times 10^{-23}$$ $$M=1, M' =4$$ $$2.3121\times 10^{-22}$$ $$1.4715\times 10^{-20}$$ $$8.9944\times 10^{-23}$$ $$9.3637 \times 10^{8}$$ $$M=1, M' =5$$ $$1.5279\times 10^{-23}$$ $$6.6134\times 10^{-20}$$ $$1.2681\times 10^{-22}$$ $$1.1247 \times 10^{9}$$ $$M=1, M' =8$$ $$3.9518\times 10^{-22}$$ $$1.5329 \times 10^{9}$$ $$1.3831 \times 10^{-21}$$ $$1.5329 \times 10^{9}$$ $$M=1, M' =10$$ $$1.7069 \times 10^{-21}$$ $$1.8313\times 10^{-4}$$ $$1.0888 \times 10^{-21}$$ $$1.7312\times 10^{9}$$ $$M=1, M' =12$$ $$5.4848\times 10^{-21}$$ $$1.0738\times 10^{-5}$$ $$1.0542 \times 10^{-21}$$ $$1.8949 \times 10^{9}$$ $$M, M'$$ $$\alpha = \frac{1}{2}$$ $$\alpha = \frac{1}{4}$$ $$\alpha = 1$$ $$\alpha = 2$$ $$M=1, M' =2 $$ $$1.2907 \times 10^{-29}$$ $$1.1500\times 10^{-23}$$ $$1.1707\times 10^{-23}$$ $$4.6001 \times 10^{-23}$$ $$M=1, M' =4$$ $$2.3121\times 10^{-22}$$ $$1.4715\times 10^{-20}$$ $$8.9944\times 10^{-23}$$ $$9.3637 \times 10^{8}$$ $$M=1, M' =5$$ $$1.5279\times 10^{-23}$$ $$6.6134\times 10^{-20}$$ $$1.2681\times 10^{-22}$$ $$1.1247 \times 10^{9}$$ $$M=1, M' =8$$ $$3.9518\times 10^{-22}$$ $$1.5329 \times 10^{9}$$ $$1.3831 \times 10^{-21}$$ $$1.5329 \times 10^{9}$$ $$M=1, M' =10$$ $$1.7069 \times 10^{-21}$$ $$1.8313\times 10^{-4}$$ $$1.0888 \times 10^{-21}$$ $$1.7312\times 10^{9}$$ $$M=1, M' =12$$ $$5.4848\times 10^{-21}$$ $$1.0738\times 10^{-5}$$ $$1.0542 \times 10^{-21}$$ $$1.8949 \times 10^{9}$$ Table 6 Comparison of the estimated values of $$ \tilde{J} $$ for integer orders $$ \nu =\gamma =1 $$ with different values of $$ \alpha $$ for Example 2 $$M, M'$$ $$\alpha = \frac{1}{2}$$ $$\alpha = \frac{1}{4}$$ $$\alpha = 1$$ $$\alpha = 2$$ $$M=1, M' =2 $$ $$1.2907 \times 10^{-29}$$ $$1.1500\times 10^{-23}$$ $$1.1707\times 10^{-23}$$ $$4.6001 \times 10^{-23}$$ $$M=1, M' =4$$ $$2.3121\times 10^{-22}$$ $$1.4715\times 10^{-20}$$ $$8.9944\times 10^{-23}$$ $$9.3637 \times 10^{8}$$ $$M=1, M' =5$$ $$1.5279\times 10^{-23}$$ $$6.6134\times 10^{-20}$$ $$1.2681\times 10^{-22}$$ $$1.1247 \times 10^{9}$$ $$M=1, M' =8$$ $$3.9518\times 10^{-22}$$ $$1.5329 \times 10^{9}$$ $$1.3831 \times 10^{-21}$$ $$1.5329 \times 10^{9}$$ $$M=1, M' =10$$ $$1.7069 \times 10^{-21}$$ $$1.8313\times 10^{-4}$$ $$1.0888 \times 10^{-21}$$ $$1.7312\times 10^{9}$$ $$M=1, M' =12$$ $$5.4848\times 10^{-21}$$ $$1.0738\times 10^{-5}$$ $$1.0542 \times 10^{-21}$$ $$1.8949 \times 10^{9}$$ $$M, M'$$ $$\alpha = \frac{1}{2}$$ $$\alpha = \frac{1}{4}$$ $$\alpha = 1$$ $$\alpha = 2$$ $$M=1, M' =2 $$ $$1.2907 \times 