Let \(t_{n}=n\tau\), \(n=0,1,\ldots,N\), \(T=\tau N\), \(N\in\mathbb{Z}^{+}\), and \(x_{j}=a+jh\), \(j=1,0,\ldots,M+1\), \(h=(ba)/M\), \(M\in\mathbb{Z}^{+}\). On this timespace lattice, we set about deriving the desired exponential Bspline collocation method for Eqs. (1.1)(1.3).
Discretization of Caputo derivative
We recall the definitions of fractional derivatives. Given a smooth enough \(f(x,t)\), the αth Caputo derivative is defined by
$$ {^{C}_{0}}D^{\alpha}_{t}f(x,t)= \frac{1}{\Gamma(m\alpha)} \int^{t}_{0}\frac{\partial^{m} f(x,\xi)}{\partial\xi^{m}}\frac{d\xi}{(t\xi )^{1+\alpham}}, $$
(3.1)
and the αth RiemannLiouville type derivative is defined by
$$ {^{\mathrm{RL}}_{0}}D^{\alpha}_{t}f(x,t)= \frac{1}{\Gamma(m\alpha)}\frac{\partial ^{m}}{\partial t^{m}} \int^{t}_{0}\frac{f(x,\xi)\,d\xi}{(t\xi)^{1+\alpham}}, $$
(3.2)
where \(m1<\alpha<m\), \(m\in\mathbb{N}\) is not less than 1. In common sense, (3.1) owns merits in handling the initialvalued problems, and thereby is utilized in time in most instances. (3.1), (3.2) interconvert into each other through
$$ {^{C}_{0}}D^{\alpha}_{t}f(x,t)={^{\mathrm{RL}}_{0}}D^{\alpha}_{t}f(x,t) \sum^{m1}_{l=0}\frac {f^{(l)}(x,0)t^{l\alpha}}{\Gamma(l+1\alpha)}. $$
(3.3)
They are equal when \(f^{(k)}(x,0)=0\), \(k=0,1,\ldots,m1\), are fixed; we refer the readers to [15, 16] for deeper insight. An effective approximation for Caputo derivative can be derived by rewriting Eq. (3.3) and using proper schemes to discretize (3.2), i.e.,
$$ {^{C}_{0}}D^{\alpha}_{t}f(x,t_{n}) \approx\frac{1}{\tau^{\alpha}}\sum_{k=0}^{n} \omega^{q,\alpha}_{k}f(x,t_{nk}) \frac{1}{\tau^{\alpha}}\sum ^{m1}_{l=0}\sum^{n}_{k=0} \frac{\omega ^{q,\alpha}_{k} f^{(l)}(x,0)t_{nk}^{l}}{l!}, $$
(3.4)
with several sets of coefficients \(\omega^{q,\alpha}_{k}\), \(q=1,2,3,4,5\), (see [37]). Let \(\omega^{\alpha}_{k}=\omega^{1,\alpha}_{k}\). Then
$$ \omega^{\alpha}_{k}=(1)^{k}\binom{\alpha}{k}=\frac{\Gamma{(k\alpha)}}{\Gamma {(\alpha)}\Gamma{(k+1)}}, \quad k=0,1,2,\ldots $$
(3.5)
in which case the scheme is the one given by Gorenflo et al. [38]. On imposing \(0<\alpha<1\), (3.4) simply reduces to
$$ {^{C}_{0}}D^{\alpha}_{t}f(x,t_{n})= \frac{1}{\tau^{\alpha}}\sum_{k=0}^{n}\omega ^{q,\alpha}_{k}f(x,t_{nk}) \frac{1}{\tau^{\alpha}}\sum _{k=0}^{n}\omega^{q,\alpha}_{k}f(x,0)+{R_{q}( \tau)}, $$
(3.6)
with the truncated error \(R_{q}(\tau)\) satisfying \(R_{q}(\tau)=O(\tau ^{q})\), \(q=1,2,3,4,5\).
