Discretization of Caputo derivative and semi-discrete scheme
First, we obtain the semi-discrete scheme for (1.1)–(1.3). The discretization of J̅ is performed with a constant time step \(\tau =\frac{T}{N}\), where \(N \in \mathcal{N}^{*}\). Denote \(t_{n}=n \tau \) for \(N=0:N\). Let \(u^{n}=u(x,t_{n})\). For a discrete function \(\{u^{n}\}_{n=0}^{N+1}\), we provide some preliminaries concerning the approximation of the time fractional derivative \(\partial _{t}^{\alpha} u(x,t)\) with \(1<\alpha <2\). A Caputo derivative approximation formula (CDAF) for \(\partial _{t}^{\alpha} u(x,t_{n+1})\) with \(1<\alpha <2\) can be defined as a linear combination of the discrete second time derivatives \(\{\partial ^{2} u^{j}\}_{j=1}^{n+1}\) [47]
$$\begin{aligned} \partial _{t}^{\alpha}u^{n+1}= \frac{\tau ^{2-\alpha}}{\Gamma (3-\alpha )} \sum _{j=0}^{n}b_{j} \partial ^{2}u^{n+1-j}+\mathcal{R}_{1}^{n+1}(u), \end{aligned}$$
(2.1)
where
$$ b_{j}=(j+1)^{2-\alpha}-j^{2-\alpha},$$
and \(\mathcal{R}_{1}^{n+1}(u)\) is the local truncation error such that
$$ \bigl\vert \mathcal{R}_{1}^{n+1}(u) \bigr\vert \leq C_{u} \tau ^{3-\alpha} \text{or} \mathcal{R}_{1}^{n+1}(u)=O \bigl( \tau ^{3-\alpha}\bigr).$$
The following lemma summarizes some properties of the coefficients \(b_{j}\) which will be used in this paper.
Lemma 2.1
(See [47]. Properties of the coefficients \(b_{j}^{n+1}\))
For any \(1 < \alpha <2\), the coefficients of \(b_{j}^{n+1}\) satisfy the following properties:
-
\(b_{j}>0\), \(j=0,1,\ldots ,n\),
-
\(1=b_{0}>b_{1}>\cdots >b_{n}\) and \(b_{n} \rightarrow 0\) as \(n\rightarrow \infty \),
-
\(\sum_{j=0}^{n}(b_{j}-b_{j+1})=1\).
Substituting (2.1) into (1.1) gives
$$\begin{aligned}& a(\alpha ,\tau ) \bigl(u^{n+1}-2u^{n}+u^{n-1} \bigr)+a(\alpha ,\tau )\sum_{j=1}^{n} b_{j}\bigl(u^{n+1-j}-2u^{n-j}+u^{n-1-j}\bigr) \\& \quad =\Delta u^{n+1}(x)-u^{n+1}(x)+ f^{n+1}(x)+R^{n+1}(u), \end{aligned}$$
(2.2)
where \(a(\alpha ,\tau )=\frac{1}{\tau ^{\alpha}\Gamma (3-\alpha )}\) and \(f^{n+1}(x)=f(x,t_{n+1})\).
Replacing \(u^{n+1}\) by the approximate solution \(U^{n+1}\), we can obtain the following semi-discrete problem.
