In contrast to the articles [1, 3], we will give a strict and accurate error analysis in this section. Firstly, we recall two useful lemmas.
Lemma 5.1
Suppose that function
\(x(u)\), \(u\in[0,1)\)satisfies the bounded condition and
\(e(u)=x(u)-x_{m}(u)\), where
\(x_{m}(u)\)is m approximations of HWs of
\(x(u)\), then
$$ \Vert e \Vert _{L^{2}([0,1))}^{2}= \int_{0}^{1}e^{2}(u)\,du \leq O \bigl(h^{2} \bigr). $$
(19)
Proof
See [1]. □
Lemma 5.2
Suppose that the function
\(x(u,v)\)satisfying the bounded condition is defined on
\(\mathbf{D}=[0,1)\times[0,1)\)and
\(e(u,v)=x(u,v)-x_{m}(u,v)\), where
\(x_{m}(u,v)\)is m approximations of HWs of
\(x(u,v)\), then
$$ \Vert e \Vert _{L^{2}(\mathbf{D})}^{2}= \int_{0}^{1} \int_{0}^{1}e^{2}(u,v)\,du\,dv \leq O \bigl(h^{2} \bigr). $$
(20)
Proof
See [1]. □
Secondly, let \(e(v)=x(v)-x_{m}(v)\), where \(x_{m}(v)\), \(x_{0_{m}}(v)\), \(k_{m}(u,v)\), and \(r_{m}(u,v)\) are m approximations of Haar wavelets of \(x(v)\), \(x_{0}(v)\), \(k(u,v)\), and \(r(u,v)\), respectively.
$$ \begin{aligned}[b]e(v) = {}& x(v)-x_{m}(v) \\ ={} & x_{0}(v)-x_{0_{m}}(v) \\ & + \int_{0}^{v} \bigl[{k(u,v)} {\sigma \bigl(x(u) \bigr)}-k_{m}(u,v){\sigma \bigl(x_{m}(u) \bigr)} \bigr]\,du \\ & + \int_{0}^{v} \bigl[{r(u,v)} {\rho \bigl(x(u) \bigr)}-{r_{m}(u,v)} {\rho \bigl(x_{m}(u) \bigr)} \bigr] \,dB(u).\end{aligned} $$
(21)
Lastly, the main convergence theorem is proved.
Theorem 5.1
Suppose that analytic functionsσandρsatisfy the following conditions:
- (i)
\(|\sigma(x)-\sigma(y)| \leq l_{1}|x-y|\), \(| \rho(x)-\rho(y)|\leq l_{3} |x-y|\);
- (ii)
\(|\sigma(x)|\leq l_{2}\), \(|\rho(y)|\leq l_{4}\);
- (iii)
\(|k(u,v)|\leq l_{5}\), \(|r(u,v)|\leq l_{6}\),
where
\(x,y\in\mathbb{R}\), constant
\(l_{i}>0\), \(i=1, 2, \ldots, 6\). Then
$$\begin{aligned} \int_{0}^{T} \mathbb{E} \bigl( \bigl\vert e_{m}(v) \bigr\vert ^{2} \bigr)\,dv= \int _{0}^{T}\mathbb{E} \bigl( \bigl\vert x(v)-x_{m}(v) \bigr\vert ^{2} \bigr)\,dv\leq O \bigl(h^{2} \bigr), \quad T\in[0,1). \end{aligned}$$
Proof
For (21), we have
$$\begin{aligned} \mathbb{E} \bigl( \bigl\vert e_{m}(v) \bigr\vert ^{2} \bigr) \leq& 3 \biggl[ \mathbb{E} \bigl( \bigl\vert x_{0}(v)-x_{0_{m}}(v) \bigr\vert ^{2} \bigr) \\ &{}+ \mathbb{E} \biggl( \biggl\vert \int_{0}^{v} \bigl({k(u,v)} {\sigma \bigl(x(u) \bigr)}-{r_{m}(u,v)} {\sigma \bigl(x_{m}(u) \bigr)} \bigr)\,du \biggr\vert ^{2} \biggr) \\ &{}+ \mathbb{E} \biggl( \biggl\vert \int_{0}^{v} \bigl({r(u,v)} {\rho \bigl(x(u) \bigr)}-{r_{m}(u,v)} {\rho \bigl(x_{m}(u) \bigr)} \bigr) \,dB(u) \biggr\vert ^{2} \biggr) \biggr]. \end{aligned}$$