Lemma 3.1
The coefficients
\(\omega^{\alpha}_{k}\)
defined in (3.5) fulfill

(a)
\(\omega^{\alpha}_{0}=1\), \(\omega^{\alpha}_{k}< 0\), \(\forall k\geq1\),

(b)
\(\sum_{k=0}^{\infty}\omega^{\alpha}_{k}=0\), \(\sum_{k=0}^{n1}\omega^{\alpha}_{k}>0\).
Proof
See references [15, 39] for details. □
A fully discrete exponential Bspline based scheme
Define \(V_{M+3}=\textrm{span}\{B_{1}(x),B_{0}(x),\ldots ,B_{M}(x),B_{M+1}(x)\}\) over the interval \([a,b]\) referred to as an \((M+3)\)dimensional exponential spline space. Then an approximate solution to Eqs. (1.1)(1.3) is sought on \(V_{M+3}\) in the form
$$ u_{N}(x,t)=\sum_{j=1}^{M+1} \alpha_{j}(t)B_{j}(x), $$
(3.7)
with the unknown weights \(\{\alpha_{j}(t)\}_{j=1}^{M+1}\) yet to be determined by some certain restrictions. Discretizing Eq. (1.1) by using (3.6) in time, we have
$$\begin{gathered} \omega^{q,\alpha}_{0}u(x,t_{n})\tau^{\alpha}\kappa\frac{\partial^{2} u(x,t_{n})}{\partial x^{2}}\\\quad= \sum_{k=1}^{n1} \omega^{q,\alpha}_{k}u(x,t_{nk})+\sum _{k=0}^{n1}\omega ^{q,\alpha}_{k}u(x,0)+ \tau^{\alpha}f(x,t_{n})+{\tau^{\alpha}R_{q}( \tau)}.\end{gathered} $$
Let \(\alpha^{n}_{j}=\alpha_{j}(t_{n})\). On replacing \(u(x,t)\) by \(u_{N}(x,t)\) and imposing the following collocation and boundary conditions
$$\begin{gathered} \omega^{q,\alpha}_{0}u_{N}(x_{j},t_{n}) \tau^{\alpha}\kappa\frac{\partial^{2} u_{N}(x_{j},t_{n})}{\partial x^{2}} \\\quad= \sum_{k=1}^{n1} \omega^{q,\alpha}_{k}u_{N}(x_{j},t_{nk})+ \sum_{k=0}^{n1}\omega^{q,\alpha}_{k}u_{N}(x_{j},0)+ \tau^{\alpha}f(x_{j},t_{n}), \\ u_{N}(x_{0},t_{n})=g_{1}(t_{n}), \qquad u_{N}(x_{M},t_{n})=g_{2}(t_{n}),\end{gathered} $$
at each nodal point \(x_{j}\), \(j=0,1,\ldots,M\), we obtain
$$ A\alpha^{n}_{j1}+A' \alpha^{n}_{j}+A\alpha^{n}_{j+1} =\sum _{k=1}^{n1}\omega^{q,\alpha}_{k}P^{nk}_{j}+ \sum_{k=0}^{n1}\omega ^{q,\alpha}_{k}P_{j}^{0}+R^{n}_{j}, $$
(3.8)
and the boundary sets
$$\begin{aligned}& \frac{sph}{2(phcs)}\alpha^{n}_{1}+\alpha^{n}_{0}+ \frac {sph}{2(phcs)}\alpha^{n}_{1}=g_{1}^{n}, \end{aligned}$$
(3.9)
$$\begin{aligned}& \frac{sph}{2(phcs)}\alpha^{n}_{M1}+\alpha^{n}_{M}+ \frac {sph}{2(phcs)}\alpha^{n}_{M+1}=g_{2}^{n}, \end{aligned}$$
(3.10)
owing to (3.7) and (2.1)(2.3) with
$$\begin{gathered} A=\tau^{\alpha}\kappa p^{2}s+\omega^{q,\alpha}_{0}(sph), \qquad A'=2\tau ^{\alpha}\kappa p^{2}s+2 \omega^{q,\alpha}_{0}(phcs), \\ P^{m}_{j}=(sph)\alpha^{m}_{j1}+2(phcs) \alpha^{m}_{j}+(sph)\alpha ^{m}_{j+1}, \qquad R^{n}_{j}=2\tau^{\alpha}(phcs)f_{j}^{n}, \end{gathered}$$
