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
$$ biglvert mathcal{R}_{1}^{n+1}(u) bigrvert leq C_{u} tau ^{3-alpha} text{or} mathcal{R}_{1}^{n+1}(u)=O bigl( tau ^{3-alpha}bigr).$$
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 (nrightarrow 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}-2tau phi _{2}(x)) and find (U^{n+1}) ((n=0,1,2,ldots ,N-1)) such that
$$begin{aligned} textstylebegin{cases} a(alpha ,tau )(U^{n+1}-2U^{n}+U^{n-1})+a(alpha ,tau )sum_{j=1}^{n}b_{j}(U^{n+1-j}-2U^{n-j}+U^{n-1-j}) \ quad =Delta U^{n+1}(x)-U^{n+1}(x)+ f^{n+1}(x), \ U^{n+1}|_{x in partial Omega}=0,quad -1leq n leq N-1. end{cases}displaystyle end{aligned}$$
For the convenience of discussion, define the linear operator L as follows:
$$begin{aligned} textrm{L} (*)=textstylebegin{cases} ((2a(alpha ,tau )+1)-Delta ) (*),& n=0, \ ((a(alpha ,tau )+1)-Delta )(*),& 1leq n leq N-2. end{cases}displaystyle end{aligned}$$
Therefore, a semi-discrete problem can be converted into the following equivalent:
$$begin{aligned} textrm{L} U^{n+1}(x)=F^{n+1}(x),quad 0leq n leq N-1, end{aligned}$$
(2.3)
where
$$ F^{n+1}(x)=textstylebegin{cases} -a(alpha ,tau )(-2phi _{1}(x)-2tau phi _{2}(x))+ f^{1}(x),quad n=0, \ -a(alpha ,tau )(-2U^{n}+U^{n-1})-a(alpha ,tau )sum_{j=1}^{n}b_{j}(U^{n+1-j} \ quad {}-2U^{n-j}+U^{n-1-j})+ f^{n+1}(x),quad 1leq n leq N-2. end{cases} $$
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 (xin 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
$$begin{aligned} boldsymbol{mathsf{H}}_{p}[0,b]=bigl{ u|u(x),u^{prime }(x),u^{prime prime }(x)in AC[0,b], u^{prime prime prime }(x) in L^{2}[0,b],u(0)=u(b)=0, xin [0,b]bigr} . end{aligned}$$
The inner product in (boldsymbol{mathsf{H}}_{p}[0,b]) is in the form
$$begin{aligned} langle u,vrangle _{boldsymbol{mathsf{H}}_{p}}=u(0)v(0)+u(b)v(b)+u^{prime }(0)v^{prime }(0)+ int _{a}^{b}u^{prime prime prime }(x)v^{prime prime prime }(x),dx, end{aligned}$$
(2.5)
and the norm (|u|_{boldsymbol{mathsf{H}}}) is defined by
$$begin{aligned} Vert u Vert _{boldsymbol{mathsf{H}}_{p}}=sqrt{langle u,urangle _{mathsf{H}}}, end{aligned}$$
(2.6)
where (u,vin 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]
$$begin{aligned} K_{y}(x)= textstylebegin{cases} frac{1}{120b^{2}}(b-x)y(120bx+x(6b^{2}x-120-4bx^{2}+x^{3})y \ quad {}-5bxy^{3}+(b+x)y^{4}), quad y< x, \ frac{1}{120b^{2}}(b-y)x(120by+y(6b^{2}y-120-4by^{2}+y^{3})x \ quad {}-5byx^{3}+(b+y)x^{4}), quad ygeq x. end{cases}displaystyle end{aligned}$$
(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]):
$$begin{aligned} textrm{L} U^{n+1}(x)=F^{n+1}(x),quad 0leq n leq N-1, end{aligned}$$
where
$$ F^{n+1}(x)=textstylebegin{cases} -a(alpha ,tau )(-2phi _{1}(x)-2tau phi _{2}(x))+ f^{1}(x),quad n=0, \ -a(alpha ,tau )(-2U^{n}+U^{n-1})-a(alpha ,tau )sum_{j=1}^{n}b_{j}(U^{n+1-j} \ quad {}-2U^{n-j}+U^{n-1-j})+ f^{n+1}(x),quad 1leq n leq N-2, end{cases} $$
