New Trends in Applying LRM to Nonlinear Ill-Posed Equations

Abstract

Tautenhahn (2002) studied the Lavrentiev regularization method (LRM) to approximate a stable solution for the ill-posed nonlinear equation $\kappa \left(u\right)=v$, where $\kappa :D\left(\kappa \right)\subseteq X⟶X$ is a nonlinear monotone operator and X is a Hilbert space. The operator in the example used in Tautenhahn’s paper was not a monotone operator. So, the following question arises. Can we use LRM for ill-posed nonlinear equations when the involved operator is not monotone? This paper provides a sufficient condition to employ the Lavrentiev regularization technique to such equations whenever the operator involved is non-monotone. Under certain assumptions, the error analysis and adaptive parameter choice strategy for the method are discussed. Moreover, the developed theory is applied to two well-known ill-posed problems—inverse gravimetry and growth law problems.

1. Introduction

Consider the ill-posed nonlinear equation [1]

$$\kappa \left(u\right)=v,$$

(1)

where $\kappa :D\left(\kappa \right)\subseteq X⟶X,$  X is a Hilbert space. The norm and inner product on X are $\parallel ·\parallel$ and $\langle ·,·\rangle ,$ respectively. Suppose ∃$\widehat{u}\in D\left(\kappa \right)$, such that $\kappa \left(\widehat{u}\right)=v$ and $v^{\delta }\in X$, such that

$$\parallel v-v^{\delta }\parallel \le \delta .$$

The standard regularization method to approximate $\widehat{u}$ of (1) is Tikhonov regularization (see [2,3,4,5,6,7,8,9,10,11] for the latest work on regularization methods for (1)), wherein the regularised approximation $u_{\alpha }^{\delta }$ is the minimizer of the Tikhonov’s functional

$$J_{\alpha }(u)= {\parallel \kappa (u)-v^{\delta }\parallel }^2+ \alpha {\parallel u-u_0\parallel }^2,$$

where $u_0\in X$ is the initial guess and $\alpha >0$ is the regularization parameter.

It is known that if $\kappa$ is monotone [1,12,13,14,15], i.e.,

$$\langle \kappa \left(u\right)-\kappa \left(v\right),u-v\rangle \ge 0\forall u,v\in D\left(\kappa \right),$$

then one can use LRM to approximate the exact solution $\widehat{u}$. Recall that [1,13,14,15], in LRM for (1), with a unique solution $u_{\alpha }$, (which consists) of

$$\kappa \left(u\right)+\alpha (u-u_0)=v$$

(2)

is considered an approximation for $\widehat{u}$ with $u_0$ as an initial guess for $\widehat{u}.$

As (1) is ill-posed, and in most practical problems, the available data are $v^{\delta }$ (the noisy data), one has to deal with the solution $u_{\alpha }^{\delta }$ of

$$\kappa \left(u\right)+\alpha (u-u_0)=v^{\delta }.$$

(3)

This unique solution $u_{\alpha }^{\delta }$ is observed to be a good approximation for $\widehat{u}$, provided that the regularization parameter $\alpha$ is chosen appropriately. The existence and uniqueness of the solution are due to the Minty–Browder theorem [1,12].

To obtain the error bounds on the distance $\parallel u_{\alpha }^{\delta }-\widehat{u}\parallel ,$ one needs some additional smoothness assumptions on $\widehat{u}$ with respect to the operator ${\kappa }^{\prime }\left(\widehat{u}\right)$ or ${\kappa }^{\prime }\left(u_0\right).$ In the literature, various source conditions are used, including the well-known Holder-type conditions [1,13,14,16]. Our study considers one such assumption, as given in Assumption 2.

The regularization theory for linear ill-posed problems is already well developed, thus directing our attention to the more interesting nonlinear case. One can see a detailed overview in [17] for the linear case.

The theoretical analysis of LRM for (1) with a monotone operator $\kappa$ has been well studied by many authors [1,13,14,15,18,19,20,21]. However, the examples of operators used in some of the papers are not monotone. For example, in [1], Tautenhahn considered the following example to illustrate the theoretical results, assuming the operator in (1) was monotone, which it actually was not. It was proven in [22] that the operator $\kappa$ involved was not monotone.

Example 1.

Consider the problem of identifying $u\left(t\right),t\in (0,1)$ in the initial value problem

$${\displaystyle \frac{dv}{dt}}=u\left(t\right)v\left(t\right),v\left(0\right)=c,t\in (0,1),c>0,$$

(4)

when the given noisy data are $v^{\delta }\left(t\right)\in L^2(0,1).$ The problem can be reformulated as an ill-posed operator equation $\kappa \left(u\right)=v$ with

$$\left[\kappa \left(u\right)\right]\left(t\right)=c{e}^{{\int }_0^{t}u\left(\theta \right)d\theta },u\in L^2(0,1),t\in (0,1).$$

(5)

In [23], the following example was used to illustrate the theoretical results obtained for the Lavrentiev regularization method when the operator in (1) was monotone, but the operator $\kappa$ in the following example (inverse gravimetry problem [24]) was also not monotone.

Example 2.

The operator equation

$$\kappa \left(w\right)\equiv -\int {\int }_{\left(\right[0,m]\times [0,m\left]\right)}{\displaystyle \frac{1}{{[{(\theta -{\theta }^{\prime })}^2+{(\vartheta -{\vartheta }^{\prime })}^2+w^2({\theta }^{\prime },{\vartheta }^{\prime })]}^{1/2}}}d{\theta }^{\prime }d{\vartheta }^{\prime }=q(\theta ,\vartheta ),$$

(6)

where $q(\theta ,\vartheta )=\Delta g(\theta ,\vartheta )+\kappa \left(H\right),$ $\Delta g(\theta ,\vartheta )$ is the gravitational field (anomalous), H is the asymptotic horizontal plane and $\kappa :H^{\prime }([0,m]\times [0,m])\subseteq L^2([0,m]\times [0,m])\to L^2([0,m]\times [0,m])$ where $w_0(\theta ,\vartheta )=H$ is constant, is an ill-posed equation (details in [15,24]).

Note that Equations (2) and (3) have unique solutions when $\kappa$ is monotone. So, the question is, under what condition do Equations (2) and (3) have unique solutions when the operator $\kappa$ in (1) is not monotone.

The goal of this paper is to provide the conditions under which one can utilize the Lavrentiev regularization method to solve (1), when the operator $\kappa$ is not monotone.

Here, Section 2 gives the main results and Section 3 consists of the convergence analysis. The adaptive parameter choice strategy is discussed in Section 4 and the algorithm is given in Section 5, numerical examples and conclusions in Section 6 and Section 7, respectively.

2. Main Results

In the first theorem of this section, we prove that ${\kappa }^{\prime }\left(u_0\right)+\alpha I$ is invertible. We use the notation $B(u,\rho )=\{v\in X: \parallel v-u\parallel <\rho \}$ for $u\in X$ and $\rho >0$, and $\overline{B}(u,\rho )$ to denote the closure of $B(u,\rho )$.

Theorem 1.

Suppose that for every ${ϵ}_h>0,$ there exists a positive self-adjoint operator $A_h$, such that

$$\parallel {\kappa }^{\prime }\left(u_0\right)-A_h\parallel \le {ϵ}_h$$

for some $u_0\in B(\widehat{u},r_0^{\prime })$ and for some $r_0^{\prime }>0.$ Then, ${({\kappa }^{\prime }\left(u_0\right)+\alpha I)}^{-1}$ exists for fixed $\alpha \in (\max \{2{ϵ}_h,\delta \},\overline{\alpha })$ for some $\overline{\alpha }>\max \{2{ϵ}_h,\delta \}.$ Furthermore,

$$\parallel \alpha {({\kappa }^{\prime }\left(u_0\right)+\alpha I)}^{-1}\parallel \le 2$$

(7)

and

$$\parallel {({\kappa }^{\prime }\left(u_0\right)+\alpha I)}^{-1}{\kappa }^{\prime }\left(u_0\right)\parallel \le 3.$$

(8)

Proof. 