10^{-29}$$ $$1.1500\times 10^{-23}$$ $$1.1707\times 10^{-23}$$ $$4.6001 \times 10^{-23}$$ $$M=1, M' =4$$ $$2.3121\times 10^{-22}$$ $$1.4715\times 10^{-20}$$ $$8.9944\times 10^{-23}$$ $$9.3637 \times 10^{8}$$ $$M=1, M' =5$$ $$1.5279\times 10^{-23}$$ $$6.6134\times 10^{-20}$$ $$1.2681\times 10^{-22}$$ $$1.1247 \times 10^{9}$$ $$M=1, M' =8$$ $$3.9518\times 10^{-22}$$ $$1.5329 \times 10^{9}$$ $$1.3831 \times 10^{-21}$$ $$1.5329 \times 10^{9}$$ $$M=1, M' =10$$ $$1.7069 \times 10^{-21}$$ $$1.8313\times 10^{-4}$$ $$1.0888 \times 10^{-21}$$ $$1.7312\times 10^{9}$$ $$M=1, M' =12$$ $$5.4848\times 10^{-21}$$ $$1.0738\times 10^{-5}$$ $$1.0542 \times 10^{-21}$$ $$1.8949 \times 10^{9}$$ Table 7 Comparison of the estimated values of $$ \tilde{J} $$ for orders $$ \nu =\gamma =0.8 $$ with $$ \alpha = \frac{1}{2}, 1 $$ for Example 2 $$M, M' $$ $$\alpha = \frac{1}{2} $$ $$ \alpha = 1$$ $$M=1, M' =1 $$ $$ 0 $$ $$ 2.95\times 10^{-30} $$ $$M=1, M' =3 $$ $$ 2.17\times 10^{-25}-1.11\times 10^{-30}i $$ $$ 6.12\times 10^{-24}+6.71\times 10^{-29}i $$ $$M=1, M' =5 $$ $$ 8.59\times 10^{-16}-2.63\times 10^{-24}i $$ $$ 6.71\times 10^{-23}-4.97\times 10^{-31}i $$ $$M=1, M' =7 $$ $$ 2.95\times 10^{-22}+2.55\times 10^{-30}i $$ $$ 5.90\times 10^{-23}-6.50\times 10^{-31}i $$ $$M=1, M' =9 $$ $$ 2.49\times 10^{-21}-5.08\times 10^{-29}i $$ $$ 5.06\times 10^{-22}+2.29\times 10^{-29}i $$ $$M=1, M' =11$$ $$5.84\times 10^{-21}+1.11\times 10^{-29}i $$ $$2.47\times 10^{-23}+1.70\times 10^{-29}i$$ $$M, M' $$ $$\alpha = \frac{1}{2} $$ $$ \alpha = 1$$ $$M=1, M' =1 $$ $$ 0 $$ $$ 2.95\times 10^{-30} $$ $$M=1, M' =3 $$ $$ 2.17\times 10^{-25}-1.11\times 10^{-30}i $$ $$ 6.12\times 10^{-24}+6.71\times 10^{-29}i $$ $$M=1, M' =5 $$ $$ 8.59\times 10^{-16}-2.63\times 10^{-24}i $$ $$ 6.71\times 10^{-23}-4.97\times 10^{-31}i $$ $$M=1, M' =7 $$ $$ 2.95\times 10^{-22}+2.55\times 10^{-30}i $$ $$ 5.90\times 10^{-23}-6.50\times 10^{-31}i $$ $$M=1, M' =9 $$ $$ 2.49\times 10^{-21}-5.08\times 10^{-29}i $$ $$ 5.06\times 10^{-22}+2.29\times 10^{-29}i $$ $$M=1, M' =11$$ $$5.84\times 10^{-21}+1.11\times 10^{-29}i $$ $$2.47\times 10^{-23}+1.70\times 10^{-29}i$$ Table 7 Comparison of the estimated values of $$ \tilde{J} $$ for orders $$ \nu =\gamma =0.8 $$ with $$ \alpha = \frac{1}{2}, 1 $$ for Example 2 $$M, M' $$ $$\alpha = \frac{1}{2} $$ $$ \alpha = 1$$ $$M=1, M' =1 $$ $$ 0 $$ $$ 2.95\times 10^{-30} $$ $$M=1, M' =3 $$ $$ 2.17\times 10^{-25}-1.11\times 10^{-30}i $$ $$ 6.12\times 10^{-24}+6.71\times 10^{-29}i $$ $$M=1, M' =5 $$ $$ 8.59\times 10^{-16}-2.63\times 10^{-24}i $$ $$ 6.71\times 10^{-23}-4.97\times 10^{-31}i $$ $$M=1, M' =7 $$ $$ 2.95\times 10^{-22}+2.55\times 10^{-30}i $$ $$ 5.90\times 10^{-23}-6.50\times 10^{-31}i $$ $$M=1, M' =9 $$ $$ 2.49\times 10^{-21}-5.08\times 10^{-29}i $$ $$ 5.06\times 10^{-22}+2.29\times 10^{-29}i $$ $$M=1, M' =11$$ $$5.84\times 10^{-21}+1.11\times 10^{-29}i $$ $$2.47\times 10^{-23}+1.70\times 10^{-29}i$$ $$M, M' $$ $$\alpha = \frac{1}{2} $$ $$ \alpha = 1$$ $$M=1, M' =1 $$ $$ 0 $$ $$ 2.95\times 10^{-30} $$ $$M=1, M' =3 $$ $$ 2.17\times 10^{-25}-1.11\times 10^{-30}i $$ $$ 6.12\times 10^{-24}+6.71\times 10^{-29}i $$ $$M=1, M' =5 $$ $$ 8.59\times 10^{-16}-2.63\times 10^{-24}i $$ $$ 6.71\times 10^{-23}-4.97\times 10^{-31}i $$ $$M=1, M' =7 $$ $$ 2.95\times 10^{-22}+2.55\times 10^{-30}i $$ $$ 5.90\times 10^{-23}-6.50\times 10^{-31}i $$ $$M=1, M' =9 $$ $$ 2.49\times 10^{-21}-5.08\times 10^{-29}i $$ $$ 5.06\times 10^{-22}+2.29\times 10^{-29}i $$ $$M=1, M' =11$$ $$5.84\times 10^{-21}+1.11\times 10^{-29}i $$ $$2.47\times 10^{-23}+1.70\times 10^{-29}i$$ Table 8 Comparison of the estimated values of $$ \tilde{J} $$ for orders $$ \nu =0.75, \gamma =0.8 $$ with $$ \alpha =\frac{1}{2}, 1 $$ for Example 2 $$M, M'$$ $$ \alpha = \frac{1}{2} $$ $$ \alpha = 1$$ $$M=1, M' =1 $$ $$ 3.70\times 10^{-22} $$ $$ 2.17 \times 10^{-21} $$ $$M=1, M' =3 $$ $$ 9.02\times 10^{-18}+3.28\times 10^{-25}i $$ $$ 3.31\times 10^{-23}-7.25\times 10^{-31}i $$ $$M=1, M' =5 $$ $$ 4.78\times 10^{-22}+2.88\times 10^{-29}i $$ $$ 1.69\times 10^{-23}+1.92\times 10^{-29}i $$ $$M=1, M' =7 $$ $$ 5.76\times 10^{-22}+1.18\times 10^{-29}i $$ $$ 2.00\times 10^{-22}-3.66\times 10^{-38}i $$ $$M=1, M' =9 $$ $$ 3.66\times 10^{-22}-7.02\times 10^{-29}i $$ $$ 2.91\times 10^{-22}+4.23\times 10^{-29}i $$ $$M=1, M' =11$$ $$3.90\times 10^{-21}-3.25\times 10^{-31}i$$ $$1.68 \times 10^{-22}-3.84\times 10^{-30}i$$ $$M, M'$$ $$ \alpha = \frac{1}{2} $$ $$ \alpha = 1$$ $$M=1, M' =1 $$ $$ 3.70\times 10^{-22} $$ $$ 2.17 \times 10^{-21} $$ $$M=1, M' =3 $$ $$ 9.02\times 10^{-18}+3.28\times 10^{-25}i $$ $$ 3.31\times 10^{-23}-7.25\times 10^{-31}i $$ $$M=1, M' =5 $$ $$ 4.78\times 10^{-22}+2.88\times 10^{-29}i $$ $$ 1.69\times 10^{-23}+1.92\times 10^{-29}i $$ $$M=1, M' =7 $$ $$ 5.76\times 10^{-22}+1.18\times 10^{-29}i $$ $$ 2.00\times 10^{-22}-3.66\times 10^{-38}i $$ $$M=1, M' =9 $$ $$ 3.66\times 10^{-22}-7.02\times 10^{-29}i $$ $$ 2.91\times 10^{-22}+4.23\times 10^{-29}i $$ $$M=1, M' =11$$ $$3.90\times 10^{-21}-3.25\times 