In this section, we present a review of the notations and preliminary definitions of the RKHS theory. Additionally, we demonstrate how to solve FVIDE (1)–(2) using the RKHS method. Accordingly, we build orthonormal function systems of the space \(W_{2}^{3} [ 0,1 ] \oplus W_{2}^{3} [ 0,1 ] \) based on the Gram–Schmidt orthogonalization process.
Definition 4.1
(see [24])
Let H be a Hilbert space. A function \(K:\varOmega\times \varOmega \rightarrow \mathbb{R} \) is a reproducing kernel of H if the following conditions are satisfied:
-
(i)
for each \(x\in\varOmega\), \(K(\cdot,x)\in H\);
-
(ii)
for each \(x\in\varOmega\) and \(\phi\in H\), \(\langle \phi ,K(\cdot,x) \rangle=\phi(x)\), which is called the reproducing property.
Definition 4.2
(see [24])
The Hilbert space \(W_{2}^{m} [ a,b ] \) is defined as \(W_{2}^{m} [a,b ]= \{ y(t) |y (t ), y'(t),\ldots,y^{(m-1)}(t)\mbox{ are absolutely continuous }, y^{(m)}(t)\in L^{2}[a,b]\mbox{ and } y(a)=y'(a)=\cdots= y^{(m-1)}(a)=0\mbox{ whenever }m\neq1 \}\). Whilst the inner product and the norm in \(W_{2}^{m} [ a,b ] \) are defined by
$$ \bigl\langle y_{1} ( t ) ,y_{2}(t) \bigr\rangle _{W_{2}^{m}}=\sum_{i=0}^{m-1}y_{1}^{ ( i ) } ( a ) y_{2}^{ ( i ) } ( a ) + \int_{a}^{b}y_{1}^{{(m)}}(t)y_{2}^{{(m)}}(t) \,dt $$
(4)
and \(\Vert y_{1}(t) \Vert _{W_{2}^{m}}=\sqrt{ \langle y_{1}(t),y_{1}(t) \rangle_{W_{2}^{m}}}\), where \(y_{1}\), \(y_{2}\in W_{2}^{m} [ a,b ] \).
Definition 4.3
The Hilbert space \(W_{2}^{m} [ a,b ] \oplus W_{2}^{m} [ a,b ] \), \(m=1,2,3,\ldots,n\), can be defined as \(W_{2}^{m} [ a,b ] \oplus W_{2}^{m} [ a,b ] =\{y=(y_{1},y_{2})^{T}\mid y_{1},y_{2}\in W_{2}^{m} [ a,b ] \}\). Accordingly, the inner product and the norm in \(W_{2}^{m} [ a,b ] \oplus W_{2}^{m} [ a,b ] \) are built, respectively, by \(\langle y ( t ) ,z(t) \rangle _{W_{2}^{m}\oplus W_{2}^{m}}=\sum_{i=1}^{2} \langle y_{i} ( t ) ,z_{i}(t) \rangle_{W_{2}^{m}}\) and \(\Vert y(t) \Vert _{W_{2}^{m}\oplus W_{2}^{m}}=\sqrt{\sum_{i=1}^{2} \Vert y_{i}(t) \Vert _{W_{2}^{m}}^{2}}\).
