In this section, we will give the error estimate of the quasi-boundary value regularization method under the *a priori* parameter choice rule. For \(\gamma>0\), let \(D^{\gamma}(\Omega)\) be the set of all function \(\psi\in L^{2}(\Omega)\) defined by

$$ {\Vert\psi\Vert}_{D^{\gamma}(\Omega)}=\Biggl(\sum _{n=1}^{\infty }n^{2\gamma}{\bigl\vert ( \psi, \varphi_{n} ) \bigr\vert }^{2}\Biggr)^{\frac{1}{2}}< \infty. $$

(4.1)

### Lemma 4.1

*For any*
\(q>0\), \(0<\mu<1\), *and*
\(n\geq1 >0\), *we have the following inequality*:

$$ A(n)=\frac{\mu n^{2-q}{\pi}^{2}}{C_{1}+\mu n^{2}{\pi}^{2}}\leq \textstyle\begin{cases} \frac{1}{2}(((2-q)C_{1})/({\pi}^{2}q))^{1-\frac{q}{2}}{\pi}^{2}q\mu ^{\frac{q}{2}}, &0< q< 2,\\ ({\pi}^{2}\mu)/C_{1}, &q\geq2, 1< n< \frac{1}{\mu},\\ {\pi}^{2}\mu^{q-1}, &q\geq2, n\geq\frac{1}{\mu}. \end{cases} $$

(4.2)

### Proof

For \(0< q<2\), we can easily see

$$ \lim_{n\rightarrow0}A(n)=0 \quad\mbox{and} \quad \lim _{n\rightarrow \infty}A(n)=0, $$

then we infer

$$ \sup_{n\geq1}A(n)\leq A\bigl(n^{*}\bigr), $$

where \(n^{*}\) is the root of \(A'(n)=0\), and \(n^{*}=\sqrt{\frac {(2-q)C_{1}}{\mu\pi^{2}q}}\).

So

$$ A(n)\leq A\bigl(n^{*}\bigr)=\frac{1}{2}\biggl( \frac{(2-q)C_{1}}{{\pi}^{2}q}\biggr)^{1-\frac {q}{2}}{\pi}^{2}q\mu^{\frac{q}{2}}. $$

For \(q\geq2\) and \(1< n<\frac{1}{\mu}\), we have

$$ A(n)=\frac{\mu n^{2-q}{\pi}^{2}}{C_{1}+\mu n^{2}{\pi}^{2}}< \frac{{\pi }^{2}\mu}{C_{1}}. $$

For \(q\geq2\) and \(n\geq\frac{1}{\mu}\), we get

$$ A(n)=\frac{\mu n^{2-q}{\pi}^{2}}{C_{1}+\mu n^{2}{\pi}^{2}}< {\pi}^{2}\mu^{q-1}. $$

□

### Lemma 4.2

*For any*
\(0<\mu<1\), *and*
\(n\geq1\), *we have the following inequality*:

$$ B(n)=\frac{n^{2}}{C_{1}+\mu n^{2}{\pi}^{2}}\leq\frac{1}{\mu{\pi}^{2}}. $$

(4.3)

The proof is very easy and we omit it here.

The main result of this section is the following.

### Theorem 4.1

*Assume an a priori bound is imposed as follows*:

$$ {\Vert p \Vert}_{D^{q}(\Omega)}\leq E, $$

(4.4)

*where*
\(q>0\)
*and*
\(E>0\)
*are two constants*. *Suppose the a priori condition* (4.4) *and the noises data assumption* (1.3) *hold*. *We have an estimate as follows*:

$$ \begin{aligned}[b] &\mathbb{E} {\bigl\Vert \tilde{p}_{\mu,M}(x)-p(x) \bigr\Vert }^{2} \\ &\quad \leq\frac{R_{\mathrm{max}}^{2}}{M {\mu}^{2}}+ \textstyle\begin{cases} \frac{1}{4}{\pi}^{4}q^{2}(((2-q)C_{1})/(q{\pi}^{2}))^{2-q}{\mu }^{q}E^{2}, &0< q< 2,\\ ({\pi}^{4}/C_{1}^{2}){\mu}^{2}E^{2}, &q\geq2,1< n< \frac{1}{\mu},\\ {\pi}^{4}{\mu}^{2q-2}E^{2}, &q\geq2,n\geq\frac{1}{\mu}. \end{cases}\displaystyle \end{aligned} $$

