Numerical scheme for Erdélyi–Kober fractional diffusion equation using Galerkin–Hermite method

Abstract

The aim of this work is to devise and analyse an accurate numerical scheme to solve Erdélyi–Kober fractional diffusion equation. This solution can be thought as the marginal pdf of the stochastic process called the generalized grey Brownian motion (ggBm). The ggBm includes some well-known stochastic processes: Brownian motion, fractional Brownian motion, and grey Brownian motion. To obtain a convergent numerical scheme we transform the fractional diffusion equation into its weak form and apply the discretization of the Erdélyi–Kober fractional derivative. We prove the stability of the solution of the semi-discrete problem and its convergence to the exact solution. Due to the singularity in the time term appearing in the main equation, the proposed method converges slower than first order. Finally, we provide a numerical analysis of the full-discrete problem using orthogonal expansion in terms of Hermite functions.

Appendix

In the following section, we present a rigorous proof of Theorem 1.

Proof of Theorem 1

The asymptotic relation for Erdélyi–Kober fractional integral operator $I_{\alpha /\beta }^{0,\beta }$ was given in [46], therefore we will not provide a detailed proof of it here.

Since we know the order of discretization of Erdélyi–Kober fractional integral operator, we proceed to find the order of discrete representations of the operator $D_{\alpha /\beta }^{\beta -1,1-\beta }$. At the beginning, let us consider the operator $G_{\alpha /\beta }^{\beta -1,1-\beta }$. Note that Erdélyi–Kober fractional differential operator can be rewritten as

$$\begin{aligned} D_{\alpha /\beta }^{\beta -1,1-\beta }\psi (t_n)&= \beta I_{\alpha /\beta }^{0,\beta }\psi (t_n)+\frac{\beta }{\alpha }t_n\frac{d}{d t}I_{\alpha /\beta }^{0,\beta }\psi (t_n) =\beta L_{\alpha /\beta }^{0,\beta }\psi (t_n) +\beta R_n\\&\quad + \frac{\beta }{\alpha }t_n\frac{d}{d t}I_{\alpha /\beta }^{0,\beta }\psi (t_n)\\&= \beta L_{\alpha /\beta }^{0,\beta }\psi (t_n) +\beta R_n + \frac{\beta }{\alpha } t_n {\overline{\partial }} L_{\alpha /\beta }^{0,\beta }\psi (t_n) + \frac{\beta }{\alpha }{\widehat{R}}_n, \end{aligned}$$

where $R_n$ is a discretization error of the Erdélyi–Kober fractional integral operator (2.6) and ${\widehat{R}}_n= t_n (\frac{d}{d t}I_{\alpha /\beta }^{0,\beta }\psi (t_n) - {\overline{\partial }}I_{\alpha /\beta }^{0,\beta }\psi (t_n)+ {\overline{\partial }}I_{\alpha /\beta }^{0,\beta }\psi (t_n)- {\overline{\partial }}L_{\alpha /\beta }^{0,\beta }\psi (t_n))={\widehat{R}}_{n,1}+{\widehat{R}}_{n,2}$. The first difference, ${\widehat{R}}_{n,1}$, can be estimated in a standard way

$$\begin{aligned} {\widehat{R}}_{n,1}&= \frac{d}{d t}I_{\alpha /\beta }^{0,\beta }\psi (t_n) - \frac{1}{k}(I_{\alpha /\beta }^{0,\beta }\psi (t_n)-I_{\alpha /\beta }^{0,\beta }\psi (t_{n-1}))\\&= \frac{1}{n} \frac{d^2}{d t^2} I_{\alpha /\beta }^{0,\beta }\psi (t^{*})= \frac{\varGamma \bigl (\frac{2 \beta }{\alpha }+1\bigr )}{\varGamma \bigl (\frac{2 \beta }{\alpha }+\beta +1\bigr )} \psi ''(\sigma ^* t^*) \frac{1}{n}, \end{aligned}$$

where $\sigma ^* \in (0,1)$ and $t^* \in [t_{n-1},t_n]$. However, to estimate the order of the second term in ${\widehat{R}}_n$, we need to investigate this term carefully. Note that using again the Mean Value Theorem for Integrals and Sums, and the relation $c_{n,i}<c_{n-1,i}$ for $i\in \{1,2,\ldots ,n-1\}$ we have