As $A_h$ is positive self adjoint operator, $A_h+\alpha I$ is invertible for any fixed $\alpha >0.$

Suppose $\alpha \in (\max \{2{ϵ}_h,\delta \},\overline{\alpha })$ for some $\overline{\alpha }>\max \{2{ϵ}_h,\delta \}.$ Then,

$$\begin{array}{ccc}& & \parallel {(A_h+\alpha I)}^{-1}[({\kappa }^{\prime }\left(u_0\right)+\alpha I)-(A_h+\alpha I)]\parallel \hfill \\ & =& \parallel {(A_h+\alpha I)}^{-1}[{\kappa }^{\prime }\left(u_0\right)-A_h]\parallel \hfill \\ & \le & {\displaystyle \frac{{ϵ}_h}{\alpha }}\le {\displaystyle \frac{{ϵ}_h}{2{ϵ}_h}}={\displaystyle \frac{1}{2}}<1.\hfill \end{array}$$

Therefore, ${\kappa }^{\prime }\left(u_0\right)+\alpha I$ is invertible by the Banach lemma on invertible operators [25,26] and

$$\parallel {({\kappa }^{\prime }\left(u_0\right)+\alpha I)}^{-1}(A_h+\alpha I)\parallel \le {\displaystyle \frac{1}{1-\frac{1}{2}}}=2$$

or

$$\parallel {({\kappa }^{\prime }\left(u_0\right)+\alpha I)}^{-1}\parallel \le 2\parallel {(A_h+\alpha I)}^{-1}\parallel ={\displaystyle \frac{2}{\alpha }}.$$

Thus,

$$\parallel \alpha {({\kappa }^{\prime }\left(u_0\right)+\alpha I)}^{-1}\parallel \le {\displaystyle \frac{2\alpha }{\alpha }}\le 2$$

and

$$\begin{array}{ccc}\hfill \parallel {({\kappa }^{\prime }\left(u_0\right)+\alpha I)}^{-1}{\kappa }^{\prime }\left(u_0\right)\parallel & =& \parallel {\left({\kappa }^{\prime }\right(u_0)+\alpha I)}^{-1}({\kappa }^{\prime }\left(u_0\right)+\alpha I-\alpha I\parallel \hfill \\ & \le & \parallel {\left({\kappa }^{\prime }\right(u_0)+\alpha I)}^{-1}({\kappa }^{\prime }\left(u_0\right)+\alpha I)\parallel \hfill \\ & & +\parallel \alpha {({\kappa }^{\prime }\left(u_0\right)+\alpha I)}^{-1}\parallel \hfill \\ & \le & 3.\hfill \end{array}$$

□

Remark 1.

Note that, as ${\kappa }^{\prime }\left(u_0\right)+\alpha I$ is invertible for all $\alpha >0$ (as ${\kappa }^{\prime }(·)$ is positive type or sectorial [12,27]) in Example (1), one can apply the theory developed in this paper to the Example (1). Observe that if ${\kappa }^{\prime }(·)$ is a positive type, then the condition $\alpha \in (\max \{2{ϵ}_h,\delta \},\overline{\alpha })$ is not required.

In what follows, we assume the following:

Assumption 1

([1,13,16]). There exists an element $\phi (u,u_0,v)\in X$ and a constant $k_0>0$, such that ∀$u\in D\left(\kappa \right)$ and $v\in X,$

$$\begin{array}{ccc}\hfill [{\kappa }^{\prime }\left(u\right)-{\kappa }^{\prime }\left(u_0\right)]v& =& {\kappa }^{\prime }\left(u_0\right)\phi (u,u_0,v),\hfill \\ \hfill \parallel \phi (u,u_0,v)\parallel & \le & k_0\parallel v\parallel \parallel u-u_0\parallel .\hfill \end{array}$$

3. Approximating Sequence

For each $n=1,2,\dots$, define the sequence $\left(u_{n,\alpha }^{\delta }\right)$ as

$$\begin{array}{ccc}\hfill u_{0,\alpha }^{\delta }& :=& u_0,\hfill \\ \hfill u_{n+1,\alpha }^{\delta }& =& u_{n,\alpha }^{\delta }-{(A_h+\alpha I)}^{-1}[\kappa \left(u_{n,\alpha }^{\delta }\right)+\alpha (u_{n,\alpha }^{\delta }-u_0)-v^{\delta }].\hfill \end{array}$$

(9)

We require some parameters to prove $\left(u_{n,\alpha }^{\delta }\right)$ converges. Let $k_0\le {\displaystyle \frac{1}{24}},$

$$r_0^{\prime }={\displaystyle \frac{\sqrt{9+12k_0\left(\frac{1}{24k_0}-1\right)}-3}{3k_0}},$$

$$\gamma ={\displaystyle \frac{3}{4}}{k}_0{r}_0^{\prime 2}+{\displaystyle \frac{3}{2}}{r}_0^{\prime }+1,$$

$$r={\displaystyle \frac{\frac{1}{2}+\sqrt{\frac{1}{4}-6k_0\gamma }}{3k_0}}$$

and

$$\sigma ={\displaystyle \frac{1+3k_{0}r}{2}}.$$

Remark 2.

For the above $r_0^{\prime },$ we have $\gamma \le {\displaystyle \frac{1}{24k_0}}$ and $\sigma <1$ and $\gamma <{\displaystyle \frac{\gamma }{1-\sigma }}\le r.$

Theorem 2.

Let Assumption 1 be satisfied. Then, sequence $\left(u_{n,\alpha }^{\delta }\right)$ is well defined and $\left(u_{n,\alpha }^{\delta }\right)\in B(u_0,r),$ $\forall n\ge 0.$ Also, the sequence $\left(u_{n,\alpha }^{\delta }\right)$ is Cauchy in $B(u_0,r)$ and thus convergent to some $u_{\alpha }^{\delta }\in \overline{B(u_0,r)}$, satisfying $\kappa \left(u_{\alpha }^{\delta }\right)+\alpha (u_{\alpha }^{\delta }-u_0)=v^{\delta }.$

Further,

$$\begin{array}{c}\hfill \parallel u_{n,\alpha }^{\delta }-u_{\alpha }^{\delta }\parallel \le {\displaystyle \frac{{\sigma }^n}{1-\sigma }}\gamma .\end{array}$$

(10)

Proof. 

We employ mathematical induction to prove $u_{n,\alpha }^{\delta }\in B(u_0,r),$ $\forall n\ge 0.$

From Assumption 1 and as $\kappa \left(\widehat{u}\right)=v,$ we have

$$\begin{array}{ccc}\hfill \parallel u_{1,\alpha }^{\delta }-u_0\parallel & =& \parallel {(A_h+\alpha I)}^{-1}[\kappa \left(u_0\right)-v^{\delta }]\parallel \hfill \\ & =& \parallel {(A_h+\alpha I)}^{-1}[\kappa \left(u_0\right)-\kappa \left(\widehat{u}\right)+v-v^{\delta }]\parallel \hfill \\ & \le & \parallel {(A_h+\alpha I)}^{-1}{\int }^1\left({\kappa }^{\prime }(\widehat{u}+t(u_0-\widehat{u})-{\kappa }^{\prime }\left(u_0\right))dt(u_0-\widehat{u})\right)\parallel \hfill \\ & & +\parallel {(A_h+\alpha I)}^{-1}[{\kappa }^{\prime }\left(u_0\right)-A_h+A_h](u_0-\widehat{u})\parallel \hfill \\ & & +\parallel {(A_h+\alpha I)}^{-1}(v-v^{\delta })\parallel \hfill \\ & \le & \parallel {(A_h+\alpha I)}^{-1}{\kappa }^{\prime }\left(u_0\right)\parallel {\int }_0^1\phi (\widehat{u}+t(u_0-\widehat{u},u_0,u_0-\widehat{u})dt\hfill \\ & & +\parallel {(A_h+\alpha I)}^{-1}[{\kappa }^{\prime }\left(u_0\right)-A_h+A_h](u_0-\widehat{u})\parallel \hfill \\ & & +\parallel {(A_h+\alpha I)}^{-1}(v-v^{\delta })\parallel \hfill \end{array}$$

$$\begin{array}{ccc}& \le & \parallel {(A_h+\alpha I)}^{-1}[{\kappa }^{\prime }\left(u_0\right)-A_h+A_h]\parallel {\int }_0^1{k}_0|1-t|\parallel u_0-\widehat{u}{\parallel }^{2}dt\hfill \\ & & +\parallel {(A_h+\alpha I)}^{-1}[{\kappa }^{\prime }\left(u_0\right)-A_h+A_h](u_0-\widehat{u})\parallel \hfill \\ & & +\parallel {(A_h+\alpha I)}^{-1}(v-v^{\delta })\parallel \hfill \\ & \le & \left({\displaystyle \frac{{ϵ}_h}{\alpha }}+1\right){\displaystyle \frac{k_0}{2}}{r}_0^{\prime 2}+\left({\displaystyle \frac{{ϵ}_h}{\alpha }}+1\right)r_0^{\prime }+{\displaystyle \frac{\delta }{\alpha }}\hfill \\ & \le & {\displaystyle \frac{3}{4}}{k}_0{r}_0^{\prime 2}+{\displaystyle \frac{3}{2}}{r}_0^{\prime }+1\hfill \\ & =& \gamma \le r.\hfill \end{array}$$

Next, we assume that $u_{k,\alpha }^{\delta }\in B(u_0,r),$ for some k. Then,

$$\begin{array}{ccc}\hfill \parallel u_{k+1,\alpha }^{\delta }-u_0\parallel & \le & \parallel u_{k+1,\alpha }^{\delta }-u_{k,\alpha }^{\delta }\parallel + \parallel u_{k,\alpha }^{\delta }-u_{k-1,\alpha }^{\delta }\parallel +\dots \hfill \\ & & +\parallel u_{1,\alpha }^{\delta }-u_0\parallel \hfill \\ & \le & ({\sigma }^k+{\sigma }^{k-1}+\dots +1)\gamma \le {\displaystyle \frac{\gamma }{1-\sigma }}\le r.\hfill \end{array}$$

Therefore, $u_{k+1,\alpha }^{\delta }\in B(u_0,r)$ and induction yields our assertion.

