A filter method for inverse nonlinear sideways heat equation

Abstract

In this paper, we study a sideways heat equation with a nonlinear source in a bounded domain, in which the Cauchy data at $x = \mathcal {X}$ are given and the solution in $0 \le x < \mathcal {X}$ is sought. The problem is severely ill-posed in the sense of Hadamard. Based on the fundamental solution to the sideways heat equation, we propose to solve this problem by the filter method of degree α, which generates a well-posed integral equation. Moreover, we show that its solution converges to the exact solution uniformly and strongly in $\mathscr {L}^{p}(\omega,\mathcal {X};\mathscr {L}^{2}(\mathbb {R}))$, $\omega\in [0,\mathcal {X})$ under a priori assumptions on the exact solution. The proposed regularized method is illustrated by numerical results in the final section.

1 Introduction

In this paper, we determine the surface temperature $u( x,t)$ for $0 \leq x< \mathcal {X}$ from the known temperature measurements $u(\mathcal {X},t)=\phi(t)$ and heat-flux measurement $\frac{\partial u}{\partial x} (\mathcal {X},t)=\psi(t)$ when $u(x,t)$ satisfies the following system:

$$\begin{aligned} \textstyle\begin{cases} \frac{\partial u}{\partial t}=\frac{\partial^{2} u}{\partial x^{2}} + f(u)G(x,t;u),& (x,t) \in(0,\mathcal {X}) \times\mathbb {R}, \\ u(x,t)|_{t\to\pm\infty} = 0, & ( x,t) \in(0,\mathcal {X}) \times \mathbb {R},\\ u(\mathcal {X},t)=\phi(t),& t \in\mathbb {R}, \\ \frac{\partial u}{\partial x}(\mathcal {X},t)=\psi(t),& t \in\mathbb {R}, \end{cases}\displaystyle \end{aligned}$$

(1.1)

where $\phi,\psi\in\mathscr {L}^{2}(\mathbb {R})$ are given functions. The source terms $f(u)$, $G(u)$ are globally Lipschitz functions satisfying (2.15a) and (2.15b), respectively.

The problem called the inverse nonlinear sideways heat equation (INSHE for short) is a model of a problem where one wants to determine the temperature on both sides of a thick wall, but where one side is inaccessible to measurements. In many dynamic heat transfer situations, one wishes to determine the temperature on the surface of a body, where the surface itself is inaccessible for measurements. The physical situation at the surface may be unsuitable for attaching a sensor, or the accuracy of a surface measurement may be seriously impaired by the presence of the sensor. Typical practical applications are the estimation of the heat flux and the temperature at the surface of the body under investigation, e.g., re-entry vehicles, calorimeter-type instrumentation, and combustion chambers [1, 2, 5, 11, 13, 15, 16, 19, 20]. In such cases, one is restricted to interior measurements, and from these one wishes to compute the surface temperature.

Cannon (1984) [4] considered the direct problem for the homogeneous heat equation in the quarter plane ($x \geq0$, $t \geq0$):

$$\begin{aligned} \textstyle\begin{cases} \frac{\partial u}{\partial t}=\frac{\partial^{2} u}{\partial x^{2}},& x \geq0, t \geq0, \\ u(x,0) = 0, & x \geq0,\\ u(0,t)=\phi(t),& t\geq0. \end{cases}\displaystyle \end{aligned}$$

(1.2)

The functions $\phi(\cdot)$ and $u(x, \cdot)$ are to be in $\mathscr {L}^{2}(\mathbb {R})$ (ϕ and u vanish for $t < 0$). The author proved that, for each $\phi\in \mathscr {L}^{2}(\mathbb {R})$, (1.2) has a unique solution u with $u( x, \cdot) \in\mathscr {L}^{2}(\mathbb {R})$ for each $x \geq0$.

Fredrik Berntsson (1999) [3] considered the sideways heat equation

$$\begin{aligned} \textstyle\begin{cases} \frac{\partial u}{\partial t}=\kappa\frac{\partial^{2} u}{\partial x^{2}},& 0< x < 1, t \geq0, \\ u(x,0) = 0, & 0 < x < 1, \\ u(1,t)=g(t),& t\geq0 \\ \frac{\partial u}{\partial x}(1,t)=h(t),& t\geq0. \end{cases}\displaystyle \end{aligned}$$

(1.3)

The author used the spectral method to solve problem (1.3). Error estimates for the regularized solution were derived, and a procedure for selecting an appropriate regularization parameter was given.

In recent years, linear homogeneous problem (1.1), i.e., $f(u) G(x,t;u) =0$, has been researched by many authors, and various numerical methods have been proposed, e.g., the boundary element Tikhonov regularization method (Lesnic et al. (1996) [9]), the conjugate gradient method (Hao (2012) [8]), the difference regularization method (Xiong et al. (2006a) [17]), the ”optimal filtering” method (Seidman & Elden (1990) [14]), the Fourier method (Xiong et al. 2006b [18]), the quasi-reversibility method (Elden (1987) [6], Liu & Wei (2013) [10]), the wavelet, wavelet-Galerkin, and the spectral regularization methods (Elden et al. (2000) [7], Reginska & Elden (1997) [12]), to mention only a few.

