The direct problems for elliptic equations (Dirichlet, Neumann or mixed boundary value problems) extensively studied. However, in some practical problems, the whole boundary data often can not be known, we only know the noisy data on a part of the boundary or at some interior points of the concerned domain, which will lead to some inverse problems.

This paper considers the following inverse problem of nonlinear elliptic equation

$ \begin{align} \left\{ \begin{aligned} &u_{xx}+u_{yy} = f(x, y, u(x, y)), & 0< x<\pi, 0<y<T, \\ &u(x, 0) = \varphi(x), &0\leq x\leq\pi, \\ &u_y(x, 0) = 0, &0\leq x\leq\pi, \\ &u(0, y) = u(\pi, y) = 0, &0\leq y\leq T, \end{aligned} \right. \end{align} $

(1.1)

where $ \varphi(x)\in L^2(0, \pi) $ is known function, $ f:\mathbb{R}\times \mathbb{R}\times L^2(0, \pi)\rightarrow L^2(0, \pi) $ is an uniform Lipschitz continuous function, i.e., existing $ L>0 $ independent of $ x, y\in\mathbb{R} $, $ u, v\in L^2(0, \pi) $, such that

$ \begin{equation} \|f(x, y, u)-f(x, y, v)\|\leq L\|u-v\|. \end{equation} $

(1.2)

Our assignment is to determine $ u(\cdot, y) $ from (1.1)–(1.2).

This problem is called the Cauchy problem. As $ f(x, y, u) = -k^2u $, it is the Helmholtz equation that has important applications in acoustic, hydrodynamic and electromagnetic waves [1, 2], etc. If $ f $ is independent of $ u $, it becomes as the Poisson equation. If setting $ f(x, y, u) = \sin u $, we deduce that the classical nonlinear elliptic sine-Gordon equation which mainly appears in the theory of Josephson effects, superconductors, spin waves in ferromagnets, see [3, 4].

Problem(1.1) is ill-posed in the sense that the solution (even if it exists) does not depend continuously on the given Cauchy data [5, 6]. It causes great difficulty in doing the numerical calculations, thus some special regularization techniques are required to stabilize numerical computations, see [6, 7]. Note that, as $ f = 0 $, the problem is the Cauchy problem of Laplace equation, many regularization methods were presented to study it in past years, such as the quasi-reversibility method [8-11], Tikhonov method [12], discretization method [13, 14], the fundamental solution method [15], non-local boundary value problem method [16, 17], etc. In 2014–2018, [18–21] respectively used modified boundary Tikhonov-type, Fourier truncated, filtering function and Generalized Tikhonov-type regularization methods to solve the abstract Cauchy problem of semi-linear elliptic equation in the general bounded domain. On the other similar works for the Cauchy problem of nonlinear elliptic equation, please see [22-28].

In 2012, [29] considered the Cauchy problem of Laplace equation

$ \begin{align} \left\{ \begin{aligned} &u_{xx}+u_{yy} = 0, \quad 0 < x < \pi, \quad 0< y < T, \\ & u(0, y) = u(\pi, y) = 0, \quad 0 \leq y \leq T, \\ &u(x, 0) = \varphi (x), \quad 0 \leq x \leq \pi, \\ &u_y(x, 0) = 0, \quad 0 \leq x \leq \pi, \end{aligned} \right. \end{align} $

(1.3)

and for $ p\geq1 $, we defined the regularization solution as

$ \begin{align} u_{\alpha}^{\delta}(x, y) = \left\{ \begin{aligned} &\sum \limits_{n = 1}^{\infty}\frac{\cosh{\left(ny\right)}}{1+\alpha n^p\sinh {\left(nT\right)}} c_n^\delta X_n(x), \quad p \text{ is odd}, \\ & \sum \limits_{n = 1}^{\infty}\frac{\cosh{\left(ny\right)}}{1+\alpha n^p\cosh {\left(nT\right)}} c_n^\delta X_n(x), \quad p \text{ is even}, \\ \end{aligned} \right. \end{align} $

(1.4)

i.e., the so-called improved non-local boundary value problem method, where $ c_n^\delta = \int_0^\pi \varphi^\delta(\xi) $ $ X_n(\xi)d\xi $, $ X_n(x): = \sqrt{\frac{2}{\pi}}\sin{(nx)}(n\ge 1 $), and $ \{X_n(x)\}_{n = 1}^{\infty} $ is an orthonormal basis of $ L^2(0, \pi) $, $ \varphi^{\delta} $ denotes the measured data and satisfies $ \|\varphi^{\delta}-\varphi\| \leq \delta $, $ \| \cdot\| $ denotes the $ L^2 $-norm and $ \delta>0 $ is a noise level, $ \alpha $ is the regularization parameter. Inspired by this method, in the present paper we investigate the nonlinear problem (1.1)–(1.2) and adopting a similar technique to overcome its ill-poseness (see Section 2), this is an extension to the work in [29].

The article is constructed as below. Section $ 2 $ uses our method to treat problem (1.1) and proves the well-posedness of regularized problem. In Section $ 3 $, the convergence estimate for this method is derived. An iterative scheme is proposed to calculate the regularization solution and make some numerical verifications in Section $ 4 $. Some conclusions are given in Section $ 5 $.