Scheme I
Given \(U^{0}=\phi _{1}(x)\), \(U^{-1}=U^{1}-2\tau \phi _{2}(x)\) and find \(U^{n+1}\) (\(n=0,1,2,\ldots ,N-1\)) such that
$$\begin{aligned} \textstyle\begin{cases} a(\alpha ,\tau )(U^{n+1}-2U^{n}+U^{n-1})+a(\alpha ,\tau )\sum_{j=1}^{n}b_{j}(U^{n+1-j}-2U^{n-j}+U^{n-1-j}) \\ \quad =\Delta U^{n+1}(x)-U^{n+1}(x)+ f^{n+1}(x), \\ U^{n+1}|_{x \in \partial \Omega}=0,\quad -1\leq n \leq N-1. \end{cases}\displaystyle \end{aligned}$$
For the convenience of discussion, define the linear operator L as follows:
$$\begin{aligned} \textrm{L} (*)=\textstyle\begin{cases} ((2a(\alpha ,\tau )+1)-\Delta ) (*),& n=0, \\ ((a(\alpha ,\tau )+1)-\Delta )(*),& 1\leq n \leq N-2. \end{cases}\displaystyle \end{aligned}$$
Therefore, a semi-discrete problem can be converted into the following equivalent:
$$\begin{aligned} \textrm{L} U^{n+1}(x)=F^{n+1}(x),\quad 0\leq n \leq N-1, \end{aligned}$$
(2.3)
where
$$ F^{n+1}(x)=\textstyle\begin{cases} -a(\alpha ,\tau )(-2\phi _{1}(x)-2\tau \phi _{2}(x))+ f^{1}(x),\quad n=0, \\ -a(\alpha ,\tau )(-2U^{n}+U^{n-1})-a(\alpha ,\tau )\sum_{j=1}^{n}b_{j}(U^{n+1-j} \\ \quad {}-2U^{n-j}+U^{n-1-j})+ f^{n+1}(x),\quad 1\leq n \leq N-2. \end{cases} $$
A pseudo-spectral kernel-based method
Now, we employ a pseudo-spectral method based on RK to discrete the space direction and obtain a full-discrete scheme of (2.3). To obtain this, we need some notations and preliminaries.
We now give background material and preliminaries, which are used in the following sections. Recall that a real reproducing kernel Hilbert space (RKHS) on a nonempty abstract set Ω is a particular type of a real Hilbert space H of functions that satisfies the following additional property (called reproducing kernel property): for each \(x\in \Omega \), there exists \(K(x,\cdot)\in \mathsf{H}\) (\(R:\Omega \times \Omega \longrightarrow \mathbf{R} \)) such that, for every \(u \in \mathsf{H}\), one has
$$\begin{aligned} u(x)=\bigl(u(\cdot),K(x,\cdot)\bigr)_{\mathsf{H}},\quad \forall u \in \mathsf{H}, \forall x \in \Omega . \end{aligned}$$
(2.4)
Definition 2.2
(See [40])
A Hilbert space H of real functions on a set Ω is called an RKHS if there exists an RK \(K(x,\cdot)\) of H.
Theorem 2.3
(See [40])
Suppose that \(\boldsymbol{\mathsf{H}}\) is an RKHS with RK \(K:\Omega \times \Omega \longrightarrow \mathbf{R} \). Then \(K(x,\cdot)\) is positive definite. Moreover, \(K(x,\cdot)\) is strictly positive definite if and only if the point evaluation functionals \(\bigl\{ \scriptsize{ \begin{array}{l@{\quad}l} I_{x}: \boldsymbol{\mathsf{H}}\longrightarrow \mathbf{R}, \\ I_{x}(u)=u(x) \end{array}}\) are linearly independent in \(\boldsymbol{\mathsf{H}}^{*}\), where \(\boldsymbol{\mathsf{H}}^{*}\) is the space of bounded linear functionals on \(\boldsymbol{\mathsf{H}}\).
Definition 2.4
(See [40]. One-dimensional RKHS)
The inner product space \(\boldsymbol{\mathsf{H}}_{p}[0,b]\) for a function u is defined as
$$\begin{aligned} \boldsymbol{\mathsf{H}}_{p}[0,b]=\bigl\{ u|u(x),u^{\prime }(x),u^{\prime \prime }(x)\in AC[0,b], u^{\prime \prime \prime }(x) \in L^{2}[0,b],u(0)=u(b)=0, x\in [0,b]\bigr\} . \end{aligned}$$
The inner product in \(\boldsymbol{\mathsf{H}}_{p}[0,b]\) is in the form
$$\begin{aligned} \langle u,v\rangle _{\boldsymbol{\mathsf{H}}_{p}}=u(0)v(0)+u(b)v(b)+u^{\prime }(0)v^{\prime }(0)+ \int _{a}^{b}u^{\prime \prime \prime }(x)v^{\prime \prime \prime }(x)\,dx, \end{aligned}$$
(2.5)
and the norm \(\|u\|_{\boldsymbol{\mathsf{H}}}\) is defined by
$$\begin{aligned} \Vert u \Vert _{\boldsymbol{\mathsf{H}}_{p}}=\sqrt{\langle u,u\rangle _{\mathsf{H}}}, \end{aligned}$$
(2.6)
where \(u,v\in \boldsymbol{\mathsf{H}}_{p}[0,b]\).