10^{-31}i$$ $$1.68 \times 10^{-22}-3.84\times 10^{-30}i$$ Table 8 Comparison of the estimated values of $$ \tilde{J} $$ for orders $$ \nu =0.75, \gamma =0.8 $$ with $$ \alpha =\frac{1}{2}, 1 $$ for Example 2 $$M, M'$$ $$ \alpha = \frac{1}{2} $$ $$ \alpha = 1$$ $$M=1, M' =1 $$ $$ 3.70\times 10^{-22} $$ $$ 2.17 \times 10^{-21} $$ $$M=1, M' =3 $$ $$ 9.02\times 10^{-18}+3.28\times 10^{-25}i $$ $$ 3.31\times 10^{-23}-7.25\times 10^{-31}i $$ $$M=1, M' =5 $$ $$ 4.78\times 10^{-22}+2.88\times 10^{-29}i $$ $$ 1.69\times 10^{-23}+1.92\times 10^{-29}i $$ $$M=1, M' =7 $$ $$ 5.76\times 10^{-22}+1.18\times 10^{-29}i $$ $$ 2.00\times 10^{-22}-3.66\times 10^{-38}i $$ $$M=1, M' =9 $$ $$ 3.66\times 10^{-22}-7.02\times 10^{-29}i $$ $$ 2.91\times 10^{-22}+4.23\times 10^{-29}i $$ $$M=1, M' =11$$ $$3.90\times 10^{-21}-3.25\times 10^{-31}i$$ $$1.68 \times 10^{-22}-3.84\times 10^{-30}i$$ $$M, M'$$ $$ \alpha = \frac{1}{2} $$ $$ \alpha = 1$$ $$M=1, M' =1 $$ $$ 3.70\times 10^{-22} $$ $$ 2.17 \times 10^{-21} $$ $$M=1, M' =3 $$ $$ 9.02\times 10^{-18}+3.28\times 10^{-25}i $$ $$ 3.31\times 10^{-23}-7.25\times 10^{-31}i $$ $$M=1, M' =5 $$ $$ 4.78\times 10^{-22}+2.88\times 10^{-29}i $$ $$ 1.69\times 10^{-23}+1.92\times 10^{-29}i $$ $$M=1, M' =7 $$ $$ 5.76\times 10^{-22}+1.18\times 10^{-29}i $$ $$ 2.00\times 10^{-22}-3.66\times 10^{-38}i $$ $$M=1, M' =9 $$ $$ 3.66\times 10^{-22}-7.02\times 10^{-29}i $$ $$ 2.91\times 10^{-22}+4.23\times 10^{-29}i $$ $$M=1, M' =11$$ $$3.90\times 10^{-21}-3.25\times 10^{-31}i$$ $$1.68 \times 10^{-22}-3.84\times 10^{-30}i$$ 7. Conclusion In the present work, a new scheme for two-dimensional FOCPs is introduced. In this method, general formulations for the Riemann–Liouville fractional integral operators of generalized fractional-order Bernoulli and generalized fractional-order Legendre have been derived. First the given problem is transformed into an equivalent variational problem, then the variational problem is solved approximately by utilizing the generalized fractional-order Bernoulli–Legendre functions, Riemann-Liouville fractional integral operators, two-dimensional LG quadrature rule and Newton’s iterative formula for solving the non-linear system of equations. Illustrative examples are given to demonstrate the validity and applicability of the proposed method. Moreover, only a small number of generalized fractional-order Bernoulli–Legendre functions is needed to obtain a satisfactory result. 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