On the basis of Lipschitz continuity, Itô isometry, and Cauchy–Schwarz inequality, it yields
$$\begin{aligned} \mathbb{E} \bigl( \bigl\vert e_{m}(v) \bigr\vert ^{2} \bigr) \leq& 3 \biggl[ \mathbb{E} \bigl( \bigl\vert x_{0}(v)-x_{0_{m}}(v) \bigr\vert ^{2} \bigr) \\ &{}+ \mathbb{E} \biggl( \int_{0}^{v} \bigl\vert {k(u,v)} {\sigma \bigl(x(u) \bigr)}-{k_{m}(u,v)} {\sigma \bigl(x_{m}(u) \bigr)} \bigr\vert ^{2}\,du \biggr) \\ &{}+ \mathbb{E} \biggl( \int_{0}^{v} \bigl\vert {r(u,v)} {\rho \bigl(x(u) \bigr)}-{r_{m}(u,v)} {\rho \bigl(x_{m}(u) \bigr)} \bigr\vert ^{2}\,du \biggr) \biggr] \\ = & 3 \biggl[\mathbb{E} \bigl( \bigl\vert x_{0}(v)-x_{0_{m}}(v) \bigr\vert ^{2} \bigr) \\ &{}+ \int_{0}^{v}\mathbb{E} \bigl( \bigl\vert {k(u,v)} \bigl( \sigma \bigl(x(u) \bigr)-\sigma \bigl(x_{m}(u) \bigr) \bigr)\\ &{} + \sigma \bigl(x_{m}(u) \bigr) \bigl({k(u,v)}-{k_{m}(u,v)} \bigr) \bigr\vert ^{2} \bigr)\,du \\ &{}+ \int_{0}^{v}\mathbb{E} \bigl( \bigl\vert {r(u,v)} \bigl(\rho \bigl(x(u) \bigr)-\rho \bigl(x_{m}(u) \bigr) \bigr) \\ &{}+\rho \bigl(x_{m}(u) \bigr) \bigl( {r(u,v)}-{r_{m}(u,v)} \bigr) \bigr\vert ^{2} \bigr)\,du \biggr] \\ \leq& 3 \biggl[ \bigl\vert x_{0}(v)-x_{0_{m}}(v) \bigr\vert ^{2} \\ &{}+ 2{l_{1}}^{2}{l_{5}}^{2} \int_{0}^{v}\mathbb{E} \bigl( \bigl\vert e_{m}(u) \bigr\vert ^{2} \bigr)\,du +2{l_{2}}^{2} \int_{0}^{v} \bigl\vert {k(u,v)}-{k_{m}(u,v)} \bigr\vert ^{2}\,du \\ &{}+ 2{l_{3}}^{2}{l_{6}}^{2} \int_{0}^{v}\mathbb{E} \bigl( \bigl\vert e_{m}(u) \bigr\vert ^{2} \bigr)\,du +2{l_{4}}^{2} \int_{0}^{v} \bigl\vert {r(u,v)}-{r_{m}(u,v)} \bigr\vert ^{2}\,du \biggr]. \end{aligned}$$
Then we can get
$$\begin{aligned} \mathbb{E} \bigl( \bigl\vert e_{m}(v) \bigr\vert ^{2} \bigr) \leq{}& 3 \biggl[ \bigl\vert x_{0}(v)-x_{0_{m}}(v) \bigr\vert ^{2}+2{l_{2}}^{2} \int _{0}^{v} \bigl\vert {k(u,v)}-{k_{m}(u,v)} \bigr\vert ^{2}\,du \\ & + 2{l_{4}}^{2} \int_{0}^{v} \bigl\vert {r(u,v)}-{r_{m}(u,v)} \bigr\vert ^{2}\,du \biggr] \\ & +6 \bigl({l_{1}}^{2}{l_{5}}^{2}+{l_{3}}^{2}{l_{6}}^{2} \bigr) \int_{0}^{v}\mathbb{E} \bigl( \bigl\vert e_{m}(u) \bigr\vert ^{2} \bigr)\,du,\end{aligned} $$
or