The space \(\boldsymbol{\mathsf{H}}_{p}[0,b]\) is an RKHS and the RK \(K_{y}(x)\) can be denoted by [40]
$$\begin{aligned} K_{y}(x)= \textstyle\begin{cases} \frac{1}{120b^{2}}(b-x)y(120bx+x(6b^{2}x-120-4bx^{2}+x^{3})y \\ \quad {}-5bxy^{3}+(b+x)y^{4}), \quad y< x, \\ \frac{1}{120b^{2}}(b-y)x(120by+y(6b^{2}y-120-4by^{2}+y^{3})x \\ \quad {}-5byx^{3}+(b+y)x^{4}), \quad y\geq x. \end{cases}\displaystyle \end{aligned}$$
(2.7)
With the help of pseudo-spectral method based on RK, we will illustrate how to derive the numerical solution. Now, we will give the representation of a numerical solution to the semi-discrete problem (2.3) in the RKHS \(\boldsymbol{\mathsf{H}}_{p}[0,b]\). Let \(\mathcal{B}_{M}=\{x_{j}\}_{j = 1}^{M}\) be a distinct subset of Ω̅. We consider the finite-dimensional space
$$\begin{aligned} \mathcal{U}_{M}=\operatorname{\textbf{span}}\bigl\{ \psi _{j}(x)=K_{y}(x)|_{y=x_{j}}, x_{j} \in \mathcal{B}_{M}\bigr\} \subset \boldsymbol{\mathsf{H}}_{p}[0,b], \end{aligned}$$
where \(K_{y}(x)\) is the RK constructed in \(\boldsymbol{\mathsf{H}}_{p}[0,b]\).
The semi-discrete problem can be written into following equivalent form \(\boldsymbol{\mathsf{H}}_{p}[0,b]\) to \(C[0,b]\):
$$\begin{aligned} \textrm{L} U^{n+1}(x)=F^{n+1}(x),\quad 0\leq n \leq N-1, \end{aligned}$$
where
$$ F^{n+1}(x)=\textstyle\begin{cases} -a(\alpha ,\tau )(-2\phi _{1}(x)-2\tau \phi _{2}(x))+ f^{1}(x),\quad n=0, \\ -a(\alpha ,\tau )(-2U^{n}+U^{n-1})-a(\alpha ,\tau )\sum_{j=1}^{n}b_{j}(U^{n+1-j} \\ \quad {}-2U^{n-j}+U^{n-1-j})+ f^{n+1}(x),\quad 1\leq n \leq N-2, \end{cases} $$
and \(F^{n+1}\in C[0,b]\) as \(u^{k}\in \boldsymbol{\mathsf{H}}_{p}[0,b]\).
An approximant \(U_{M}^{n+1}\) to \(U^{n+1}\) can be obtained by calculating a truncated series based on trial functions as follows:
(2.8)
To determine the interpolation coefficients \(\{\alpha ^{n+1}_{j}\}_{j=1}^{M}\), the set of collocation conditions is used by applying (2.3) to \(\mathcal{B}_{M}\). Thus
$$ \lambda _{i}\bigl[U_{M}^{n+1} \bigr]:=\textrm{L} U_{M}^{n+1}(x_{i})=\sum _{j=1}^{M} \alpha ^{n+1}_{j} \textrm{L}\psi _{j}(x_{i})=F^{n+1}(x_{i}),\quad i=1,2, \ldots ,M, $$
(2.9)
where the functional \(\lambda _{i}\), (\(1\leq i \leq M\)) is defined by applying the differential operator followed by a point evaluation at \(x_{i} \in \mathcal{B}_{n}\). In general, a single set \(\Lambda _{M}:=\{\lambda _{i}\}_{i=1}^{M}\) of functionals contains several types of differential operators.
The arising collocation matrix K is unsymmetric and has the ij-entries:
$$ \textbf{K}_{ij}:=\lambda _{i}[\psi _{j}]=\lambda ^{x}_{i} K_{y}(x)|_{y=x_{j}},\quad 1 \leq i,j \leq M, $$
(2.10)
where the subscript x in \(\lambda ^{x}_{j}\) indicates that \(\lambda ^{x}_{j}\) applies to the function of x.