(4.5)

*As*
\(0<\mu<1\)
*and*
\(\lim_{M\rightarrow+\infty} \frac{1}{M \mu^{2}}=0\), *we obtain*

$$ \mathbb{E} {\bigl\Vert \tilde{p}_{\mu,M}(x)-p(x) \bigr\Vert }^{2} \textit{ is of order } \textstyle\begin{cases} \max(\frac{1}{M\mu^{2}}, \mu^{q}), &0< q< 2,\\ \max(\frac{1}{M\mu^{2}}, \mu^{2}), &q\geq2,1< n< \frac{1}{\mu},\\ \max(\frac{1}{M\mu^{2}}, \mu^{2q-2}) &q\geq2,n\geq\frac{1}{\mu}. \end{cases} $$

(4.6)

### Proof

Applying (2.16) and (3.4), we can get

$$ \begin{aligned}[b] \tilde{p}_{\mu,M}(x)-p(x) &= \sum_{n=1}^{M}\frac{\frac{1}{M}\sum_{k=1}^{M}\varepsilon _{k}\sigma_{k}E_{\alpha,1}(-n^{2}\pi^{2}T^{\alpha})}{E_{\alpha ,1}(-n^{2}{\pi}^{2}T^{\alpha})[\mu+E_{\alpha,1}(-n^{2}\pi^{2}T^{\alpha })]} \varphi_{n}(x) \\ &\quad {}-\sum_{n=1}^{\infty}\frac{[g_{n}-F_{n}(T)T^{\alpha }E_{\alpha,\alpha+1}(-n^{2}{\pi}^{2}T^{\alpha})]\mu}{E_{\alpha ,1}(-n^{2}{\pi}^{2}T^{\alpha})[\mu+E_{\alpha,1}(-n^{2}\pi^{2}T^{\alpha })]} \varphi_{n}(x). \end{aligned} $$

(4.7)

Using Parseval’s equality, we obtain

$$ \begin{aligned}[b] \bigl\Vert \tilde{p}_{\mu,M}(x)-p(x) \bigr\Vert ^{2} &\leq\sum_{n=1}^{M} \biggl[\frac{\frac{1}{M}\sum_{k=1}^{M}\varepsilon_{k}\sigma_{k}}{\mu+E_{\alpha,1}(-n^{2}{\pi }^{2}T^{\alpha})}\biggr]^{2} \\ &\quad {}+\sum_{n=1}^{\infty}\biggl[\frac{[g_{n}-F_{n}(T)T^{\alpha }E_{\alpha,\alpha+1}(-n^{2}{\pi}^{2}T^{\alpha})]\mu}{E_{\alpha ,1}(-n^{2}{\pi}^{2}T^{\alpha})[\mu+E_{\alpha,1}(-n^{2}\pi^{2}T^{\alpha})]} \biggr]^{2}. \end{aligned} $$

We use \(\mathbb{E}(\varepsilon_{j}\varepsilon_{l})=0\) (\(j\neq l\)), and \(\mathbb{E}\varepsilon_{j}^{2}=1\), \(j=1,2,\ldots,M\). Then we obtain

$$ \begin{aligned}[b] \mathbb{E} {\bigl\Vert \tilde{p}_{\mu,M}(x)-p(x) \bigr\Vert }^{2} &\leq\underbrace{\sum _{n=1}^{M}\frac{\frac{1}{M^{2}}\sum_{k=1}^{M}\mathbb{E}\varepsilon^{2}_{k}\sigma^{2}_{k}}{(\mu+E_{\alpha ,1}(-n^{2}{\pi}^{2}T^{\alpha}))^{2}}}_{D_{1}} \\ &\quad {}+\underbrace{\sum_{n=1}^{\infty}\biggl[ \frac {[g_{n}-F_{n}(T)T^{\alpha}E_{\alpha,\alpha+1}(-n^{2}{\pi}^{2}T^{\alpha })]\mu}{E_{\alpha,1}(-n^{2}{\pi}^{2}T^{\alpha})[\mu+E_{\alpha ,1}(-n^{2}\pi^{2}T^{\alpha})]}\biggr]^{2}}_{D_{2}}. \end{aligned} $$