Adopting the similar method as in [19], we can derive that the unique solution to problem (1.1) satisfies the nonlinear integral equation

$ \begin{equation} u(x, y) = \sum\limits_{n = 1}^\infty \left(\cosh(ny)\varphi_n+\int_0^{y}\frac{\sinh(n(y-\tau))}{n}f_n(u)(\tau)d\tau\right) \sin(nx), \end{equation} $

(2.1)

where

$ \begin{equation} \varphi_n = \frac{2}{\pi}\int_0^\pi\varphi(x)\sin(nx)dx, \quad f_n(u)(y) = \frac{2}{\pi}\int_0^{\pi}f(x, y, u(x, y))\sin(nx)dx. \end{equation} $

(2.2)

(2.1) means that $ \cosh(ny) $, $ \sinh(n(y-\tau))/n\rightarrow\infty $ ($ n\rightarrow\infty $), so in order to restore the stability of problem (1.1), we must eliminate the high frequency parts of two functions to design the regularization solution of original problem.

Based on the analysis above and considering the compatibility of physical dimension, when $ p\geq1 $ is even, we approximate the nonlinear problem (1.1) as

$ \begin{align} \left\{ \begin{aligned} &(u^\delta_\alpha)_{xx}+(u^\delta_\alpha)_{yy} = \sum\limits_{n = 1}^\infty\frac{1/\cosh(nT)}{\alpha n^p+1/\cosh(nT)}f_n(u^\delta_\alpha)(y)\sin(nx), &0< x<\pi, 0<y<T, \\ &u^\delta_\alpha(x, 0) = \sum\limits_{n = 1}^\infty\frac{1/\cosh(nT)}{\alpha n^p+1/\cosh(nT)}\varphi^\delta_n\sin(nx), &0\leq x\leq\pi, \\ &(u^\delta_\alpha)_y(x, 0) = 0, &0\leq x\leq\pi, \\ &u^\delta_\alpha(0, y) = u^\delta_\alpha(\pi, y) = 0, &0\leq y\leq T, \end{aligned} \right. \end{align} $

(2.3)

when $ p\geq1 $ is odd, it is regularized by

$ \begin{align} \left\{ \begin{aligned} &(u^\delta_\alpha)_{xx}+(u^\delta_\alpha)_{yy} = \sum\limits_{n = 1}^\infty\frac{1/\sinh(nT)}{\alpha n^p+1/\sinh(nT)}f_n(u^\delta_\alpha)(y)\sin(nx), &0< x<\pi, 0<y<T, \\ &u^\delta_\alpha(x, 0) = \sum\limits_{n = 1}^\infty\frac{1/\sinh(nT)}{\alpha n^p+1/\sinh(nT)}\varphi^\delta_n\sin(nx), &0\leq x\leq\pi, \\ &(u^\delta_\alpha)_y(x, 0) = 0, &0\leq x\leq\pi, \\ &u^\delta_\alpha(0, y) = u^\delta_\alpha(\pi, y) = 0, &0\leq y\leq T, \end{aligned} \right. \end{align} $

(2.4)

where

$ \begin{equation} \varphi^\delta_n = \frac{2}{\pi}\int_0^\pi\varphi^\delta(x)\sin(nx)dx, \quad f_n(u^\delta_\alpha)(y) = \frac{2}{\pi}\int_0^{\pi}f(x, y, u^\delta_\alpha(x, y))\sin(nx)dx, \end{equation} $

(2.5)

and $ \alpha>0 $ is the regularization parameter, the measured data $ \varphi^\delta\in L^2(0, \pi) $ satisfies $ \|\varphi^\delta-\varphi\|\leq\delta $, $ \delta>0 $ is the error level, $ \|\cdot\| $ denotes $ L^2 $-norm. In fact, it is easily verified that the solution to problem (2.3) satisfies the nonlinear integral equation

$ \begin{eqnarray} &&u^\delta_\alpha(x, y)\\ = &&\sum\limits_{n = 1}^\infty \frac{1/\cosh(nT)}{\alpha n^p+1/\cosh(nT)}\left(\cosh(ny)\varphi^\delta_n+\int_0^{y} \frac{\sinh(n(y-\tau))}{n}f_n(u^\delta_\alpha)(\tau)d\tau\right)\sin(nx), \end{eqnarray} $

(2.6)

the solution of problem (2.4) satisfies the nonlinear integral equation

$ \begin{equation} u^\delta_\alpha(x, y) = \sum\limits_{n = 1}^\infty \frac{1/\sinh(nT)}{\alpha n^p+1/\sinh(nT)}\left(\cosh(ny)\varphi^\delta_n+\int_0^{y} \frac{\sinh(n(y-\tau))}{n}f_n(u^\delta_\alpha)(\tau)d\tau\right)\sin(nx), \end{equation} $

(2.7)

where $ \varphi^\delta_n $ and $ f_n(u^\delta_\alpha)(y) $ are given by (2.5).

According to the regularization theory, in order to ensure that the regularization solution is a stable approximation to exact solution, we need proof the existence, uniqueness, and stability for the solution of (2.6) and (2.7).