Therefore the unknown coefficients \(\alpha ^{n+1}_{j}\), \(j=1,2,\ldots ,M\), can be obtained by solving the following system:
$$\begin{aligned}& \textbf{K}[\alpha ]^{n+1}=\textbf{F}^{n+1}, \end{aligned}$$
where
$$\begin{aligned}& [\alpha ]^{n+1}=\bigl[\alpha _{1}^{n+1},\alpha _{2}^{n+1},\ldots ,\alpha _{M}^{n+1} \bigr]^{T},\\& \textbf{F}^{n+1}=\bigl[F^{n+1}(x_{1}),F^{n+1}(x_{2}), \ldots ,F^{n+1}(x_{M})\bigr]^{T}, \end{aligned}$$
and
(2.11)
We know that
$$\begin{aligned} \textbf{U}^{n+1}=\textbf{A}[\alpha ]^{n+1}, \end{aligned}$$
(2.12)
where
$$\begin{aligned} \textbf{A}=[A_{ij}]_{M \times M},\qquad A_{ij}=\psi _{j}(x_{i}) \end{aligned}$$
and
$$\begin{aligned} \textbf{U}^{n+1}=\bigl[U^{n+1}(x_{1}),U^{n+1}(x_{2}), \ldots ,U^{n+1}(x_{M})\bigr]^{T}. \end{aligned}$$
The following matrix vector form is achieved by differentiating (2.12) with respect to x and evaluating it at the point girds \(x_{i}\in \mathcal{B}_{M}\):
$$\begin{aligned} \Delta \textbf{U}^{n+1}=\textbf{A}_{xx}[\alpha ]^{n+1}, \end{aligned}$$
where
$$\begin{aligned} \Delta \textbf{U}^{n+1}=\bigl[\Delta U^{n+1}(x_{1}), \Delta U^{n+1}(x_{2}), \ldots ,\Delta U^{n+1}(x_{M}) \bigr]^{T} \end{aligned}$$
and
$$\begin{aligned} \textbf{A}_{xx}=[A_{xx,ij}]_{M \times M},\qquad A_{xx,ij}= \frac{\partial ^{2}\psi _{j}}{\partial x^{2}}|_{x=x_{i}}. \end{aligned}$$
Now, from \(\textbf{U}^{n+1}=\textbf{A}[\alpha ]^{n+1}\) we have
$$\begin{aligned}{} [\alpha ]^{n+1}=\textbf{A}^{-1}\textbf{U}^{n+1}, \end{aligned}$$
and therefore
$$\begin{aligned} \Delta \textbf{U}^{n+1}=\textbf{A}_{xx} \textbf{A}^{-1}\textbf{U}^{n+1}. \end{aligned}$$
(2.13)
Now, by using (2.13), we can write
$$\begin{aligned} \textbf{K} \textbf{U}^{n+1}=\textbf{F}^{n+1},\quad 0 \leq n \leq N-2, \end{aligned}$$
where
$$\begin{aligned} \textbf{K}=\textstyle\begin{cases} ((2a(\alpha ,\tau )+1)\textbf{I}-\textbf{A}_{xx}\textbf{A}^{-1} ) ,& n=0, \\ ((a(\alpha ,\tau )+1)\textbf{I}-\textbf{A}_{xx}\textbf{A}^{-1} ),& 1\leq n \leq N-2, \end{cases}\displaystyle \end{aligned}$$
(2.14)
and
$$ \textbf{F}^{n+1}=\textstyle\begin{cases} a(\alpha ,\tau )(2\Phi _{1}+2\tau \Phi _{2})+ \textbf{f}^{1},\quad n=0, \\ -a(\alpha ,\tau )(-2\textbf{U}^{n}+\textbf{U}^{n-1})-a(\alpha ,\tau ) \sum_{j=1}^{n}b_{j}(\textbf{U}^{n+1-j} \\ \quad {}-2\textbf{U}^{n-j}+\textbf{U}^{n-1-j})+ \textbf{f}^{n+1}, \quad 1\leq n \leq N-2, \end{cases} $$
in which
$$\begin{aligned} \textbf{f}^{n+1}=\bigl[ f^{n+1}(x_{1}), f^{n+1}(x_{2}),\ldots ,f^{n+1}(x_{M}) \bigr]^{T} \end{aligned}$$
and
$$\begin{aligned} \Phi _{j}=\bigl[ \phi _{j}(x_{1}), \phi _{j}(x_{2}),\ldots ,\phi _{j}(x_{M}) \bigr]^{T},\quad j=1,2. \end{aligned}$$
Nonsingularity of the collocation matrix
Lemma 2.5
(See [40])
Let \(K_{y}(x)\) be the reproducing kernel of the space \(\boldsymbol{\mathsf{H}}_{p}[0,b]\), then
$$ \frac{\partial ^{i+j}K_{y}(x)}{\partial x^{i} \partial y^{j}}, \quad 0 \leq i+j \leq 2, $$
(2.15)
is absolutely continuous with respect to x and y.
