### 4.1 MRKHS solution of GBT equation along with boundary conditions

In this section, we present a brief description for the notations and preliminary definitions of the MRKHS theory. Additionally, we explain how to solve GBT with boundary conditions (1.2)–(1.4) using the MRKHS method. Accordingly, we construct orthonormal function systems of the space based on the process orthogonalization of Gram–Schmidt.

Consider the GBT equation within CFFD

$$ \textstyle\begin{cases} a_{1}(\xi ) \omega ''(\xi )+ a_{2}(\xi ) {}^{CF}\mathcal{D}_{0}^{ \frac{3}{2}}\omega (\xi )+ a_{3}(\xi )\omega '(\xi )\\ \quad {}+ a_{4}(\xi ){}^{CF} \mathcal{D}_{0}^{\frac{1}{2}}\omega (\xi )+ a_{5}(\xi )\omega (\xi )=h( \xi ) , \\ \omega (0)=\mu _{1} ;\qquad \omega (X)=\mu _{2} . \end{cases} $$

(4.1)

Now, to apply our technique to GBT equation (4.1) on the Hilbert space \(\mathcal{H}_{2}[0,X]\), we consider a linear differential operator defined as follows:

$$ \textstyle\begin{cases} \mathcal{L}:\mathcal{H}_{3}[0,X] \longrightarrow \mathcal{H}_{1}[0,X] , \\ \mathcal{L}\omega (\xi )= a_{1}(\xi ) \omega ''(\xi )+ a_{2}(\xi ) {}^{CF} \mathcal{D}_{0}^{\frac{3}{2}}\omega (\xi )+ a_{3}(\xi )\omega '(\xi )\\ \hphantom{\mathcal{L}\omega (\xi )= }{}+ a_{4}(\xi ){}^{CF}\mathcal{D}_{0}^{\frac{1}{2}}\omega (\xi )+ a_{5}( \xi )\omega (\xi ). \end{cases} $$

(4.2)

Thus, using the simple transform \(z(\xi ):=(\omega (\xi )-(\mu _{2}-\mu _{1})\xi +\mu _{1})\), the GBT with the boundary condition equation (4.1) can be equivalently converted to the form

$$ \textstyle\begin{cases} \mathcal{L}z(\xi )=h(\xi ), \quad \xi \in [0, X] , \\ z(0)=0,\qquad z(X)=0 . \end{cases} $$

(4.3)

### Theorem 4.1

*The differential operator* \(\mathcal{L}\) *from* \(\mathcal{H}_{3}[0,X] \) *into* \(\mathcal{H}_{1}[0,X]\) *is bounded and linear*. *Hence*, \(\mathcal{L}\) *is continuous*.

### Proof

In order to prove that \(\mathcal{L}\) is a bounded operator, it is enough to find \(M>0\) such that \(\frac{\|\mathcal{L}z(\xi )\|_{\mathcal{H}_{1}}}{\|z(\xi )\|_{\mathcal{H}_{3}}} \leq M\). By the definition of the inner product for \(m=1\) in (3.2) on the Hilbert space \(\mathcal{H}_{1}[0,X]\), we have

$$ \bigl\Vert \mathcal{L}z(\xi ) \bigr\Vert _{\mathcal{H}_{1}}^{2}= \bigl\langle \mathcal{L}z( \xi ) ,\mathcal{L}z(\xi )\bigr\rangle _{\mathcal{H}_{1}}= \bigl[\mathcal{L}z(0)\bigr]^{2}+ \int ^{X}_{0}\bigl| (\mathcal{L}z)'(\xi )\bigr| ^{2}\,d\xi . $$

On the other hand, using the reproducing property of the MRKHS and by the Cauchy–Schwarz inequality, we can write

$$ \begin{aligned} \bigl| (\mathcal{L}z)^{(i)}(\xi )\bigr| &= \bigl\vert \bigl\langle z(\xi ),\bigl(\mathcal{L}V^{ \{3\}}_{\xi } \bigr)^{(i)}(\xi )\bigr\rangle _{\mathcal{H}_{3}} \bigr\vert \\ & \leq \bigl\Vert \bigl(\mathcal{L}V^{\{3\}}_{\xi } \bigr)^{(i)}(\xi ) \bigr\Vert _{\mathcal{H}_{3}} \bigl\Vert z(\xi ) \bigr\Vert _{\mathcal{H}_{3}} \\ & \leq M_{\{i\}} \bigl\Vert z(\xi ) \bigr\Vert _{\mathcal{H}_{3}},\quad i=0,1. \end{aligned} $$

(4.4)

Hence,

$$ \begin{aligned} \bigl\Vert \mathcal{L}z(\xi ) \bigr\Vert _{\mathcal{H}_{1}}^{2} & \leq \biggl[M_{\{0\}}^{2}+ \int ^{X}_{0}M_{\{1\}}^{2}\,d\xi \biggr] \bigl\Vert z(\xi ) \bigr\Vert ^{2}_{\mathcal{H}_{3}} \\ & \leq \bigl(M_{\{0\}}^{2}+ XM_{\{1\}}^{2} \bigr) \bigl\Vert z(\xi ) \bigr\Vert ^{2}_{\mathcal{H}_{3}} \\ & \leq M^{2} \bigl\Vert z(\xi ) \bigr\Vert ^{2}_{\mathcal{H}_{3}}, \end{aligned} $$

(4.5)

where \(M=\sqrt{(M_{\{0\}}^{2}+XM_{\{1\}}^{2})}\).