Now, let $ x>0 $, $ 0\leq\tau\leq y\leq T $, we define two functions

$ \begin{equation} h(x) = \frac{1}{\alpha x+e^{-xT}}, \quad g(x) = \frac{e^{(y-\tau-T)x}}{\alpha x+e^{-Tx}}. \end{equation} $

(2.8)

Noting that when $ \alpha<T $, $ h(x) $ attain unique maximum at the point $ x_0 $ such that

$ \begin{equation} h(x)\leq h(x_0) = h\left(\frac{\ln(T/\alpha)}{T}\right) = \frac{T}{\alpha(1+\ln(T/\alpha))}, \end{equation} $

(2.9)

and it is also proved that

$ \begin{equation} g(x)\leq T^{\frac{y-\tau}{T}}(\alpha\ln(T/\alpha))^{\frac{\tau-y}{T}}, \end{equation} $

(2.10)

clearly, when $ \tau = 0 $, we have

$ \begin{equation} g(x) = \frac{e^{(y-T)x}}{\alpha x+e^{-Tx}}\leq T^{\frac{y}{T}}(\alpha\ln(T/\alpha))^{-\frac{y}{T}}. \end{equation} $

(2.11)

Theorem 2.1 Let $ \varphi^\delta\in L^2(0, \pi) $, $ f $ satisfies (1.2), then problem (2.6) exists a unique solution $ u^{\delta}_{\alpha}\in C([0, T]; L^2(0, \pi)) $.

Proof For $ w\in C([0, T]; L^2(0, \pi)) $, we define the operator $ \Gamma(w)(\cdot, y) $ as

$ \begin{equation} \Gamma(w)(\cdot, y) = \sum\limits_{n = 1}^\infty \left(\frac{\cosh(ny)\varphi^\delta_n} {1+\alpha n^{p}\cosh(nT)}+\int_0^{y} \frac{\sinh(n(y-\tau))f_n(w)(\tau)}{n\left(1+\alpha n^{p}\cosh(nT)\right)}d\tau\right)\sin(nx), \end{equation} $

(2.12)

then for $ w, v \in C([0, T];L^2(0, \pi)) $, $ q\geq1 $, we prove the following result

$ \begin{equation} \left\||\Gamma^q(w)(\cdot, y)-\Gamma^q(v)(\cdot, y)|\right\|\leq \frac{(LC_{\alpha}T)^q}{\sqrt{q!}}\left\||w-v|\right\|, \end{equation} $

(2.13)

where $ C_{\alpha} = T/\alpha $, $ \||\cdot|\| $ denotes the sup norm in $ C([0, T];L^2(0, \pi)) $. First, we use the induction method to derive the following estimate

$ \begin{equation} \|\Gamma^q(w)(\cdot, y)-\Gamma^q(v)(\cdot, y)\|\leq (LC_{\alpha})^q\frac{y^{q/2}T^{q/2}}{\sqrt{q!}}\left\||w-v|\right\|, \end{equation} $

(2.14)

For $ q = 1 $, from (2.12), (2.9) and (1.2), and $ e^{(y-\tau-T)}\leq1 $, we can get

$ \begin{eqnarray} &&\|\Gamma(w)(\cdot, y)-\Gamma(v)(\cdot, y)\|^2\\ & = & \|\sum\limits_{n = 1}^\infty\int_0^{y}\frac{\sinh(n(y-\tau))}{n (1+\alpha n^{p}\cosh(nT) )} (f_n(w)(\tau)-f_n(v)(\tau) )d\tau \sin(nx)\|^2\\ &\leq&\frac{\pi}{2}\sum\limits_{n = 1}^\infty (\int_0^{y}\frac{\sinh(n(y-\tau))}{n (1+\alpha n^{p}\cosh(nT) )} (f_n(w)(\tau)-f_n(v)(\tau) )d\tau )^2\\ &\leq&\frac{\pi}{2}\sum\limits_{n = 1}^\infty\int_0^{y} (\frac{e^{n(y-\tau-T)}} {\alpha n+e^{-nT}} )^2d\tau\int_0^{y} (f_n(w)(\tau)-f_n(v)(\tau) )^2d\tau\\ &\leq&\frac{\pi}{2}\sum\limits_{n = 1}^\infty\int_0^{y} (\frac{1} {\alpha n+e^{-nT}} )^2d\tau\int_0^{y} (f_n(w)(\tau)-f_n(v)(\tau) )^2d\tau\\ &\leq& \frac{\pi}{2}y (\frac{T}{\alpha (1+\ln(T/\alpha) )} )^2 \int_0^{y}\sum\limits_{n = 1}^\infty (f_n(w)(\tau)-f_n(v)(\tau) )^2d\tau\\ &\leq& TC_\alpha^2 \int_0^{y} \|f(\cdot, \tau, w(\cdot, \tau))-f(\cdot, \tau, v(\cdot, \tau)) \|^2d\tau\\ &\leq& L^2TC_\alpha^2 \int_0^{y} \|w(\cdot, \tau)-v(\cdot, \tau) \|^2d\tau\\ &\leq& (LC_\alpha)^2yT \||w-v| \|^2. \end{eqnarray} $

When $ q = i $, suppose that

$ \begin{equation} \|\Gamma^i(w)(\cdot, y)-\Gamma^i(v)(\cdot, y)\|^2\leq (LC_{\alpha})^{2i}\frac{y^{i}T^{i}}{i!} \||w-v| \|^2, \end{equation} $

