Cauchy problem for the Laplace equation in a radially symmetric hollow cylinder

Abstract

In this paper, an axisymmetric Cauchy problem for the Laplace equation in an unbounded hollow cylinder is considered. The Cauchy data are given on the inside surface of the cylinder, and the solution on the whole domain is sought. We propose a Fourier method with a priori and a posteriori parameter choice rules to solve this ill-posed problem. It is shown that the approximate solutions are stably convergent to the exact ones with explicit error estimates. A further comparison in the numerical aspects demonstrates the effectiveness and accuracy of the presented methods.

1 Introduction

The Cauchy problem for the Laplace equation is an old yet persistent problem arising in many practical applications, and the general form is

$$ \textstyle\begin{cases} \triangle u=0, \quad x\in\Omega, \\ u|_{\Gamma}=g, \\ \frac{\partial u}{\partial n}|_{\Gamma}=h, \end{cases} $$

(1.1)

where Ω is a domain in $\mathbb{R}^{n}$, $\Gamma\subset \partial\Omega$ is part of the boundary, g and h are given functions in $L^{2}(\Gamma)$, and the solution u is sought in the whole domain Ω. This problem arises in many practical contexts, for example, in the problem of electrical prospecting, u denotes the potential of the electrostatic field artificially created in the interior of the Earth. We have $\triangle u=0$, $x\in\Omega$, where $u|_{\Gamma}$ is the magnitude of the potential u and $\frac{\partial u}{\partial n}|_{\Gamma}$ is the intensity of the potential, both measured at the accessible surface Γ of the Earth. In general, this problem is ill-posed since for some Cauchy data g and h there is not a solution, and even if there exists a solution it does not always depend continuously on the data. Therefore, several regularization methods have been presented to solve it such as the quasi-reversibility method [1, 2], the boundary element method [3, 4], the Fourier regularization method [5, 6], the central difference regularization method [7], the mollification method [8], etc. However, most of the results are in two dimensions. For the high dimensional case, both theoretical analysis and numerical computation are very difficult. In [9], the authors transfer high dimensional Cauchy problem for Laplace equation into moment problem, and then construct a series of polynomial functions to approximate solutions of the moment problem. In [10], a quasi-boundary-value method together with left-preconditioned generalized minimum residual method are proposed to deal with an ill-posed Cauchy problem for a 3D elliptic partial differential equation with variable coefficients.

In this paper, we consider the problem of an extension of the field potential specified on the inside surface of a hollow cylinder into space, and it is reduced to the axisymmetric Cauchy problem for the Laplace equation. This problem is involved in practical calculations of various electron optic systems. The hollow cylinder case is interesting since the hole leaves spaces for the measurement devices or devices that generate electric or magnetic fields. For example, electric fields with rotational symmetry are usually generated by electrodes in the shape of cylinders, cups and diaphragms. In recent years, Lu et al. [11] applied an analytical approach to study the transient heat conduction in a composite hollow cylinder. Cheng et al. [12] studied the inverse heat conduction problem in a hollow spherically symmetric domain. Marin and Marinescu [13, 14] investigated the existence, uniqueness and the asymptotic partition of total energy for the solutions of the initial boundary value problem within the context of the thermoelasticity of initially stressed bodies, and further considered micropolar thermoelastic body occupying a prismatic cylinder [15]. Şeremet and Şeremet [16] presented new steady-state Green’s functions for displacements and thermal stresses for plane problem within a rectangular region, and the proposed technique could be extended to many 3D problems. More detailed descriptions of the model of hollow cylinders can be found in [17].

In this paper, suppose the considered 3D hollow cylinder domain is regular, and the internal and external radii are denoted $r_{0}$ and R, respectively. Under the assumption that the inside surface of the cylinder is composed of insulating materials, it makes most sense to use cylindrical coordinates to transfer problem (1.1) into the following problem:

$$ \textstyle\begin{cases} u_{rr}+\frac{1}{r}u_{r}+u_{zz}=0, & r_{0}< r< R, z\in\mathbb{R}, \\ u(r_{0},z)=g(z), & z\in\mathbb{R}, \\ u_{r}(r_{0},z)=0, & z\in\mathbb{R}, \end{cases} $$

(1.2)

where $r=\sqrt{x^{2}+y^{2}}$, and $g(z)$ is a known potential distribution along the inside surface.

In practice, the data $g(z)$ is often obtained by the instrument installed inside the hollow cylinder, and there exist unavoidable errors. We assume that instead of exact data $g(z)\in L^{2}(\mathbb {R})$, only a noisy data $g^{\delta}(z)\in L^{2}(\mathbb{R})$ with

$$ \bigl\Vert g(z)-g^{\delta}(z)\bigr\Vert \leq\delta $$

(1.3)

is available. $\delta>0$ represents the ‘noise level’ and $\|\cdot\|$ denotes the $L^{2}$-norm.

We further assume that

$$ u(r,\cdot)\in L^{2}(\mathbb{R}),\quad \mbox{for each }r \in(r_{0}, R), $$

(1.4)

and the following a priori bound holds:

$$ \bigl\Vert u(R,\cdot)\bigr\Vert \leq E, $$

(1.5)

where E is a positive constant.

This paper is organized as follows. In Section 2, we present the expression of the solution and analyze the ill-posedness of problem (1.2). The a priori and a posteriori parameter choice rules which yield error estimates of Hölder type are suggested in Section 3. In Section 4, some numerical examples are given to illustrate the validity of the theoretical results. Finally, Section 5 ends this paper with a short conclusion.