$$\begin{aligned} {\widehat{R}}_{n,2}&= \frac{t_n}{k}\sum \limits _{i=1}^{n-1} \frac{\alpha }{\beta \varGamma (\beta )}\int \limits _{i-1}^{i}(f(n,s)-f(n-1,s))(\psi (i k)-\psi (s k))\, d s\\&\quad + \frac{t_n}{k}\frac{\alpha }{\beta \varGamma (\beta )}\int \limits _{n-1}^{n} f(n,s)(\psi (n k)-\psi (s k))\, d s\\&= t_n \psi '(x k) \sum \limits _{i=1}^{n-1} \frac{\alpha }{\beta \varGamma (\beta )}\int \limits _{i-1}^{i}(f(n,s)-f(n-1,s))(i-s)\, d s\\&\quad + t_n \psi '(x_n k)\frac{\alpha }{\beta \varGamma (\beta )}\int \limits _{n-1}^{n}f(n,s)(n-s)\, d s, \end{aligned}$$

where $i-1\le s<\sigma _i <i$, $0<x<n-1,\ n-1<x_n<n$ and

$$\begin{aligned} f(n, s):= \frac{1}{n}\biggl (\frac{s}{n}\biggr )^{\frac{\alpha }{\beta }-1} \biggl (1-\biggl (\frac{s}{n}\biggr )^{\frac{\alpha }{\beta }}\biggr )^{\beta -1}. \end{aligned}$$

Furthermore, performing a straightforward calculation, we get

$$\begin{aligned} C_n:=\sum \limits _{i=1}^{n} \frac{\alpha }{\beta \varGamma (\beta )}\int \limits _{i-1}^{i}f(n,s) (i-s) \, d s\le \sum \limits _{i=1}^{n} \frac{\alpha }{\beta \varGamma (\beta )}\int \limits _{i-1}^{i}f(n,s) \, d s=\frac{1}{\beta \varGamma (\beta )}. \end{aligned}$$

Moreover, note that $C_n$ can also be written in the following way

$$\begin{aligned} C_n= \sum \limits _{i=0}^{n-1} \frac{1}{\beta \varGamma (\beta )} \biggl (1-\biggl (\frac{i}{n}\biggr )^{\alpha /\beta }\biggr )^\beta - \frac{n \varGamma \bigl (\frac{\beta }{\alpha }+1\bigr )}{\varGamma \bigl (\frac{\beta }{\alpha }+\beta +1\bigr )}. \end{aligned}$$

(6.1)

Next, we use the Intermediate Value Theorem to get

$$\begin{aligned} \begin{aligned}&\alpha \biggl (\psi '(x k) \sum \limits _{i=1}^{n-1} \int \limits _{i-1}^{i}f(n,s) (i-s)\, d s + \psi '(x_n k) \int \limits _{n-1}^{n} f(n,s)(n-s)\, d s \biggr )\\&\quad = \beta \varGamma (\beta )\psi '(x^* k) C_n, \end{aligned} \end{aligned}$$

(6.2)

where $x^*\in (x, x_n)$. Finally, we have

$$\begin{aligned} \begin{aligned} {\widehat{R}}_{n,2}&= t_n (\psi '(x^* k) C_n - \psi '(x k) C_{n-1}) \\&= t_n \psi '(x^* k) (C_n-C_{n-1})+ t_n C_n (\psi '(x^* k)-\psi '(x k))\\&= t_n \psi '(x^* k) (C_n-C_{n-1})+ t_n C_n (x^*-x) k \psi ''(x^{* *} k), \end{aligned} \end{aligned}$$

where $x^{* *}\in (x,x^*)$. Let us notice that the difference $|x-x^*|$ does not increase when n becomes larger: $\psi$ is continuously differentiable on the interval $[0,t_n]$ and due to the fact that when increasing n the contribution of the integral in the second term in (6.2) to $C_n$ becomes negligible. Next, we use (6.1) to obtain