Next, for all \(z,\eta \in \mathcal{H}_{3}[0,X] \), we have \(\|\mathcal{L}(z+\eta )-\mathcal{L}(z)\|_{\mathcal{H}_{1}}= \| \mathcal{L}(\eta )\|_{\mathcal{H}_{1}}\leq M\|\eta \|_{\mathcal{H}_{3}}\). Letting \(\eta \rightarrow 0\) implies that \(\mathcal{L}\) is continuous. □

We construct an orthonormal function system of \(\mathcal{H}_{3}[0,X]\) as follows: Put \(\Phi _{i}(\cdot)=V^{\{1\}}_{\xi _{i}}(\cdot)\) and \(\Psi _{i}(\cdot)=\mathcal{L}^{\star }\Phi _{i}(\cdot)\), where \(\mathcal{L}^{\star }\) is the adjoint operator of \(\mathcal{L}\) and \(\{\xi \}_{i=1}^{\infty }\) is a dense set on \([0, X]\).

The orthonormal system \(\{\widehat{\Psi }_{i}(\xi )\}_{i=1}^{\infty }\) of the space \(\mathcal{H}_{3}[0,X]\) can be generated from the well-known Gram–Schmidt orthogonalization process as follows:

$$ \begin{aligned} \widehat{\Psi }_{i}(\xi )&=\sum ^{i}_{k=1} \sigma _{ik}\Psi _{i1}(\xi ),\quad i=1,2,\ldots, \end{aligned} $$

(4.6)

where \(\sigma _{ik}>0\) are the orthogonalization coefficients.

### Theorem 4.2

*For* (4.3), *if* \(\{\xi _{i}\}_{i=1}^{\infty }\) *is a dense set on* [0, *X*], *then the orthogonal function system* \(\{ \Psi _{i}(\xi ) \}_{i=1}^{\infty }\) *is complete in* \(\mathcal{H}_{3}[0,X]\).

### Proof

Note that

$$ \Psi _{i}(\xi )=\bigl\langle \mathcal{L}^{\star }\Phi _{i}(\tau ) ,V^{\{3\}}_{ \xi _{i}}(\tau ) \bigr\rangle = \bigl\langle \Phi _{i}(\tau ) ,\mathcal{L}_{ \tau }V^{\{3\}}_{\xi _{i}}( \tau ) \bigr\rangle =\mathcal{L}_{\tau }V^{\{3\}}_{ \xi _{i}}( \tau )\in \mathcal{H}_{1}[0,X]. $$

(4.7)

Now, for each \(z(\xi )\in \mathcal{H}_{3}[0,X]\), suppose \(\langle z(\xi ),\Psi _{i}(\xi ) \rangle =0\), thus we have

$$ \begin{aligned} \bigl\langle z(\xi ),\Psi _{i}(\xi ) \bigr\rangle &=\bigl\langle z(\xi ),\mathcal{L}^{ \star }\Phi _{i}(\xi ) \bigr\rangle \\ & =\bigl\langle \mathcal{L}z(\xi ),\Phi _{i}(\xi ) \bigr\rangle \\ & =\mathcal{L}z(\xi _{i})=0. \end{aligned} $$

(4.8)

Because \(\{\xi _{i}\}_{i=1}^{\infty }\) is dense in [0, X] and \(\mathcal{L}\) is continuous, we get \(\mathcal{L}z(\xi _{i})=0\). Using the existence of the inverse operator \(\mathcal{L}^{-1}\), we conclude that \(z(\xi )=0\). So the proof is complete. □

### Theorem 4.3

*If* \(\{\xi _{i}\}_{i=1}^{\infty }\) *is a dense subset on* [0, *X*] *and the analytic solution* \(z(\xi )\) *of* (4.1) *is unique*, *then the solution* \(z(\xi )\) *can be represented in the following form*:

$$ \begin{aligned} z(\xi )&= \sum _{i=1}^{\infty }\sum_{k=1}^{i} \sigma _{ik}h(\xi _{k})) \widehat{\Psi }_{i}(\xi ). \end{aligned} $$

(4.9)

### Proof

For each \(z(\xi )\in \mathcal{H}_{3}[0,X]\), \(z(\xi )\) can be extended in the Fourier series \(\sum^{\infty }_{i=1}\langle z(\xi ), \widehat{\Psi }_{i}(\xi ) \rangle \widehat{\Psi }_{i}(\xi ) \) about the orthonormal function system \(\{ \Psi _{i}(\xi ) \}_{i=1}^{\infty }\) of \(\mathcal{H}_{3}[0,X]\). Moreover, the series \(\sum^{\infty }_{i=1}\langle z(\xi ),\widehat{\Psi }_{i}(\xi ) \rangle \) is uniformly convergent in the Hilbert space \(\mathcal{W}_{2}[0,X]\). So, we have

