In this section, we will give two convergence estimates under an a priori regularization parameter choice rule and an a posteriori regularization parameter choice rule, respectively.
An a priori regularization parameter choice rule
Theorem 4.1
Let
\(f(x)\)
given by (3.6) be the exact solution of problem (1.1). Let
\(f^{m,\delta }(x)\)
given by (3.12) be the regularization Landweber iterative approximation solution. Choosing the regularization parameter
\(m=[z]\), where
$$ z=\biggl(\frac{E}{\delta }\biggr)^{\frac{4}{p+2}}, $$
(4.1)
then we obtain the following error estimate:
$$ \bigl\Vert f^{m,\delta }(\cdot)-f(\cdot)\bigr\Vert \leq C_{2}E^{\frac{2}{p+2}} \delta^{\frac{p}{p+2}}, $$
(4.2)
where
\([z]\)
denotes the largest integer less than or equal to
z, and
\(C_{2}=\sqrt{a}+(\frac{4p}{a})^{\frac{p}{4}}\)
is a constant dependent on
a, p.
Proof
Due to the triangle inequality, we know
$$\begin{aligned}& \bigl\Vert f^{m,\delta }(\cdot)-f(\cdot)\bigr\Vert \\& \quad =\Biggl\Vert \sum_{n=1}^{\infty }\frac{1-(1-a\mu_{n}^{2})^{m}}{\mu_{n}}g_{n}^{\delta }X_{n}(x)- \sum_{n=1}^{\infty}\mu_{n}^{-1}g_{n}X_{n}(x) \Biggr\Vert \\& \quad \leq \Biggl\Vert \sum_{n=1}^{\infty } \frac{1-(1-a\mu_{n}^{2})^{m}}{\mu_{n}}g_{n}^{\delta }X_{n}(x)-\sum _{n=1}^{\infty }\frac{1-(1-a\mu_{n}^{2})^{m}}{\mu_{n}}g_{n}X_{n}(x) \Biggr\Vert \\& \qquad {}+\Biggl\Vert \sum_{n=1}^{\infty } \frac{1-(1-a\mu_{n}^{2})^{m}}{\mu_{n}}g_{n}X_{n}(x)-\sum _{n=1}^{\infty }\mu_{n}^{-1}g_{n}X_{n}(x) \Biggr\Vert \\& \quad =\bigl\Vert f^{m,\delta }(\cdot)-f^{m}(\cdot)\bigr\Vert + \bigl\Vert f^{m}(\cdot)-f(\cdot)\bigr\Vert . \end{aligned}$$
From condition (1.2) and (3.12), we have
$$\begin{aligned}& \bigl\Vert f^{m,\delta }(\cdot)-f^{m}(\cdot)\bigr\Vert ^{2} \\& \quad =\Biggl\Vert \sum_{n=1}^{\infty } \frac{1-(1-a\mu_{n}^{2})^{m}}{\mu_{n}}g_{n}^{\delta }X_{n}(x)-\sum _{n=1}^{\infty}\frac{1-(1-a\mu_{n}^{2})^{m}}{\mu_{n}}g_{n}X_{n}(x) \Biggr\Vert ^{2} \\& \quad =\Biggl\Vert \sum_{n=1}^{\infty } \frac{1-(1-a\mu_{n}^{2})^{m}}{\mu_{n}}\bigl(g_{n}^{\delta }-g_{n} \bigr)X_{n}(x)\Biggr\Vert ^{2} \\& \quad \leq \sup_{n\geq 1}H(n)^{2}\delta^{2}, \end{aligned}$$
where \(H(n):=\frac{1-(1-a\mu_{n}^{2})^{m}}{\mu_{n}}\).