$$\begin{aligned} \begin{aligned} \frac{\beta \varGamma (\beta )}{\alpha }(C_n-C_{n-1})=&\sum \limits _{i=0}^{n-1} \frac{1}{\alpha }\biggl (1-\biggl (\frac{i}{n}\biggr )^{\alpha /\beta }\biggr )^\beta -\frac{1}{\alpha }\biggl (1-\biggl (\frac{i}{n-1}\biggr )^{\alpha /\beta }\biggr )^\beta \\&-\frac{\beta \varGamma (\beta ) \varGamma \bigl (1+\frac{\beta }{\alpha }\bigr )}{\alpha \varGamma \bigl (\frac{\beta }{\alpha }+\beta +1\bigr )}\\ =&\sum \limits _{i=0}^{n-1} \biggl [ \frac{1}{\alpha }\biggl (1-\biggl (\frac{i}{n}\biggr )^{\alpha /\beta }\biggr )^\beta -\frac{1}{\alpha }\biggl (1-\biggl (\frac{i}{n-1}\biggr )^{\alpha /\beta }\biggr )^\beta \\&\ \qquad -\frac{i}{n}\int \limits _{\frac{i}{n}}^{\frac{i+1}{n}} s^{\frac{\alpha }{\beta }-1}(1-s^{\frac{\alpha }{\beta }})^{\beta -1} \, d s\biggr ] \\&+ \sum \limits _{i=0}^{n-1} \frac{i}{n}\int \limits _{\frac{i}{n}}^{\frac{i+1}{n}} s^{\frac{\alpha }{\beta }-1}(1-s^{\frac{\alpha }{\beta }})^{\beta -1}\biggl (\frac{i}{n}-s\biggr ) \, d s =S_1 + S_2. \end{aligned} \end{aligned}$$

Note that the absolute value of the second term can be easily bounded from above as follows

$$\begin{aligned} |S_2|&\le \sum \limits _{i=0}^{n-1}\int \limits _{\frac{i}{n}}^{\frac{i+1}{n}} s^{\frac{\alpha }{\beta }-1}(1-s^{\frac{\alpha }{\beta }})^{\beta -1} \biggl (s-\frac{i}{n}\biggr )\, d s\\&\le \frac{1}{n} \sum \limits _{i=0}^{n-1} \int \limits _{\frac{i}{n}}^{\frac{i+1}{n}} s^{\frac{\alpha }{\beta }-1}(1-s^{\frac{\alpha }{\beta }})^{\beta -1} \, d s=\frac{1}{\alpha } \frac{1}{n}, \end{aligned}$$

where the last inequality was obtained using $0 \le n s-i\le 1$. Considering now $S_1$, we evaluate the integral for each i to get

$$\begin{aligned} S_1&= \sum \limits _{i=1}^n \biggl [ \frac{1}{\alpha }\biggl (1-\biggl (\frac{i}{n}\biggr )^{\alpha /\beta }\biggr )^\beta -\frac{1}{\alpha }\biggl (1-\biggl (\frac{i}{n-1}\biggr )^{\alpha /\beta }\biggr )^\beta \\&\quad -\frac{i}{n} \biggl (\frac{1}{\alpha }\biggl (1-\biggl (\frac{i}{n}\biggr )^{\alpha /\beta }\biggr )^{\beta }-\frac{1}{\alpha }\biggl (1-\biggl (\frac{i+1}{n}\biggr )^{\alpha /\beta }\biggr )^{\beta }\biggr )\biggr ] =\sum \limits _{i=1}^n \theta _i. \end{aligned}$$

Next, for each $i\in \{1,2,\ldots ,n-2\}$ we expand $\theta _i$ in the Taylor series for large n with fixed i/n

$$\begin{aligned} \theta _i= & {} -\frac{1}{2 \beta n^2}\biggl (1-\biggl (\frac{i}{n}\biggr )^{\alpha /\beta }\biggr )^{\beta -2} \biggl (\frac{i}{n}\biggr )^{\frac{\alpha }{\beta }-1} \biggl ( \beta \biggl (\alpha \biggl (\frac{i}{n}-1\biggr )+\frac{i}{n}+1\biggr ) \biggl (\frac{i}{n}\biggr )^{\alpha /\beta }\\&+ \alpha -\beta -\frac{\alpha i}{n}-\frac{\beta i}{n}\biggr )\\&+{\mathcal {O}}\biggl (\frac{(1-\bigl (\frac{i}{n}\bigr ))^{\beta -2} \bigl (\frac{i}{n}\bigr )^{\frac{\alpha }{\beta }-2}}{n^3}\biggr ). \end{aligned}$$