2 Expression of the solution

For $f(z)\in L^{1}(\mathbb{R})$, $\hat{f}(\xi)$ denotes its Fourier transform, which is defined by

$$\hat{f}(\xi):=\frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}}f(z)e^{-i\xi z}\,dz. $$

Thus by using Fourier transform, the problem (1.2) is transformed into the following initial problem (2.1) of ODE in frequency domain:

$$ \textstyle\begin{cases} \hat{u}_{rr}+\frac{1}{r}\hat{u}_{r}-\xi^{2}\hat{u}=0,& r_{0}< r< R, \xi\in\mathbb{R}, \\ \hat{u}(r_{0},\xi)=\hat{g}(\xi), & \xi\in\mathbb{R}, \\ \hat{u}_{r}(r_{0},\xi)=0, & \xi\in\mathbb{R}. \end{cases} $$

(2.1)

Lemma 2.1

The solution of problem (2.1) is given by

$$ \hat{u}(r,\xi)=r_{0}|\xi|\Phi(r,\xi)\hat{g}(\xi),\quad r\in [r_{0},R], \xi\in\mathbb{R}, $$

(2.2)

where

$$ \Phi(r,\xi)=I_{0}\bigl(r\vert \xi \vert \bigr)K_{1}\bigl(r_{0}\vert \xi \vert \bigr)+K_{0}\bigl(r\vert \xi \vert \bigr)I_{1} \bigl(r_{0}\vert \xi \vert \bigr), $$

(2.3)

and $I_{0}(\cdot)$, $I_{1}(\cdot)$, $K_{0}(\cdot)$, $K_{1}(\cdot)$ denote the modified Bessel function.

Proof

From [18], we know that the modified Bessel equation has two linearly independent solutions $I_{0}$ and $K_{0}$, then the general solution of equation in problem (2.1) is

$$ \hat{u}(r,\xi)=C_{1}(\xi)I_{0}\bigl(r\vert \xi \vert \bigr)+C_{2}(\xi)K_{0}\bigl(r\vert \xi \vert \bigr), \quad r\in [r_{0},R], \xi\in\mathbb{R}. $$

Combining the boundary conditions in (2.1) with the properties $I_{0}'(x)=I_{1}(x)$, $K_{0}'(x)=-K_{1}(x)$, and $I_{0}(x)K_{1}(x)+I_{1}(x)K_{0}(x)=\frac{1}{x}$ for $x>0$, we get equation (2.2) of the solution to problem (2.1). □

Note that if $r_{0}=0$, it is a special case to problem (1.2) and the corresponding solution is

$$ u(r,z)=\frac{1}{\sqrt{2\pi}} \int_{\xi\in\mathbb{R}}e^{iz\xi }I_{0}(r\xi) \hat{g}^{\delta}(\xi)\,d\xi. $$

There is only one modified Bessel function $I_{0}$ in this expression, and this case has been discussed in [19].

In order to get a better understanding of the property of solution (2.2), it is necessary to list some important properties of function $\Phi(r,\xi)$. The following lemma establishes the relationship between $\Phi(r,\xi)$ and some basic elementary functions.

Lemma 2.2

For $\xi\neq0$, there exist positive constants $C_{1}$ and $C_{2}$ such that the following inequalities hold:

$$ C_{1}\leq\frac{|\xi|\Phi(r,\xi)}{e^{(r-r_{0})|\xi|}}\leq C_{2}. $$

(2.4)

Proof

Case 1: $|\xi|\geq1$.

According to [18], the ‘asymptotic expansions for large arguments’ of modified Bessel functions $I_{0}$, $I_{1}$, $K_{0}$, and $K_{1}$ are as follows:

$$ I_{\nu}(x)\sim\frac{1}{\sqrt{2\pi}}\frac{e^{x}}{\sqrt{x}},\qquad K_{\nu}(x)\sim\sqrt{\frac{\pi}{2}}\frac{e^{-x}}{\sqrt{x}},\quad \nu=0,1, $$

(2.5)

and then, combining with the continuous property of Bessel functions, we know that there exist four pairs of positive constants ${c}_{\nu}$, $c_{\nu}'$, $d_{\nu}$, and $d_{\nu}'$ ($\nu=0,1$), such that

$$ c_{\nu}\frac{e^{x}}{\sqrt{x}}\leq I_{\nu}(x)\leq c_{\nu}' \frac {e^{x}}{\sqrt{x}},\qquad d_{\nu} \frac{e^{-x}}{\sqrt{x}}\leq K_{\nu}(x)\leq d_{\nu}' \frac {e^{-x}}{\sqrt{x}}, \quad x\geq r_{0}. $$

(2.6)

Furthermore, on denoting $\mu_{1}=c_{0}d_{1}$ and $\mu _{2}=c_{0}'d_{1}'+c_{1}'d_{0}'$, then the following inequalities are straightforward calculations by using equation (2.3) and the above inequalities (2.6):

$$ \frac{\mu_{1}e^{(r-r_{0})|\xi|}}{|\xi|}\leq\Phi(r,\xi)\leq\frac {\mu_{2}e^{(r-r_{0})|\xi|}}{|\xi|}, \quad \mbox{for } |\xi| \geq1. $$

(2.7)

Case 2: $0<|\xi|<1$.