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}), ((1leq 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
$$begin{aligned} textbf{K}=textstylebegin{cases} ((2a(alpha ,tau )+1)textbf{I}-textbf{A}_{xx}textbf{A}^{-1} ) ,& n=0, \ ((a(alpha ,tau )+1)textbf{I}-textbf{A}_{xx}textbf{A}^{-1} ),& 1leq n leq N-2, end{cases}displaystyle end{aligned}$$
(2.14)
and
$$ textbf{F}^{n+1}=textstylebegin{cases} a(alpha ,tau )(2Phi _{1}+2tau Phi _{2})+ textbf{f}^{1},quad n=0, \ -a(alpha ,tau )(-2textbf{U}^{n}+textbf{U}^{n-1})-a(alpha ,tau ) sum_{j=1}^{n}b_{j}(textbf{U}^{n+1-j} \ quad {}-2textbf{U}^{n-j}+textbf{U}^{n-1-j})+ textbf{f}^{n+1}, quad 1leq n leq N-2, end{cases} $$
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
$$begin{aligned} sum_{j=1}^{M}c_{j} lambda ^{x}_{j} K_{y}(x)|_{y=x_{i}}=0, quad igeq 1. end{aligned}$$
(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
$$begin{aligned} 0 =&Biggllangle U(cdot),sum_{j=1}^{M}c_{j} lambda ^{x}_{j} K(x,cdot)Biggrrangle _{ boldsymbol{mathsf{H}}_{p}} \ =&sum _{j=1}^{M}c_{j}lambda ^{x}_{j} bigllangle U(cdot), K(x,cdot)bigrrangle _{boldsymbol{mathsf{H}}_{p}} \ =&sum_{j=1}^{M}c_{j}lambda _{j}[U], quad forall U in boldsymbol{mathsf{H}}_{p}[0,b], end{aligned}$$
(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
$$begin{aligned} biglvert U^{n+1}(x)-U_{M}^{n+1}(x) bigrvert leq biglVert U^{n+1} bigrVert _{boldsymbol{mathsf{H}}_{p}} BigglVert K(x,cdot)-sum_{i=1}^{M}overline{ psi}_{i}(x)psi _{i} BiggrVert _{ boldsymbol{mathsf{H}}_{p}}. end{aligned}$$
(2.20)
Proof
Using the reproducing property, we have
$$begin{aligned} U_{M}^{n+1}(x) =&sum_{i=1}^{M}U^{n+1}(x_{i}) overline{psi}_{i}(x) \ =&sum_{i=1}^{M}bigllangle U^{n+1},psi _{i}bigrrangle _{ boldsymbol{mathsf{H}}_{p}}overline{ psi}_{i}(x) \ =& Biggllangle U^{n+1},sum_{i=1}^{M} overline{psi}_{i}(x)psi _{i} Biggrrangle _{boldsymbol{mathsf{H}}_{p}}. end{aligned}$$
(2.21)
Thus
$$begin{aligned} biglvert U^{n+1}(x)-U_{M}^{n+1}(x) bigrvert =& Bigglvert Biggllangle U^{n+1},K(x,cdot)-sum _{i=1}^{M} overline{psi}_{i}(x)psi _{i}Biggrrangle _{boldsymbol{mathsf{H}}_{p}} Biggrvert \ leq& biglVert U^{n+1} bigrVert _{boldsymbol{mathsf{H}}_{p}} BigglVert K(x,cdot)-sum_{i=1}^{M} overline{ psi}_{i}(x)psi _{i} BiggrVert _{boldsymbol{mathsf{H}}_{p}}. end{aligned}$$
(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
$$ Vert U Vert _{L^{2}[0,b]} leq d max _{x_{j} in mathcal{B}_{n}} biglvert U(x_{j}) bigrvert +ch^{3} bigglVert frac{d^{3}U}{dx^{3}} biggrVert _{L^{2}[0,b]},quad h=sup_{x in [0,b]}min_{x_{j} in mathcal{B}_{M}} Vert x-x_{j} Vert , $$
(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:
$$begin{aligned} biglVert U^{n+1}-U_{M}^{n+1} bigrVert _{L^{2}[0,b]} leq ch^{3} biglVert U^{n+1} bigrVert _{ boldsymbol{mathsf{H}}_{p}}, end{aligned}$$
(2.24)
where c is a real constant.