Lemma 2.6
Let \(\{x_{j}\}_{j=1}^{\infty}\) be dense in the domain \([0,b]\) and the set of functions \(\{\lambda ^{x}_{j} K(x,\cdot)\}_{j=1}^{M}\) be linearly independent on the reproducing kernel space \(\boldsymbol{\mathsf{H}}_{p}[0,b]\). Then the set of vectors \(\{( \lambda ^{x}_{j} K_{y}(x)|_{y=x_{1}},\lambda ^{x}_{j} K_{y}(x)|_{y=x_{2}}, \ldots )^{T}\}_{j=1}^{M}\) is linearly independent.
Proof
If \(\{c_{j}\}_{j=1}^{M}\) satisfies \(\sum_{j=1}^{M}c_{j}( \lambda ^{x}_{j} K_{y}(x)|_{y=x_{1}},\lambda ^{x}_{j} K_{y}(x)|_{y=x_{2}},\ldots )^{T}=0\), one can deduce that
$$\begin{aligned} \sum_{j=1}^{M}c_{j} \lambda ^{x}_{j} K_{y}(x)|_{y=x_{i}}=0, \quad i\geq 1. \end{aligned}$$
(2.16)
From Lemma 2.5 it is clear that the functions \(\lambda ^{x}_{j} K(x,\cdot)\) for \(\lambda _{j} \in \Lambda _{M}\) are continuous. Furthermore, note that \(\{x_{i}\}_{i \geq 1}\) is a dense set. Therefore \(\sum_{j=1}^{M}c_{j} \lambda ^{x}_{j} K(x,\cdot)=0\), which implies that \(c_{j}=0\) (\(j=1,2,\ldots ,M\)). This completes the proof. □
From Lemma 2.6 we can extract the following theorem.
Theorem 2.7
Let the set of functions \(\{\lambda ^{x}_{j} K(x,\cdot)\}_{j=1}^{M}\) be linearly independent on \(\boldsymbol{\mathsf{H}}_{p}[0,b]\). Then there exist m points \(\mathcal{B}_{n}=\{x_{j}\}_{j=1}^{M}\) such that the collocation matrix K is nonsingular.
Lemma 2.8
Let the set of functionals \(\{\lambda _{j}\}_{j=1}^{M}\) be linearly independent on \(\boldsymbol{\mathsf{H}}_{p}[0,b]\). Then the set of functions \(\{\lambda ^{x}_{j} K(x,\cdot)\}_{j=1}^{M}\) is linearly independent.
Proof
If \(\{c_{j}\}_{j=1}^{M}\) satisfies \(\sum_{j=1}^{M}c_{j}\lambda ^{x}_{j} K(x,\cdot)=0\), then we get
$$\begin{aligned} 0 =&\Biggl\langle U(\cdot),\sum_{j=1}^{M}c_{j} \lambda ^{x}_{j} K(x,\cdot)\Biggr\rangle _{ \boldsymbol{\mathsf{H}}_{p}} \\ =&\sum _{j=1}^{M}c_{j}\lambda ^{x}_{j} \bigl\langle U(\cdot), K(x,\cdot)\bigr\rangle _{\boldsymbol{\mathsf{H}}_{p}} \\ =&\sum_{j=1}^{M}c_{j}\lambda _{j}[U], \quad \forall U \in \boldsymbol{\mathsf{H}}_{p}[0,b], \end{aligned}$$
(2.17)
which implies that \(c_{j}=0\) (\(j=1,2,\ldots ,M\)), and this completes the proof. □
From Lemma 2.8 and Theorem 2.7, we can derive the following theorem.
Theorem 2.9
Let the set of functionals \(\{\lambda _{j}\}_{j=1}^{M}\) be linearly independent on \(\boldsymbol{\mathsf{H}}_{p}[0,b]\). Then there exist n points \(\mathcal{B}_{n}=\{x_{j}\}_{j=1}^{n}\) such that the collocation matrix K is nonsingular.