Based on equation (2.3), we have

$$ \Phi(r,\xi)\sim\frac{1}{r_{0}|\xi|},\quad \mbox{for } |\xi | \rightarrow0. $$

(2.8)

For ease of use, an alternative form of (2.8) is as follows:

$$ \Phi(r,\xi)\sim\frac{e^{(r-r_{0})|\xi|}}{r_{0}|\xi|}, \quad \mbox{for } |\xi|\rightarrow0, $$

and combining with the continuous property of $\Phi(r,\xi)$ on $[r_{0},R]\times(0,1)$, we know that there exist two positive constants $\mu_{3}$ and $\mu_{4}$ such that the following inequalities hold:

$$ \frac{\mu_{3}e^{(r-r_{0})|\xi|}}{|\xi|}\leq\Phi(r,\xi)\leq\frac {\mu_{4}e^{(r-r_{0})|\xi|}}{|\xi|}, \quad \mbox{for } 0< | \xi|< 1. $$

If we take $C_{1}=\min(\mu_{1},\mu_{3})$, and $C_{2}=\max(\mu_{2},\mu_{4})$, then for $\xi\neq0$ inequalities (2.4) are obtained. □

By the Parseval equality, we know

$$ \bigl\Vert u(r,\cdot)\bigr\Vert ^{2} = \int_{\mathbb{R}}\bigl\vert r_{0}\xi\Phi(r,\xi)\hat{g}( \xi) \bigr\vert ^{2}\,d\xi. $$

(2.9)

Combining with (1.4), (2.9), and Lemma 2.2, we know that the Fourier transform $\hat{g}(\xi)$ of the exact data $g(z)$ must decay rapidly for $|\xi|\rightarrow\infty$ in order to ensure the convergence of (2.9). However, for the noisy data $g^{\delta}(z)$, its Fourier transform may not possess such a property, and the noisy perturbation will be multiplied by the diverging factor $|\xi\Phi(r,\xi)|$.

3 Fourier method and error estimates

Since the ill-posedness of problem (1.2) is caused by the high frequency perturbation of the noisy data, it is reasonable to stabilize the problem by eliminating high frequencies of the noisy data directly from the solution. This is the so-called Fourier method, it was put forward first by Lars Eldén et al. to deal with the inverse heat conduction problem [20]. Afterwards, this method has been successfully applied to deal with various inverse problems, e.g. the problem of a numerical pseudodifferential operator [21], the problem of numerical analytic continuation [22], the Cauchy problem for the Helmholtz equation [23], etc. However, for the Cauchy problem in a hollow cylinder, there are few efficient numerical methods, especially with a posteriori regularization parameter choice rule. In the following, we attempt to solve this problem by using Fourier method together with both a priori and a posteriori parameter choice rules.

According to [20], we eliminate all high frequencies from the solution and consider (2.2) only for $|\xi|<\xi _{\max}$, i.e., define the Fourier regularization solution of problem (1.2) as

$$ v_{\xi_{\max}}^{\delta}(r,z)=\frac{1}{\sqrt{2\pi}} \int_{\mathbb {R}}e^{iz\xi}r_{0}|\xi|\Phi(r,\xi) \hat{g}^{\delta}(\xi)\chi_{\xi _{\max}}\,d\xi, $$

(3.1)

where $\xi_{\max}>0$ is the regularization parameter to be determined, and χ is the characteristic function. In the following, we will establish error estimates between the exact solution and its regularization approximations.

3.1 A priori parameter choice rule

Theorem 3.1

Assume the conditions (1.3)-(1.5) hold. The regularized solution of problem (1.2) is defined by (3.1). If we take the regularization parameter $\xi_{\max}^{*}$ to be

$$ \xi_{\max}^{*}= \frac{1}{R-r_{0}}\ln\frac{E}{\delta}, $$

(3.2)

then the following Hölder stability holds:

$$ \bigl\Vert u(r,\cdot)-v_{\xi_{\max}^{*}}^{\delta}(r,\cdot) \bigr\Vert \leq \bigl(r_{0}+C_{1}^{-1} \bigr)C_{2}E^{\frac{r-r_{0}}{R-r_{0}}}\delta^{\frac{R-r}{R-r_{0}}},\quad r_{0}< r< R. $$

(3.3)

Proof

For the exact data $g(z)$, we define

$$ v_{\xi_{\max}}(r,z)=\frac{1}{\sqrt{2\pi}} \int_{\mathbb {R}}e^{iz\xi}r_{0}|\xi|\Phi(r,\xi) \hat{g}(\xi)\chi_{\xi_{\max }}\,d\xi. $$

Combining with (2.2) and (3.1), we have

$$\begin{aligned} \bigl\Vert u(r,\cdot)-v_{\xi_{\max}}^{\delta}(r,\cdot)\bigr\Vert \leq&\bigl\Vert u(r,\cdot )-v_{\xi_{\max}}(r,\cdot)\bigr\Vert +\bigl\Vert v_{\xi_{\max}}(r,\cdot)-v_{\xi _{\max}}^{\delta}(r,\cdot)\bigr\Vert \\ \leq& \biggl( \int_{{\vert \xi \vert >\xi_{\max}}}\bigl\vert r_{0}\xi\Phi(r,\xi)\hat {g} \bigr\vert ^{2}\,d\xi \biggr)^{\frac{1}{2}} \\ &{}+ \biggl( \int_{\vert \xi \vert \leq\xi_{\max }}\bigl\vert r_{0}\xi\Phi(r,\xi) \bigl( \hat{g}-\hat{g}^{\delta}\bigr)\bigr\vert ^{2}\,d\xi \biggr)^{\frac{1}{2}} \\ :=&I_{1}+I_{2}. \end{aligned}$$

We will estimate $I_{1}$ and $I_{2}$ separately.