$$ \begin{aligned} z(\xi ) &=\sum^{\infty }_{i=1} \bigl\langle z(\xi ),\widehat{\Psi }_{i}(\xi ) \bigr\rangle \widehat{ \Psi }_{i}(\xi ) \\ & =\sum_{i=1}^{\infty }\sum _{k=1}^{i}\sigma _{ik}\bigl\langle z(\xi ), \Psi _{k}(\xi )\bigr\rangle \widehat{\Psi }_{i}(\xi ) \\ & =\sum_{i=1}^{\infty }\sum _{k=1}^{i}\sigma _{ik}\bigl\langle \mathcal{L}z( \xi ),\Phi _{k1}(\xi ) \bigr\rangle \widehat{\Psi }_{i}(\xi ) \\ & = \sum_{i=1}^{\infty }\sum _{k=1}^{i}\sigma _{ik}\mathcal{L}z(\xi _{k}) \widehat{\Psi }_{i}(\xi ) \\ &=\sum_{i=1}^{\infty }\sum _{k=1}^{i}\sigma _{ik}h(\xi _{k}) \widehat{\Psi }_{i}(\xi ). \end{aligned} $$

(4.10)

□

Now, the approximate solution can be obtained by truncating the series in (4.9) as follows:

$$ \begin{aligned} z(\xi )&= \sum _{i=1}^{n}\sum_{k=1}^{i} \sigma _{ik}h(\xi _{k})) \widehat{\Psi }_{i}(\xi ). \end{aligned} $$

(4.11)

In the following theorem, we prove that the error that results when approximating the solution in (4.9) by the form (4.11) is decreasing to zero.

### Theorem 4.4

*Let* \(E_{n}=\|z-z_{n}\|_{\mathcal{H}_{3}}\), *where* *z*, \(z_{n}\) *are respectively the exact and the approximate solution of* (4.1) *represented in* (4.9) *and* (4.11), *then the error* \(E_{n}\) *decreases monotonously in the sense of* \(\|\cdot\|_{\mathcal{H}_{3}}\), *and* \(E_{n}\rightarrow 0\) *as* \(n\rightarrow \infty \).

### Proof

We have

$$ \begin{aligned} E_{n}^{2} &= \Vert z-z_{n} \Vert ^{2}_{\mathcal{H}_{3}} \\ & = \Biggl\Vert \sum_{i=n+1}^{\infty }\sum _{k=1}^{i}\sigma _{ik}h(\xi _{k}) \widehat{\Psi }_{i}(\xi ) \Biggr\Vert ^{2}_{\mathcal{H}_{3}} \\ &= \Biggl\Vert \sum_{i=n+1}^{\infty }A_{i} \widehat{\Psi }_{i}(\xi ) \Biggr\Vert ^{2}_{ \mathcal{H}_{3}} \\ &= \sum_{i=n+1}^{\infty }(A_{i})^{2} \end{aligned} $$

and

$$ \begin{aligned} E_{n-1}^{2} = \Vert z-z_{n-1} \Vert ^{2}_{\mathcal{H}_{3}}=\sum _{i=n-1}^{ \infty }(A_{i})^{2}=(A_{n-1})^{2}+ \sum_{i=n}^{\infty }(A_{i})^{2}. \end{aligned} $$

Note that \(E_{n-1}^{2} > E_{n}^{2} \), so we conclude that the error \(E_{n} \) is monotone decreasing in the sense of \(\|\cdot\|^{2}_{\mathcal{H}_{3}}\), and because \(\sum^{\infty }_{i=1}\langle z(\xi ),\widehat{\Psi }_{i}(\xi ) \rangle \widehat{\Psi }_{i}(\xi ) \) is convergent, \(E_{n}\rightarrow 0\) as \(n\rightarrow \infty \). Hence, the proof is complete. □

### 4.2 MRKHS solution of GBT equation along with initial conditions

Now, we give some notations and preliminary definitions of the MRKHS theory. We then explain how to reformulate the GBT under the initial conditions (1.2)–(1.3) into an equivalent system of first-order fractional differential equations and how to implement our method to solve this system, with highlighting the relationship between the solution of the system and the solution of GBT. Accordingly, we construct an orthonormal function system of the space based on the Gram–Schmidt orthogonalization process.

Consider the following GBT with initial conditions:

$$ \textstyle\begin{cases} a_{1}(\xi ) \omega ''(\xi )+ a_{2}(\xi ) {}^{CF}\mathcal{D}_{xi}^{ \frac{3}{2}}\omega (\xi )+ a_{3}(\xi )\omega '(\xi )\\ \quad {}+ a_{4}(\xi ){}^{CF} \mathcal{D}_{\xi }^{\frac{1}{2}}\omega (\xi )+ a_{5}(\xi )\omega (\xi )=h( \xi ), & \\ \omega (0)=\mu _{1},\qquad \omega ^{\prime }(0)=\mu _{2}. & \end{cases} $$

(4.12)

We can obtain an equivalent form of (4.12) by homogenizing the initial conditions using a transformation given by the following formula \(z(\xi ):=\omega (\xi )-(\mu _{2}\xi +\mu _{1})\)

$$ \textstyle\begin{cases} a_{1}(\xi ) z''(\xi )+ a_{2}(\xi ) {}^{CF}\mathcal{D}_{\xi }^{ \frac{3}{2}}z(\xi )+ a_{3}(\xi )z'(\xi )\\ \quad {}+ a_{4}(\xi ){}^{CF}\mathcal{D}_{ \xi }^{\frac{1}{2}}z(\xi )+ a_{5}(\xi )z(\xi )=H(\xi ),& \\ z(0)=0,\qquad z^{\prime }(0)=0, & \end{cases} $$

(4.13)

where \(H(\xi )=h(\xi )-(a_{3}(\xi )\mu _{2}+2\mu _{2} a_{4}(\xi )(1-e^{-\xi })+a_{5}( \xi )(\mu _{2}\xi +\mu _{1}))\).