Suppose $u_{n+1,\alpha }^{\delta },u_{n,\alpha }^{\delta }\in B(u_0,r),n>0.$ Then, by (9),

$$\begin{array}{ccc}\hfill u_{n+1,\alpha }^{\delta }-u_{n,\alpha }^{\delta }& =& u_{n,\alpha }^{\delta }-u_{n-1,\alpha }^{\delta }-{(A_h+\alpha I)}^{-1}[\kappa \left(u_{n,\alpha }^{\delta }\right)-\kappa \left(u_{n-1,\alpha }^{\delta }\right)\hfill \\ & & +\alpha (u_{n,\alpha }^{\delta }-u_{n-1,\alpha }^{\delta })]\hfill \\ & =& {(A_h+\alpha I)}^{-1}((A_h+\alpha I)(u_{n,\alpha }^{\delta }-u_{n-1,\alpha }^{\delta })\hfill \\ & & -(\kappa \left(u_{n,\alpha }^{\delta }\right)-\kappa \left(u_{n-1,\alpha }^{\delta }\right)+\alpha (u_{n,\alpha }^{\delta }-u_{n-1,\alpha }^{\delta })))\hfill \\ & =& {(A_h+\alpha I)}^{-1}(A_h-{\int }_0^1{\kappa }^{\prime }(u_{n-1,\alpha }^{\delta }+t(u_{n,\alpha }^{\delta }-u_{n-1,\alpha }^{\delta }))dt)\hfill \\ & & \times (u_{n,\alpha }^{\delta }-u_{n-1,\alpha }^{\delta })\hfill \\ & =& {(A_h+\alpha I)}^{-1}(A_h-{\kappa }^{\prime }\left(u_0\right)\hfill \\ & & +{\int }_0^1[{\kappa }^{\prime }\left(u_0\right)-{\kappa }^{\prime }(u_{n-1,\alpha }^{\delta }+t(u_{n,\alpha }^{\delta }-u_{n-1,\alpha }^{\delta }))]dt\hfill \\ & & \times (u_{n,\alpha }^{\delta }-u_{n-1,\alpha }^{\delta }).\hfill \end{array}$$

From Assumption 1, we obtain

$$\begin{array}{ccc}\hfill \parallel u_{n+1,\alpha }^{\delta }-u_{n,\alpha }^{\delta }\parallel & \le & {\displaystyle \frac{{ϵ}_h}{\alpha }}\parallel u_{n,\alpha }^{\delta }-u_{n-1,\alpha }^{\delta }\parallel + \parallel {(A_h+\alpha I)}^{-1}{\kappa }^{\prime }\left(u_0\right)\hfill \\ & & {\int }_0^1\phi (u_{n-1,\alpha }^{\delta }+t(u_{n,\alpha }^{\delta }-u_{n-1,\alpha }^{\delta }),u_0,u_{n,\alpha }^{\delta }-u_{n-1,\alpha }^{\delta })dt\parallel \hfill \\ & \le & {\displaystyle \frac{{ϵ}_h}{\alpha }}\parallel u_{n,\alpha }^{\delta }-u_{n-1,\alpha }^{\delta }\parallel + \parallel {(A_h+\alpha I)}^{-1}[{\kappa }^{\prime }\left(u_0\right)-A_h+A_h]\hfill \\ & & {\int }_0^1\phi (u_{n-1,\alpha }^{\delta }+t(u_{n,\alpha }^{\delta }-u_{n-1,\alpha }^{\delta }),u_0,u_{n,\alpha }^{\delta }-u_{n-1,\alpha }^{\delta })dt\parallel \hfill \\ & \le & {\displaystyle \frac{{ϵ}_h}{\alpha }}\parallel u_{n,\alpha }^{\delta }-u_{n-1,\alpha }^{\delta }\parallel +({\displaystyle \frac{{ϵ}_h}{\alpha }}+1)k_{0}r\parallel u_{n,\alpha }^{\delta }-u_{n-1,\alpha }^{\delta }\parallel \hfill \\ & & (\mathrm{b}\mathrm{e}\mathrm{c}\mathrm{a}\mathrm{u}\mathrm{s}\mathrm{e}{u}_{n-1,\alpha }^{\delta }+t(u_{n,\alpha }^{\delta }-u_{n-1,\alpha }^{\delta })\in B(u_0,r))\hfill \\ & \le & {\displaystyle \frac{1+3k_{0}r}{2}}\parallel u_{n,\alpha }^{\delta }-u_{n-1,\alpha }^{\delta }\parallel \hfill \\ \hfill & =& \sigma \parallel u_{n,\alpha }^{\delta }-u_{n-1,\alpha }^{\delta }\parallel .\hfill \end{array}$$

(11)

Let $n,m\in \mathbb{N}.$ Consider,

$$\begin{array}{ccc}\hfill \parallel u_{n+m,\alpha }^{\delta }-u_{n,\alpha }^{\delta }\parallel & \le & \parallel u_{n+m,\alpha }^{\delta }-u_{n+m-1,\alpha }^{\delta }\parallel + \parallel u_{n+m-1,\alpha }^{\delta }-u_{n+m-2,\alpha }^{\delta }\parallel +\dots \hfill \\ & & +\parallel u_{n+1,\alpha }^{\delta }-u_{n,\alpha }^{\delta }\parallel \hfill \\ & \le & ({\sigma }^{n+m-1}+{\sigma }^{n+m-2}+\dots +{\sigma }^n)\gamma \hfill \\ & \le & {\displaystyle \frac{{\sigma }^n}{1-\sigma }}\gamma .\hfill \end{array}$$

(12)

Thus, $\left(u_{n,\alpha }^{\delta }\right)$ is Cauchy in $B(u_0,r).$ Therefore $\left(u_{n,\alpha }^{\delta }\right)$ converges to $u_{\alpha }^{\delta }\in \overline{B(u_0,r)}.$ As $n\to \infty$ in (9), we obtain $\kappa \left(u_{\alpha }^{\delta }\right)+\alpha (u_{\alpha }^{\delta }-u_0)=v^{\delta }.$

Further, taking $m\to \infty$ in (12), we obtain

$$\begin{array}{c}\hfill \parallel u_{n,\alpha }^{\delta }-u_{\alpha }^{\delta }\parallel \le {\displaystyle \frac{{\sigma }^n}{1-\sigma }}\gamma .\end{array}$$

(13)

□

In the following, an additional assumption is given.

Assumption 2.

∃ is a constant $\rho >0$ and an element $\xi \in D\left(\kappa \right)$ with $\parallel \xi \parallel \le \rho$, such that $u_0-\widehat{u}={\left({\kappa }^{\prime }{\left(u_0\right)}^{*}{\kappa }^{\prime }\left(u_0\right)\right)}^{\omega }\xi ,$ $0<\omega \le 1.$

We define some parameters to be used in the proof of the results that follow.

Hereafter, we assume

$$r_0={\displaystyle \frac{3-\sqrt{8+6k_0}}{3k_0}},k_0<{\displaystyle \frac{1}{6}},$$

$$R_1={\displaystyle \frac{(1-3k_0{r}_0)-\sqrt{1-18k_0{r}_0+9k_0^2{r}_0^2}}{3k_0}}$$

and

$$R_2={\displaystyle \frac{(1-3k_0{r}_0)-\sqrt{1-18k_0{r}_0+9k_0^2{r}_0^2-6k_0}}{3k_0}}.$$

Then, one can see that $R_1<R_2.$ Let $R=\max \{R_2,r+r_0\}.$

Remark 3.

When $\delta =0,$ $\left(u_{n,\alpha }^{\delta }\right)$ converges to $u_{\alpha }$, which satisfies $\kappa \left(u_{\alpha }\right)+\alpha (u_{\alpha }-u_0)=v.$ Further, one can observe that $u_{n,\alpha }^{\delta }\in B(\widehat{u},R),$ for all $n=0,1,2,\dots .$

Theorem 3.