By Lemma 2.1, we get
$$ \frac{1-(1-a\mu_{n}^{2})^{m}}{\mu_{n}}\leq \sqrt{am}, $$
(4.3)
i.e.,
$$ H(n)\leq \sqrt{am}. $$
(4.4)
So
$$ \bigl\Vert f^{m,\delta }(\cdot)-f^{m}(\cdot)\bigr\Vert \leq \sqrt{am}\delta. $$
(4.5)
On the other hand, using (3.7), we have
$$\begin{aligned} \bigl\Vert f^{m}(\cdot)-f(\cdot)\bigr\Vert ^{2} =& \Biggl\Vert \sum_{n=1}^{\infty } \frac{1-(1-a\mu_{n}^{2})^{m}}{\mu_{n}}g_{n}X_{n}(x)-\sum _{n=1}^{\infty }\mu_{n}^{-1}g_{n}X_{n}(x) \Biggr\Vert ^{2} \\ =&\Biggl\Vert \sum_{n=1}^{\infty } \frac{(1-a\mu_{n}^{2})^{m}}{\mu_{n}}g_{n}X_{n}(x)\Biggr\Vert ^{2} \\ =&\Biggl\Vert \sum_{n=1}^{\infty }\bigl(1-a \mu_{n}^{2}\bigr)^{m}\bigl(n^{2}+k^{2} \bigr)^{-p/2}\bigl(n^{2}+k^{2}\bigr)^{p/2}f_{n}X_{n}(x) \Biggr\Vert ^{2} \\ \leq &\sup_{n\geq 1}Q(n)^{2}E^{2}, \end{aligned}$$
where \(Q(n):=(1-a(-\frac{1-e^{-\sqrt{n^{2}+k^{2}}}}{n^{2}+k^{2}})^{2})^{m}(n ^{2}+k^{2})^{-\frac{p}{2}}\).
Using Lemma 2.2 and Lemma 2.4, we have
$$ Q(n)\leq \biggl(1-\frac{a}{4(n^{2}+k^{2})^{2}}\biggr)^{m}\bigl(n^{2}+k^{2} \bigr)^{- \frac{p}{2}}. $$
(4.6)
Let \(t:=n^{2}+k^{2}\),
$$ F(t):=\biggl(1-\frac{a}{(2t)^{2}}\biggr)^{m}t^{-\frac{p}{2}}. $$
(4.7)
Let \(t_{0}\) satisfy \(F'(t_{0})=0\), and we easily get
$$ t_{0}=\biggl(\frac{a(4m+p)}{4p}\biggr)^{\frac{1}{2}}. $$
(4.8)
Thus
$$\begin{aligned} F( t_{0}) =&\biggl(1-\frac{p}{4m+p}\biggr)^{m}\biggl( \frac{a(4m+p)}{4p}\biggr)^{-\frac{p}{4}} \\ \leq & \biggl(\frac{4p}{a(m+1)}\biggr)^{\frac{p}{4}}, \end{aligned}$$
(4.9)
i.e.,
$$ F(t)\leq \biggl(\frac{4p}{a}\biggr)^{\frac{p}{4}}(m+1)^{-\frac{p}{4}}. $$
(4.10)
Thus we obtain
$$ Q(n)\leq \biggl(\frac{4p}{a}\biggr)^{\frac{p}{4}}(m+1)^{-\frac{p}{4}}. $$
(4.11)
Hence
$$ \bigl\Vert f^{m}(\cdot)-f(\cdot)\bigr\Vert \leq \biggl( \frac{4p}{a}\biggr)^{\frac{p}{4}}(m+1)^{- \frac{p}{4}}E. $$
(4.12)
Combining (4.5) and (3.12), we choose \(m=[z]\), and get
$$ \bigl\Vert f^{m,\delta }(\cdot)-f(\cdot)\bigr\Vert \leq C_{2}E^{\frac{2}{p+2}} \delta^{\frac{p}{p+2}}, $$
(4.13)
where \(C_{2}:=\sqrt{a}+(\frac{4p}{a})^{\frac{p}{4}}\).
We complete the proof of Theorem 4.1. □
An a posteriori regularization parameter choice rule
We consider the a posteriori regularization parameter choice in the Morozov discrepancy, and construct regularization solution sequences \(f^{m,\delta }(x)\) by Landweber iterative method. Assume \(\tau >1\) is a given fixed constant. Stop the algorithm at the first occurrence of \(m=m(\delta)\in \mathbb{N}_{0}\) with
$$ \bigl\Vert Kf^{m,\delta }(\cdot)-g^{\delta }(\cdot)\bigr\Vert \leq \tau \delta, $$
(4.14)
where \(\Vert g^{\delta }\Vert \geq \tau \delta\).
Lemma 4.2
Let
\(\gamma (m)=\Vert Kf^{m,\delta }(\cdot)-g^{\delta }(\cdot)\Vert \), then we have the following conclusions:
-
(a)
\(\gamma (m)\)
is a continuous function;
-
(b)
\(\lim_{m\rightarrow 0}\gamma (m)=\Vert g^{\delta }\Vert \);
-
(c)
\(\lim_{m\rightarrow +\infty }\gamma (m)=0\);
-
(d)
\(\gamma (m)\)
is a strictly decreasing function, for any
\(m\in (0,+\infty)\).