For the first term $I_{1}$, due to (1.5), we have

$$\begin{aligned} I_{1} =& \biggl( \int_{|\xi|>\xi_{\max}}\bigl\vert r_{0}\xi\Phi(r,\xi)\hat {g} \bigr\vert ^{2}\,d\xi \biggr)^{\frac{1}{2}}\leq E\sup _{|\xi|>\xi_{\max }} \biggl[\frac{\Phi(r,\xi)}{\Phi(R,\xi)} \biggr] \\ \leq& \frac{C_{2}E}{C_{1}}\sup_{|\xi|>\xi_{\max}}e^{-(R-r)|\xi|}\leq \frac{C_{2}E}{C_{1}}e^{-(R-r)\xi_{\max}}. \end{aligned}$$

For the second term $I_{2}$, due to (1.3), we know that

$$\begin{aligned} I_{2} =& \biggl( \int_{|\xi|\leq\xi_{\max}}\bigl\vert r_{0}\xi\Phi(r,\xi) \bigl(\hat {g}-\hat{g}^{\delta}\bigr)\bigr\vert ^{2}\,d\xi \biggr)^{\frac{1}{2}} \\ \leq& \delta\sup_{|\xi|\leq\xi_{\max}}r_{0}|\xi|\Phi(r,\xi )\leq C_{2}r_{0}\delta\sup_{|\xi|\leq\xi_{\max}} e^{(r-r_{0})|\xi |}\leq C_{2}r_{0}\delta e^{(r-r_{0})\xi_{\max}}. \end{aligned}$$

Combining $I_{1}$ and $I_{2}$, we have

$$ \bigl\Vert u(r,\cdot)-v_{\xi_{\max}}^{\delta}(r,\cdot)\bigr\Vert \leq\frac {C_{2}E}{C_{1}}e^{-(R-r)\xi_{\max}}+C_{2}r_{0}\delta e^{(r-r_{0})\xi_{\max}}. $$

If we replace $\xi_{\max}$ by $\xi_{\max}^{*}$ defined by (3.2), the final estimate is obtained as (3.3). The proof is completed. □

3.2 A posteriori parameter choice rule

Set

$$ \rho(\xi_{\max}):=\bigl\Vert v_{\xi_{\max}}^{\delta}(r_{0},z)-g^{\delta}(z) \bigr\Vert . $$

(3.4)

From equation (3.1) of the regularized solution $v_{\xi_{\max}}^{\delta}$, it is easy to see that

$$\begin{aligned} \rho(\xi_{\max})=\bigl\Vert \hat{g}^{\delta}\chi_{\xi_{\max}}- \hat {g}^{\delta}\bigr\Vert = \biggl( \int_{|\xi|>\xi_{\max}}\bigl\vert \hat{g}^{\delta }(\xi)\bigr\vert ^{2}\,d\xi \biggr)^{\frac{1}{2}}. \end{aligned}$$

Lemma 3.1

The function $\rho(\xi_{\max})$ satisfies

• 1. $\rho(\xi_{\max})$ is a continuous and decreasing function on $(0,\infty)$,

• 2. $\lim_{\xi_{\max}\rightarrow\infty}\rho(\xi_{\max})=0$,

• 3. $\lim_{\xi_{\max}\rightarrow0}\rho(\xi_{\max})=\| g^{\delta}\|$.

According to the discrepancy principle, we will take the solution $\xi _{\max}$ of the equation

$$ \rho(\xi_{\max})=\tau\delta $$

(3.5)

to be the regularization parameter, where $\tau>1$ is a constant. In practice, we always have $\|g^{\delta}\|>\delta$, otherwise, $v_{\xi _{\max}}^{\delta}\equiv0$ would be an acceptable approximation to u. Therefore, for an appropriate constant $\tau>1$, equation (3.5) is always solvable, and if solution of (3.5) is not unique, the $\xi_{\max}$ will be understood as the minimal solution of the equation.

For the choice rule of regularization parameter, we have a range estimate for $\xi_{\max}$ given by the following lemma.

Lemma 3.2

Assume that conditions (1.3)-(1.5) hold, if $\xi_{\max}$ is taken as the solution of equation (3.5), then the following inequality holds:

$$ e^{(R-r_{0})\xi_{\max}}\leq\frac{E}{C_{1}r_{0}(\tau-1)\delta}. $$

(3.6)

Proof

It is easy to observe that

$$\begin{aligned} \|\hat{g}\chi_{\xi_{\max}}-\hat{g}\| =& \biggl( \int_{\vert \xi \vert >\xi _{\max}}\bigl\vert \hat{g}(\xi)\bigr\vert ^{2} \,d\xi \biggr)^{\frac{1}{2}} \\ =& \biggl( \int_{\vert \xi \vert >\xi_{\max}}\bigl\vert r_{0}\xi\Phi(R,\xi)\hat{g}(\xi )\bigr\vert ^{2}r_{0}^{-2}\xi^{-2} \Phi^{-2}(R,\xi)\,d\xi \biggr)^{\frac {1}{2}} \\ \leq&E\sup_{\vert \xi \vert >\xi_{\max}}\bigl[r_{0}\vert \xi \vert \Phi(R,\xi)\bigr]^{-1}\leq \frac{E}{C_{1}r_{0}}e^{-(R-r_{0})\xi_{\max}}. \end{aligned}$$

(3.7)

In view of equation (3.5) and the triangle inequality, we also have

$$\begin{aligned} \Vert \hat{g}\chi_{\xi_{\max}}-\hat{g}\Vert =&\bigl\Vert (1-\chi_{\xi_{\max }})\hat{g}\bigr\Vert =\bigl\Vert (1-\chi_{\xi_{\max}}) \bigl(\hat{g}-\hat{g}^{\delta }+\hat{g}^{\delta}\bigr)\bigr\Vert \\ \geq&\bigl\Vert (1-\chi_{\xi_{\max}})\hat{g}^{\delta}\bigr\Vert - \bigl\Vert (1-\chi_{\xi _{\max}}) \bigl(\hat{g}-\hat{g}^{\delta}\bigr) \bigr\Vert \\ \geq&(\tau-1)\delta. \end{aligned}$$