Let us notice that for sufficiently large n we can bound $|\theta _i|$ from above as follows

$$\begin{aligned} \begin{aligned} |\theta _i|&\le C \frac{1}{2 \beta n^2}\biggl (1-\biggl (\frac{i}{n}\biggr )^{\alpha /\beta }\biggr )^{\beta -2} \biggl (\frac{i}{n}\biggr )^{\frac{\alpha }{\beta }-1} \biggl |\biggl ( \beta \biggl (\alpha \biggl (\frac{i}{n}-1\biggr )+\frac{i}{n}+1\biggr ) \biggl (\frac{i}{n}\biggr )^{\alpha /\beta }\\&\quad + \alpha -\beta -\frac{ i}{n}(\alpha +\beta )\biggr )\biggr |. \end{aligned} \end{aligned}$$

The formula on the right hand side of the above can be written in the shorter form, $C\cdot W\!\left( \frac{i}{n}\right) /n^2$, where C does not depend on n. Moreover, it is easy to note that

$$\begin{aligned} \frac{1}{n} \sum \limits _{i=1}^{n-2} W\left( \frac{i}{n}\right) \xrightarrow {n\rightarrow \infty } \int \limits _{0}^{1}W(x)\, d x, \end{aligned}$$

where convergence is the result of the definition of a Riemann sum with the set $\{i/n:1\le i\le n-2\}$ as a partition of (0, 1) and the integrability of the function under the integral sign. The function W(x) has two singularities: at $x=0$ and $x=1$. Performing the standard calculation, we get $W(x)=(\alpha -\beta )x^{\alpha /\beta -1}+{\mathcal {O}}(x^{2\alpha /\beta -1})$ as $x\rightarrow 0$ and $W(x)=\alpha (1+\beta )(1-x)^{\beta -1}+{\mathcal {O}}((1-x)^\beta )$ as $x\rightarrow 1$. Since $\alpha ,\beta >0$, the function W(x) is integrable around its singular points. Furthermore, it is easy to see that

$$\begin{aligned} \theta _0=\theta _{n-1}={\mathcal {O}}\biggl (\frac{1}{n}\biggr ), \quad \mathrm {as}\ n\rightarrow \infty . \end{aligned}$$

Based on the above considerations, we get

$$\begin{aligned} C_n-C_{n-1}={\mathcal {O}}\biggl (\frac{1}{n}\biggr ),\quad \mathrm {as} n\rightarrow \infty . \end{aligned}$$

Finally, we have

$$\begin{aligned} |t_n{\overline{\partial }}I_{\alpha /\beta }^{0,\beta }\psi (t_n) - t_n{\overline{\partial }}L_{\alpha /\beta }^{0,\beta }\psi (t_n)|\le t_n \psi '(x^* k) \frac{C_1}{n}+t_n^2 C_n C_2 \psi ''(x^{* *}k) \frac{1}{n}, \end{aligned}$$

where $C_1$ and $C_2$ do not depend on n. Taking into account the estimates of all components in ${\widehat{R}}_n$, we further obtain

$$\begin{aligned} \begin{aligned} |{\widehat{R}}_n|&= \biggl |t_n \biggl (\frac{d}{d t}I_{\alpha /\beta }^{0,\beta }\psi (t_n) - {\overline{\partial }}I_{\alpha /\beta }^{0,\beta }\psi (t_n)+ {\overline{\partial }}I_{\alpha /\beta }^{0,\beta }\psi (t_n)- {\overline{\partial }}L_{\alpha /\beta }^{0,\beta }\psi (t_n)\biggr )\biggr |\\&\le \frac{\beta }{\alpha }\rho _1 |\psi ''(\sigma ^* t^*)| \frac{t_n}{n} + |\psi '(x^* k)| \frac{C_1 t_n}{n}+ |\psi ''(x^{* *}k)| \frac{ C_2 C_n t_n^2}{\beta \varGamma (\beta ) n}\\&\le \frac{1}{n}\biggl [ \frac{\beta t_n}{\alpha }\rho _1 |\psi ''(t^{*})| + t_n C_1 \psi '(t^{* *})|+\frac{C_2 C_n t_n^2 }{\beta \varGamma (\beta )} |\psi ''(t^{* * *})|\biggr ], \end{aligned} \end{aligned}$$

where $t^{*},t^{* *},t^{* * *} \in (0,t_n)$. The above estimate together with $R_n$ yields (2.7).