Theorem 4.4
(see [25])
The space
\(W_{2}^{m} [ a,b ] \)
is a complete reproducing kernel space. That is, for each fixed
\(t\in [ a,b ] \), there exists
\(R_{t} ( s ) \in W_{2}^{m} [ a,b ] \)
such that
\(\langle y(s),R_{t} ( s ) \rangle _{W_{2}^{m}}=y(t)\)
for any
\(y(t)\in W_{2}^{m} [ a,b ] \)
and any
\(t\in [ a,b ] \). The reproducing kernel
\(R_{t} ( s ) \)
can be written as
$$ R_{t} ( s ) = \textstyle\begin{cases} \sum_{i=0}^{2m-1}p_{i}(t)s^{i}, & s\leq t,\\ \sum_{i=0}^{2m-1}q_{i}(t)s^{i}, & s>t.\end{cases} $$
(5)
The representation of the reproducing kernel function \(R_{t} ( s ) \) in \(W_{2}^{3} [ 0,1 ] \), using Mathematica 7.0 software package, is provided by
$$ R_{t} ( s ) =\textstyle\begin{cases} 1+\frac{1}{12}t^{2}s^{2}(3+t)+ts(1-\frac{1}{24}t^{4})+\frac{1}{120}t^{5}, & s\leq t, \\ 1+\frac{1}{12}t^{2}s^{2}(3+s)+ts(1-\frac{1}{24}s^{4})+\frac{1}{120}s^{5}, & s>t.\end{cases} $$
(6)
To apply our technique on the Hilbert space \(W_{2}^{3} [ a,b ] \oplus W_{2}^{3} [ a,b ] \), we define the linear invertible operator \(L:W_{2}^{3} [ a,b ] \oplus W_{2}^{3} [ a,b ] \longrightarrow W_{2}^{1} [ a,b ] \oplus W_{2}^{1} [ a,b ] \) as \(Ly(t)=y^{\prime\prime}(t)\) such that and \(y(t)=(y_{1,\alpha}(t),y_{2,\alpha}(t))^{T}\). Thus, Eqs. (1) and (2) can be converted into the form
$$ Ly(t)=G \bigl( t,f(t),h \bigl(y(t) \bigr) \bigr) , $$
(7)
subject to the fuzzy initial conditions
$$ y(a)=\gamma_{1},\qquad y^{\prime}(a)=\gamma_{2}, $$
(8)
where \(h(y(t))=\int_{a}^{t}k(t,s)g(y(t))\,ds\), \(y(t)\in W_{2}^{3} [ a,b ] \oplus W_{2}^{3} [ a,b ] \), and \(f(t)\in W_{2}^{1} [ a,b ] \oplus W_{2}^{1} [ a,b ] \). Here, \(G ( t,f(t),h(y(t)) ) \) is \(G ( t,f_{1,\alpha }(t),f_{2,\alpha}(t),h(y_{1,\alpha}(t)),h(y_{2,\alpha}(t)) ) \).
Lemma 4.5
\(L:W_{2}^{3} [ a,b ] \oplus W_{2}^{3} [ a,b ] \longrightarrow W_{2}^{1} [ a,b ] \oplus W_{2}^{1} [ a,b ] \)
is a bounded linear operator.
Proof
Firstly, it is easy to prove that L is a linear operator from \(W_{2}^{3} [ a,b ] \oplus W_{2}^{3} [ a,b ] \) into \(W_{2}^{1} [ a,b ] \oplus W_{2}^{1} [ a,b ] \). Secondly, we need to prove that \(\Vert Ly \Vert _{W_{2}^{1}\oplus W_{2}^{1}}\leq\xi \Vert y \Vert _{W_{2}^{3}\oplus W_{2}^{3}}\), where \(\xi >0\). For each \(y\in W_{2}^{3} [ a,b ] \oplus W_{2}^{3} [ a,b ] \), we have