Error analysis
Suppose that \(\mathcal{B}_{M}=\{x_{i}\}_{i = 1}^{M}\) and \(\mathcal{U}_{M}=\operatorname{Span}\{\psi _{1},\psi _{2},\ldots , \psi _{M}\}\). Applying the Gram–Schmidt orthogonalization process to \(\{\psi _{1},\psi _{2},\ldots , \psi _{M}\}\), we can obtain
$$\begin{aligned} \overline{\psi}_{i}(x)=\sum _{k=1}^{i}\beta _{ik}\psi _{k}(x),\quad ( \beta _{ii}>0,i=1,2,\ldots ,M). \end{aligned}$$
(2.18)
Therefore \(\{\overline{\psi}_{1},\overline{\psi}_{2},\ldots , \overline{\psi}_{n} \}\) is an orthonormal basis for \(\mathcal{U}_{M}\).
Therefore, we can write the interpolant \(U_{M}^{n+1}(x)\) to \(U^{n+1}\) at \(\mathcal{B}_{n}\) in the following form:
$$ U^{n+1}(x)\approx U_{M}^{n+1}(x)= \sum_{i=1}^{m}U^{n+1}(x_{i}) \overline{\psi}_{i}(x). $$
(2.19)
Theorem 2.10
Suppose that \(U_{M}^{n+1}(x) \in \boldsymbol{\mathsf{H}}_{p}[0,b]\) and \(U^{n+1}(x)\) are the approximate solution and exact solution for (2.3), respectively. Then, for any \(U^{n+1}(x) \in \boldsymbol{\mathsf{H}}_{p}[0,b]\), we have
$$\begin{aligned} \bigl\vert U^{n+1}(x)-U_{M}^{n+1}(x) \bigr\vert \leq \bigl\Vert U^{n+1} \bigr\Vert _{\boldsymbol{\mathsf{H}}_{p}} \Biggl\Vert K(x,\cdot)-\sum_{i=1}^{M}\overline{ \psi}_{i}(x)\psi _{i} \Biggr\Vert _{ \boldsymbol{\mathsf{H}}_{p}}. \end{aligned}$$
(2.20)
Proof
Using the reproducing property, we have
$$\begin{aligned} U_{M}^{n+1}(x) =&\sum_{i=1}^{M}U^{n+1}(x_{i}) \overline{\psi}_{i}(x) \\ =&\sum_{i=1}^{M}\bigl\langle U^{n+1},\psi _{i}\bigr\rangle _{ \boldsymbol{\mathsf{H}}_{p}}\overline{ \psi}_{i}(x) \\ =& \Biggl\langle U^{n+1},\sum_{i=1}^{M} \overline{\psi}_{i}(x)\psi _{i} \Biggr\rangle _{\boldsymbol{\mathsf{H}}_{p}}. \end{aligned}$$
(2.21)
Thus
$$\begin{aligned} \bigl\vert U^{n+1}(x)-U_{M}^{n+1}(x) \bigr\vert =& \Biggl\vert \Biggl\langle U^{n+1},K(x,\cdot)-\sum _{i=1}^{M} \overline{\psi}_{i}(x)\psi _{i}\Biggr\rangle _{\boldsymbol{\mathsf{H}}_{p}} \Biggr\vert \\ \leq& \bigl\Vert U^{n+1} \bigr\Vert _{\boldsymbol{\mathsf{H}}_{p}} \Biggl\Vert K(x,\cdot)-\sum_{i=1}^{M} \overline{ \psi}_{i}(x)\psi _{i} \Biggr\Vert _{\boldsymbol{\mathsf{H}}_{p}}. \end{aligned}$$
(2.22)
Thus, the proof is completed. □
Lemma 2.11
(See [48])
Suppose that \(u \in C^{3}[0,b]\) and \(\mathcal{B}_{M}=\{x_{i}\}_{i = 1}^{M} \subset [0,b] \) is a distinct subset of \([0,b]\), then
$$ \Vert U \Vert _{L^{2}[0,b]} \leq d \max _{x_{j} \in \mathcal{B}_{n}} \bigl\vert U(x_{j}) \bigr\vert +ch^{3} \biggl\Vert \frac{d^{3}U}{dx^{3}} \biggr\Vert _{L^{2}[0,b]},\quad h=\sup_{x \in [0,b]}\min_{x_{j} \in \mathcal{B}_{M}} \Vert x-x_{j} \Vert , $$
(2.23)
where c and d are real constants.