Let $\overline{\alpha }>0$ be a fixed number, such that $\overline{\alpha }>\max \{2{ϵ}_h,\delta +{ϵ}_h\}.$ Let $\alpha \in (\max \{2{ϵ}_h,\delta +{ϵ}_h\},\overline{\alpha })$ and $u_0$, such that

$$\parallel u_0-\widehat{u}\parallel =r_0.$$

Then, $u_{\alpha },u_{\alpha }^{\delta }\in B(\widehat{u},R).$

Proof. 

As

$$\kappa \left(u_{\alpha }\right)+\alpha (u_{\alpha }-u_0)=v,$$

(14)

we have $\kappa \left(u_{\alpha }\right)-\kappa \left(\widehat{u}\right)+\alpha (u_{\alpha }-u_0)=0$ or

$${\int }_0^1{\kappa }^{\prime }(\widehat{u}+t(u_{\alpha }-\widehat{u}))dt(u_{\alpha }-\widehat{u})+\alpha (u_{\alpha }-u_0)=0.$$

That is,

$$\begin{array}{ccc}\hfill ({\kappa }^{\prime }\left(u_0\right)+\alpha I)(u_{\alpha }-\widehat{u})& =& {\kappa }^{\prime }\left(u_0\right)(u_{\alpha }-\widehat{u})\hfill \\ & & -{\int }_0^1{\kappa }^{\prime }(\widehat{u}+t(u_{\alpha }-\widehat{u}))dt(u_{\alpha }-\widehat{u})+\alpha (u_0-\widehat{u})\hfill \end{array}$$

and hence

$$\begin{array}{ccc}\hfill u_{\alpha }-\widehat{u}& =& {({\kappa }^{\prime }\left(u_0\right)+\alpha I)}^{-1}{\int }_0^1[{\kappa }^{\prime }\left(u_0\right)-{\kappa }^{\prime }(\widehat{u}+t(u_{\alpha }-\widehat{u}))]dt(u_{\alpha }-\widehat{u})\hfill \\ \hfill & & +{({\kappa }^{\prime }\left(u_0\right)+\alpha I)}^{-1}\alpha (u_0-\widehat{u}).\hfill \end{array}$$

(15)

Thus, from Assumption 1, we obtain

$$\begin{array}{ccc}\hfill \parallel u_{\alpha }-\widehat{u}\parallel & \le & 3k_0\parallel u_{\alpha }-\widehat{u}\parallel \left[{\displaystyle \frac{\parallel u_{\alpha }-\widehat{u}\parallel }{2}}+\parallel \widehat{u}-u_0\parallel \right]+2\parallel u_0-\widehat{u}\parallel \hfill \\ & \le & {\displaystyle \frac{3k_0}{2}}{\parallel u_{\alpha }-\widehat{u}\parallel }^2+ 3k_0{r}_0\parallel u_{\alpha }-\widehat{u}\parallel +2r_0.\hfill \end{array}$$

This implies

$$\parallel u_{\alpha }-\widehat{u}\parallel \le R_1<R,$$

so $u_{\alpha }\in B(\widehat{u},R).$

Next, note that $\kappa \left(u_{\alpha }^{\delta }\right)+\alpha (u_{\alpha }^{\delta }-u_0)=v^{\delta }$ or

$$\kappa \left(u_{\alpha }^{\delta }\right)-\kappa \left(\widehat{u}\right)+\alpha (u_{\alpha }^{\delta }-u_0)=v^{\delta }-v.$$

This implies

$${\int }_0^1{\kappa }^{\prime }(\widehat{u}+t(u_{\alpha }^{\delta }-\widehat{u}))dt(u_{\alpha }^{\delta }-\widehat{u})+\alpha (u_{\alpha }^{\delta }-\widehat{u})=v^{\delta }-v+\alpha (u_0-\widehat{u})$$

so,

$$\begin{array}{ccc}\hfill ({\kappa }^{\prime }\left(u_0\right)+\alpha I)(u_{\alpha }^{\delta }-\widehat{u})& =& {\kappa }^{\prime }\left(u_0\right)(u_{\alpha }^{\delta }-\widehat{u})-{\int }_0^1{\kappa }^{\prime }(\widehat{u}+t(u_{\alpha }^{\delta }-\widehat{u}))dt(u_{\alpha }^{\delta }-\widehat{u})\hfill \\ & & +\alpha (u_0-\widehat{u})+(v^{\delta }-v)\hfill \end{array}$$

hence

$$\begin{array}{ccc}\hfill u_{\alpha }^{\delta }-\widehat{u}& =& {({\kappa }^{\prime }\left(u_0\right)+\alpha I)}^{-1}{\int }_0^1[{\kappa }^{\prime }\left(u_0\right)-{\kappa }^{\prime }(\widehat{u}+t(u_{\alpha }^{\delta }-\widehat{u}))]dt(u_{\alpha }^{\delta }-\widehat{u})\hfill \\ & & +{({\kappa }^{\prime }\left(u_0\right)+\alpha I)}^{-1}\alpha (u_0-\widehat{u})+{({\kappa }^{\prime }\left(u_0\right)+\alpha I)}^{-1}(v^{\delta }-v).\hfill \end{array}$$

Thus, from Assumption 1, we obtain

$$\begin{array}{ccc}\hfill \parallel u_{\alpha }^{\delta }-\widehat{u}\parallel & \le & 3k_0\parallel u_{\alpha }^{\delta }-\widehat{u}\parallel \left[{\displaystyle \frac{\parallel u_{\alpha }^{\delta }-\widehat{u}\parallel }{2}}+\parallel \widehat{u}-u_0\parallel \right]+2\parallel u_0-\widehat{u}\parallel +{\displaystyle \frac{\delta }{\alpha -{ϵ}_h}}\hfill \\ & \le & {\displaystyle \frac{3k_0}{2}}{\parallel u_{\alpha }^{\delta }-\widehat{u}\parallel }^2+ 3k_0{r}_0\parallel u_{\alpha }^{\delta }-\widehat{u}\parallel +2r_0+1\hfill \\ & & (\mathrm{b}\mathrm{e}\mathrm{c}\mathrm{a}\mathrm{u}\mathrm{s}\mathrm{e}{\displaystyle \frac{\delta }{\alpha -{ϵ}_h}}<1).\hfill \end{array}$$

This implies

$$\parallel u_{\alpha }^{\delta }-\widehat{u}\parallel \le R_2<R,$$

so $u_{\alpha }^{\delta }\in B(\widehat{u},R).$ □

The next theorem provides an estimate for $\parallel \widehat{u}-u_{\alpha }\parallel$ and $\parallel \widehat{u}-u_{\alpha }^{\delta }\parallel .$ The following Lemma is considered in its proof.

Lemma 1

([28]). Suppose that $\mathcal{A}:X\to X$ is a linear bounded operator with $\mathcal{A}={\mathcal{A}}^{*}\ge 0.$ If $p>0$ and $a>0$, then, for any linear bounded operator ${\mathcal{A}}_h:X\to X$ with ${\mathcal{A}}_h={\mathcal{A}}_h^{*}\ge 0,$ $\parallel {\mathcal{A}}_h\parallel \le a$, we have

$$\begin{array}{c}\hfill \parallel {\mathcal{A}}^p-{\mathcal{A}}_h^p\parallel \le a_p{\parallel \mathcal{A}-{\mathcal{A}}_h\parallel }^{\min \{p,1\}}.\end{array}$$

(16)

Here, $a_p={\displaystyle \frac{4}{\pi }},$ if $p\le 1$ and $p\mapsto a_p$ is bounded in $(0,p_0]$ for any $p_0>0.$

Theorem 4.

Let $\overline{\alpha } >0$ be a fixed number, such that $\overline{\alpha }>\max \{2{ϵ}_h,\delta +{ϵ}_h\}.$ Let $\alpha \in (\max \{2{ϵ}_h,\delta +{ϵ}_h\},\overline{\alpha })$ and $u_0$, such that

$$\parallel u_0-\widehat{u}\parallel =r_0.$$

Let $u_{\alpha }^{\delta }$ and $u_{\alpha }$ be the same as in Theorem 2 and Remark 3, respectively. Suppose Assumption 1 and Assumption 2 hold and $6k_{0}R<1.$ Then,

(a) 

$$\begin{array}{ccc}\hfill \parallel \widehat{u}-u_{\alpha }\parallel & \le & c_1{\alpha }^{\omega },\hfill \end{array}$$

(17)

where $c_1={\displaystyle \frac{2\rho }{1-3k_{0}R}}\left[{\displaystyle \frac{4}{\pi }}(\parallel {\kappa }^{\prime }\left(u_0\right)\parallel + \parallel A_h{\parallel )}^{\omega }+{\parallel A_h\parallel }^{\omega }\right].$

(b) 

$$\parallel u_{\alpha }^{\delta }-u_{\alpha }\parallel \le c_2{\displaystyle \frac{\delta }{\alpha }},$$

(18)

where $c_2={\displaystyle \frac{2}{1-6k_{0}R}}.$

(c) 

$\parallel u_{\alpha }^{\delta }-\widehat{u}\parallel \le \max \{c_1,c_2\}({\alpha }^{\omega }+{\displaystyle \frac{\delta }{\alpha }}).$

In particular, for $\alpha ={\delta }^{{\displaystyle \frac{1}{\omega +1}}},$ we have

$$\begin{array}{c}\hfill \parallel u_{\alpha }^{\delta }-\widehat{u}\parallel =O({\delta }^{{\displaystyle \frac{\omega }{\omega +1}}}).\end{array}$$

(19)

Proof. 