$$\begin{aligned} \bigl\Vert Ly(t) \bigr\Vert _{W_{2}^{1}\oplus W_{2}^{1}}^{2} ={}&\sum _{i=1}^{2} \bigl\Vert L_{i}y_{i,\alpha}(t) \bigr\Vert _{W_{2}^{1}}^{2}= \bigl\Vert L_{1}y_{1,\alpha}(t) \bigr\Vert _{W_{2}^{1}}^{2}+ \bigl\Vert L_{2}y_{2,\alpha}(t) \bigr\Vert _{W_{2}^{1}}^{2} \\ ={}& \bigl\langle L_{1}y_{1,\alpha}(t),L_{1}y_{1,\alpha}(t) \bigr\rangle _{W_{2}^{1}}+ \bigl\langle L_{2}y_{2,\alpha}(t),L_{2}y_{2,\alpha }(t) \bigr\rangle _{W_{2}^{1}} \\ ={}& \bigl[L_{1}y_{1,\alpha}(a) \bigr]^{2}+ \int_{a}^{b} \biggl[\frac{d}{dt} \bigl(L_{1}y_{1,\alpha}(t) \bigr) \biggr]^{2}\,dt+ \bigl[L_{2}y_{2,\alpha}(a) \bigr]^{2} \\ & {} + \int_{a}^{b} \biggl[\frac{d}{dt} \bigl(L_{2}y_{2,\alpha}(t) \bigr) \biggr]^{2}\,dt. \end{aligned}$$
By the reproducing property of \(R_{t} ( s ) \), then \(y_{1,\alpha }(t)= \langle y_{1,\alpha}(s),R_{t} ( s ) \rangle _{W_{2}^{3}}\) and \(y_{2,\alpha}(t)= \langle y_{2,\alpha }(s), R_{t} ( s ) \rangle_{W_{2}^{3}}\). Thus,
$$\begin{aligned} &L_{1}y_{1,\alpha}(t)=\bigl\langle y_{1,\alpha}(s),L_{1}R_{t} ( s ) \bigr\rangle _{W_{2}^{3}}, \qquad L_{2}y_{2,\alpha }(t)= \bigl\langle y_{2,\alpha}(s),L_{2}R_{t} ( s ) \bigr\rangle _{W_{2}^{3}}, \\ &\frac{d}{dt} \bigl( L_{1}y_{1,\alpha}(t) \bigr) = \biggl\langle y_{1,\alpha}(s),\frac{d}{dt} \bigl(L_{1}R_{t} (s ) \bigr) \biggr\rangle _{W_{2}^{3}}, \\ &\frac{d}{dt} \bigl(L_{2}y_{2,\alpha}(t) \bigr) = \biggl\langle y_{2,\alpha}(s),\frac{d}{dt} \bigl(L_{2}R_{t} (s ) \bigr) \biggr\rangle _{W_{2}^{3}}. \end{aligned}$$
From the continuity of \(R_{t} ( s ) \) on \([a,b]\), we have
$$\begin{aligned} &\bigl\vert L_{1}y_{1,\alpha}(t) \bigr\vert = \bigl\vert \bigl\langle y_{1,\alpha}(s),L_{1}R_{t} (s ) \bigr\rangle _{W_{2}^{3}} \bigr\vert \leq \Vert y_{1,\alpha} \Vert _{W_{2}^{3}} \bigl\Vert L_{1}R_{t} (s ) \bigr\Vert _{W_{2}^{3}}\leq k_{1} \Vert y_{1,\alpha} \Vert _{W_{2}^{3}}, \\ &\bigl\vert L_{2}y_{2,\alpha}(t) \bigr\vert = \bigl\vert \bigl\langle y_{2,\alpha}(s),L_{2}R_{t} (s ) \bigr\rangle _{W_{2}^{3}} \bigr\vert \leq \Vert y_{2,\alpha} \Vert _{W_{2}^{3}} \bigl\Vert L_{2}R_{t} (s ) \bigr\Vert _{W_{2}^{3}}\leq k_{2} \Vert y_{2,\alpha} \Vert _{W_{2}^{3}}, \\ &\biggl\vert \frac{d}{dt} \bigl(L_{1}y_{1,\alpha}(t) \bigr) \biggr\vert = \biggl\vert \biggl\langle y_{1,\alpha}(s), \frac {d}{dt} \bigl(L_{1}R_{t} (s ) \bigr) \biggr\rangle _{W_{2}^{3}} \biggr\vert \leq \Vert y_{1,\alpha} \Vert _{W_{2}^{3}} \biggl\Vert \frac{d}{dt} \bigl(L_{1}R_{t} (s ) \bigr) \biggr\Vert _{W_{2}^{3}} \\ &\phantom{\biggl\vert \frac{d}{dt} \bigl(L_{1}y_{1,\alpha}(t) \bigr) \biggr\vert }\leq k_{3} \Vert y_{1,\alpha} \Vert _{W_{2}^{3}}, \\ &\biggl\vert \frac{d}{dt} \bigl(L_{2}y_{2,\alpha}(t) \bigr) \biggr\vert = \biggl\vert \biggl\langle y_{2,\alpha}(s), \frac {d}{dt} \bigl(L_{2}R_{t} (s ) \bigr) \biggr\rangle _{W_{2}^{3}} \biggr\vert \leq \Vert y_{2,\alpha} \Vert _{W_{2}^{3}} \biggl\Vert \frac{d}{dt} \bigl(L_{2}R_{t} (s ) \bigr) \biggr\Vert _{W_{2}^{3}} \\ &\phantom{\biggl\vert \frac{d}{dt} \bigl(L_{2}y_{2,\alpha}(t) \bigr) \biggr\vert }\leq k_{4} \Vert y_{2,\alpha} \Vert _{W_{2}^{3}}. \end{aligned}$$
