Numerical stability of Grünwald–Letnikov method for time fractional delay differential equations

Abstract

This paper is concerned with the numerical stability of time fractional delay differential equations (F-DDEs) based on Grünwald–Letnikov (GL) approximation for the Caputo fractional derivative. In particular, we focus on the numerical stability region and the Mittag–Leffler stability. Using the boundary locus technique, we first derive the exact expression of the numerical stability region in the parameter plane, and show that the fractional backward Euler scheme is not $\tau (0)$-stable, which is different from the backward Euler scheme for integer DDE models. Secondly, we prove the numerical Mittag–Leffler stability for the numerical solutions provided that the parameters fall into the numerical stability region, by employing the singularity analysis of generating function. Our results show that the numerical solutions of F-DDEs are completely different from the classical integer order DDEs, both in terms of $\tau (0)$-stability and the long-time decay rate. Numerical examples are given to confirm the theoretical results.

Appendices

Proof of Lemma 4.1

Proof of Lemma 4.1

For (i), we need to show that

$$\begin{aligned} 2^{\alpha }<2\frac{[(1-\alpha )\pi ]^{\alpha }\sin (\alpha \pi /2)}{\sin (\alpha \pi )} =\frac{[(1-\alpha )\pi ]^{\alpha }}{\sin (\pi (1-\alpha )/2)}, \end{aligned}$$

by Lemma 2.2. This is equivalent to

$$\begin{aligned} \sin \left( \frac{(1-\alpha )\pi }{2}\right) <\left[ \frac{(1-\alpha )\pi }{2}\right] ^{\alpha }. \end{aligned}$$

When $\frac{(1-\alpha )\pi }{2}>1$, the inequality clearly holds. If $\frac{(1-\alpha )\pi }{2}\in (0, 1]$, then

$$\begin{aligned} \sin \left( \frac{(1-\alpha )\pi }{2}\right) <\frac{(1-\alpha )\pi }{2} \le \left[ \frac{(1-\alpha )\pi }{2}\right] ^{\alpha }\quad \text {as}\quad \alpha \in (0, 1). \end{aligned}$$

(ii). By Lemma 2.2, we only need to show that the straight lines are below the boundary,

$$\begin{aligned} 2^{\alpha }k^{\alpha }\ge \frac{[(1-\alpha )\pi ]^{\alpha }}{\sin (\pi (1-\alpha )/2)},~~k\ge 3. \end{aligned}$$

(A.1)

It suffices to show that

$$\begin{aligned} \begin{aligned} \sin \left( \frac{\pi (1-\alpha )}{2}\right) - \left[ \frac{(1-\alpha )\pi }{2k} \right] ^{\alpha }\ge 0. \end{aligned} \end{aligned}$$

(A.2)

It’s worth noting that the function $h_1(x)= \frac{\sin x}{x}$ is decreasing with $\frac{2}{\pi }\le h_1(x)\le 1$ and it is concave on $[0, \pi /2]$ (note that the sign of the second order derivative is determined by the sign of $(1-\frac{x^2}{2})\frac{\sin x}{x}-\cos x$, which is negative for $x\in [0, \pi /2]$).

For $k\ge 5$, since $\sin (\pi (1-\alpha )/2)\ge \frac{\pi }{2}(1-\alpha )\frac{2}{\pi }$, it suffices to show that

$$\begin{aligned} \frac{\pi }{2}(1-\alpha )\frac{2}{\pi }- \left[ \frac{(1-\alpha )\pi }{2k}\right] ^{\alpha }\ge 0. \end{aligned}$$

The latter is equivalent to show that (setting $\beta =1-\alpha$) $h_2(\beta )=\beta \log \beta -(1-\beta )\log \left( \frac{\pi }{2k} \right) \ge 0.$ The derivative $h'_2(\beta )$ is negative for $k\ge 5$ and the value at $\beta =1$ is zero. Hence, the inequality (A.2) holds for $k\ge 5$.

If $k=3$, consider directly (setting $\beta =1-\alpha$) that $g(\beta ):=\log \left( \sin (\frac{\pi }{2}\beta ) \right) +(\beta -1)\log \left( \frac{\beta \pi }{6}\right) .$ As $\beta \rightarrow 1^-$, $g(\beta )$ tends to 0. Hence, to show $g(\beta )>0$ for $\beta \in (0,1)$, we only need to show that the first order derivative

$$\begin{aligned} g'(\beta )=\frac{\pi }{2}\frac{\cos (\pi \beta /2)}{\sin (\pi \beta /2)}+\log \left( \frac{\pi }{6}\right) +\log (\beta )+(\beta -1)/\beta \end{aligned}$$

is always negative on $\beta \in (0, 1)$.