By (15),

$$\begin{array}{ccc}\hfill u_{\alpha }-\widehat{u}& =& {({\kappa }^{\prime }\left(u_0\right)+\alpha I)}^{-1}{\int }_0^1[{\kappa }^{\prime }\left(u_0\right)-{\kappa }^{\prime }(\widehat{u}+t(u_{\alpha }-\widehat{u}))]dt(u_{\alpha }-\widehat{u})\hfill \\ & & +{({\kappa }^{\prime }\left(u_0\right)+\alpha I)}^{-1}\alpha (u_0-\widehat{u})\hfill \end{array}$$

hence, from Assumption 1, we have

$$\begin{array}{ccc}\hfill \parallel u_{\alpha }-\widehat{u}\parallel & \le & \parallel \alpha {({\kappa }^{\prime }\left(u_0\right)+\alpha I)}^{-1}(u_0-\widehat{u})\parallel +3k_{0}R\parallel u_{\alpha }-\widehat{u}\parallel \hfill \end{array}$$

or

$$\begin{array}{ccc}\hfill (1-3k_{0}R)\parallel u_{\alpha }-\widehat{u}\parallel & \le & \parallel \alpha {({\kappa }^{\prime }\left(u_0\right)+\alpha I)}^{-1}(u_0-\widehat{u})\parallel .\hfill \end{array}$$

(20)

Further, note that

$$\begin{array}{ccc}\hfill \parallel \alpha {({\kappa }^{\prime }\left(u_0\right)+\alpha I)}^{-1}(u_0-\widehat{u})\parallel & \le & \parallel \alpha [{({\kappa }^{\prime }\left(u_0\right)+\alpha I)}^{-1}-{(A_h+\alpha I)}^{-1}](u_0-\widehat{u})\parallel \hfill \\ \hfill & & +\parallel \alpha {(A_h+\alpha I)}^{-1}(u_0-\widehat{u})\parallel \hfill \\ \hfill & =& \parallel \alpha {({\kappa }^{\prime }\left(u_0\right)+\alpha I)}^{-1}(A_h-{\kappa }^{\prime }\left(u_0\right))\hfill \\ \hfill & & \times {(A_h+\alpha I)}^{-1}(u_0-\widehat{u})\parallel \hfill \\ \hfill & & +\parallel \alpha {(A_h+\alpha I)}^{-1}(u_0-\widehat{u})\parallel \hfill \\ \hfill & \le & \left({\displaystyle \frac{{ϵ}_h}{\alpha -{ϵ}_h}}+1\right)\parallel \alpha {(A_h+\alpha I)}^{-1}(u_0-\widehat{u})\parallel \hfill \\ \hfill & \le & 2\parallel \alpha {(A_h+\alpha I)}^{-1}(u_0-\widehat{u})\parallel .\hfill \end{array}$$

(21)

Now, from Assumption 2 and Lemma 1, we have

$$\begin{array}{ccc}\hfill \parallel \alpha {(A_h+\alpha I)}^{-1}(u_0-\widehat{u})\parallel & =& \parallel \alpha {(A_h+\alpha I)}^{-1}{\left({\kappa }^{\prime }{\left(u_0\right)}^{*}{\kappa }^{\prime }\left(u_0\right)\right)}^{\omega }\xi \parallel \hfill \\ & \le & \parallel \alpha {(A_h+\alpha I)}^{-1}[{\left({\kappa }^{\prime }{\left(u_0\right)}^{*}{\kappa }^{\prime }\left(u_0\right)\right)}^{\omega }-{\left(A_h^{*}{A}_h\right)}^{\omega }]\xi \parallel \hfill \\ & & +\parallel \alpha {(A_h+\alpha I)}^{-1}{\left(A_h^{*}{A}_h\right)}^{\omega }\xi \parallel \hfill \\ & \le & {\displaystyle \frac{4\rho }{\pi }}{\parallel {\kappa }^{\prime }{\left(u_0\right)}^{*}{\kappa }^{\prime }\left(u_0\right)-A_h^{*}{A}_h\parallel }^{\omega }\parallel \xi \parallel \hfill \\ & & +\parallel \alpha {(A_h+\alpha I)}^{-1}{A}_h^{\omega }{A}_h^{\omega }\xi \parallel \hfill \\ & =& {\displaystyle \frac{4\rho }{\pi }}[\parallel {\kappa }^{\prime }{\left(u_0\right)}^{*}({\kappa }^{\prime }\left(u_0\right)-A_h)+({\kappa }^{\prime }{\left(u_0\right)}^{*}-A_h^{*})A_h{\parallel ]}^{\omega }\parallel \xi \parallel \hfill \\ & & +\parallel \alpha {(A_h+\alpha I)}^{-1}{A}_h^{\omega }{A}_h^{\omega }\xi \parallel \hfill \\ & \le & \rho \left[\right(\parallel {\kappa }^{\prime }\left(u_0\right)\parallel + \parallel A_h{\parallel )}^{\omega }{\displaystyle \frac{4{ϵ}_h^{\omega }}{\pi }} + \parallel A_h{\parallel }^{\omega }{\alpha }^{\omega }].\hfill \end{array}$$

(22)

Thus, from (20), (21), (22) and the fact that ${ϵ}_h<\alpha ,$ we have

$$\parallel u_{\alpha }-\widehat{u}\parallel \le {\displaystyle \frac{2\rho }{1-3k_{0}R}}\left[{\displaystyle \frac{4}{\pi }}(\parallel {\kappa }^{\prime }\left(u_0\right)\parallel + \parallel A_h{\parallel )}^{\omega }+{\parallel A_h\parallel }^{\omega }\right]{\alpha }^{\omega }.$$

This completes the proof of (17).

Next, as

$$\begin{array}{c}\hfill \kappa \left(u_{\alpha }^{\delta }\right)+\alpha (u_{\alpha }^{\delta }-u_0)=v^{\delta },\end{array}$$

(23)

by (14) and (23), we have

$$\kappa \left(u_{\alpha }^{\delta }\right)-\kappa \left(u_{\alpha }\right)+\alpha (u_{\alpha }^{\delta }-u_{\alpha })=v^{\delta }-v.$$

So, from the mean value theorem

$$\begin{array}{ccc}\hfill \left[{\int }_0^1{\kappa }^{\prime }(u_{\alpha }+t(u_{\alpha }^{\delta }-u_{\alpha }))dt+\alpha I\right](u_{\alpha }^{\delta }-u_{\alpha })& =& v^{\delta }-v\hfill \end{array}$$

or

$$\begin{array}{ccc}\hfill ({\kappa }^{\prime }\left(u_0\right)+\alpha I)(u_{\alpha }^{\delta }-u_{\alpha })& =& {\int }_0^1[{\kappa }^{\prime }\left(u_0\right)-{\kappa }^{\prime }(u_{\alpha }+t(u_{\alpha }^{\delta }-u_{\alpha }))]dt(u_{\alpha }^{\delta }-u_{\alpha })\hfill \\ & & +v^{\delta }-v.\hfill \end{array}$$

Therefore, from Assumption 1, we have

$$\begin{array}{ccc}\hfill \parallel u_{\alpha }^{\delta }-u_{\alpha }\parallel & \le & \parallel {\int }_0^1{({\kappa }^{\prime }\left(u_0\right)+\alpha I)}^{-1}[{\kappa }^{\prime }\left(u_0\right)-{\kappa }^{\prime }(u_{\alpha }+t(u_{\alpha }^{\delta }-u_{\alpha }))]dt\hfill \\ & & \times (u_{\alpha }^{\delta }-u_{\alpha })\parallel + \parallel {({\kappa }^{\prime }\left(u_0\right)+\alpha I)}^{-1}(v^{\delta }-v)\parallel \hfill \\ & \le & 3k_0\parallel u_{\alpha }^{\delta }-u_{\alpha }\parallel \left[\parallel u_{\alpha }-u_0\parallel +{\displaystyle \frac{\parallel u_{\alpha }^{\delta }-u_0\parallel + \parallel u_{\alpha }-u_0\parallel }{2}}\right]+{\displaystyle \frac{\delta }{\alpha -{ϵ}_h}}\hfill \\ & \le & 6k_{0}r\parallel u_{\alpha }^{\delta }-u_{\alpha }\parallel +{\displaystyle \frac{\delta }{\alpha -{ϵ}_h}}\hfill \end{array}$$

and hence

$$\begin{array}{ccc}\hfill \parallel u_{\alpha }^{\delta }-u_{\alpha }\parallel & \le & {\displaystyle \frac{1}{1-6k_{0}R}}{\displaystyle \frac{\delta }{\alpha -{ϵ}_h}}\hfill \\ & \le & {\displaystyle \frac{1}{1-6k_{0}R}}{\displaystyle \frac{2\delta }{\alpha }}.\hfill \end{array}$$

This proves (18); (c) and (19) follows from (17) and (18). □

Theorem 2 and Theorem 4 lead to the following theorem.