(3.8)

Combining (3.7) with (3.8), we have

$$ \frac{E}{C_{1}r_{0}}e^{-(R-r_{0})\xi_{\max}}\geq(\tau-1)\delta, $$

and therefore (3.6) holds. □

Lemma 3.3

Assume that conditions (1.3)-(1.5) hold, then we have

$$\begin{aligned}& \bigl\Vert v_{\xi_{\max}}^{\delta}(r,\cdot)-u(r,\cdot)\bigr\Vert \leq C\bigl\Vert v_{\xi_{\max }}^{\delta}(r_{0}, \cdot)-u(r_{0},\cdot)\bigr\Vert ^{\frac{R-r}{R-r_{0}}} \bigl\Vert v_{\xi_{\max}}^{\delta}(R,\cdot)-u(R,\cdot)\bigr\Vert ^{\frac {r-r_{0}}{R-r_{0}}}, \\& \quad r_{0}< r< R, \end{aligned}$$

(3.9)

where C is a constant independent on δ and E.

Proof

Let the index $\alpha=\frac{R-r}{R-r_{0}}$, it can be deduced from Lemma 2.2 that there exists a positive constant C such that

$$\bigl(r_{0}\vert \xi \vert \Phi(R,\xi)\bigr)^{\alpha} \frac{\Phi(r,\xi)}{\Phi(R,\xi )}\leq C,\quad \mbox{for } \vert \xi \vert >0. $$

Then we have the following estimate:

$$\begin{aligned}& \bigl\Vert v_{\xi_{\max}}^{\delta}(r,\cdot)-u(r,\cdot)\bigr\Vert \\& \quad = \bigl\Vert r_{0}\xi\Phi (r,\xi)\hat{g}^{\delta} \chi_{\xi_{\max}}-r_{0}\xi\Phi(r,\xi)\hat {g}\bigr\Vert \\& \quad = \biggl[ \int_{\xi\in\mathbb{R}}\bigl(r_{0}\vert \xi \vert \Phi(r,\xi) \bigr)^{2}\bigl(\hat {g}^{\delta}\chi_{\xi_{\max}}-\hat{g} \bigr)^{2}\,d\xi \biggr]^{\frac {1}{2}} \\& \quad = \biggl[ \int_{\xi\in\mathbb{R}}\bigl(\hat{g}^{\delta}\chi_{\xi _{\max}}- \hat{g}\bigr)^{2\alpha}\bigl(r_{0}\vert \xi \vert \Phi(R,\xi) \bigl(\hat{g}^{\delta }\chi_{\xi_{\max}}-\hat{g}\bigr)\bigr)^{2-2\alpha} \\& \qquad {}\times \bigl(r_{0}\vert \xi \vert \Phi(R,\xi )\bigr)^{2\alpha} \frac{\Phi^{2}(r,\xi)}{\Phi^{2}(R,\xi)}\,d\xi \biggr]^{\frac{1}{2}} \\& \quad \leq C\bigl\Vert v_{\xi_{\max}}^{\delta}(r_{0}, \cdot)-u(r_{0},\cdot)\bigr\Vert ^{\frac {R-r}{R-r_{0}}} \bigl\Vert v_{\xi_{\max}}^{\delta}(R,\cdot)-u(R,\cdot)\bigr\Vert ^{\frac{r-r_{0}}{R-r_{0}}}. \end{aligned}$$

Thus the proof is completed. □

Lemma 3.4

Assume that conditions (1.3)-(1.5) hold. If $\xi_{\max}^{*}$ is taken as the solution of equation (3.5), then we have

$$\begin{aligned}& \bigl\Vert u(r_{0},\cdot)-v_{\xi_{\max}^{*}}^{\delta}(r_{0}, \cdot)\bigr\Vert \leq(1+\tau )\delta, \end{aligned}$$

(3.10)

$$\begin{aligned}& \bigl\Vert u(R,\cdot)-v_{\xi_{\max}^{*}}^{\delta}(R,\cdot)\bigr\Vert \leq C^{*} E, \end{aligned}$$

(3.11)

where $C^{*}$ is a constant independent on δ and E.

Proof

From noise level (1.3) and the choice rule for $\xi _{\max}^{*}$, the first inequality is easy to obtain,

$$ \bigl\Vert u(r_{0},\cdot)-v_{\xi_{\max}^{*}}^{\delta}(r_{0}, \cdot)\bigr\Vert \leq\bigl\Vert g-g^{\delta}\bigr\Vert +\bigl\Vert g^{\delta}-v_{\xi_{\max}^{*}}^{\delta}(r_{0},\cdot)\bigr\Vert \leq(1+\tau)\delta. $$

For the second inequality, we have

$$\begin{aligned}& \bigl\Vert u(R,\cdot)-v_{\xi_{\max}^{*}}^{\delta}(R,\cdot)\bigr\Vert \\& \quad \leq \bigl\Vert u(R,\cdot)-v_{\xi_{\max}^{*}}(R,\cdot)\bigr\Vert +\bigl\Vert v_{\xi_{\max}^{*}}(R,\cdot)-v_{\xi_{\max}^{*}}^{\delta}(R,\cdot)\bigr\Vert \\& \quad \leq E+ \biggl[ \int_{|\xi|\leq\xi_{\max}^{*}}\bigl(r_{0}\xi\Phi(R,\xi ) \bigl(\hat{g}- \hat{g}^{\delta}\bigr)\bigr)^{2}\,d\xi \biggr]^{\frac{1}{2}} \\& \quad \leq E+\delta\sup_{|\xi|\leq\xi_{\max}^{*}}r_{0}|\xi|\Phi(R,\xi) \\& \quad \leq E+C_{2}r_{0}\delta e^{(R-r_{0})\xi_{\max}^{*}}. \end{aligned}$$