Thus,
$$\begin{aligned} \bigl\Vert Ly(t) \bigr\Vert _{W_{2}^{1}\oplus W_{2}^{1}}^{2} \leq {}&k_{1}^{2} \Vert y_{1,\alpha} \Vert _{W_{2}^{3}}^{2}+(b-a)k_{3}^{2} \Vert y_{1,\alpha} \Vert _{W_{2}^{3}}^{2}+k_{2}^{2} \Vert y_{2,\alpha} \Vert _{W_{2}^{3}}^{2} +(b-a)k_{4}^{2} \Vert y_{2,\alpha} \Vert _{W_{2}^{3}}^{2} \\ ={}& \bigl(k_{1}^{2}+(b-a)k_{3}^{2} \bigr) \Vert y_{1,\alpha} \Vert _{W_{2}^{3}}^{2}+ \bigl(k_{2}^{2}+(b-a)k_{4}^{2} \bigr) \Vert y_{2,\alpha } \Vert _{W_{2}^{3}}^{2} \\ \leq{}&\xi^{2} \bigl( \Vert y_{1,\alpha} \Vert _{W_{2}^{3}}^{2}+ \Vert y_{2,\alpha} \Vert _{W_{2}^{3}}^{2} \bigr) \\ ={}&\xi^{2}\sum_{i=1}^{2} \Vert y_{i,\alpha} \Vert _{W_{2}^{3}}^{2} \\ ={}&\xi^{2} \Vert y \Vert _{{W_{2}^{3}\oplus W_{2}^{3}}}^{2}, \end{aligned}$$
where \(\xi^{2}=\max\{ k_{1}^{2}+(b-a)k_{3}^{2},k_{2}^{2}+(b-a)k_{4}^{2}\}\). The proof is complete. □
Next, to construct an orthogonal system of functions \(\{ \psi _{ij}(t) \} _{(i,j)=(1,1)}^{(\infty,2)}\) of the space \(W_{2}^{3} [ a,b ] \oplus W_{2}^{3} [ a,b ] \), we let \(\psi _{ij}(t)=L^{\ast}\varPhi_{ij}(t)\), where is the adjoint operator of L and \(\varPhi_{ij}(t)=(\varPhi _{i1}(t),\varPhi_{i2}(t))^{T}\). From now on the orthonormal system \(\{\overline{\psi}_{ij} ( t ) \}_{(i,j)=(1,1)}^{(\infty,2)}\) in the space \(W_{2}^{3} [ a,b ] \oplus W_{2}^{3} [ a,b ] \) can be derived from the Gram–Schmidt orthogonalization process of \(\{\psi{}_{ij} (t )\}_{(i,j)=(1,1)}^{(\infty ,2)}\) as follows:
$$ \overline{\psi}_{ij} ( t ) =\sum_{l=1}^{i} \sum_{k=1}^{j}\beta_{lk}^{ij} \psi_{lk}(t), $$
(9)
where \(\beta_{lk}^{ij}\) are orthogonalization coefficients given by
$$\begin{aligned} &\beta_{11}^{ij}=\frac{1}{ \Vert \psi_{11} \Vert },\qquad \beta _{lk}^{ij}= \Biggl( \Vert \psi_{lk} \Vert ^{2}-\sum_{p=1}^{l-1} \bigl\langle \psi_{lk}(t),\overline{\psi}_{lp}(t) \bigr\rangle ^{2} \Biggr) ^{-\frac{1}{2}}\quad (l=k\neq1), \\ &\beta_{lk}^{ij}=\frac{-\sum_{p=k}^{l-1} \langle\psi _{lk}(t),\overline{\psi}_{lp}(t) \rangle\beta_{pk}^{ij}}{\sqrt{ \Vert \psi_{lk} \Vert ^{2}-\sum_{p=1}^{l-1} \langle\psi _{lk}(t),\overline{\psi}_{lp}(t) \rangle^{2}}}\quad (l>k). \end{aligned}$$
The main result of this article is the following theorems, which give the exact expression of the solution of FVIDE (7)–(8) in the space \(W_{2}^{3} [ a,b ] \oplus W_{2}^{3} [ a,b ] \) and the convergence of our method.