Lemma 4.3
For fixed
\(\tau >1\), and Landweber’s iteration method with stopping rule (4.14), we can see that the regularization parameter
\(m=m(\delta,g^{\delta })\in \mathbb{N}_{0}\)
satisfies
$$ m\leq \biggl(\frac{2(p+2)}{a}\biggr) \biggl(\frac{E}{(\tau -1)\delta } \biggr)^{\frac{4}{p+2}}. $$
(4.15)
Proof
From (3.12), we have the representation
$$ R_{m}g=\sum_{n=1}^{\infty } \frac{1-(1-a\mu_{n}^{2})^{m}}{\mu_{n}}g _{n}X_{n}(x) $$
(4.16)
for every \(g\in H^{2}(\Omega)\) and thus
$$\begin{aligned} \Vert KR_{m}g-g\Vert ^{2} =&\Biggl\Vert \sum _{n=1}^{\infty }\bigl(1-\bigl(1-a \mu_{n}^{2}\bigr)^{m}\bigr)g_{n}X_{n}(x)- \sum_{n=1}^{\infty }g_{n}X_{n}(x) \Biggr\Vert ^{2} \\ =&\Biggl\Vert \sum_{n=1}^{\infty }-\bigl(1-a \mu_{n}^{2}\bigr)^{m}g_{n}X_{n}(x) \Biggr\Vert ^{2} \\ =&\sum_{n=1}^{\infty }\bigl(1-a \mu_{n}^{2}\bigr)^{2m}g_{n}^{2}. \end{aligned}$$
From \(\vert 1-a\mu_{n}^{2}\vert <1\), we conclude that \(\Vert KR_{m-1}-I\Vert \leq 1\).
We know that m is the minimum value that satisfies \(\Vert KR_{m}g^{\delta }-g^{\delta }\Vert =\Vert Kf^{m,\delta }-g^{\delta }\Vert \leq \tau \delta \). Hence
$$\begin{aligned} \Vert KR_{m-1}g-g\Vert \geq &\bigl\Vert KR_{m-1}g^{\delta }-g^{\delta } \bigr\Vert -\bigl\Vert (KR_{m-1}-I) \bigl(g-g^{\delta }\bigr) \bigr\Vert \\ \geq &\tau \delta -\Vert KR_{m-1}-I\Vert \delta \\ \geq &(\tau -1)\delta. \end{aligned}$$
On the other hand, using (3.7), we obtain
$$\begin{aligned} \Vert KR_{m-1}g-g\Vert =&\Biggl\Vert \sum _{n=1}^{\infty }\bigl(1-\bigl(1-a\mu_{n}^{2} \bigr)^{m-1}\bigr)g_{n}X_{n}(x)-\sum _{n=1}^{\infty }g_{n}X_{n}(x)\Biggr\Vert \\ =&\Biggl\Vert \sum_{n=1}^{\infty }-\bigl(1-a \mu_{n}^{2}\bigr)^{m-1}(g,X_{n})\Biggr\Vert \\ =&\Biggl\Vert \sum_{n=1}^{\infty }\bigl(1-a \mu_{n}^{2}\bigr)^{m-1}\mu_{n} \bigl(n^{2}+k^{2}\bigr)^{-\frac{p}{2}}f_{n} \bigl(n^{2}+k^{2}\bigr)^{\frac{p}{2}}X_{n}(x)\Biggr\Vert \\ \leq &\sup_{n\geq 1}\bigl\vert \bigl(1-a\mu_{n}^{2} \bigr)^{m-1}\mu_{n}\bigl(n^{2}+k^{2} \bigr)^{-\frac{p}{2}}\bigr\vert E. \end{aligned}$$
Let
$$ S(n):=\bigl(1-a\mu_{n}^{2}\bigr)^{m-1} \mu_{n}\bigl(n^{2}+k^{2}\bigr)^{-\frac{p}{2}}, $$
(4.17)
so
$$ (\tau -1)\delta \leq S(n)E. $$
(4.18)
Using Lemma 2.4, we have
$$ S(n)\leq \biggl(1-a\biggl(\frac{1}{2t}\biggr)^{2} \biggr)^{m-1}t^{-\frac{p}{2}-1}. $$
(4.19)
Let
$$ G(t)=\biggl(1-a\biggl(\frac{1}{2t}\biggr)^{2} \biggr)^{m-1}t^{-\frac{p}{2}-1}, $$
(4.20)
suppose \(t_{\ast }\) satisfy \(G'(t_{\ast })=0\), we easily get
$$ t_{\ast }=\biggl(\frac{a(4m+p-2)}{4(p+2)}\biggr)^{\frac{1}{2}}, $$
(4.21)
so
$$\begin{aligned} G(t_{\ast }) =&\biggl(1-\frac{4(p+2)}{4m+p-2}\biggr)^{m-1}\biggl( \frac{a(4m+p-2)}{4(p+2)}\biggr)^{-\frac{p+2}{4}} \\ \leq &\biggl(\frac{2(p+2)}{ma}\biggr)^{\frac{p+2}{4}}. \end{aligned}$$
Combining (4.18) with (4.21), we obtain (4.15). □
Theorem 4.4
Let
\(f(x)\)
given by (3.6) be the exact solution of problem (1.1). Let
\(f^{m,\delta }(x)\)
given by (3.12) be the Landweber iterative regularization approximation solution. The regularization parameter is given by (4.14), then we have the following error estimate:
$$ \bigl\Vert f^{m,\delta }(\cdot)-f(\cdot)\bigr\Vert \leq \bigl(C_{1}(\tau +1)^{ \frac{2}{p+2}}+C_{3} \bigr)E^{\frac{2}{p+2}}\delta^{\frac{p}{p+2}}, $$
(4.22)
where
\(C_{3}=(2(p+2))(\frac{1}{\tau -1})^{\frac{2}{p+2}}\)
is a constant.