(2.15)

then for $ q = i+1 $, by (2.15), and use the similar process, we obtain that

$ \begin{eqnarray} &&\|\Gamma^{i+1}(w)(\cdot, y)-\Gamma^{i+1}(v)(\cdot, y)\|^2\\ & = & \|\sum\limits_{n = 1}^\infty\int_0^{y}\frac{\sinh(n(y-\tau))}{n (1+\alpha n^{p}\cosh(nT) )} (f_n(\Gamma^i(w))(\tau)-f_n(\Gamma^i(v))(\tau) )d\tau \sin(nx) \|^2\\ &\leq&\frac{\pi}{2}\sum\limits_{n = 1}^\infty (\int_0^{y}\frac{\sinh(n(y-\tau))}{n (1+\alpha n^{p}\cosh(nT) )} (f_n(\Gamma^i(w))(\tau)-f_n(\Gamma^i(v))(\tau) )d\tau )^2\\ & \leq& TC_\alpha^2 \int_0^{y} \|f(\cdot, \tau, \Gamma^i(w)(\cdot, \tau))-f(\cdot, \tau, \Gamma^i(v)(\cdot, \tau)) \|^2d\tau\\ &\leq& L^2TC_\alpha^2 \int_0^{y} \|\Gamma^i(w)(\cdot, \tau)-\Gamma^i(v)(\cdot, \tau) \|^2d\tau \leq(LC_{\alpha})^{2i+2}\frac{y^{i+1}T^{i+1}}{(i+1)!} \||w-v| \|^2. \end{eqnarray} $

By the induction principle, we have

$ \begin{equation} \|\Gamma^q(w)(\cdot, y)-\Gamma^q(v)(\cdot, y)\|\leq (LC_{\alpha})^q\frac{y^{q/2}T^{q/2}}{\sqrt{q!}}\left\||w-v|\right\|. \end{equation} $

(2.16)

Subsequently, it is clearly derived that

$ \begin{equation} \left\||\Gamma^q(w)(\cdot, y)-\Gamma^q(v)(\cdot, y)|\right\|\leq \frac{(LC_{\alpha}T)^q}{\sqrt{q!}}\left\||w-v|\right\|. \end{equation} $

(2.17)

Consider the operator $ \Gamma: C([0, T];L^2(0, \pi))\rightarrow C([0, T];L^2(0, \pi)) $, and we know

$ \begin{equation} \lim\limits_{q\rightarrow\infty}\frac{(LC_{\alpha}T)^q}{\sqrt{q!}} = 0. \end{equation} $

(2.18)

It exists a positive integer constant $ q_0 $ and satisfies $ 0<\frac{(LC_{\alpha}T)^{q_0}}{\sqrt{q_0!}}<1 $, then $ \Gamma^{q_0} $ is a contraction, this indicates that $ \Gamma^{q_0}(w) = w $ has a unique solution $ u^{\delta}_{\alpha}\in C([0, T]; L^2(0, \pi)) $. Note that $ \Gamma\left(\Gamma^{q_0}\left(u^{\delta}_{\alpha}\right)\right) = \Gamma\left(u^{\delta}_{\alpha}\right) $, hence $ \Gamma^{q_0}\left(G\left(u^{\delta}_{\alpha}\right)\right) = \Gamma\left(u^{\delta}_{\alpha}\right) $. By the uniqueness of the fixed point, we obtain that $ \Gamma\left(u^{\delta}_{\alpha}\right) = u^{\delta}_{\alpha} $, so $ \Gamma(w) = w $ exists a unique solution $ u^{\delta}_{\alpha}\in C([0, T];L^2(0, \pi)) $.

Theorem 2.2 Let $ u^\delta_{\alpha1} $ and $ u^\delta_{\alpha2} $ be the solutions of problem (2.6) corresponding to the noisy data $ \varphi^\delta_1 $ and $ \varphi^\delta_2 $, respectively, then for $ \alpha<T $, we have

$ \begin{equation} \|u^\delta_{\alpha1}(\cdot, y)-u^\delta_{\alpha2}(\cdot, y)\|\leq C_0\left(\alpha\ln(T/\alpha)\right)^{-\frac{y}{T}}\left\|\varphi_1^\delta-\varphi_2^\delta\right\| \end{equation} $

(2.19)

with $ T_0 = \max\{T, T^{\frac{y}{T}}, T^{\frac{y-\tau}{T}}\} $, $ C_0 = \sqrt{8T^2_0\left(1+2yL^2T^3_0e^{2yL^2T^3_0}\right)} $.

Proof From equation (2.6), we have

$ \begin{equation} u^\delta_{\alpha1}(x, y) = \sum\limits_{n = 1}^\infty \frac{1/\cosh(nT)}{\alpha n^p+1/\cosh(nT)}\left(\cosh(ny)\varphi^\delta_{1n} +\int_0^{y}\frac{\sinh(n(y-\tau))}{n}f_n(u^\delta_{\alpha1})(\tau)d\tau\right)\sin(nx), \end{equation} $

(2.20)

$ \begin{equation} u^\delta_{\alpha2}(x, y) = \sum\limits_{n = 1}^\infty \frac{1/\cosh(nT)}{\alpha n^p+1/\cosh(nT)}\left(\cosh(ny)\varphi^\delta_{2n} +\int_0^{y}\frac{\sinh(n(y-\tau))}{n}f_n(u^\delta_{\alpha2})(\tau)d\tau\right)\sin(nx), \end{equation} $

(2.21)

where $ \varphi^\delta_{\mu n} = \frac{2}{\pi}\int_0^\pi\varphi_\mu^\delta(x)\sin(nx)dx $, $ \mu = 1, 2 $.