Let us now consider the operator $K_{\alpha /\beta }^{\beta -1,1-\beta }$. Similarly to the above considerations, the integral part of $D_{\alpha /\beta }^{\beta -1,1-\beta }$ is approximated using $\beta L_{\alpha /\beta }^{0,\beta }$. Now, let us examine solely the differentiation part of the analysed operator. Performing an appropriate transformation, we have

$$\begin{aligned} \begin{aligned}&\sum \limits _{i=1}^{n} d_{n,i} n (\psi (t_i)- \psi (t_{i-1})) - \frac{\beta \varGamma (\beta )}{\alpha } t_n \frac{d}{d t} I_{\alpha /\beta }^{0,\beta }\psi (t_n) \\&\quad = \biggl ( \sum \limits _{i=1}^{n}\int \limits _{\frac{i-1}{n}}^{\frac{i}{n}}\tau ^{\frac{\alpha }{\beta }}\bigl (1-\tau ^{\frac{\alpha }{\beta }}\bigr )^{\beta -1} n (\psi (t_i)-\psi (t_{i-1}))\, d \tau \\&\qquad -\frac{\beta \varGamma (\beta )}{\alpha } t_n \frac{d}{d t} \int \limits _{0}^{1}\tau ^{\frac{\alpha }{\beta }-1}\bigl (1-\tau ^{\frac{\alpha }{\beta }}\bigr )^{\beta -1}\psi (\tau t_n)\, d \tau \biggr )\\&\quad = \biggl ( \sum \limits _{i=1}^{n}\int \limits _{\frac{i-1}{n}}^{\frac{i}{n}}\tau ^{\frac{\alpha }{\beta }}\bigl (1-\tau ^{\frac{\alpha }{\beta }}\bigr )^{\beta -1}\biggl [ n (\psi (t_i) -\psi (t_{i-1})) - \frac{d}{d \tau }\psi (\tau t_n)\biggr ]\, d \tau \biggr ). \end{aligned} \end{aligned}$$

We use Taylor series expansion for $\psi (x t_n)$ at the point $x=\tau$ and obtain

$$\begin{aligned} \psi (x t_n)=\psi (\tau t_n)+ \frac{d }{d x}\psi (\tau t_n) (x-\tau ) + \frac{t_n^2}{2} \psi ''(\xi t_n) (x-\tau )^2,\quad \xi \in (\tau ,x). \end{aligned}$$

Therefore, the expression in the square bracket under the integral sign can be rewritten in the following way

$$\begin{aligned} \begin{aligned} n (\psi (t_i)-\psi (t_{i-1}))- \frac{d}{d \tau }\psi (\tau t_n)&= n \frac{t_n^2}{2}\biggl ( \psi ''(\xi _i t_n) \biggl (\frac{i}{n}-\tau \biggr )^2\\&\quad -\psi ''(\xi _{i-1} t_n) \biggl (\frac{i-1}{n}-\tau \biggr )^2\biggr ), \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \bigl | \frac{\alpha }{\beta \varGamma (\beta )} \sum \limits _{i=1}^{n}&d_{n,i} n (\psi (t_i)-\psi (t_{i-1})-t_n \frac{d}{d t} I_{\alpha /\beta }^{0,\beta }\psi (t_n) \bigr | \le \rho _2 \frac{t_n^2 |\psi ''(\sigma ^* t_n)|}{n}, \end{aligned} \end{aligned}$$

where $\rho _2= \varGamma \bigl (1+\frac{\beta }{\alpha }\bigr )/\varGamma \bigl (\frac{\beta }{\alpha }+\beta +1\bigr )$ and $\sigma ^*\in (0,1)$. The above inequalities end the proof.