Theorem 5.

Let $\overline{\alpha }>0$ be a fixed number, such that $\overline{\alpha }>\max \{2{ϵ}_h,\delta +{ϵ}_h\},$ $\alpha \in (\max \{2{ϵ}_h,\delta +{ϵ}_h\},\overline{\alpha })$ and let $u_0$, such that

$$\parallel u_0-\widehat{u}\parallel =r_0.$$

Let $u_{\alpha }^{\delta }$ and $u_{\alpha }$ be the same as in Theorem 2 and Remark 3, respectively. Suppose Assumption 1 and Assumption 2 hold and $6k_{0}R<1.$ Let

$$n_{\alpha ,\delta }:=\min \{m\in \mathbb{N}:\alpha {\sigma }^m\le \delta \}.$$

(24)

Then,

$$\parallel u_{n_{\alpha ,\delta },\alpha }^{\delta }-\widehat{u}\parallel \le \tilde{c}({\displaystyle \frac{\delta }{\alpha }}+{\alpha }^{\omega }),$$

(25)

where $\tilde{c}=\max \{c_1,c_2,{\displaystyle \frac{\gamma }{1-\sigma }}\}.$

One should choose $\alpha$, such that ${\alpha }^{1+\omega }=\delta .$ But, such a choice is not possible as $\omega$ is unknown. Therefore, a strategy for the parameter choice is to be considered. Here, the adaptive parameter choice introduced by Pereverzyev and Schock in [29] is adopted.

4. Adaptive Choice and Stopping Rule

In our study, the regularization parameter $\alpha (>0)$ is chosen and the error bound is deduced based on an adaptive scheme, which is not only independent of $\omega$ but also of the regularization method involved. This was introduced by Pereverzyev and Schock in [29] (also see [20]). Here, different regularization parameters ${\alpha }_i$ are usually chosen from a finite set $\Delta =\{{\alpha }_0,{\alpha }_1,{\alpha }_2,\dots ,{\alpha }_n\}$, with ${\alpha }_0<{\alpha }_1<{\alpha }_2<\dots <{\alpha }_n$ and, then, the corresponding $u_{n_{{\alpha }_i,\delta },{\alpha }_i}^{\delta }$ are studied.

Theorem 6.

Assume the conditions in Theorem 5 hold. Further, assume ∃$i\in \{0,1,2,\dots ,n\}$ such that ${\alpha }_i^{\omega }\le {\displaystyle \frac{\delta }{{\alpha }_i}}$ and for some $\mu >1,$

$$\begin{array}{c}\hfill {\alpha }_i=:{\mu }^i{\alpha }_0\forall i=1,2,\dots ,n,\mathrm{w}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}{\alpha }_0=C\delta ,\end{array}$$

for some constant $C.$ Let

$$\begin{array}{ccc}\hfill n_{{\alpha }_i,\delta }& =& \min \{m:{\alpha }_i{\sigma }^m\le \delta \},\hfill \\ \hfill l& =& \max \{i:{\alpha }_i^{\omega }\le {\displaystyle \frac{\delta }{{\alpha }_i}}\}<n\hfill \end{array}$$

and

$$\begin{array}{c}\hfill k=\max \{i: \parallel u_{n_{{\alpha }_i,\delta },{\alpha }_i}^{\delta }-u_{n_{{\alpha }_j,\delta },{\alpha }_j}^{\delta }\parallel \le {\displaystyle \frac{4\tilde{c}\delta }{{\alpha }_j}},j=0,1,2,\dots ,i\}.\end{array}$$

Then, $l\le k$ and

$$\begin{array}{c}\hfill \parallel \widehat{u}-u_{n_{{\alpha }_k,\delta },{\alpha }_k}^{\delta }\parallel \le 6\tilde{c}\mu {\delta }^{{\displaystyle \frac{\omega }{\omega +1}}}=O({\delta }^{{\displaystyle \frac{\omega }{\omega +1}}}),0<\omega \le 1.\end{array}$$

Proof. 

First, we assert that $l\le k.$ To prove this, it is enough to show that for each $i=1,2,\dots ,n,$ ${\alpha }_i^{\omega }\le {\displaystyle \frac{\delta }{{\alpha }_i}}$ implies $\parallel u_{n_{{\alpha }_i,\delta },{\alpha }_i}^{\delta }-u_{n_{{\alpha }_j,\delta },{\alpha }_j}^{\delta }\parallel \le {\displaystyle \frac{4\tilde{c}\delta }{{\alpha }_j}},\forall j=0,1,2,\dots ,i.$

Let $j\le i.$ Consider

$$\begin{array}{ccc}\hfill \parallel u_{n_{{\alpha }_i,\delta },{\alpha }_i}^{\delta }-u_{n_{{\alpha }_j,\delta },{\alpha }_j}^{\delta }\parallel & \le & \parallel u_{n_{{\alpha }_i,\delta },{\alpha }_i}^{\delta }-\widehat{u}\parallel + \parallel u_{n_{{\alpha }_j,\delta },{\alpha }_j}^{\delta }-\widehat{u}\parallel \hfill \\ & \le & \tilde{c}\left({\displaystyle \frac{\delta }{{\alpha }_i}}+{\alpha }_i^{\omega }\right)+\tilde{c}\left({\displaystyle \frac{\delta }{{\alpha }_j}}+{\alpha }_j^{\omega }\right)\hfill \\ & \le & \tilde{c}\left({\displaystyle \frac{\delta }{{\alpha }_i}}+{\displaystyle \frac{\delta }{{\alpha }_i}}\right)+\tilde{c}\left({\displaystyle \frac{\delta }{{\alpha }_j}}+{\displaystyle \frac{\delta }{{\alpha }_j}}\right)\hfill \\ & \le & 4\tilde{c}{\displaystyle \frac{\delta }{{\alpha }_j}}.\hfill \end{array}$$

Hence, the assertion.

Next, we fix ${\alpha }_k$ and as $l\le k,$ one can observe that

$$\begin{array}{ccc}\hfill \parallel \widehat{u}-u_{n_{{\alpha }_k,\delta },{\alpha }_k}^{\delta }\parallel & \le & \parallel \widehat{u}-u_{n_{{\alpha }_l,\delta },{\alpha }_l}^{\delta }\parallel + \parallel u_{n_{{\alpha }_l,\delta },{\alpha }_l}^{\delta }-u_{n_{{\alpha }_k,\delta },{\alpha }_k}^{\delta }\parallel \hfill \\ & \le & \tilde{c}\left({\alpha }_l^{\omega }+{\displaystyle \frac{\delta }{{\alpha }_l}}\right)+4\tilde{c}{\displaystyle \frac{\delta }{{\alpha }_k}}\hfill \\ & \le & \tilde{c}\left({\displaystyle \frac{\delta }{{\alpha }_l}}+{\displaystyle \frac{\delta }{{\alpha }_l}}\right)+4\tilde{c}{\displaystyle \frac{\delta }{{\alpha }_l}}\hfill \\ & \le & 6\tilde{c}{\displaystyle \frac{\delta }{{\alpha }_l}}.\hfill \end{array}$$

(26)

Note that the quantity ${\alpha }^{\omega }+{\displaystyle \frac{\delta }{\alpha }}$ attains its least value for the choice $\alpha :={\alpha }_{\delta }$, such that ${\alpha }_{\delta }^{\omega }={\displaystyle \frac{\delta }{{\alpha }_{\delta }}}.$

We have ${\alpha }_{\delta }\le {\alpha }_{l+1}\le \mu {\alpha }_l.$ This gives, ${\displaystyle \frac{\delta }{{\alpha }_l}}\le {\displaystyle \frac{\mu \delta }{{\alpha }_{\delta }}}=\mu {\alpha }_{\delta }^{\omega }=\mu {\delta }^{{\displaystyle \frac{\omega }{\omega +1}}}.$

Thus,

$$\begin{array}{ccc}\hfill \parallel \widehat{u}-u_{n_{{\alpha }_k,\delta },{\alpha }_k}^{\delta }\parallel & \le & 6\tilde{c}\mu {\delta }^{{\displaystyle \frac{\omega }{\omega +1}}}\hfill \\ & =& O({\delta }^{{\displaystyle \frac{\omega }{\omega +1}}})\mathrm{w}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}0<\omega \le 1.\hfill \end{array}$$

(27)

□

5. Algorithm

The algorithm associated with the adaptive parameter choice strategy in Theorem (6) is as follows:

• Choose ${\alpha }_0=C\delta$ and $\mu >1.$
• Choose ${\alpha }_i={\mu }^i{\alpha }_0,i=0,1,2,\dots ,n.$

(i)

Set $i=0.$

(ii)

Choose the minimum n, such that ${\sigma }^n\le {\displaystyle \frac{\delta }{{\alpha }_i}}$ (say $n_{{\alpha }_i,\delta }$).