Combining with Lemma 3.2, we have

$$\begin{aligned} \bigl\Vert u(R,\cdot)-v_{\xi_{\max}^{*}}^{\delta}(R,\cdot)\bigr\Vert \leq& E+\frac {C_{2}E}{C_{1}(\tau-1)} \\ =&C^{*}E, \end{aligned}$$

where $C^{*}=1+\frac{C_{2}}{C_{1}(\tau-1)}$. The proof of this lemma is completed. □

Theorem 3.2

Assume that conditions (1.3)-(1.5) hold. Further suppose that $\delta<\|g^{\delta}\|$. Choose $\tau>1$ such that $0<\tau\delta<\|g^{\delta}\|$. If $\xi_{\max}^{*}$ is taken as the solution of equation (3.5), then the following error estimates are satisfied:

$$ \bigl\Vert v_{\xi_{\max}^{*}}^{\delta}(r,\cdot)-u(r,\cdot)\bigr\Vert \leq C \bigl(C^{*}E\bigr)^{\frac{r-r_{0}}{R-r_{0}}}\bigl[(1+\tau)\delta\bigr]^{\frac {R-r}{R-r_{0}}}, \quad r_{0}< r< R, $$

(3.12)

where C and $C^{*}$ are the same as in Lemmas 3.3 and 3.4, respectively.

The proposition of Theorem 3.2 follows immediately from Lemmas 3.3 and 3.4.

4 Numerical experiment

In this section, we present the numerical implementation of the Fourier method with a priori and a posteriori parameter choice rule, respectively. The fast Fourier transform and the inverse fast Fourier transform are used to compute the approximate solutions, and all computations are done in Matlab 7.0. In the computation, we always take $R=1$, $r_{0}=0.1$, and we consider the problem in domain $\{ 0.1< r<1, -10<z<10\}$. The numbers of grids on the $(r,z)$ domain are denoted M and K. In practical applications, the data $g(z)$ is obtained by measurement and there are inevitable errors. Thus in our experiment, we will consider the noisy data $g^{\delta}$ created by

$$ g^{\delta}=g\bigl(1+\varepsilon \operatorname{rand}\bigl( \operatorname{size}(g)\bigr)\bigr), $$

where $g=(g(z_{1}),g(z_{2}),\ldots,g(z_{K}))$, $z_{k}=-10+\frac {20(k-1)}{K-1}$, $k=1,2,\ldots,K$, and the noise level

$$ \delta=\bigl\Vert g-g^{\delta}\bigr\Vert _{L^{2}}=\sqrt {\frac{20}{K}\sum_{k=1}^{K} \bigl(g(z_{k})-g^{\delta}(z_{k})\bigr)^{2}}. $$

The function ‘$\operatorname{rand}(\cdot)$’ generates arrays of uniformly distributed random numbers.

For the a priori parameter choice rule, the regularization parameter depends on both the a priori bound and the noise level. The a priori bound is computed according to

$$ E=\sqrt{\frac{20}{K}\sum_{k=1}^{K} \bigl\vert f(z_{k})\bigr\vert ^{2}}. $$

To test the accuracy of the computed approximations, we use the relative root mean square error (RES), which is defined as

$$ \operatorname{RES}\bigl(v_{N}^{\delta}(r,\cdot)\bigr)= \frac{\sqrt{\sum_{k=1}^{K}(u(r,z_{k})-v_{N}^{\delta}(r,z_{k}))^{2}}}{ \sqrt{\sum_{k=1}^{K}u^{2}(r,z_{k})}}. $$

(4.1)

We will consider two examples where there are no exact solutions. The data function $g(z)$ is obtained by solving the following direct problem:

$$\begin{aligned} \textstyle\begin{cases} u_{rr}+\frac{1}{r}u_{r}+u_{zz}=0, & r_{0}< r< R, z\in\mathbb{R}, \\ u_{r}(r_{0},z)=0, & z\in\mathbb{R}, \\ u(R,z)=f(z), & z\in\mathbb{R}, \end{cases}\displaystyle \end{aligned}$$

(4.2)

where $f(z)$ is selected to be a function with some interesting features. The well-posed problem (4.2) could then be solved by the standard five-point difference scheme and the corresponding data function $g(z)$ can be computed.

Example 1

$f(z)=e^{-z^{2}}$, $z\in\mathbb {R}$.

Tables 1 and 2 show the effect of increasing M and K on accuracy. From these tables, we find that M and K have small influence on the results when they become large. That is to say, the degree of ill-posedness of numerical problems does not increase with the refinement of the mesh used. Thus we shall always take $M=31$, $K=109$ in the numerical experiment.

Table 1 Relative errors RES with $\pmb{r=0.2}$ , $\pmb{\varepsilon=0.001}$ , $\pmb{M=31}$ for Example 1

Table 2 Relative errors RES with $\pmb{r=0.2}$ , $\pmb{\varepsilon=0.001}$ , $\pmb{K=109}$ for Example 1

Table 3 lists the relative root mean square error (RES) between the exact solution and its approximations with different perturbations in the data. The approximations are obtained by the Fourier method with both the a priori and the a posteriori parameter choice rule. For the a priori Fourier method, the regularization parameter is selected according to Theorem 3.1. For the a posteriori Fourier method, we choose $\xi_{\max}^{*}$ as the solution of equation (3.5), where τ is some constant greater than unity, which can be taken heuristically to be 1.1. Note that the regularization parameter chosen by the a posteriori rule is only dependent on noise level δ, and computing accuracy is improved. This table shows that the smaller r, the better the computed results, which is consistent with the theoretical result in Theorems 3.1 and 3.2. The suggested method is still stable for higher noise levels on the data, and the smaller ε, the more accurate the approximations.