For $ n\geq1 $, from (2.20), (2.21), (2.9), (2.10) and (1.2), we get

$ \begin{eqnarray*} &&\|u^\delta_{\alpha1}(\cdot, y)-u^\delta_{\alpha2}(\cdot, y)\|^2\\\nonumber & = &\frac{\pi}{2}\sum^\infty_{n = 1} \|\frac{\cosh(ny)/\cosh(nT)}{\alpha n^p+1/\cosh(nT)} (\varphi^\delta_{1n}-\varphi^\delta_{2n} ) +\int_0^{y}\frac{\sinh(n(y-\tau))/\cosh(nT)}{n (\alpha n^p+1/\cosh(nT) )}(f_n(u^\delta_{\alpha1})(\tau)-f_n(u^\delta_{\alpha2})(\tau))d\tau \|^2\\\nonumber &\leq& \pi\sum^\infty_{n = 1} (\frac{\cosh(ny)/\cosh(nT)}{\alpha n^p+1/\cosh(nT)} )^2 (\varphi^\delta_{1n}-\varphi^\delta_{2n} )^2\\ &&+\pi\sum^\infty_{n = 1}\frac{1}{n^2} (\int_0^{y}\frac{\sinh(n(y-\tau))/\cosh(nT)}{\alpha n^p+1/\cosh(nT)}(f_n(u^\delta_{\alpha1})(\tau)-f_n(u^\delta_{\alpha2})(\tau))d\tau )^2\\ &\leq& 4\pi\sum^\infty_{n = 1} (\frac{e^{n(y-T)}}{\alpha n+e^{-nT}} )^2 (\varphi^\delta_{1n}-\varphi^\delta_{2n} )^2 +\pi\sum^\infty_{n = 1}\frac{1}{n^2} (\int_0^{y}\frac{e^{n(y-\tau-T)}}{\alpha n+e^{-nT}}(f_n(u^\delta_{\alpha1})(\tau)-f_n(u^\delta_{\alpha2})(\tau))d\tau )^2\\\nonumber &\leq&8T^{\frac{2y}{T}}(\alpha\ln(T/\alpha))^{-\frac{2y}{T}}\|\varphi^\delta_1-\varphi^\delta_2\|^2 +\pi\sum^\infty_{n = 1}\int_0^{y} (\frac{e^{n(y-\tau-T)}}{\alpha n+e^{-nT}} )^2d\tau\int_0^{y} \|f_n(u^\delta_{\alpha1})(\tau)-f_n(u^\delta_{\alpha2})(\tau)) \|^2d\tau \end{eqnarray*} $

$ \begin{eqnarray*} &\leq&8T^{\frac{2y}{T}}(\alpha\ln(T/\alpha))^{-\frac{2y}{T}}\|\varphi^\delta_1-\varphi^\delta_2\|^2\\\nonumber &&+2yL^2T^{\frac{2y-2\tau}{T}}(\alpha\ln(T/\alpha))^{-\frac{2y}{T}}\int_0^{y}(\alpha\ln(T/\alpha))^{\frac{2\tau}{T}} \|u^\delta_{\alpha1}(\cdot, \tau)-u^\delta_{\alpha2}(\cdot, \tau)\|^2d\tau. \end{eqnarray*} $

Then

$ \begin{eqnarray*} &&(\alpha\ln(T/\alpha))^{\frac{2y}{T}}\|u^\delta_{\alpha1}(\cdot, y)-u^\delta_{\alpha2}(\cdot, y)\|^2\\\nonumber &\leq&8T^2_0\|\varphi^\delta_1-\varphi^\delta_2\|^2 +2L^2T^3_0\int_0^{y}(\alpha\ln(T/\alpha))^{\frac{2\tau}{T}} \|u^\delta_{\alpha1}(\cdot, \tau)-u^\delta_{\alpha2}(\cdot, \tau)\|^2d\tau\nonumber. \end{eqnarray*} $

Using Gronwall's inequality [30], it can be obtained that

$ \begin{equation} (\alpha\ln(T/\alpha))^{\frac{2y}{T}}\|u^\delta_{\alpha1}(\cdot, y)-u^\delta_{\alpha2}(\cdot, y)\|^2\leq8T^2_0 (1+2yL^2T^3_0e^{2yL^2T^3_0} ) \|\varphi^\delta_1-\varphi^\delta_2 \|^2. \end{equation} $

(2.22)

From (2.22), estimate (2.19) can be established.