Theorem 4.6
Suppose
\(\{t_{i}\}_{i=1}^{\infty}\)
is dense in
\([a,b]\), if
\(y(t)\in W_{2}^{3} [ a,b ] \oplus W_{2}^{3} [ a,b ] \)
is the solution of FVIDE (7)–(8), then
$$ y(t)=\sum_{i=1}^{\infty }\sum _{j=1}^{2}\sum_{l=1}^{i} \sum_{k=1}^{j}\beta _{lk}^{ij}G_{k} \bigl( t_{l},f(t_{l}),h \bigl(y(t_{l}) \bigr) \bigr) \overline {\psi}_{ij} ( t ) , $$
(10)
which is a convergent series in the sense of
\(\Vert \cdot \Vert _{W_{2}^{3}\oplus W_{2}^{3}}\).
Proof
Firstly, we need to prove that \(\{\psi_{ij} ( t ) \}_{(i,j)=(1,1)}^{(\infty,2)}\) is the complete system in \(W_{2}^{3} [ a,b ] \oplus W_{2}^{3} [ a,b ] \) as follows:
$$\begin{aligned} \psi_{ij}(t) &=L^{\ast}\varPhi_{ij}(t)= \bigl\langle L^{\ast}\varPhi _{ij}(s),R_{t} ( s ) \bigr\rangle _{W_{2}^{3}\oplus W_{2}^{3}} \\ &= \bigl\langle \varPhi_{ij}(s),L_{s}R_{t} ( s ) \bigr\rangle _{W_{2}^{1}\oplus W_{2}^{1}} \\ &=L_{s}R_{t} ( s ) |_{s=t_{i}}\in W_{2}^{3} [ a,b ] \oplus W_{2}^{3} [ a,b ] . \end{aligned}$$
On the other hand, for each \(y\in W_{2}^{3} [ a,b ] \oplus W_{2}^{3} [ a,b ] \), let \(\langle y ( t ) ,\psi_{ij} ( t ) \rangle_{W_{2}^{3}\oplus W_{2}^{3}}=0\). One has
$$\begin{aligned} \bigl\langle y ( t ) ,\psi_{ij} ( t ) \bigr\rangle _{W_{2}^{3}\oplus W_{2}^{3}} &= \bigl\langle y_{1,\alpha}(t),\psi _{i1} ( t ) \bigr\rangle _{W_{2}^{3}}+ \bigl\langle y_{2,\alpha }(t),\psi_{i2} ( t ) \bigr\rangle _{W_{2}^{3}} \\ &= \bigl\langle y_{1,\alpha}(t),L_{1}^{\ast} \varPhi_{i1}(t) \bigr\rangle _{W_{2}^{3}}+ \bigl\langle y_{2,\alpha}(t),L_{2}^{\ast}\varPhi _{i2}(t) \bigr\rangle _{W_{2}^{3}} \\ &= \bigl\langle L_{1}y_{1,\alpha}(t),\varPhi_{i1}(t) \bigr\rangle _{W_{2}^{1}}+ \bigl\langle L_{2}y_{2,\alpha}(t), \varPhi_{i2}(t) \bigr\rangle _{W_{2}^{1}} \\ &=L_{1}y_{1,\alpha}(t_{i})+L_{2}y_{2,\alpha}(t_{i}) \\ &=Ly(t_{i}). \end{aligned}$$
Thus, \(Ly ( t ) =0\) because \(\{ t_{i} \} _{i=1}^{\infty}\)is dense on \([ a,b ] \). Hence, \(y ( t ) =0\) because of the existence of \(L^{-1}\). Secondly, since \(\{\psi_{ij} ( t ) \}_{(i,j)=(1,1)}^{(\infty,2)}\) is the complete system in \(W_{2}^{3} [ a,b ] \oplus W_{2}^{3} [ a,b ] \) and from Eq. (9), then the sequence \(\{\overline{\psi}_{ij} ( t ) \}_{(i,j)=(1,1)}^{(\infty,2)}\) is the complete orthonormal system in \(W_{2}^{3} [ a,b ] \oplus W_{2}^{3} [ a,b ] \). Thirdly, using Fourier series expansion about \(\{ \overline{\psi}_{ij} ( t ) \}_{(i,j)=(1,1)}^{(\infty ,2)}\), we have