Table 3 Comparison of relative errors for Example 1

Figure 1 is the comparison of exact solution and its approximations at different values of the radius $r=0.2, 0.6, 0.9$. In Figure 1(a1)-(c1), the perturbation is $\varepsilon=0.005$. In Figure 1(a2)-(c2), the perturbation is $\varepsilon=0.05$.

Figure 1

Comparisons of exact solution and its approximations at $\pmb{r=0.2, 0.6, 0.9}$ for Example 1 . The results for $\varepsilon=0.005$ are (a1)-(c1) and the results for $\varepsilon=0.05$ are (a2)-(c2).

Figure 2 gives the corresponding comparison of exact solution and its approximations in terms of $u(\cdot, z)$ distributions at different constant $z=-10, 0, 2, 5$ for $\varepsilon=0.05$.

Figure 2

Comparisons of exact solution and its approximations at $\pmb{z=-10, 0, 2, 5}$ with $\pmb{\varepsilon=0.05}$ for Example 1 .

From Figures 1 and 2, and Table 3, we see that there is almost no difference for the a posteriori and a priori Fourier method with exact a priori bound E when r is relatively small. However, with the increase of r, the numerical effect of the a posteriori Fourier method is better than the a priori one. It is generally known that a priori bound E has a great influence on the accuracy of regularized solutions computed by a priori method, and a wrong a priori bound may lead to bad regularized solutions. This is just the weakness of the a priori parameter choice rule. The following example will also confirm this matter.

Example 2

$f(z)=\sin\frac{\pi z}{10}$, $z\in\mathbb{R}$.

Table 4 lists the RES between the exact solution and its approximations with different perturbations in the data for Example 2.

Table 4 Comparison of relative errors for Example 2

Figure 3 is a comparison of the exact solution and its approximations at different values of the radius $r=0.2, 0.6, 0.9$ for Example 2. In Figure 3(a1)-(c1), the perturbation is $\varepsilon=0.005$. In Figure 3(a2)-(c2), the perturbation is $\varepsilon=0.05$.

Figure 3

Comparisons of exact solution and its approximations at $\pmb{r=0.2, 0.6, 0.9}$ with $\pmb{\varepsilon=0.05}$ for Example 2 .

The explanation for Table 4 and Figure 3 is similar to that for Example 1, but it is worth noting that numerical results for the a posteriori Fourier method are more accurate at $r=0.9$. The reason for this phenomenon is mainly the that regularization parameter selected by a posteriori choice rule depends only on the noise level δ, and is not related to other factors. Table 5 gives a discussion as regards the impact of bound E on the relative error of the regularized approximation for Example 2, and we also see that a wrong constant E may lead to bad regularized solutions.

Table 5 The impact of a priori bound E on the relative errors for Example 2 at $\pmb{r=0.9}$ with $\pmb{\varepsilon=10^{-2}}$

For linear ill-posed problems defined on a ‘strip’ or ‘cylinder’ domain, the Fourier method is the most simple and a very effective regularization method. We repeated the computations of Examples 1 and 2 using the modified Tikhonov regularization method. The advantage of this method is that explicit error estimate for some specific problems could be obtained. The expression of the modified Tikhonov regularized solution and the corresponding error estimate are listed in the appendix.

Figure 4 shows the comparison between the a posteriori Fourier method and the modified Tikhonov method on $r=0.9$ with different perturbations. From this figure, it is easy to see that the Fourier method is much stable and better for larger r and ε.

Figure 4

Comparisons of exact solution and its approximations by using different methods at $\pmb{r=0.9}$ . The results for Example 1 are (a1) and (a2) with $\varepsilon=0.005$ and $\varepsilon=0.05$, respectively. The results for Example 2 are (b1) and (b2) with $\varepsilon=0.005$ and $\varepsilon=0.05$, respectively.

5 Conclusion

In this paper, we have applied the Fourier method together with a priori parameter choice rule and a posteriori parameter choice rule to solve the Cauchy problem for the Laplace equation in a hollow cylinder domain. The Hölder type error estimates between the exact solution and its approximation are obtained. As for any a priori regularization method, the choice of the regularization parameter usually depends on both the a priori bound and the noise level. In general, the a priori bound cannot be known exactly in practice, and working with a wrong a priori bound may lead to bad regularization solution. The advantage of the a posteriori method is that one does not need to know the smoothness and the a priori bound of the unknown solution. The numerical results also show that the Fourier method with a posteriori parameter choice rule is much stable than the one with a priori parameter choice rule for larger r and ε. However for the a posteriori method, some important information as regards the solution is concealed and hidden for the discrepancy principle, such that the theoretical analysis of the convergence rate is rather difficult obtain for some problems. The related theory is particularly worthy of further development.

Appendix

In order to compare the results with Fourier method, we repeat the computations of Examples 1 and 2 using the modified Tikhonov regularization method. Define the approximation as follows:

$$ v_{\alpha}^{\delta}(r,z)=\frac{1}{\sqrt{2\pi}} \int_{\mathbb {R}}\frac{e^{i\xi z}r_{0}|\xi|\Phi(r,\xi)}{1+{\alpha}^{2}[r_{0}|\xi |\Phi(R,\xi)]^{2}}\hat{g}^{\delta}(\xi)\,d\xi. $$

(A.1)

Before giving the explicit error estimate, we present first the following lemma which is crucial for the error estimate.