Before starting to describe the approximate RKHS scheme, we find it appropriate to rewrite the equivalent GBT equation (4.13) in the form of a system of fractional differential equations (SFDE) of first order by setting \(z(\xi )=z_{1}(\xi )\) and \(z_{1}'(\xi )=z_{2}(\xi )\). In this sense, the equivalent SFDE that we design has the form

$$ \begin{gathered} z'_{1}(\xi )= z_{2}(\xi ), \\ a_{1}(\xi )z'_{2}(\xi )+ a_{2}(\xi ){}^{CF}\mathcal{D}_{0}^{\frac{1}{2}}z_{2}( \xi )= H(\xi )- a_{3}(\xi )z_{1}(\xi ), \end{gathered} $$

(4.14)

equipped with the initial conditions

$$ z_{1}(0)=0 ;\qquad z_{2}(0)=0. $$

(4.15)

It is obvious that GBT equation (4.13) with its original initial conditions is equivalent to SFDE (4.14) with the new initial conditions (4.15) in the following sense: whenever \(Z=(z_{1},z_{2})^{T}\) with \(z_{1}\in \mathcal{H}_{2}[0,X]\) is a solution of SFDE (4.14) associated with initial conditions (4.15), the solution \(z := z_{1}\) solves GBT equation (4.13). Also, whenever \(z\in \mathcal{H}_{2}[0,X]\) is a solution to GBT equation (4.13) with original initial conditions, the vector of solutions \(Z:=(z_{1},z_{2})^{T}:= (z,z')^{T}\) solves SFDE (4.14) equipped with initial conditions (4.15).

Now, we begin by applying the MRKHS approach to solve the SFDE by defining differential operators as

$$ \mathcal{L}_{1},\mathcal{L}_{2}:\mathcal{H}_{2}[0,X] \longrightarrow \mathcal{H}_{1}[0,X], $$

(4.16)

whereas \(\mathcal{L}_{1}z_{1}(\xi )=z'_{1}(\xi )\) and \(\mathcal{L}_{2}z_{2}(\xi )= a_{1}(\xi )z'_{2}(\xi )+ a_{2}(\xi ){}^{CF} \mathcal{D}_{\xi }^{\frac{1}{2}}z_{2}(\xi )\). Put \mathcal{L}=\left(\begin{array}{cc}{\mathcal{L}}_{1}& 0\\ 0& {\mathcal{L}}_{2}\end{array}\right)\in {\mathcal{W}}_{2}[0,X], \(H_{1}(\xi ,Z(\xi ))=z_{2}(\xi )\), \(H_{2}(\xi ,Z(\xi ))= a_{3}(\xi )z_{1}(\xi )+H(\xi )\), and \(H(\xi ,Z(\xi ))= (H_{1}(\xi ,Z(\xi )),H_{2}(\xi ,Z(\xi )))^{T}\).

Consequently, the initial value GBT equation can be converted into the form

$$ \mathcal{L}Z(\xi )=H\bigl(\xi ,Z(\xi )\bigr) , $$

equipped with the initial conditions

$$ Z(\mathbf{0})=\mathbf{0}. $$

### Lemma 4.5

*The differential operators* \(\mathcal{L}_{1},\mathcal{L}_{2}:\mathcal{H}_{2}[0,X]\longrightarrow \mathcal{H}_{1}[0,X]\) *are linear and bounded operators*. *Consequently*, *the operator* \(\mathcal{L}:\mathcal{W}_{2}[0,X] \longrightarrow \mathcal{W}_{1}[0,X]\) *is also linear and bounded*.

### Proof

The proof is divided into two parts; first, we prove that \(\mathcal{L}_{j}:\mathcal{H}_{2}[0,X]\longrightarrow \mathcal{H}_{1}[0,X]\), \(j=1,2\), are bounded and linear. The linearity is obvious since both integer order and Caputo–Fabrizio derivatives are linear.