Theorem 2.12
If \(U^{n+1}_{M}\) is the approximate solution of (2.3) in the space \(\boldsymbol{\mathsf{H}}_{p}\). Then the following relation holds:
$$\begin{aligned} \bigl\Vert U^{n+1}-U_{M}^{n+1} \bigr\Vert _{L^{2}[0,b]} \leq ch^{3} \bigl\Vert U^{n+1} \bigr\Vert _{ \boldsymbol{\mathsf{H}}_{p}}, \end{aligned}$$
(2.24)
where c is a real constant.
Proof
According to Lemma 2.11, we have
$$\begin{aligned}& \bigl\Vert U^{n+1}-U_{M}^{n+1} \bigr\Vert _{L^{2}[0,b]} \\& \quad \leq d \max_{x_{j} \in \mathcal{B}_{n}} \bigl\vert U^{n+1}(x_{j})-U_{M}^{n+1}(x_{j}) \bigr\vert \\& \qquad {} +ch^{3} \biggl\Vert \frac{d^{3}U^{n+1}}{dx^{3}}- \frac{d^{3}U_{M}^{n+1}}{dx^{3}} \biggr\Vert _{L^{2}[0,b]} \\& \quad \leq ch^{3} \bigl\Vert U^{n+1}-U_{M}^{n+1} \bigr\Vert _{\boldsymbol{\mathsf{H}}_{p}}, \end{aligned}$$
(2.25)
where d and c are constants.
We know that
$$\begin{aligned} \bigl\Vert U^{n+1} \bigr\Vert ^{2}_{\boldsymbol{\mathsf{H}}_{p}}= \bigl\Vert U^{n+1}-U_{M}^{n+1} \bigr\Vert ^{2}_{ \boldsymbol{\mathsf{H}}_{p}}+ \bigl\Vert U_{M}^{n+1} \bigr\Vert ^{2}_{\boldsymbol{\mathsf{H}}_{p}}+2 \langle U-U_{M},U_{M} \rangle _{\boldsymbol{\mathsf{H}}_{p}}. \end{aligned}$$
Since
$$\begin{aligned} \langle U-U_{M},U_{M}\rangle _{\boldsymbol{\mathsf{H}}_{p}} =&\Biggl\langle U-U_{M}, \sum_{j=1}^{M} \alpha ^{n+1}_{j}\psi _{j}\Biggr\rangle _{ \boldsymbol{\mathsf{H}}_{p}} \\ =&\sum_{j=1}^{M}\alpha ^{n+1}_{j}\bigl\langle U-U_{M},K(\cdot ,x_{j}) \bigr\rangle _{ \boldsymbol{\mathsf{H}}_{p}} \\ =&\sum_{j=1}^{M} \alpha ^{n+1}_{j} (U-U_{M}) (x_{j})=0, \end{aligned}$$
then
$$\begin{aligned} \bigl\Vert U^{n+1} \bigr\Vert ^{2}_{\boldsymbol{\mathsf{H}}_{p}}= \bigl\Vert U^{n+1}-U_{M}^{n+1} \bigr\Vert ^{2}_{ \boldsymbol{\mathsf{H}}_{p}}+ \bigl\Vert U_{M}^{n+1} \bigr\Vert ^{2}_{\boldsymbol{\mathsf{H}}_{p}}. \end{aligned}$$
Therefore, we have
$$\begin{aligned} \bigl\Vert U^{n+1}-U_{M}^{n+1} \bigr\Vert _{\boldsymbol{\mathsf{H}}_{p}} \leq \bigl\Vert U^{n+1} \bigr\Vert _{ \boldsymbol{\mathsf{H}}_{p}}. \end{aligned}$$
(2.26)
Now, from (2.25) and (2.26), we can obtain
$$\begin{aligned}& \bigl\Vert U^{n+1}-U_{M}^{n+1} \bigr\Vert _{L^{2}[0,b]} \leq ch^{3} \bigl\Vert U^{n+1} \bigr\Vert _{ \boldsymbol{\mathsf{H}}_{p}}. \end{aligned}$$
(2.27)
Thus, the proof is completed. □