where \(m=0,1,\ldots,n1\). As a result, using Eqs. (3.9)(3.10) to remove the unknown variables \(\alpha^{n}_{1}\), \(\alpha^{n}_{M+1}\) in Eq. (3.8) when \(j=0\), M, the above system admits a linear system of algebraic equations of size \((M+1)\times (M+1)\) as follows:
$$ \textbf{A}\boldsymbol{\alpha}^{n}=\sum _{k=1}^{n1}\omega^{q,\alpha }_{k} \textbf{B}\boldsymbol{\alpha}^{nk} +\sum_{k=0}^{n1} \omega^{q,\alpha}_{k}\textbf{B}\boldsymbol{\alpha }^{0}+ \textbf{F}^{n}, \quad{q=1,2,3,4,5,} $$
(3.11)
where
$$\begin{gathered} \textbf{A}=\left ( \textstyle\begin{array}{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c} 2\tau^{\alpha}\kappa p^{3}hs(c1) & 0 & & & \\ A & A' & A & & \\ & \ldots& \ldots& \ldots& \\ & & \ldots& \ldots& \ldots\\ & & A & A' & A \\ & & & 0 & 2\tau^{\alpha}\kappa p^{3}hs(c1) \ \end{array}\displaystyle \right ), \\\textbf{B}=\left ( \textstyle\begin{array}{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c} 0 & 0 & & & \\ sph & 2(phcs) & sph & & \\ & \ldots& \ldots& \ldots& \\ & & \ldots& \ldots& \ldots\\ & & sph & 2(phcs) & sph \\ & & & 0 & 0 \end{array}\displaystyle \right ), \\\boldsymbol{\alpha}^{m}=\left ( \textstyle\begin{array}{c} \alpha^{m}_{0} \\ \alpha^{m}_{1} \\ \vdots\\ \alpha^{m}_{M1} \\ \alpha^{m}_{M} \end{array}\displaystyle \right ), \qquad\textbf{F}^{n}=(phcs)\left ( \textstyle\begin{array}{c} 2\tau^{\alpha}(sph)f^{n}_{0}+d^{n}_{0} \\ 2\tau^{\alpha}f^{n}_{1} \\ \vdots\\ 2\tau^{\alpha}f^{n}_{M1} \\ 2\tau^{\alpha}(sph)f^{n}_{M}+d^{n}_{M} \end{array}\displaystyle \right ), \end{gathered}$$
in which \(m=0,1,\ldots,n\) and \(d^{n}_{0}\), \(d^{n}_{M}\) are as follows:
$$\begin{gathered} d^{n}_{0}=2(sph)\sum_{k=0}^{n1} \omega^{q,\alpha}_{k}g_{1}^{nk}+2(sph)\sum _{k=0}^{n1}\omega^{q,\alpha}_{k} \varphi_{0}+2\tau^{\alpha}\kappa p^{2}sg^{n}_{1}, \\ d^{n}_{M}=2(sph)\sum_{k=0}^{n1} \omega^{q,\alpha}_{k}g_{2}^{nk}+2(sph)\sum _{k=0}^{n1}\omega^{q,\alpha}_{k} \varphi_{M}+2\tau^{\alpha}\kappa p^{2}sg^{n}_{2}. \end{gathered}$$
The unknown weights \(\boldsymbol{\alpha}^{n}\) depend on \(\boldsymbol {\alpha}^{nk}\), \(k=0,1,\ldots,n\), at their previous time levels and are found via a recursive style; once \(\boldsymbol{\alpha}^{n}\) is obtained, \(\alpha ^{n}_{1}\), \(\alpha^{n}_{M+1}\) can further be determined with the help of Eqs. (3.9)(3.10). On the other hand, A is an \((M+1)\times(M+1)\) tridiagonal matrix, therefore the system can be performed by the wellknown Thomas algorithm, which simply needs the arithmetic operation cost \(O(M+1)\).