Proof
Using triangle inequality, we obtain
$$ \bigl\Vert f^{m,\delta }(\cdot)-f^{m}(\cdot)\bigr\Vert \leq \bigl\Vert f^{m,\delta }(\cdot)-f^{m}(\cdot)\bigr\Vert +\bigl\Vert f^{m}(\cdot)-f(\cdot)\bigr\Vert . $$
(4.23)
By (4.5) and Lemma 4.3, we get
$$\begin{aligned} \bigl\Vert f^{m,\delta }(\cdot)-f^{m}(\cdot)\bigr\Vert \leq& \sqrt{am}\delta \\ \leq& C _{4}E^{\frac{2}{p+2}}\delta^{\frac{p}{p+2}}, \end{aligned}$$
(4.24)
where \(C_{3}=(2(p+2))^{\frac{1}{2}}(\frac{1}{\tau -1})^{\frac{2}{p+2}}\).
For the second part of the right side of (4.24), we have
$$\begin{aligned}& K\bigl(f^{m}(\cdot)-f(\cdot)\bigr) \\& \quad =\sum _{n=1}^{\infty }-\bigl(1-a\mu_{n}^{2} \bigr)^{m}g _{n}X_{n}(x) \\& \quad =\sum_{n=1}^{\infty }-\bigl(1-a \mu_{n}^{2}\bigr)^{m}\bigl(g_{n}-g_{n}^{\delta } \bigr)X _{n}(x)+\sum_{n=1}^{\infty }- \bigl(1-a\mu_{n}^{2}\bigr)^{m}g_{n}^{\delta }X_{n}(x). \end{aligned}$$
Using (1.2) and (4.15), we have
$$ \bigl\Vert K\bigl(f^{m}(\cdot)-f(\cdot)\bigr)\bigr\Vert \leq ( \tau +1)\delta. $$
(4.25)
Due to
$$\begin{aligned} \bigl\Vert f^{m}(\cdot)-f(\cdot)\bigr\Vert _{H^{p}} =& \Biggl(\sum_{n=1}^{\infty }-\bigl(1-a\mu _{n}^{2}\bigr)^{2m}f_{n}^{2} \bigl(n^{2}+k^{2}\bigr)^{p}\Biggr)^{\frac{1}{2}} \\ \leq &\Biggl(\sum_{n=1}^{\infty }f_{n}^{2} \bigl(n^{2}+k^{2}\bigr)^{p}\Biggr)^{\frac{1}{2}} \\ \leq & E, \end{aligned}$$
and using Theorem 3.1, we have
$$ \bigl\Vert f^{m}(\cdot)-f(\cdot)\bigr\Vert \leq C_{1}(\tau +1)^{\frac{p}{p+2}}E^{ \frac{2}{p+2}}\delta^{\frac{p}{p+2}}. $$
(4.26)
Therefore,
$$ \bigl\Vert f^{m,\delta }(\cdot)-f(\cdot)\bigr\Vert \leq \bigl(C_{1}(\tau +1)^{ \frac{p}{p+2}}+C_{3} \bigr)E^{\frac{2}{p+2}}\delta^{\frac{p}{p+2}}. $$
(4.27)
This completes the proof of Theorem 4.4. □