Theorem 3.1 Suppose that $ u $ is the solution of problem (1.1) and $ u^\delta_\alpha $ is the solution of problem (2.6). Let the measured data $ \varphi^\delta $ satisfy $ \|\varphi^\delta-\varphi\|\leq\delta $, and the exact solution $ u $ satisfy

$ \begin{equation} \frac{\pi}{2}\sum\limits^{\infty}_{n = 1}n^{2p}e^{2n(T-y)} |\langle u(\cdot, y), \sin(nx)\rangle |^2\leq E^2 , \end{equation} $

(3.1)

and the regularization parameter $ \alpha $ is chosen as

$ \begin{equation} \alpha = \delta, \end{equation} $

(3.2)

then for fixed $ 0<y\leq T $ and $ \delta<T $, we have the following convergence estimate

$ \begin{equation} \|u^\delta_\alpha(\cdot, y)-u(\cdot, y)\|\leq C_2\delta^{1-\frac{y}{T}}(\ln(T/\delta))^{-\frac{y}{T}}, \end{equation} $

(3.3)

where $ C_2 = C_0+C_1 $, $ C_1 = \sqrt{32T^2_0E^2 (1+2yL^2T^3_0e^{2yL^2T^3_0} )} $.

Proof Denoting $ u_\alpha $ as the corresponding solution of problem (2.6) with the exact data $ \varphi $. Using the triangle inequality, we have

$ \begin{equation} \|u^\delta_\alpha-u\|\leq\|u^\delta_\alpha-u_\alpha\|+\|u_\alpha-u\|. \end{equation} $

(3.4)

From Theorem 2.2, we get

$ \begin{eqnarray} \|u^\delta_\alpha(\cdot, y)-u_\alpha(\cdot, y)\| \leq C_0 \left(\alpha\ln(T/\alpha)\right)^{-\frac{y}{T}} \left\|\varphi^\delta-\varphi\right\|. \end{eqnarray} $

(3.5)

By (2.1), (2.6), (2.10), (2.11), (3.1), and use the inequality $ e^{t}/2\leq\cosh(t)\leq e^{t} $ for $ t>0 $, it can be derived that

$ \begin{eqnarray*} &&\|u_\alpha(\cdot, y)-u(\cdot, y)\|^2\\ &\leq&\frac{\pi}{2}\sum^\infty_{n = 1}\|\frac{\alpha n^p}{\alpha n^p+1/\cosh(nT)}(\varphi_n\cosh(ny)+\frac{1}{n}\int_0^{y}\sinh(n(y-\tau))f_n(u)(\tau)d\tau)\\\nonumber &&+\frac{1}{n}\int_0^{y}\frac{\sinh(n(y-\tau))/\cosh(nT)}{\alpha n^p+1/\cosh(nT)}(f_n(u_{\alpha})(\tau)-f_n(u)(\tau))d\tau\|^2\\\nonumber &\leq& \pi\sum^\infty_{n = 1}\left(\frac{\alpha\cosh(ny)/\cosh(nT)}{\alpha n^p+1/\cosh(nT)}\right)^2\left(\frac{\cosh(nT)}{\cosh(ny)}\right)^2n^{2p}\left(\varphi_n\cosh(ny)+\frac{1}{n}\int_0^{y} \sinh(n(y-\tau))f_n(u)(\tau)d\tau\right)^2 \end{eqnarray*} $

$ \begin{eqnarray*} &&+\frac{1}{n^2}\left(\int_0^{y}\frac{\sinh(n(y-\tau))/\cosh(nT)}{\alpha n^p+1/\cosh(nT)}(f_n(u_{\alpha})(\tau)-f_n(u)(\tau))d\tau\right)^2\\\nonumber &\leq& 16\pi\sum^\infty_{n = 1}\left(\frac{ e^{n(y-T)}}{\alpha n+e^{-nT}}\right)^2\alpha^2n^{2p}e^{2n(T-y)}\left(\varphi_n\cosh(ny)+\frac{1}{n}\int_0^{y} \sinh(n(y-\tau))f_n(u)(\tau)d\tau\right)^2\\\nonumber &&+\pi\sum^\infty_{n = 1}\frac{1}{n^2}\left(\int_0^{y}\frac{e^{n(y-\tau-T)}}{\alpha n+e^{-nT}}(f_n(u_{\alpha})(\tau)-f_n(u)(\tau))d\tau\right)^2\\\nonumber &\leq&32T^{\frac{2y}{T}}(\alpha\ln(T/\alpha))^{-\frac{2y}{T}}\alpha^2E^2 +\pi\sum^\infty_{n = 1}\int_0^{y}\left(\frac{e^{n(y-\tau-T)}}{\alpha n+e^{-nT}}\right)^2d\tau\int_0^{y}\left\|f_n(u_{\alpha})(\tau)-f_n(u)(\tau))\right\|^2d\tau\\\nonumber &\leq&32T^{\frac{2y}{T}}(\alpha\ln(T/\alpha))^{-\frac{2y}{T}}\alpha^2E^2 +2yL^2T^{\frac{2y-2\tau}{T}}(\alpha\ln(T/\alpha))^{-\frac{2y}{T}}\int_0^{y}(\alpha\ln(T/\alpha))^{\frac{2\tau}{T}} \|u_{\alpha}(\cdot, \tau)-u(\cdot, \tau)\|^2d\tau. \end{eqnarray*} $