(4.8)

From Lemma 2.3, we know that

$$ \frac{C_{1}}{n^{2}\pi^{2}}\leq E_{\alpha,1}\bigl(-n^{2} \pi^{2}T^{\alpha }\bigr)\leq\frac{C_{2}}{n^{2}\pi^{2}}. $$

(4.9)

Since \(\sigma_{k}< R_{\mathrm{max}}\) and Lemma 4.2, we estimate \(M_{1}\) as follows:

$$ \begin{aligned}[b] D_{1} &\leq \frac{R_{\mathrm{max}}^{2}{\pi}^{4}}{M}\Bigl(\sup_{n\in\mathbb{N}}B(n)\Bigr)^{2}\leq \frac {R_{\mathrm{max}}^{2}}{M{\mu}^{2}}. \end{aligned} $$

(4.10)

By (2.14), (4.4), (4.9) and Lemma 4.1, we obtain

$$ \begin{aligned}[b] D_{2} &=\sum _{n=1}^{\infty}p_{n}^{2}\biggl[ \frac{\mu n^{2}{\pi }^{2}}{C_{1}+\mu n^{2}{\pi}^{2}}\biggr]^{2} \\ &=\sum_{n=1}^{\infty}p_{n}^{2}n^{2q}n^{-2q} \biggl[\frac{\mu n^{2}{\pi}^{2}}{C_{1}+\mu n^{2}{\pi}^{2}}\biggr]^{2} \\ &\leq E^{2}\Bigl(\sup_{n\in\mathbb{N}}A(n) \Bigr)^{2} \\ &\leq \textstyle\begin{cases} \frac{1}{4}{\pi}^{4}q^{2}(\frac{(2-q)C_{1}}{q{\pi}^{2}})^{2-q}{\mu }^{q}E^{2}, &0< q< 2,\\ \frac{{\pi}^{4}}{C_{1}^{2}}{\mu}^{2}E^{2}, &q\geq2,1< n< \frac{1}{\mu},\\ {\pi}^{4}{\mu}^{2q-2}E^{2}, &q\geq2,n\geq\frac{1}{\mu}. \end{cases}\displaystyle \end{aligned} $$

(4.11)

Combining (4.10) and (4.11) , we can easy get the conclusion. □

### Remark 4.1

By choosing \(\mu=(\frac{1}{M})^{\frac {1}{q+2}}\), and by (4.6), in the case \(0< q<2\), we can conclude that

$$ \mathbb{E} {\bigl\Vert \tilde{p}_{\mu,M}(x)-p(x) \bigr\Vert }^{2} \mbox{ is of order } \biggl(\frac{1}{M}\biggr)^{\frac{q}{q+2}}. $$

### Remark 4.2

By choosing \(\mu=(\frac{1}{M})^{\frac{1}{4}}\), and by (4.6), in the case \(q\geq2\), \(1< n<\frac{1}{\mu}\), we can conclude that

$$ \mathbb{E} {\bigl\Vert \tilde{p}_{\mu,M}(x)-p(x) \bigr\Vert }^{2} \mbox{ is of order } \biggl(\frac{1}{M}\biggr)^{\frac{1}{2}}. $$

### Remark 4.3

By choosing \(\mu=(\frac{1}{M})^{\frac {1}{2q}}\), and by (4.6), in the case \(q\geq2\), \(n\geq\frac{1}{\mu }\), we can conclude that

$$ \mathbb{E} {\bigl\Vert \tilde{p}_{\mu,M}(x)-p(x)\bigr\Vert }^{2} \mbox{ is of order } \biggl(\frac{1}{M}\biggr)^{\frac{2q-2}{2q}}. $$