For boundedness, let \(z\in \mathcal{H}_{2}[0,X]\), then

$$ \Vert \mathcal{L}_{j}z_{j} \Vert ^{2}_{\mathcal{H}_{1}}= \langle \mathcal{L}_{j}z_{j}, \mathcal{L}_{j}z_{j} \rangle _{ \mathcal{H}_{1}}= \int _{0}^{X} \bigl( (\mathcal{L}_{j}z_{j}) (\tau )\bigr)^{2} +\bigl( (\mathcal{L}_{j}z_{j})'( \tau )\bigr)^{2} \,d\tau . $$

Using the reproducing property of \(V^{\lbrace 2\rbrace }_{\xi }(\tau )\), we can write \(z(\xi )= \langle z(\cdot), V^{\lbrace 2\rbrace }_{\xi }(\cdot) \rangle _{\mathcal{H}_{2}}\) and

$$ z'(\xi )= \biggl\langle z(\cdot), \frac{d}{d\xi }V^{\lbrace 2\rbrace }_{\xi }( \cdot) \biggr\rangle _{\mathcal{H}_{2}}, $$

and

$$ a_{1}(\xi )z'(\xi )+ a_{2}(\xi ){}^{CF}\mathcal{D}_{\xi }^{\frac{1}{2}}z( \xi )= \biggl\langle z( \cdot),a_{1}(\xi ) \frac{d}{d\xi } V^{\lbrace 2 \rbrace }_{\xi }( \cdot)+ a_{2}(\xi ){}^{CF}\mathcal{D}_{\xi }^{\frac{1}{2}}V^{ \lbrace 2\rbrace }_{\xi }( \cdot) \biggr\rangle _{\mathcal{H}_{2}}. $$

Applying the Schwarz inequality and using the fact that *z*, \(a_{1}\), \(a_{2}\) are continuous over \([0,X]\) and \({}^{CF}\mathcal{D}_{\xi }^{\frac{1}{2}}V^{\lbrace 2\rbrace }_{\xi }\) is continuous and uniformly bounded, we get

$$\begin{aligned}& \bigl\vert (\mathcal{L}_{1}z) (\xi ) \bigr\vert = \bigl\vert z'(\xi ) \bigr\vert = \biggl\vert \biggl\langle z(\cdot), \frac{d}{d\xi }V^{\lbrace 2\rbrace }_{\xi }(\cdot) \biggr\rangle \biggr\vert _{\mathcal{H}_{2}} \leq \Vert z \Vert _{ \mathcal{H}_{2}} \biggl\Vert \frac{d}{d\xi }V^{\lbrace 2\rbrace }_{\xi } \biggr\Vert _{\mathcal{H}_{2}} \leq \Upsilon _{11} \Vert z \Vert _{ \mathcal{H}_{2}}, \\& \bigl\vert (\mathcal{L}_{1}z)'(\xi ) \bigr\vert = \biggl\vert \biggl\langle z(\cdot), \frac{d^{2}}{d\xi ^{2}}V^{\lbrace 2\rbrace }_{\xi }( \cdot) \biggr\rangle \biggr\vert _{\mathcal{H}_{2}} \leq \Vert z \Vert _{\mathcal{H}_{2}} \biggl\Vert \frac{d^{2}}{d\xi ^{2}}V^{\lbrace 2\rbrace }_{\xi } \biggr\Vert _{\mathcal{H}_{2}} \leq \Upsilon _{12} \Vert z \Vert _{\mathcal{H}_{2}}, \\& \begin{aligned} \bigl\vert \bigl(\mathcal{L}_{2}z(\cdot)\bigr) (\xi ) \bigr\vert &= \biggl\langle z, a_{1}(\cdot) \frac{d}{d\xi } V^{\lbrace 2\rbrace }_{\xi }(\cdot)+ a_{2}(\cdot){}^{CF} \mathcal{D}_{\xi }^{\frac{1}{2}}V^{\lbrace 2\rbrace }_{\xi }(\cdot) \biggr\rangle _{\mathcal{H}_{2}} \\ &\leq \Vert z \Vert _{\mathcal{H}_{2}} \biggl\Vert a_{1} \frac{d}{d\xi } V^{\lbrace 2\rbrace }_{\xi }+ a_{2}{}^{CF} \mathcal{D}_{\xi }^{\frac{1}{2}}V^{\lbrace 2\rbrace }_{\xi } \biggr\Vert _{ \mathcal{H}_{2}} \leq \Upsilon _{21} \Vert z \Vert _{\mathcal{H}_{2}}, \end{aligned} \end{aligned}$$

and similarly,

$$ \bigl\vert (\mathcal{L}_{2}z)'(\xi ) \bigr\vert \leq \Upsilon _{22} \Vert z \Vert _{\mathcal{H}_{2}}, $$

where \(\Upsilon _{ij} \in \Re ^{+}\), \(\forall i,j=1,2\).

Hence,

$$ \Vert \mathcal{L}_{j} z_{j} \Vert ^{2}_{\mathcal{H}_{1}} \leq \bigl( \Upsilon _{1j}^{2}+\Upsilon _{2j}^{2} \bigr)X \Vert z \Vert _{\mathcal{H}_{2}}^{2} \leq M_{j}^{2} \Vert z \Vert _{\mathcal{H}_{2}}, $$

where \(M_{j}^{2}=(\Upsilon _{1j}^{2}+\Upsilon _{2j}^{2})X\).

So, \(\Vert \mathcal{L}_{j} z_{j} \Vert _{\mathcal{H}_{1}}^{2}\leq M_{j} \Vert z \Vert _{\mathcal{H}_{2}}\).