Lemma

[24]

Let $0< x< l$, then we have

$$ \sup_{s\geq0}\frac{e^{(l-x)s}}{1+\omega^{2}e^{2ls}}\leq \omega^{\frac{x-l}{l}},\qquad \sup_{s\geq0}\frac{e^{(2l-x)s}}{1+\omega^{2}e^{2ls}}\leq \omega^{\frac{x-2l}{l}}. $$

(A.2)

Theorem

Assume that conditions (1.3)-(1.5) hold. If the regularization parameter α is selected to be

$$ \alpha=\frac{\delta}{E}, $$

(A.3)

then the Hölder type error estimate holds,

$$ \bigl\Vert u(r,z)-v_{\alpha}^{\delta}(r,z)\bigr\Vert \leq C'E^{\frac {r-r_{0}}{R-r_{0}}}\delta^{\frac{R-r}{R-r_{0}}}, $$

(A.4)

where $C'$ is a constant independent on δ and E.

Proof

For fixed $r\in(r_{0}, R)$,

$$\begin{aligned} \bigl\Vert u(r,z)-v_{\alpha}^{\delta}(r,z)\bigr\Vert =&\bigl\Vert \hat{u}(r,\xi)-v_{\alpha }^{\delta}(r,\xi)\bigr\Vert =\biggl\Vert r_{0}|\xi|\Phi(r,\xi)\hat{g}-\frac{r_{0}|\xi|\Phi(r,\xi )}{1+\alpha^{2}[r_{0}|\xi|\Phi(R,\xi)]^{2}} \hat{g}^{\delta}\biggr\Vert \\ \leq&\biggl\Vert \frac{r_{0}|\xi|\Phi(r,\xi)({\hat{g}}^{\delta}-\hat {g})}{1+\alpha^{2}[r_{0}|\xi|\Phi(R,\xi)]^{2}}\biggr\Vert + \alpha^{2}\biggl\Vert \frac{(r_{0}|\xi|)^{3}\Phi(r,\xi)\Phi^{2}(R,\xi)\hat {g}}{1+\alpha^{2}[r_{0}|\xi|\Phi(R,\xi)]^{2}}\biggr\Vert \\ =&I_{1}+\alpha^{2}I_{2}. \end{aligned}$$

Now we estimate $I_{1}$ and $I_{2}$ separately.

$$\begin{aligned}& I_{1}=\biggl\Vert \frac{r_{0}|\xi|\Phi(r,\xi)({\hat{g}}^{\delta}-\hat {g})}{1+\alpha^{2}[r_{0}|\xi|\Phi(R,\xi)]^{2}}\biggr\Vert \leq \delta\sup \frac{r_{0}|\xi|\Phi(r,\xi)}{1+\alpha^{2}[r_{0}|\xi|\Phi (R,\xi)]^{2}} \\& \hphantom{I_{1}}\leq\delta\sup\frac{C_{2}r_{0}e^{(r-r_{0})|\xi|}}{1+\alpha ^{2}C_{1}^{2}r_{0}^{2}e^{2(R-r_{0})|\xi|}} \leq\delta C_{2}r_{0}(C_{1}r_{0} \alpha)^{-\frac{r-r_{0}}{R-r_{0}}}, \\& I_{2}=\biggl\Vert \frac{(r_{0}|\xi|)^{3}\Phi(r,\xi)\Phi^{2}(R,\xi)\hat {g}}{1+\alpha^{2}[r_{0}|\xi|\Phi(R,\xi)]^{2}}\biggr\Vert =\biggl\Vert \frac{(r_{0}|\xi|)^{2}\Phi(r,\xi)\Phi(R,\xi)[r_{0}|\xi|\Phi (R,\xi)\hat{g}]}{1+\alpha^{2}[r_{0}|\xi|\Phi(R,\xi)]^{2}}\biggr\Vert \\& \hphantom{I_{2}}\leq E\sup\frac{(r_{0}|\xi|)^{2}\Phi(r,\xi)\Phi(R,\xi)}{1+\alpha ^{2}[r_{0}|\xi|\Phi(R,\xi)]^{2}}\leq E (C_{2}r_{0})^{2} \sup\frac {e^{(R+r-2r_{0})|\xi|}}{1+(C_{1}\alpha)^{2}e^{2(R-r_{0})|\xi|}} \\& \hphantom{I_{2}}\leq E (C_{2}r_{0})^{2}(C_{1} \alpha)^{\frac{2r_{0}-R-r}{R-r_{0}}}. \end{aligned}$$

From the estimate of $I_{1}$ and $I_{2}$, we have

$$ \bigl\Vert u(r,z)-v_{\alpha}^{\delta}(r,z)\bigr\Vert \leq\delta C_{2}r_{0}(C_{1}r_{0}\alpha )^{-\frac{r-r_{0}}{R-r_{0}}}+E (C_{2}r_{0})^{2}(C_{1} \alpha)^{\frac{2r_{0}-R-r}{R-r_{0}}}. $$

If we select $\alpha=\frac{\delta}{E}$, then we have

$$\begin{aligned}& \bigl\Vert u(r,z)-v_{\alpha}^{\delta}(r,z)\bigr\Vert \leq C'E^{\frac {r-r_{0}}{R-r_{0}}}\delta^{\frac{R-r}{R-r_{0}}}, \end{aligned}$$

where $C'=C_{1}^{-\frac{r-r_{0}}{R-r_{0}}}C_{2}r_{0}^{\frac {R-r}{R-r_{0}}}+C_{1}^{\frac{2r_{0}-R-r}{R-r_{0}}}C_{2}^{2}r_{0}^{2}$. The proof is completed. □