$$\begin{gathered} \mathbb{E} \bigl( \bigl\vert e_{m}(v) \bigr\vert ^{2} \bigr)\leq\beta(v)+\alpha \int_{0}^{v}\mathbb{E} \bigl( \bigl\vert e_{m}(u) \bigr\vert ^{2} \bigr)\,du, \\ \beta(v)=3 \biggl[ \bigl\vert x_{0}(v)-x_{0_{m}}(v) \bigr\vert ^{2}\\ \phantom{\beta(v)=}{}+2{l_{2}}^{2} \int _{0}^{v} \bigl\vert {k(u,v)}-{k_{m}(u,v)} \bigr\vert ^{2}\,du +2{l_{4}}^{2} \int_{0}^{v} \bigl\vert {r(u,v)}-{r_{m}(u,v)} \bigr\vert ^{2}\,du \biggr], \\ \alpha=6 \bigl({l_{1}}^{2}{l_{5}}^{2}+{l_{3}}^{2}{l_{6}}^{2} \bigr).\end{gathered} $$
Let \(f(v)=\mathbb{E} ( |e_{m}(v) |^{2} )\), we get
$$f(v)\leq\beta(v)+\alpha \int_{0}^{v}f(\tau)\,d\tau, \quad\tau\in[0,v). $$
By Gronwall’s inequality, it follows that
$$f(v)\leq\beta(v)+\alpha \int_{0}^{v}e^{\alpha(v-\tau)}\beta(\tau )\,d\tau, \quad v\in[0,1). $$
Then
$$\begin{aligned}& \int_{0}^{T}f(v)\,dv \\& \quad = \int_{0}^{T}\mathbb{E} \bigl( \bigl\vert e_{m}(v) \bigr\vert ^{2} \bigr)\,dv \\& \quad \leq \int_{0}^{T} \biggl(\beta(v)+\alpha \int_{0}^{v}e^{\alpha (v-\tau)}\beta(\tau)\,d\tau \biggr)\,dv \\& \quad = \int_{0}^{T}\beta(v)\,dv+\alpha \int_{0}^{T} \int_{0}^{v}e^{\alpha (v-\tau)}\beta(\tau)\,d\tau \,dv \\& \quad \leq \int_{0}^{T}\beta(v)\,dv+\alpha e^{\alpha T} \int_{0}^{T} \int _{0}^{v}\beta(\tau)\,d\tau \,dv \\& \quad =3 \int_{0}^{T} \bigl\vert x_{0}(v)-x_{0_{m}}(v) \bigr\vert ^{2}\,dv +6{l_{2}}^{2} \int_{0}^{T} \int_{0}^{v} \bigl\vert {k(u,v)}-{k_{m}(u,v)} \bigr\vert ^{2}\,du\,dv \\& \qquad{} + 6{l_{4}}^{2} \int_{0}^{T} \int_{0}^{v} \bigl\vert {r(u,v)}-{r_{m}(u,v)} \bigr\vert ^{2}\,du\,dv \\& \qquad{} + \alpha e^{\alpha T} \biggl[3 \int_{0}^{T} \int_{0}^{v} \bigl\vert x_{0}( \tau)-x_{0_{m}}(\tau) \bigr\vert ^{2}\,d\tau \,dv\\& \qquad{} +6{l_{2}}^{2} \int_{0}^{T} \int_{0}^{t} \int_{0}^{\tau} \bigl\vert {k(s,\tau )}-{k_{m}(s, \tau)} \bigr\vert ^{2}\,ds\,d\tau \,dt \\& \qquad{} + 6{l_{4}}^{2} \int_{0}^{T} \int_{0}^{v} \int_{0}^{\tau} \bigl\vert {r(u, \tau)}-{r_{m}(u,\tau)} \bigr\vert ^{2}\,du\,d\tau \,dv \biggr] \\& \quad =3I_{1}+6l^{2}_{2}I_{2}+6l^{2}_{4}I_{3}+ \alpha e^{\alpha T} \bigl[3I_{4}+ 6l^{2}_{2}I_{5} + 6l^{2}_{4}I_{6} \bigr]. \end{aligned}$$
By using (19) and (20), we have
$$I_{i} \leq w_{i}h^{2}, \quad i=1, 2, \ldots, 6. $$
So we can get
$$\begin{aligned} \int_{0}^{T}\mathbb{E} \bigl\vert e_{m}(v) \bigr\vert ^{2}\,dv &\leq \bigl[ \bigl(3w_{1}+6{l_{2}}^{2}w_{2}+6{l_{4}}^{2}w_{3} \bigr)+\alpha e^{\alpha T} \bigl(3w_{4}+6{l_{2}}^{2}w_{5}+6{l_{4}}^{2}w_{6} \bigr) \bigr]h^{2}\\&\leq O \bigl(h^{2} \bigr),\end{aligned} $$
where constant \(w_{i}>0\), \(i=1, 2, \ldots, 6\).
The proof is completed. □