Now, for any \(Z=(z_{1},z_{2})^{T} \in \mathcal{W}_{2}[0,X]\), we have

$$ \begin{aligned} \Vert \mathcal{L}Z \Vert _{\mathcal{W}_{1}}&=\sqrt {\sum^{2}_{i=1} \Vert \mathcal{L}_{i}z_{i} \Vert ^{2}_{\mathcal{H}_{1}}} \\ & \leq \sqrt{ M^{2}_{1} \Vert z_{1} \Vert ^{2}_{\mathcal{H}_{2}}+M^{2}_{2} \Vert z_{2} \Vert ^{2}_{\mathcal{H}_{2}}} \\ & \leq M \Vert Z \Vert _{\mathcal{W}_{2}}, \end{aligned} $$

(4.17)

where \(M=\max \{M_{1},M_{2}\}\). So, the proof is complete. □

Next, we create an orthonormal function system of \(\mathcal{W}_{2}[0,X]\) as follows: Let \(\Phi _{ij}(\cdot)=V^{\{1\}}_{\xi _{i}}(\cdot)\) and \(\Psi _{ij}(\cdot)=\mathcal{L}^{\star }\Phi _{ij}(\cdot)\) for each \(i=1,2,\ldots\) , and \(j=1,2\), where \(\mathcal{L}^{\star }\) is the adjoint operator of \(\mathcal{L}\) and \(\{\xi _{i}\}_{i=1}^{\infty }\) is a countable dense subset of \([0,X]\). Moreover, using the properties of the reproducing kernel, we find

$$ \begin{aligned} \Psi _{ij}(\xi )&=\mathcal{L^{\star }}\Phi _{ij}(\xi )= \bigl\langle \mathcal{L^{\star }}\Phi _{ij}(\tau ),V^{\{2\}}_{\xi _{i}}(\tau ) \bigr\rangle _{\mathcal{W}_{2}} =\bigl\langle \Phi _{ij}(\tau ),\mathcal{L}V^{ \{2\}}_{\xi _{i}}( \tau )\bigr\rangle _{\mathcal{W}_{1}}\\ &=\mathcal{L}V^{\{2 \}}_{\xi _{i}}(\tau )\in \mathcal{W}_{2}[0,X]. \end{aligned} $$

To build the representative form of the MRKHS solutions of SFDE (4.14) equipped with the initial conditions (4.15) in the space \(\mathcal{W}_{2}[0, X]\), we use the well-known Gram–Schmidt process that outputs an orthonormal function \(\{ (\widehat{\Psi }_{i1}(\xi ), \widehat{\Psi }_{i2}(\xi ))^{T} \}_{i=1}^{ \infty }\) of the space \(\mathcal{W}_{2}[0,X]\) constructed from \(\{ (\Psi _{i1}(\xi ),\Psi _{i2}(\xi ))^{T} \}_{i=1}^{\infty }\) such that

$$ \widehat{\Psi }_{i}(\xi )=\begin{pmatrix} \widehat{\Psi }_{i1}(\xi ) \\ \widehat{\Psi }_{i2}(\xi )) \end{pmatrix}= \begin{pmatrix} \sum_{k=1}^{i}\sigma ^{1}_{ik}\Psi _{i1}(\xi ) \\ \sum_{k=1}^{i}\sigma ^{2}_{ik}\Psi _{i2}(\xi ) \end{pmatrix}, $$

(4.18)

where \(\sigma ^{j}_{ik}\), \(j=1,2\), are the orthogonalization coefficients.

### Theorem 4.6

*If* \(\{\xi _{i}\}_{i=1}^{\infty }\) *is a dense set on* [0, *X*], *then the orthogonal function system* \(\{ (\Psi _{i1}(\xi ),\Psi _{i2}(\xi ))^{T} \}_{i=1}^{\infty }\) *is complete in* \(\mathcal{W}_{2}[0,X]\).

### Proof

For each Z(\xi )=\left(\begin{array}{c}{z}_{1}(\xi )\\ {z}_{2}(\xi )\end{array}\right)\in {\mathcal{W}}_{2}[0,X], suppose \(\langle Z(\xi ),\Psi _{ij}(\xi ) \rangle =0\). On the other hand, we have

$$ \begin{aligned} \bigl\langle Z(\xi ),\Psi _{ij}(\xi ) \bigr\rangle &=\bigl\langle Z(\xi ), \mathcal{L}^{\star }\Phi _{ij}(\xi ) \bigr\rangle \\ & = \left\langle \begin{pmatrix} z_{1}(\xi ) \\ z_{2}(\xi ) \end{pmatrix} , \begin{pmatrix} \mathcal{L}_{1}^{\star }\Phi _{i1}(\xi ) \\ \mathcal{L}_{2}^{\star }\Phi _{i2}(\xi ) \end{pmatrix} \right\rangle \\ & =\bigl\langle z_{1}(\xi ),\mathcal{L}_{1}^{\star } \Phi _{i1}(\xi ) \bigr\rangle + \bigl\langle z_{2}(\xi ), \mathcal{L}_{2}^{\star }\Phi _{i1}(\xi ) \bigr\rangle \\ & =\bigl\langle \mathcal{L}_{1}z_{1}(\xi ),\Phi _{i1}(\xi ) \bigr\rangle + \bigl\langle \mathcal{L}_{2}z_{2}( \xi ),\Phi _{i1}(\xi )\bigr\rangle \\ & = \mathcal{L}_{1}z_{1}(\xi _{i})+ \mathcal{L}_{2}z_{2}(\xi _{i}) \\ & =\mathcal{L}Z(\xi _{i})=0. \end{aligned} $$