$$\begin{aligned} y(t) &=\sum_{i=1}^{\infty}\sum _{j=1}^{2} \bigl\langle y(t),\overline{ \psi}_{ij} ( t ) \bigr\rangle _{W_{2}^{3}\oplus W_{2}^{3}}\overline{ \psi}_{ij} ( t ) \\ &=\sum_{i=1}^{\infty}\sum _{j=1}^{2} \Biggl\langle y(t),\sum _{l=1}^{i}\sum_{k=1}^{j} \beta_{lk}^{ij}\psi _{lk}(t) \Biggr\rangle _{W_{2}^{3}\oplus W_{2}^{3}}\overline{\psi}_{ij} ( t ) \\ &=\sum_{i=1}^{\infty }\sum _{j=1}^{2}\sum_{l=1}^{i} \sum_{k=1}^{j}\beta _{lk}^{ij} \bigl\langle y(t),\psi_{lk}(t) \bigr\rangle _{W_{2}^{3}\oplus W_{2}^{3}}\overline{ \psi}_{ij} ( t ) \\ &=\sum_{i=1}^{\infty }\sum _{j=1}^{2}\sum_{l=1}^{i} \sum_{k=1}^{j}\beta _{lk}^{ij} \bigl\langle y(t),L^{\ast}\varPhi_{lk}(t) \bigr\rangle _{W_{2}^{3}\oplus W_{2}^{3}}\overline{\psi}_{ij} ( t ) \\ &=\sum_{i=1}^{\infty }\sum _{j=1}^{2}\sum_{l=1}^{i} \sum_{k=1}^{j}\beta _{lk}^{ij} \bigl\langle Ly(t),\varPhi_{lk}(t) \bigr\rangle _{W_{2}^{1}\oplus W_{2}^{1}} \overline{ \psi}_{ij} ( t ) \\ &=\sum_{i=1}^{\infty }\sum _{j=1}^{2}\sum_{l=1}^{i} \sum_{k=1}^{j}\beta _{lk}^{ij} \bigl\langle G_{k} \bigl( t,f(t),h \bigl(y(t) \bigr) \bigr) ,\varPhi _{lk}(t) \bigr\rangle _{W_{2}^{1}\oplus W_{2}^{1}}\overline{\psi}_{ij} ( t ) \\ &=\sum_{i=1}^{\infty }\sum _{j=1}^{2}\sum_{l=1}^{i} \sum_{k=1}^{j}\beta _{lk}^{ij}G_{k} \bigl( t_{l},f(t_{l}),h \bigl(y(t_{l}) \bigr) \bigr) \overline {\psi}_{ij} ( t ) . \end{aligned}$$
Since \(\sum_{i=1}^{\infty }\sum_{j=1}^{2}\sum_{l=1}^{i}\sum_{k=1}^{j}\beta _{lk}^{ij}G_{k} ( t_{l},f(t_{l}),h(y(t_{l})) ) \overline{\psi}_{ij} ( t ) \) is a Fourier series in \(W_{2}^{3} [ a,b ] \oplus W_{2}^{3} [ a,b ] \) and \(W_{2}^{3} [ a,b ] \oplus W_{2}^{3} [ a,b ] \) is a Hilbert space, then it is a convergent series in the sense of \(\Vert \cdot \Vert _{W_{2}^{3}\oplus W_{2}^{3}}\). The proof is complete. □
For numerical computations, we put the initial function \(y_{0}(t_{l})=y(t_{l})\) and define the n-term numerical solution of FVIDE (7)–(8) by truncating the series given in Eq. (10) as follows:
$$ y_{n}(t)=\sum_{i=1}^{n}\sum _{j=1}^{2}\sum_{l=1}^{i}\sum_{k=1}^{j}\beta_{lk}^{ij}G_{k} \bigl( t_{l},f(t_{l}),h \bigl(y_{l-1}(t_{l}) \bigr) \bigr) \overline{\psi}_{ij} ( t ) . $$
(11)
Theorem 4.7
Suppose that
\(y(t)\)
is the solution of FVIDE (7)–(8) and
\(y_{n}(t)\)
is the approximate solution of FVIDE (7)–(8), where
\(y(t)\)
and
\(y_{n}(t)\)
are given by Eqs. (10) and (11), respectively, then
-
(i)
if \(\Vert y_{n} \Vert _{{W_{2}^{3}\oplus W_{2}^{3}}}\) is bounded and \(\{t_{i}\}_{i=1}^{n}\) is dense on \([a,b]\), then \(\Vert y_{n}-y \Vert _{{W_{2}^{3}\oplus W_{2}^{3}}}\longrightarrow0\) as \(n\rightarrow\infty\);
-
(ii)
\(\Vert y_{n}-y \Vert _{{C}}\longrightarrow0\) as \(n\rightarrow\infty\);
-
(iii)
the second derivative of \(y_{n}\) converges to the second derivative of y uniformly as \(n\rightarrow\infty\);
-
(iv)
if \(\Vert y_{n-1}-y \Vert _{{W_{2}^{3}\oplus W_{2}^{3}}}\longrightarrow0\) as \(n\rightarrow\infty\), \(\Vert y_{n-1} \Vert _{{W_{2}^{3}\oplus W_{2}^{3}}}\) is bounded, \(t_{n}\rightarrow\tau\) as \(n\rightarrow\infty\), \(G(t,f(t),h(y(t)))\) is continuous for \(t\in{}[ a,b]\) and f, h are continuous functions, then \(G(t_{n},f(t_{n}),h(y_{n-1}(t_{n})))\rightarrow G(\tau,f(\tau ),h(y(\tau)))\).