(iii)

Compute $u_{n_{{\alpha }_i,\delta },{\alpha }_i}^{\delta }$ using the iteration (9).

(iv)

If $\parallel u_{n_{{\alpha }_i,\delta },{\alpha }_i}^{\delta }-u_{n_{{\alpha }_j,\delta },{\alpha }_j}^{\delta }\parallel >{\displaystyle \frac{4\tilde{c}\delta }{{\alpha }_j}},$ $j\le i,$ then, take $k=i-1$ and return $u_{n_{{\alpha }_k,\delta },{\alpha }_k}^{\delta }.$

(v)

Else, put $i=i+1$ and get back to (ii).

6. Numerical Examples

Example 3.

Getting back to Example 2, the derivative of the operator κ at point $w_0(\theta ,\vartheta )$ is

$${\kappa }^{\prime }\left(w_0\right)h=\int {\int }_{\left(\right[0,m]\times [0,m\left]\right)}{\displaystyle \frac{w_0({\theta }^{\prime },{\vartheta }^{\prime })h({\theta }^{\prime },{\vartheta }^{\prime })}{{[{(\theta -{\theta }^{\prime })}^2+{(\vartheta -{\vartheta }^{\prime })}^2+{\left(w_0({\theta }^{\prime },{\vartheta }^{\prime })\right)}^2]}^{3/2}}}d{\theta }^{\prime }d{\vartheta }^{\prime }.$$

(28)

By applying the two-dimensional analogy of rectangle’s formula with a uniform grid for every variable in the integral Equation (6), we obtain the following system of nonlinear equations:

$$\sum _{i=1}^m\sum _{j=1}^m{\displaystyle \frac{1}{{[{({\theta }_k-{\theta }_j^{\prime })}^2+{({\vartheta }_l-{\vartheta }_i^{\prime })}^2+w^2({\theta }_j^{\prime },{\vartheta }_i^{\prime })]}^{1/2}}}\Delta \theta \Delta \vartheta =q({\theta }_k,{\vartheta }_l);$$

$(k,l=1,2,\dots ,m)$ for the unknown vector $\{w_{j,i}=w({\theta }_j,{\vartheta }_i),i,j=1,2,\dots ,m\}.$ In vector–matrix form, this system takes the form:

$${\kappa }_N\left(w_N\right)=q_N,$$

(29)

where $w_N,q_N$ are vectors of dimension $N=m^2.$

The discrete variant of the derivative ${\kappa }^{\prime }\left(w_0\right)$ has the form

$${\left\{{\kappa }_n^0{h}_n\right\}}_{k,l}=\sum _{i=1}^m\sum _{j=1}^m{\displaystyle \frac{\Delta \theta \Delta \vartheta w_0({\theta }_j^{\prime },{\vartheta }_i^{\prime })h({\theta }_j^{\prime },{\vartheta }_i^{\prime })}{{[{({\theta }_k-{\theta }_j^{\prime })}^2+{({\vartheta }_l-{\vartheta }_i^{\prime })}^2+w_0^2({\theta }_j^{\prime },{\vartheta }_i^{\prime })]}^{3/2}}},$$

(30)

where $w_0(\theta ,\vartheta )=H$ is constant, $N=m^2.$ In order to compute $w_{n,\alpha }^{\delta },$ we choose orthonormal system of box function ${\Phi }_i(t,\tau )={\zeta }_k\left(t\right){\zeta }_l\left(\tau \right),i=(k-1)m+l,k,l=1,2,3,\dots ,m,i=1,2,\dots ,N(=m^2),$ where ${\zeta }_k\left(t\right),{\zeta }_l\left(\tau \right)$ are $L_2$-orthonormalized characteristic functions of the intervals $[k-1,k),[l-1,l)$ [30], respectively, in $[0,m]\times [0,m].$

Then, there exists ${\lambda }_1^n,{\lambda }_2^n,\dots ,{\lambda }_N^n\in \mathbb{R},$ such that $w_{n,\alpha }^{\delta }\approx {\sum }_{i=1}^N{\lambda }_i^n{\Phi }_i(t,\tau ).$ Then, from (9) we have,

$$\begin{array}{ccc}\hfill (A_h+\alpha I)\sum _{i=1}^N({\lambda }_i^{n+1}-{\lambda }_i^n){\Phi }_i(t,\tau )& =& \sum _{i=1}^N{\eta }_i{\Phi }_i(t,\tau )-\sum _{i=1}^N{\kappa }_i{\Phi }_i(t,\tau )\hfill \\ & & +\alpha \sum _{i=1}^N(W_{0,i}-{\lambda }_i^n){\Phi }_i(t,\tau ),\hfill \end{array}$$

where ${\kappa }_i=\kappa {\left(w_{n,\alpha }^{\delta }\right)}_{k,l},{\eta }_i=q^{\delta }(k,l)$ and $W_{0,i}=w_0(k,l)$, $i=(k-1)m+l$, $k,l=1,2,\dots ,m,i=1,2,\dots ,N.$ Then, $w_{n+1,\alpha }^{\delta }$ is a solution of (9), if and only if $[{\lambda }^{n+1}-{\lambda }^n]={[{\lambda }_1^{n+1}-{\lambda }_1^n,{\lambda }_2^{n+1}-{\lambda }_2^n,\dots ,{\lambda }_N^{n+1}-{\lambda }_N^n]}^T$ is the unique solution of

$$[M_N+\alpha B_N][{\lambda }^{n+1}-{\lambda }^n]=B_N[\eta -{\kappa }^N+\alpha (W_0-{\lambda }^N)],$$

where $M_N=\left(\langle A_h{\Phi }_i,{\Phi }_j\rangle \right),B_N=\left(\langle {\Phi }_i,{\Phi }_j\rangle \right)i,j=1,2,\dots ,N,$

$${\kappa }^N={[{\kappa }_1,{\kappa }_2,\cdots ,{\kappa }_N]}^T,\eta ={[{\eta }_1,{\eta }_2,\cdots ,{\eta }_N]}^T,$$

$$W_0={[W_{0,1},W_{0,2},\cdots ,W_{0,N}]}^T\mathrm{a}\mathrm{n}\mathrm{d}{\lambda }^N={[{\lambda }_1^n,{\lambda }_2^n,\cdots ,{\lambda }_N^n]}^T.$$

For $m=40,{\kappa }_n^0$ is a positive symmetric matrix, with the minimal eigenvalue, ${\lambda }_{min}=1.5471\times {10}^{-18}$, and the condition number, $Cond\left({\kappa }_n^0\right)=6.4636\times {10}^{17}.$

Similarly, for $m=35,{\kappa }_n^0$ is a positive symmetric matrix, with a minimal eigenvalue, ${\lambda }_{min}=6.2379\times {10}^{-19}$ and condition number, $Cond\left({\kappa }_n^0\right)=1.6031\times {10}^{18}.$ We take

$$\begin{array}{ccc}\hfill \widehat{w}(\theta ,\vartheta )& =& 5-(exp(-{((\theta /10)-2.5)}^2\ast {((\vartheta /10)-2.5)}^2)\hfill \\ & & +3\ast exp(-{((\theta /10)+2.5)}^2\ast {((\vartheta /10)+2.5)}^2))/60,\hfill \end{array}$$

where $\widehat{w}(\theta ,\vartheta )$ is given on the domain $D=\{0\le \theta \le m,0\le \vartheta \le m\}.$ Let $\Delta \theta =\Delta \vartheta =1,N=m^2,{ϵ}_h={\displaystyle \frac{1}{m^2}},\Delta =0.25,w_0\equiv H\equiv 5$ [15,24,31,32]. We take $q^{\delta }=\kappa \left(\widehat{w}(\theta ,\vartheta )\right)+\delta$ in our computations.