Then

$ \begin{eqnarray} &&(\alpha\ln(T/\alpha))^{\frac{2y}{T}}\|u_{\alpha}(\cdot, y)-u(\cdot, y)\|^2 \\ &\leq&32T^{2}_0E^2\alpha^{2} +2L^2T^3_{0}\int_0^{y}(\alpha\ln(T/\alpha))^{\frac{2\tau}{T}} \|u_{\alpha}(\cdot, \tau)-u(\cdot, \tau)\|^2d\tau. \end{eqnarray} $

(3.6)

Using Gronwall's inequality [30], it can be derived that

$ \begin{equation} (\alpha\ln(T/\alpha))^{\frac{2y}{T}}\|u_{\alpha}(\cdot, y)-u(\cdot, y)\|^2 \leq32T^2_0E^2\left(1+2yL^2T^3_0e^{2yL^2T^3_0}\right)\alpha^2. \end{equation} $

(3.7)

i.e.,

$ \begin{eqnarray} \|u_{\alpha}(\cdot, y)-u(\cdot, y)\|\leq C_1\alpha^{1-\frac{y}{T}}(\ln(T/\alpha))^{-\frac{y}{T}}. \end{eqnarray} $

(3.8)

From (3.2), (3.4), (3.5) and (3.8), we have

$ \begin{equation} \|u^\delta_\alpha(\cdot, y)-u(\cdot, y)\| \leq (C_0+C_1)\delta^{1-\frac{y}{T}}(\ln(T/\delta))^{-\frac{y}{T}} = C_2\delta^{1-\frac{y}{T}}(\ln(T/\delta))^{-\frac{y}{T}}. \end{equation} $

(3.9)

For the case that $ p $ is odd, since the proof method is similar to the above procedure, we only give the conclusions of the stability and convergence estimate.

Theorem 3.2 Let $ u^\delta_{\alpha1} $ and $ u^\delta_{\alpha2} $ be the solutions of problem (2.7) corresponding to the data $ \varphi^\delta_1 $ and $ \varphi^\delta_2 $, respectively, then for $ \alpha<T $, we have

$ \begin{equation} \|u^\delta_{\alpha1}(\cdot, y)-u^\delta_{\alpha2}(\cdot, y)\|\leq C_3\left(\alpha\ln(T/\alpha)\right)^{-\frac{y}{T}}\left\|\varphi_1^\delta-\varphi_2^\delta\right\|, \end{equation} $

(3.10)

and suppose that $ u $ is the solution of problem (1.1) and $ u^\delta_\alpha $ is the solution of problem (2.7). Let the measured data $ \varphi^\delta $ satisfy $ \|\varphi^\delta-\varphi\|\leq\delta $, the exact solution $ u $ satisfy

$ \begin{equation} \frac{\pi}{2}\sum\limits^{\infty}_{n = 1}n^{2p}e^{2n(T-y)}\left|\langle u(\cdot, y), \sin(nx)\rangle\right|^2\leq E^2 , \end{equation} $

(3.11)

if the regularization parameter $ \alpha $ is chosen as

$ \begin{equation} \alpha = \delta, \end{equation} $

(3.12)

then for fixed $ 0<y\leq T $ and $ \delta<T $, we have the following convergence estimate

$ \begin{equation} \|u^\delta_\alpha(\cdot, y)-u(\cdot, y)\|\leq C_5\delta^{1-\frac{y}{T}}(\ln(T/\delta))^{-\frac{y}{T}}, \end{equation} $

(3.13)

where $ C_3 = \sqrt{8C_4^2T^2_0\left(1+2yL^2C_4^2T^3_0e^{2yL^2C_4^2T^3_0}\right)} $, $ C_4 = \frac{2}{1-e^{-2T}} $, and $ C_5 = C_3+C_6 $, $ C_6 = \sqrt{32T^2_0E^2\left(1+2yL^2C^2_4T^3_0e^{2yL^2C^2_4T^3_0}\right)} $.

Remark 3.3 In the research of ill-posed problems, we often impose an a-priori assumption on the exact solution to make the convergence estimate of one regularization method. For our problem, the difficulty of imposing the a-priori assumption mainly lie in the nonlinear property and the application of regularization method. In view of the factors above, here we impose a-priori assumption (3.1), (3.11), and combine with (2.10), (2.11) to proof Theorems 3.1–3.2.

Remark 3.4 Our given method can be applied to investigate the cases of nonhomogeneous Neumann data ($ u(x, 0) = 0 $, $ u_y(x, 0) = \phi(x)\neq0 $) or nonhomogeneous Dirichlet and Neumann datum ($ u(x, 0) = \varphi(x)\neq0 $, $ u_y(x, 0) = \phi(x)\neq0 $). Since the constructed procedure of regularization method is similar with the one in Subsection 2.1, here we skip it.

In this section, an iterative scheme is proposed to calculate the regularization solution and a numerical example is performed to verify the efficiency of our method. Consider the nonlinear problem

$ \begin{align} \left\{ \begin{aligned} &u_{xx}+u_{yy} = \cos(u)+g(x, y), &0<x<\pi, \quad 0<y<1, \\ &u(x, 0) = \varphi(x), &0\leq x\leq\pi, \\ &u_y(x, 0) = 0, &0\leq x\leq\pi, \\ &u(0, y) = u(\pi, y) = 0, &0\leq y\leq1. \end{aligned} \right. \end{align} $

(4.1)

It is clear that $ u(x, y) = x(x-\pi)(2+y^2) $ is an exact solution of problem (4.1), thus $ g(x, y) = 2x(x-\pi)+2(2+y^2)-\cos(x(x-\pi)(2+y^2)) $, $ \varphi(x) = u(x, 0) = 2x(x-\pi) $, and the measured data is given by $ \varphi^\delta(x) = \varphi(x)\left(1+\varepsilon (x/2-1)\right) $.