(4.19)

Because \(\{\xi _{i}\}_{i=1}^{\infty }\) is dense in [0, X] and \(\mathcal{L}\) is continuous, we get \(\mathcal{L}Z(\xi _{i})=0\). Using the existence of the inverse operator \(\mathcal{L}^{-1}\), we conclude that \(Z(\xi )=0\). So the proof is complete. □

### Theorem 4.7

*If* \(\{\xi _{i}\}_{i=1}^{\infty }\) *is a dense subset on* [0, *X*] *and the analytic solution* \(Z(\xi )\) *of SFDE* (4.14) *is unique*, *then the analytic solution* \(Z(\xi )\) *can be represented in the following form*:

$$ Z(\xi )= \begin{bmatrix} \sum_{i=1}^{\infty }\sum_{k=1}^{i}[\sigma ^{1}_{ik}H_{1}(\xi _{k},Z( \xi _{k}))+\sigma ^{2}_{ik}H_{2}(\xi _{k},Z(\xi _{k})) ] \widehat{\Psi }_{i1}(\xi ) \\ \sum_{i=1}^{\infty }\sum_{k=1}^{i}[\sigma ^{1}_{ik}H_{1}(\xi _{k},Z( \xi _{k}))+\sigma ^{2}_{ik}H_{2}(\xi _{k},Z(\xi _{k})) ] \widehat{\Psi }_{i2}(\xi ) \end{bmatrix}. $$

(4.20)

### Proof

For each \(Z(\xi )\in \mathcal{W}_{2}[0,X]\), \(Z(\xi )\) can be extended in the Fourier series \(\sum^{\infty }_{i=1}\langle Z(\xi ), \widehat{\Psi }_{i}(\xi ) \rangle \widehat{\Psi }_{i}(\xi ) \) about the orthonormal function system \(\{ (\Psi _{i1}(\xi ),\Psi _{i2}(\xi ))^{T} \}_{i=1}^{\infty }\), as \(\mathcal{W}_{2}[0,X]\). Moreover, the series \(\sum^{\infty }_{i=1}\langle Z(\xi ),\widehat{\Psi }_{i}(\xi ) \rangle \) is convergent in norm in the Hilbert space \(\mathcal{W}_{2}[0,X]\). So, we have

$$ \begin{aligned} Z(\xi ) &=\sum^{\infty }_{i=1} \bigl\langle Z(\xi ),\widehat{\Psi }_{i}(\xi ) \bigr\rangle \widehat{ \Psi }_{i}(\xi ) \\ & =\sum_{i=1}^{\infty } \left\langle \begin{pmatrix} z_{1}(\xi ) \\ z_{2}(\xi ) \end{pmatrix} , \begin{pmatrix} \widehat{\Psi }_{i1}(\xi ) \\ \widehat{\Psi }_{i2}(\xi ) \end{pmatrix} \right\rangle \widehat{\Psi }_{ij}(\xi ) \\ & =\sum_{i=1}^{\infty } \left\langle \begin{pmatrix} z_{1}(\xi ) \\ z_{2}(\xi ) \end{pmatrix} , \begin{pmatrix} \sum_{k=1}^{i}\sigma ^{1}_{ik}\Psi _{k1}(\xi ) \\ \sum_{k=1}^{i}\sigma ^{2}_{ik}\Psi _{k2}(\xi ) \end{pmatrix} \right\rangle \widehat{\Psi }_{i}(\xi ) \\ & =\sum_{i=1}^{\infty }\sum _{k=1}^{i}\sigma ^{1}_{ik}\bigl\langle z_{1}( \xi ),\Psi _{k1}(\xi )\bigr\rangle \widehat{ \Psi }_{i}(\xi ) +\sum_{i=1}^{ \infty } \sum_{k=1}^{i}\sigma ^{2}_{ik} \bigl\langle z_{2}(\xi ),\Psi _{k2}( \xi ) \bigr\rangle ] \widehat{\Psi }_{i}(\xi ) \\ & =\sum_{i=1}^{\infty }\sum _{k=1}^{i}\sigma ^{1}_{ik}\bigl\langle \mathcal{L}_{1}z_{1}(\xi ),\Phi _{k1}(\xi ) \bigr\rangle \widehat{\Psi }_{i}( \xi ) +\sum _{i=1}^{\infty }\sum_{k=1}^{i} \sigma ^{2}_{ik} \bigl\langle \mathcal{L}_{2}z_{2}( \xi ),\Phi _{i2}(\xi )\bigr\rangle \widehat{\Psi }_{i}( \xi ) \\ & = \sum_{i=1}^{\infty }\sum _{k=1}^{i}\bigl[\sigma ^{1}_{ik} \mathcal{L}_{1}z_{1}( \xi _{k})+\sigma ^{2}_{ik} \mathcal{L}_{2}z_{2}(\xi _{k})\bigr] \widehat{\Psi }_{i}(\xi ) \\ & =\begin{bmatrix} \sum_{i=1}^{\infty }\sum_{k=1}^{i}[\sigma ^{1}_{ik}H_{1}(\xi _{k},Z( \xi _{k}))+\sigma ^{2}_{ik}H_{2}(\xi _{k},Z(\xi _{k})) ] \widehat{\Psi }_{i1}(\xi ) \\ \sum_{i=1}^{\infty }\sum_{k=1}^{i}[\sigma ^{1}_{ik}H_{1}(\xi _{k},Z( \xi _{k}))+\sigma ^{2}_{ik}H_{2}(\xi _{k},Z(\xi _{k})) ] \widehat{\Psi }_{i2}(\xi ) \end{bmatrix}. \end{aligned} $$