Proof
According to Lemma 2.11, we have
$$begin{aligned}& biglVert U^{n+1}-U_{M}^{n+1} bigrVert _{L^{2}[0,b]} \& quad leq d max_{x_{j} in mathcal{B}_{n}} biglvert U^{n+1}(x_{j})-U_{M}^{n+1}(x_{j}) bigrvert \& qquad {} +ch^{3} bigglVert frac{d^{3}U^{n+1}}{dx^{3}}- frac{d^{3}U_{M}^{n+1}}{dx^{3}} biggrVert _{L^{2}[0,b]} \& quad leq ch^{3} biglVert U^{n+1}-U_{M}^{n+1} bigrVert _{boldsymbol{mathsf{H}}_{p}}, end{aligned}$$
(2.25)
where d and c are constants.
We know that
$$begin{aligned} biglVert U^{n+1} bigrVert ^{2}_{boldsymbol{mathsf{H}}_{p}}= biglVert U^{n+1}-U_{M}^{n+1} bigrVert ^{2}_{ boldsymbol{mathsf{H}}_{p}}+ biglVert U_{M}^{n+1} bigrVert ^{2}_{boldsymbol{mathsf{H}}_{p}}+2 langle U-U_{M},U_{M} rangle _{boldsymbol{mathsf{H}}_{p}}. end{aligned}$$
Since
$$begin{aligned} langle U-U_{M},U_{M}rangle _{boldsymbol{mathsf{H}}_{p}} =&Biggllangle U-U_{M}, sum_{j=1}^{M} alpha ^{n+1}_{j}psi _{j}Biggrrangle _{ boldsymbol{mathsf{H}}_{p}} \ =&sum_{j=1}^{M}alpha ^{n+1}_{j}bigllangle U-U_{M},K(cdot ,x_{j}) bigrrangle _{ boldsymbol{mathsf{H}}_{p}} \ =&sum_{j=1}^{M} alpha ^{n+1}_{j} (U-U_{M}) (x_{j})=0, end{aligned}$$
then
$$begin{aligned} biglVert U^{n+1} bigrVert ^{2}_{boldsymbol{mathsf{H}}_{p}}= biglVert U^{n+1}-U_{M}^{n+1} bigrVert ^{2}_{ boldsymbol{mathsf{H}}_{p}}+ biglVert U_{M}^{n+1} bigrVert ^{2}_{boldsymbol{mathsf{H}}_{p}}. end{aligned}$$
Therefore, we have
$$begin{aligned} biglVert U^{n+1}-U_{M}^{n+1} bigrVert _{boldsymbol{mathsf{H}}_{p}} leq biglVert U^{n+1} bigrVert _{ boldsymbol{mathsf{H}}_{p}}. end{aligned}$$
(2.26)
Now, from (2.25) and (2.26), we can obtain
$$begin{aligned}& biglVert U^{n+1}-U_{M}^{n+1} bigrVert _{L^{2}[0,b]} leq ch^{3} biglVert U^{n+1} bigrVert _{ boldsymbol{mathsf{H}}_{p}}. end{aligned}$$
(2.27)
Thus, the proof is completed. □
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