It is easy to see that $g'(\beta )<0$ for $\beta \in (0.8, 1)$ (considering only the first two terms). On the other hand, we have $g''(\beta )=\beta ^{-1}+\beta ^{-2}-(\pi /2)^2\frac{1}{\sin ^2(\pi \beta /2)}.$ This equation $g''(\beta )=0$ only has one root $\beta _*$ with $\beta \in (0, 1)$ and the root satisfies $\frac{\sin (\pi \beta _*/2)}{(\pi \beta _*/2)}=\frac{1}{\sqrt{1+\beta _*}}.$ This can be seen from the fact $h_1(x)=\frac{\sin x}{x}$ is decreasing and concave while the function $\frac{1}{\sqrt{1+x}}$ is convex on $[0, \pi /2]$. At the same time, we can check $g''(0.8)>0$ and $g''(1)<0$. Hence, we know that the root $\beta _*\in (0.8, 1)$ and $g''(\beta )$ is positive on [0, 0.8]. Together with $g'(0.8)<0$, we find that $g'(\beta )$ is negative on (0, 0.8]. Then the first derivative is always negative on $\beta \in (0, 1)$. Therefore, $k=3$ is also proved. $\square$

Proof for the properties of the $\Gamma _0$ curve

Proof of Lemma 4.3

Define $\phi =\theta /2\in \left( \frac{(1-\alpha )\pi }{2(1-\alpha /k)}, \frac{\pi }{2} \right)$. Consider that

$$\begin{aligned} h(\phi ):=\frac{\lambda _k(\theta )}{2^{\alpha }k^{\alpha }}=\frac{\sin ^{\alpha } \left( \frac{\phi }{k} \right) \sin \left( \phi +\frac{\alpha \pi }{2} -\frac{\alpha \phi }{k}\right) }{\sin (\phi )}. \end{aligned}$$

To show this function is decreasing, it is sufficient to prove that

$$\begin{aligned} \frac{d}{d\phi }\ln h(\phi )=\frac{\alpha }{k}\frac{\cos (\phi /k)}{\sin (\phi /k)}+\frac{(1-\alpha /k)\cos \left( \phi +\frac{\alpha \pi }{2}-\frac{\alpha \phi }{k}\right) }{\sin \left( \phi +\frac{\alpha \pi }{2} -\frac{\alpha \phi }{k} \right) }-\frac{\cos \phi }{\sin \phi } \le 0.\nonumber \\ \end{aligned}$$

(B.1)

Separate the middle term on the left side of the above equation and use the trigonometric function formula, the above equation can be equivalent to $\frac{\alpha }{k}\frac{\sin (\phi +\alpha \pi /2-\alpha \phi /k-\phi /k)}{\sin (\phi /k)\sin (\phi +\alpha \pi /2-\alpha \phi /k)} -\frac{\sin (\alpha \pi /2-\alpha \phi /k)}{\sin (\phi )\sin (\phi +\alpha \pi /2-\alpha \phi /k)}\le 0.$ Noting that $\sin (\phi +\alpha \pi /2-\alpha \phi /k)>0$ under the assumption $\phi \in \left( \frac{(1-\alpha )\pi }{2(1-\alpha /k)}, \frac{\pi }{2} \right)$. Hence, we need

$$\begin{aligned} \frac{\alpha \sin (\phi +\alpha \pi /2-\alpha \phi /k-\phi /k)}{\sin (\alpha \pi /2 -\alpha \phi /k)}\le \frac{k\sin (\phi /k)}{\sin \phi }. \end{aligned}$$

(B.2)

We are going to prove that (B.2) is true for $k\ge 2$. Since

$$\begin{aligned} \frac{\alpha \sin (\phi +\alpha \pi /2-\alpha \phi /k-\phi /k)}{\sin (\alpha \pi /2-\alpha \phi /k)}\le \frac{\alpha }{\sin (\alpha \pi /2-\alpha \phi /k)}\le \frac{1}{\cos (\phi /k)}, \end{aligned}$$

(B.3)

where the second inequality is due to the fact $\frac{\alpha }{\sin (\alpha \pi /2-\alpha \phi /k)}$ is increasing function for $\alpha \in (0, 1)$. By the concavity of $\sin (x)$ on $x\in (0, \pi /2)$, we know that $\sin (\phi )\le \frac{k}{2}\sin (2\phi /k )$ is true for $k\ge 2$, which implies that $\frac{1}{\cos (\phi /k)}\le \frac{k\sin (\phi /k)}{\sin (\phi )}.$ This together with (B.3) shows that the inequality (B.2) is true for $k\ge 2$. This completes the first part of the proof.

Using the above fact, to show that the line is below $\Gamma _0$, we only have to show that $\left( a_k^{(0)},-a_k^{(0)} \right)$ is above the line, where $a_k^{(0)}$ is defined in Lemma 4.2. In other words, we need

$$\begin{aligned} 2^{\alpha }k^{\alpha }\ge 2^{\alpha }k^{\alpha }\frac{\cos ^{\alpha }(\phi _1)}{\cos (\alpha \phi _1)}, \quad \text {where}\quad \phi _1=\frac{\pi (k-1)}{2(k-\alpha )}. \end{aligned}$$

(B.4)

Since the right hand side is decreasing in $\phi$, the largest value is achieved at $\phi _1=0$, which is $2^{\alpha }k^{\alpha }$. Hence, the inequality (B.4) is true. $\square$