Let $ 0 = y_0<y_1<\ldots<y_k<\ldots<y_M = 1 $ for $ k = 0, 1, 2, \ldots, M $, the regularization solution $ u^{\delta}_{\alpha}(x, y_k) $ with $ y_k = \frac{k}{M} $ can be computed by the following iteration scheme

$ \begin{equation} u^{\delta}_{\alpha}(x, y_k) = v_k(x) = w_{1, k}\sin(x)+w_{2, k}\sin(2x)+\ldots+w_{m, k}\sin(mx), \end{equation} $

(4.2)

here, for the case that $ p $ is even, $ w_{j, k} = \frac{1/\cosh(j)}{\alpha j^p+1/\cosh(j)}a_{j, k} $, when $ p $ is odd, $ w_{j, k} = \frac{1/\sinh(j)}{\alpha j^p+1/\sinh(j)}a_{j, k} $, and

$ \begin{eqnarray} &a_{j, k} = \cosh(jy_k)\varphi^\delta_{j}+\frac{2}{\pi j}\int^{y_{k}}_{y_{k-1}}\int^{\pi}_0\sinh\left(j(y_{k}-\tau)\right) \left(\cos(v_{k-1}(x))+g(x, \tau)\right)\sin(jx)dxd\tau\\ &+\frac{2}{\pi j}\int^{y_{k-1}}_{y_{k-2}}\int^{\pi}_0\sinh\left(j(y_{k}-\tau)\right) \left(\cos(v_{k-2}(x))+g(x, \tau)\right)\sin(jx)dxd\tau\\ &+\ldots+\frac{2}{\pi j}\int^{y_{2}}_{y_1}\int^{\pi}_0\sinh\left(j(y_{k}-\tau)\right) \left(\cos(v_{1}(x))+g(x, \tau)\right)\sin(jx)dxd\tau\\ &+\frac{2}{\pi j}\int^{y_{1}}_{0}\int^{\pi}_0\sinh\left(j(y_{k}-\tau)\right) \left(\cos(v_{0}(x))+g(x, \tau)\right)\sin(jx)dxd\tau, \end{eqnarray} $

(4.3)

$ \begin{equation} v_0(x) = \varphi^\delta(x) = \varphi(x)\left(1+\varepsilon (x/2-1)\right), \quad \varphi^\delta_{j} = \frac{2}{\pi}\int^\pi_{0}\varphi^\delta(x)\sin(jx)dx. \end{equation} $

(4.4)

Adopt the above algorithm, we choose $ y_k = \frac{k}{M} $ with $ M = 50 $, for $ k = 1, 2, \cdots, 50, j = 1, \cdots, m = 5 $ to compute $ u^{\delta}_{\alpha}(\cdot, y) $ at $ y = 0.4 $, $ 1 $ $ (k = 20, 50) $. In the computational procedure, the regularization $ \alpha $ is chosen by (3.2) or (3.12). For $ p = 1, 2, 3, 4 $, the numerical results for $ \varepsilon = 0.01 $ are shown in Figure 1–2, respectively. For $ p = 1, 2, 3 $, the relative root mean square errors between the exact and regularization solutions are defined by

$ \begin{equation} \epsilon(u) = \frac{\sqrt{\frac{1}{N}\sum\limits_{\ell = 1}^{N}\left(u(x_\ell) -u^\delta_\alpha(x_\ell)\right)^2}} {\sqrt{\frac{1}{N}\sum\limits_{\ell = 1}^{N}\left(u(x_\ell)\right)^2}}, \quad x_{\ell} = \frac{(\ell-1)}{N-1}\pi, \quad \ell = 1, 2, \cdots, N, \end{equation} $

(4.5)

the computed results are shown in Table 1.

Figure 1–2 and Table 1 show that our proposed method is effective and stable. Meanwhile, we note that the more small $ \varepsilon $ is, the better the calculation effect is, this is a common phenomenon in the computation of ill-posed problems. Table 1 indicates that the computational effect of our method is more better when $ p $ is an odd. Finally, we mention that the iteration scheme (4.2) can be explained as a Fourier series, since the exact solution is a polynomial function, during the procedure of computing the regularization solution, the node number $ M $ should be chosen as a larger number relatively, here we take $ M = 50 $, and according to the numerical results in [18, 19], the truncated term number $ m $ has no necessity to be taken too big, the best value should be $ 5 $ or $ 6 $, here we take it as $ 5 $.

We use a regularization method to solve a Cauchy problem of nonlinear elliptic equation. The existence, uniqueness and stability of the regularization solution are proven, under a-priori bound assumptions for the exact solution, the convergence estimates for $ 0<y\leq T $ have been obtained. Finally, an iterative scheme is proposed to calculate the regularization solution, and some numerical results indicate that our method is stable and feasible.