Here, $w_{n,{\alpha }_k}^{\delta }$ is the numerical solution obtained by method (9); the relative error of solution and residual are

$${\Delta }_1={\displaystyle \frac{\parallel \widehat{w}-w_{n,{\alpha }_k}^{\delta }\parallel }{\parallel w_{n,{\alpha }_k}^{\delta }\parallel }},{\Delta }_2={\displaystyle \frac{\parallel {\kappa }_n\left(w_{n,{\alpha }_k}^{\delta }\right)-q^{\delta }\parallel }{\parallel q^{\delta }\parallel }},$$

respectively.

Taking $k_0={\displaystyle \frac{1}{30}},$ we obtain $r_0^{\prime }=0.1662, \gamma =1.2500, r=5.0000$, $\sigma =0.7500, r_0=1.3644 ,R_1=4.1635$ and $R_2=8.6356=R.$ 

Table 1. Error and computation time.

$\mathit{\delta }$	${\mathit{\alpha }}_{\mathit{k}}$	m	${\mathbf{\Delta }}_1$	${\mathbf{\Delta }}_2$	CPU	${\mathbf{\Delta }}_1$ [33]	${\mathbf{\Delta }}_2$ [33]	CPU
					Time			Time [33]
0.01	0.0256		0.002089	0.000420	4.02	2.3448 × ${10}^{-4}$	7.0843 ×${10}^{-5}$	4.31
0.005	0.0128	40	0.002030	0.000407	3.84	2.3448 ×${10}^{-4}$	7.0871 ×${10}^{-5}$	4.19
0.002	0.0051		0.001875	0.000372	3.79	2.3448 ×${10}^{-4}$	7.0888 ×${10}^{-5}$	4.04
0.01	0.0256		0.002211	0.000491	2.44	3.0643 ×${10}^{-4}$	1.0248 ×${10}^{-4}$	2.92
0.005	0.0128	35	0.002151	0.000476	2.41	3.0643 ×${10}^{-4}$	1.0252 ×${10}^{-4}$	2.84
0.002	0.0051		0.001994	0.000438	2.37	3.064 3×${10}^{-4}$	1.0255 ×${10}^{-4}$	2.65

gives the values of ${\alpha }_k,$ the relative error, and the residual, for different δ values. Figure 1, Figure 2, Figure 3 and Figure 4 show the exact and computed solutions for different δ values. The work is further compared with the study in [33], giving values of ${\Delta }_1,$ ${\Delta }_2$ and computation time (t) in seconds in Table 1.

Example 4.

We next consider Example 1. The Fréchet derivative of κ is

$$\left[{\kappa }^{\prime }\left(u\right)h\right]\left(t\right)=\left[\kappa \left(u\right)\right]\left(t\right){\int }_0^{t}h\left(\theta \right)d\theta .$$

(31)

Then, κ is not monotone (proved in [22]), but ${\kappa }^{\prime }$ is a positive type or sectorial [12] (i.e., $\parallel {({\kappa }^{\prime }(·)+\alpha I)}^{-1}\parallel \le {\displaystyle \frac{c}{\alpha }}$ for $\alpha >0$ and $c>0$) and the spectrum of ${\kappa }^{\prime }\left(u\right)$ is the singleton set $\left\{0\right\}.$

In order to compute $u_{n,\alpha }^{\delta },$ we consider ${\Phi }_i\left(t\right),i=1,2,\dots ,m+1,$ is the characteristic function of interval $[i-1,i)$ [30] in $[0,1],$ (partitioned into m subintervals).

Then, there exists ${\lambda }_1^n,{\lambda }_2^n,\dots ,{\lambda }_{m+1}^n\in \mathbb{R},$ such that $u_{n,\alpha }^{\delta }\approx {\sum }_{i=1}^{m+1}{\lambda }_i^n{\Phi }_i\left(t\right).$ Then, from (9) we have,

$$\begin{array}{ccc}\hfill (A_h+\alpha I)\sum _{i=1}^{m+1}({\lambda }_i^{n+1}-{\lambda }_i^n){\Phi }_i\left(t\right)& =& \sum _{i=1}^{m+1}{\eta }_i{\Phi }_i\left(t\right)-\sum _{i=1}^{m+1}{\kappa }_i{\Phi }_i\left(t\right)\hfill \\ & & +\alpha \sum _{i=1}^{m+1}({\lambda }_i^0-{\lambda }_i^n){\Phi }_i\left(t\right)\hfill \end{array}$$

where ${\kappa }_i=\kappa \left(u_{n,\alpha }^{\delta }\right)\left(t_i\right)\mathrm{a}\mathrm{n}\mathrm{d}{\eta }_i=v^{\delta }\left(t_i\right),i=1,2,\dots ,m+1$ and $t_i^{\prime }s$ are the points of partition. Then, $u_{n+1,\alpha }^{\delta }$ is a solution of (3), if and only if $[{\lambda }^{n+1}-{\lambda }^n]={[{\lambda }_1^{n+1}-{\lambda }_1^n,{\lambda }_2^{n+1}-{\lambda }_2^n,\dots ,{\lambda }_{m+1}^{n+1}-{\lambda }_{m+1}^n]}^T$ is the unique solution of

$$[M+\alpha B][{\lambda }^{n+1}-{\lambda }^n]=B[\eta -{\kappa }^{m+1}+\alpha ({\lambda }^0-{\lambda }^n)]$$

where $M=\left(\langle A_h{\Phi }_i,{\Phi }_j\rangle \right),B=\left(\langle {\Phi }_i,{\Phi }_j\rangle \right)i,j=1,2,\dots ,m+1,$

$${\kappa }^{m+1}={[{\kappa }_1,{\kappa }_2,\cdots ,{\kappa }_{m+1}]}^T,\eta ={[{\eta }_1,{\eta }_2,\cdots ,{\eta }_{m+1}]}^T,$$

$${\lambda }^0={[{\lambda }_1^0,{\lambda }_2^0,\cdots ,{\lambda }_{m+1}^0]}^T\mathrm{a}\mathrm{n}\mathrm{d}{\lambda }^n={[{\lambda }_1^n,{\lambda }_2^n,\cdots ,{\lambda }_{m+1}^n]}^T.$$

For the computations, the exact solution is taken to be $\widehat{u}\left(t\right)=t,t\in (0,1),u_0\left(t\right)=0$ and $v\left(t\right)=e^{{\displaystyle \frac{t^2}{2}}}.$ We apply random noise to v to obtain $v^{\delta }.$ Let $u_{n,{\alpha }_k}^{\delta }$ be the numerical solution obtained by method (9). Taking $k_0={\displaystyle \frac{1}{30}},$ we obtain $r_0^{\prime }=0.1662,\gamma =1.2500,r=5.0000,\sigma =0.7500,r_0=1.3644,R_1=4.1635$ and $R_2=8.6356=R.$ 

Table 2. Relative error.

$\mathit{\delta }$	${\mathit{\alpha }}_{\mathit{k}}$	m	${\mathbf{\Delta }}_1$	${\mathbf{\Delta }}_2$
0.01	$1.0201\times {10}^{-4}$		$7.8823\times {10}^{-3}$	5.7002
0.001	$1.0201\times {10}^{-6}$	1000	$7.8077\times {10}^{-3}$	5.6951
0.0001	$1.0201\times {10}^{-8}$		$7.8070\times {10}^{-3}$	5.6947
0.01	$1.0201\times {10}^{-4}$		$7.8808\times {10}^{-3}$	5.6796
0.001	$1.0201\times {10}^{-6}$	800	$7.8062\times {10}^{-3}$	5.6931
0.0001	$1.0201\times {10}^{-8}$		$7.8055\times {10}^{-3}$	5.6936

gives the values of ${\alpha }_k,$ the relative error, and the residual,

$${\Delta }_1={\displaystyle \frac{\parallel \widehat{u}-u_{n,{\alpha }_k}^{\delta }\parallel }{\parallel u_{n,{\alpha }_k}^{\delta }\parallel }},{\Delta }_2={\displaystyle \frac{\parallel {\kappa }_n\left(u_{n,{\alpha }_k}^{\delta }\right)-v^{\delta }\parallel }{\parallel v^{\delta }\parallel }}$$

respectively, for different δ and m values. Figure 5, Figure 6, Figure 7, Figure 8, Figure 9 and Figure 10 show the exact and noisy data with the corresponding exact and computed solutions (C.S.), for different δ and m values.

7. Conclusions

A sufficient condition required to apply LRM to solve ill-posed nonlinear equations involving operators that are not monotone has been developed. With this theory, researchers can study a wider range of problems using Lavrentiev regularisation, irrespective of the monotonicity of the operator involved. The error estimates, convergence analysis, and adaptive parameter choice strategy are studied. The paper also provides numerical examples.