Proof
(i) From Eq. (11) and the orthonormality of \(\{\overline{\psi}_{ij} ( t ) \}_{(i,j)=(1,1)}^{(\infty ,2)}\), we have \(\Vert y_{n+1} \Vert _{W_{2}^{3}\oplus W_{2}^{3}}^{2}= \Vert y_{0} \Vert _{W_{2}^{3}\oplus W_{2}^{3}}^{2}+\sum_{i=1}^{n+1}\sum_{j=1}^{2}A_{ij}^{2}\), where \(A_{ij}=\sum_{l=1}^{i}\sum_{k=1}^{j}\beta _{lk}^{ij}G_{k} ( t_{l},f(t_{l}),h(y_{l-1}(t_{l})) ) \overline{\psi }_{ij} ( t ) \). Since \(\Vert y_{n+1} \Vert _{W_{2}^{3}\oplus W_{2}^{3}}\geq \Vert y_{n} \Vert _{W_{2}^{3}\oplus W_{2}^{3}}\) and \(\Vert y_{n} \Vert _{{W_{2}^{3}\oplus W_{2}^{3}}}\) is bounded, then \(\Vert y_{n} \Vert _{{W_{2}^{3}\oplus W_{2}^{3}}}\) is convergent and \(\{ \sum_{j=1}^{2}A_{ij}^{2} \} _{i=1}^{\infty}\in l^{2}\). If \(m>n\), then \(\Vert y_{m}-y_{n} \Vert _{W_{2}^{3}\oplus W_{2}^{3}}^{2}=\sum_{l=n+1}^{m}\sum_{j=1}^{2}A_{ij}^{2}\rightarrow0\) as \(m, n\rightarrow\infty\). But \(W_{2}^{3} [ a,b ] \oplus W_{2}^{3} [ a,b ] \) is a Hilbert space, then \(\Vert y_{n}-y \Vert _{{W_{2}^{3}\oplus W_{2}^{3}}}\longrightarrow0\) as \(n\rightarrow\infty\). More details can be found in [26].
(ii) Based on the reproducing property of \(R_{t} ( s ) \) and from (i), then
$$\begin{aligned} \bigl\vert y_{n}(t)-y(t) \bigr\vert &= \bigl\vert \bigl\langle y_{n}(s)-y(s),R_{t}(s) \bigr\rangle _{W_{2}^{3}\oplus W_{2}^{3}} \bigr\vert \\ &\leq \bigl\Vert y_{n}(s)-y(s) \bigr\Vert _{W_{2}^{3}\oplus W_{2}^{3}} \bigl\Vert \bigl( R_{t}(s) \bigr) \bigr\Vert _{W_{2}^{3}\oplus W_{2}^{3}} \\ &\leq k_{5} \Vert y_{n}-y \Vert _{W_{2}^{3}\oplus W_{2}^{3}} \rightarrow0\quad\text{as }n\rightarrow\infty, \end{aligned}$$
which means that \(\Vert y_{n}-y \Vert _{{C}}\longrightarrow0\) as \(n\rightarrow\infty\).
