Direct and integrated radial functions based quasilinearization schemes for nonlinear fractional differential equations

Abstract

In this article, two radial basis functions based collocation schemes, differentiated and integrated methods (DRBF and IRBF), are extended to solve a class of nonlinear fractional initial and boundary value problems. Before discretization, the nonlinear problem is linearized using generalized quasilinearization. An interesting proof via generalized monotone quasilinearization for the existence and uniqueness for fractional order initial value problem is given. This convergence analysis also proves quadratic convergence of the generalized quasilinearization method. Both the schemes are compared in terms of accuracy and convergence and it is found that IRBF scheme handles inherent RBF ill-condition better than corresponding DRBF method. Variety of numerical examples are provided and compared with other available results to confirm the efficiency of the schemes.

Appendix

The proof for $\widehat{T}{:}\,[v^0,w^0] \subset C[0, T]\rightarrow C[0, T]$ is given as follows. Let $y(t)=\widehat{T} x(t)$ for some $x\in [v^0,w^0]$. Hence

$$\begin{aligned} y(t)=x_0E_q(-\mu t^q)+\int _0^t(t-s)^{q-1}E_{q,q}(-\mu (t-s)^{q})(h(s,x(s))+\mu x(s))\mathrm{d}s.\nonumber \\ \end{aligned}$$

(A.1)

To show that y is a continuous function on [0, T], it is enough to show that $z(t)=\int _0^t(t-s)^{q-1}E_{q,q}(-\mu (t-s)^{q})(h(s,x(s))+\mu x(s))\mathrm{d}s$ is continuous as $E_q$ is a continuous function.

Define $g(t,s)=E_{q,q}(-\mu (t-s)^{q})(h(s,x(s))+\mu x(s))$, $(t,s)\in [0,T] \times [0,T]$ and $g_1(s)=h(s,x(s))+\mu x(s)$, $s \in [0,T]$. Let K and $K_1$ be the maximum of |g| and $|g_1|$ in their respective domains. Also $E_{q,q}(-\mu (t-s)^{q})$ is uniformly continuous for $(t,s)\in [0,T] \times [0,T]$.

Given $\epsilon >0$, there exist $\delta _1>0$ such that $|E_{q,q}(-\mu (t_1-s)^{q})-E_{q,q}(-\mu (t_2-s)^{q})| \le \frac{q\epsilon }{3K_1T^q}$ whenever $|t_1-t_2| \le \delta _1$. Further, given $\epsilon >0$ there exists a $\delta _2>0$ such that $|t_1-t_2|^q \le \frac{q\epsilon }{3K}$ whenever $|t_1-t_2| \le \delta _2$. Let $t_1 \le t_2$ in [0, T] with $|t_1-t_2|\le \delta =\min \{\delta _1,\delta _2\}$. Note that $z(t_1)-z(t_2)=I_1+I_2+I_3$ where

$$\begin{aligned} I_1= & {} \int _0^{t_1}(t_1-s)^{q-1}(E_{q,q}(-\mu (t_1-s)^{q})-E_{q,q}(-\mu (t_2-s)^{q}))g_1(s)\mathrm{d}s;~~~\\ I_2= & {} \int _0^{t_1}((t_1-s)^{q-1}-(t_2-s)^{q-1})g(t_2,s)\mathrm{d}s;~~~I_3=-\int _{t_1}^{t_2}(t_2-s)^{q-1}g(t_2,s)\mathrm{d}s \end{aligned}$$

Consequently

$$\begin{aligned} |I_1|\le & {} \int _0^{t_1}(t_1-s)^{q-1}|E_{q,q}(-\mu (t_1-s)^{q})-E_{q,q}(-\mu (t_2-s)^{q})||g_1(s)|\mathrm{d}s \nonumber \\\le & {} K_1\int _0^{t_1}(t_1-s)^{q-1}|E_{q,q}(-\mu (t_1-s)^{q})-E_{q,q}(-\mu (t_2-s)^{q})|\mathrm{d}s \nonumber \\ \text{ Thus }\,\,|I_1|\le & {} \frac{\epsilon }{3} \end{aligned}$$

(A.2)

$$\begin{aligned} |I_3|\le & {} \int _{t_1}^{t_2}(t_2-s)^{q-1}|g(t_2,s)|\mathrm{d}s~\le ~K\int _{t_1}^{t_2}(t_2-s)^{q-1}\mathrm{d}s \nonumber \\= & {} K_1\int _{0}^{t_2-t_1}\tau ^{q-1}\mathrm{d}\tau =\frac{K}{q} (t_2-t_1)^q \nonumber \\ \text{ Thus }\,\, |I_3|\le & {} \frac{\epsilon }{3} \end{aligned}$$

(A.3)

$$\begin{aligned} |I_2|\le & {} \int _0^{t_1} |(t_1-s)^{q-1}-(t_2-s)^{q-1})||g(t_2,s)|\mathrm{d}s \nonumber \\\le & {} K \int _0^{t_1}|(t_1-s)^{q-1}-(t_2-s)^{q-1}|\mathrm{d}s=K\frac{(t_2-t_1)^q}{q}+K\frac{t_1^q-t_2^q}{q} \nonumber \\ \text{ Thus }\,\, |I_2|\le & {} K\frac{(t_2-t_1)^q}{q} \le \frac{\epsilon }{3} \end{aligned}$$

(A.4)

This implies $|z(t_1)-z(t_2)| \le \epsilon$ whenever $|t_1-t_2|\le \delta$. Hence z is a continuous function on [0, T]. Consequently y is continuous in [0, T].