(4.21)

□

Moreover, if we take finitely many terms in the series representation for the analytic solution \(Z(\xi )\), we get directly the approximate solution SFDE (4.14), and it is given as the form

$$ Z_{n}(\xi )= \begin{bmatrix} \sum_{i=1}^{n}\sum_{k=1}^{i}[\sigma ^{1}_{ik}H_{1}(\xi _{k},Z(\xi _{k}))+ \sigma ^{2}_{ik}H_{2}(\xi _{k}, Z(\xi _{k})) ]\widehat{\Psi }_{i1}( \xi ) \\ \sum_{i=1}^{n}\sum_{k=1}^{i}[\sigma ^{1}_{ik}H_{1}(\xi _{k},Z(\xi _{k}))+ \sigma ^{2}_{ik}H_{2}(\xi _{k},Z(\xi _{k})) ]\widehat{\Psi }_{i2}(\xi ) \end{bmatrix} .$$

(4.22)

### Lemma 4.8

*The analytical solution of FBT equation* (4.13) *is given as*

$$ \begin{aligned} z(\xi )= \sum_{i=1}^{\infty } \sum_{k=1}^{i}\bigl[\sigma ^{1}_{ik}H_{1}\bigl( \xi _{k},Z(\xi _{k})\bigr)+\sigma ^{2}_{ik}H_{2} \bigl(\xi _{k}, Z(\xi _{k})\bigr) \bigr] \widehat{\Psi }_{i1}(\xi ). \end{aligned} $$

### Proof

From Theorem 4.6, the proof is direct. □

### Theorem 4.9

*If* \(E_{n}=\|Z-Z_{n}\|_{\mathcal{W}_{2}}\), *where* *Z*, \(Z_{n}\) *are the exact and the approximate solutions of SFDE* (4.14), *respectively*, *represented in* (4.20) *and* (4.22), *then the error* \(E_{n}\) *is monotonic decreasing in the sense of* \(\|\cdot\|_{\mathcal{W}_{2}}\), *and* \(E_{n}\rightarrow 0\) *as* \(n\rightarrow \infty \). *Consequently*, *the behavior error of the solution of the GBT equation decreases monotonously in the sense of* \(\|\cdot\|_{\mathcal{W}_{2}}\).

### Proof

We have

$$ \begin{aligned} E_{n}^{2} &= \Vert Z-Z_{n} \Vert ^{2}_{\mathcal{W}_{2}} \\ & = \left \Vert \begin{bmatrix} \sum_{i=n+1}^{\infty }\sum_{k=1}^{i}[\sigma ^{1}_{ik}H_{1}(\xi _{k},Z( \xi _{k}))+\sigma ^{2}_{ik}H_{2}(\xi _{k},Z(\xi _{k})) ] \widehat{\Psi }_{i1}(\xi ) \\ \sum_{i=n+1}^{\infty }\sum_{k=1}^{i}[\sigma ^{1}_{ik}H_{1}(\xi _{k},Z( \xi _{k}))+\sigma ^{2}_{ik}H_{2}(\xi _{k},Z(\xi _{k})) ] \widehat{\Psi }_{i2}(\xi ) \end{bmatrix} \right \Vert ^{2}_{\mathcal{W}_{2}} \\ & = \left \Vert \begin{bmatrix} \sum_{i=n+1}^{\infty }A_{i}\widehat{\Psi }_{i1}(\xi ) \\ \sum_{i=n+1}^{\infty }A_{i}\widehat{\Psi }_{i2}(\xi ) \end{bmatrix} \right \Vert ^{2}_{\mathcal{W}_{2}} \\ &= \Biggl\Vert \sum_{i=n+1}^{\infty }A_{i} \widehat{\Psi }_{i}(\xi ) \Biggr\Vert ^{2}_{ \mathcal{W}_{2}} \\ &= \sum_{i=n+1}^{\infty }(A_{i})^{2} \end{aligned} $$

and

$$ \begin{aligned} E_{n-1}^{2} = \Vert Z-Z_{n-1} \Vert ^{2}_{\mathcal{W}_{2}}=\sum _{i=n-1}^{ \infty }(A_{i})^{2}=(A_{n-1})^{2}+ \sum_{i=n}^{\infty }(A_{i})^{2}. \end{aligned} $$

Note that \(E_{n-1}^{2}> E_{n}^{2} \). So, we conclude that the error \(E_{n} \) is monotone decreasing in the sense of \(\|\cdot\|^{2}_{\mathcal{W}_{2}}\), and because \(\sum^{\infty }_{i=1}\langle Z(\xi ),\widehat{\Psi }_{i}(\xi ) \rangle \widehat{\Psi }_{i}(\xi ) \) is convergent, then \(E_{n}\rightarrow 0\) as \(n\rightarrow \infty \). So, the proof is complete. □