(iii) Since \(y_{n}\) converges to y uniformly, then
$$\begin{aligned} \bigl\vert y_{n}^{{\prime\prime}}(t)-y^{{\prime\prime}}(t) \bigr\vert &= \biggl\vert \frac{d^{2}}{dt^{2}} \bigl(y_{n}(t)-y(t) \bigr) \biggr\vert \\ &= \biggl\vert \frac{d^{2}}{dt^{2}} \bigl\langle y_{n}(s)-y(s),R_{t}(s) \bigr\rangle _{W_{2}^{3}\oplus W_{2}^{3}} \biggr\vert \\ &= \biggl\vert \biggl\langle y_{n}(s)-y(s),\frac{d^{2}}{dt^{2}} \bigl( R_{t}(s) \bigr) \biggr\rangle _{W_{2}^{3}\oplus W_{2}^{3}} \biggr\vert \\ &\leq \bigl\Vert y_{n}(s)-y(s) \bigr\Vert _{W_{2}^{3}\oplus W_{2}^{3}} \biggl\Vert \frac{d^{2}}{dt^{2}} \bigl( R_{t}(s) \bigr) \biggr\Vert _{W_{2}^{3}\oplus W_{2}^{3}} \\ &\leq k_{6} \Vert y_{n}-y \Vert _{W_{2}^{3}\oplus W_{2}^{3}} \rightarrow0\quad\text{as }n\rightarrow\infty. \end{aligned}$$
Thus, \(\Vert y_{n}^{{\prime\prime}}-y^{{\prime \prime }} \Vert _{C}\rightarrow0\) as \(n\rightarrow\infty\).
(iv) Firstly, we prove that \(y_{n-1}(t_{n})\rightarrow y(\tau )\) (\(n\rightarrow\infty\)) as follows:
$$\begin{aligned} \bigl\vert y_{n-1}(t_{n})-y(\tau) \bigr\vert &= \bigl\vert y_{n-1}(t_{n})-y_{n-1}(\tau)+y_{n-1}( \tau)-y(\tau) \bigr\vert \\ &\leq \bigl\vert y_{n-1}(t_{n})-y_{n-1}(\tau) \bigr\vert + \bigl\vert y_{n-1}(\tau)-y(\tau) \bigr\vert . \end{aligned}$$
From \(\Vert y_{n-1}-y \Vert _{{W_{2}^{3}\oplus W_{2}^{3}}}\longrightarrow0\) as \(n\rightarrow\infty\), it follows that \(\vert y_{n-1}(\tau)-y(\tau) \vert \rightarrow0\) as \(n\rightarrow\infty\). Based on the reproducing property of \(R_{t} ( s ) \), then
$$\begin{aligned} \bigl\vert y_{n-1}(t_{n})-y_{n-1}(\tau) \bigr\vert &= \bigl\vert \bigl\langle y_{n-1}(t),R_{t_{n}}(s) \bigr\rangle - \bigl\langle y_{n-1}(t),R_{\tau}(s) \bigr\rangle \bigr\vert \\ &= \bigl\vert \bigl\langle y_{n-1}(t),R_{t_{n}}(s)-R_{\tau}(s) \bigr\rangle \bigr\vert \\ &\leq \bigl\Vert y_{n-1}(t) \bigr\Vert _{{W_{2}^{3}\oplus W_{2}^{3}}} \bigl\Vert R_{t_{n}}(s)-R_{\tau}(s) \bigr\Vert _{{W_{2}^{3}\oplus W_{2}^{3}}}. \end{aligned}$$
Since the kernel function \(R_{t}(s)\) is symmetric, \(\Vert R_{t_{n}}(s)-R_{\tau}(s) \Vert _{{W_{2}^{3}\oplus W_{2}^{3}}}\rightarrow0\) as \(n\rightarrow\infty\), which means that \(\vert y_{n-1}(t_{n})-y_{n-1}(\tau) \vert \rightarrow0\) as \(n\rightarrow\infty\). Thus, \(\vert y_{n-1}(t_{n})-y(\tau) \vert \rightarrow0\) as \(n\rightarrow\infty\). Secondly, since f and h are continuous functions and \(t_{n}\rightarrow\tau\), then \(f(t_{n})\rightarrow f(\tau)\) as \(n\rightarrow\infty\) and \(h(y_{n-1}(t_{n}))\rightarrow h(y(\tau))\) as \(n\rightarrow\infty\). From the continuity of G, then \(G(t_{n},f(t_{n}),h(y_{n-1}(t_{n})))\rightarrow G(\tau,f(\tau ),h(y(\tau)))\). The proof is complete. □