The more important but challenging semilinear sideways heat equation with the heat source depends nonlinearly on the temperature, which occurs in many applications related to reaction-diffusion. The function $f (u)G(u)$ is known as a special type of locally Lipschitz function. For example, if we choose $f(u) := u$, $G(x,t;u) :=\sin u$ (individually they are globally Lipschitz), then

$$\begin{aligned}[b] \bigl\vert f(u)G(x,t;u) - f(v) G(x,t;v) \bigr\vert &= \vert u \sin u - v \sin v \vert \\ &\leq \vert u-v \vert \sin u + v \vert \sin u - \sin v \vert \\ & = \vert u-v \vert \sin u + 2v \cos\frac{u+v}{2} \sin \frac { \vert u-v \vert }{2} \\ & \leq \vert u-v \vert u + v \vert u-v \vert \leq\max\{u; v\} \vert u-v \vert .\end{aligned} $$

(1.4)

Although there are some works on the nonlinear case, the literature on the case of locally Lipschitz sources $f(u) G(x,t;u)$ is quite scarce. Our results extend problem (1.3), and we propose a new filter method to establish regularized solutions of problem (1.1) in the case of the locally Lipschitz function $f(u) G(x,t;u)$.

The paper is organized as follows. In Sect. 2, the formulation of problem and regularization methods is given. In Sect. 3, a stability estimate in $\mathscr {L}^{p}(\omega,\mathcal {X};\mathscr {L}^{2}(\mathbb {R}))$, $\omega\in[0,\mathcal {X})$ is proved under a priori condition of the exact solution and the locally Lipschitz source term. Finally, we present a numerical result to illustrate the proposed regularized method in Sect. 4.

2 Mathematical problem and mild solution of (INSHE)

For $w \in\mathscr {L}^{2}(\mathbb {R})$, we have the Fourier transform

$$\widehat{w}(\xi)= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} w(t) e^{-i\xi t} \,\mathrm {d} t, \quad\xi\in\mathbb {R}, $$

and the $\mathscr {L}^{2}$ norm of w is

$$\begin{aligned} \Vert w \Vert _{\mathscr {L}^{2}(\mathbb {R})}^{2}= \Vert \widehat{w} \Vert _{\mathscr {L}^{2}(\mathbb {R})}^{2} = \int_{-\infty}^{\infty} \bigl\vert \widehat{w}(\xi) \bigr\vert ^{2} \,\mathrm {d} \xi. \end{aligned}$$

(2.1)

Suppose that the solution of problem (1.1) is represented as a Fourier transform

$$\begin{aligned} u(x,t)= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \widehat {u}(x,\xi)e^{i\xi t} \, \mathrm {d} \xi \end{aligned}$$

(2.2)

with

$$\begin{aligned} \widehat{u}( x,\xi)= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty } u(x,t) e^{-i\xi t} \,\mathrm {d} t. \end{aligned}$$

(2.3)

Throughout this paper, we let $\mathcal {W} (x,t;u) = f(u) G(x,t;u)$, $\forall(x,t) \in(0,\mathcal {X}) \times\mathbb {R}$.

From (1.1), we have the following systems of second order ordinary equation:

$$\begin{aligned} \textstyle\begin{cases} - \widehat{\partial_{{xx}} u}(x,\xi) + i \xi \widehat{u}(x,\xi) = \widehat{\mathcal {W}}(u)( x,\xi),& \\ \widehat{u}(\mathcal {X},\xi) = \widehat{\phi}(\xi), & \\ \widehat{\partial_{x} u}(\mathcal {X},\xi) = \widehat{\psi}(\xi). & \end{cases}\displaystyle \end{aligned}$$

(2.4)

We thus have after some direct calculation

$$\begin{aligned} \widehat{u}(x,\xi) = {}&\cosh \bigl((\mathcal {X}-x) \sqrt{i\xi} \bigr) \widehat{u}(\mathcal {X},\xi) - \frac{\sinh ((\mathcal {X} - x) \sqrt{i\xi} )}{\sqrt{i\xi}} \widehat{ \partial_{x} u}(\mathcal {X},\xi) \\ & - \int_{ x}^{\mathcal {X}} \frac{\sinh ((z- x) \sqrt{i\xi} )}{\sqrt{i\xi}} \widehat{\mathcal {W}} (u ) (z,\xi) \,\mathrm {d}z, \end{aligned}$$

(2.5)

where

$$\sqrt{i\xi} = \sqrt{\frac{ \vert \xi \vert }{2}}(1+\sigma i), \quad\sigma= \operatorname{sign} (\xi), \xi\in\mathbb {R}. $$

Remark 2.1

In (2.5), for $\xi= 0$, $\frac{\sinh (y \sqrt{i\xi } )}{\sqrt{i\xi}}$ is defined as y, since

$$\lim_{\xi\to0}\frac{\sinh (y \sqrt{i\xi} )}{\sqrt {i\xi}} = y. $$

Moreover, for $\xi=0$, we have

$$\begin{aligned} \widehat{u}( x,0) &= \widehat{u}(\mathcal {X},0) - (\mathcal {X} - x) \widehat{ \partial_{x} u}(\mathcal {X},0) - \int_{ x}^{\mathcal {X}} (z-x) \widehat{\mathcal {W}} (u ) (z,0)\, \mathrm {d}z. \end{aligned}$$

From (2.5), the exact form of u is given by

$$\begin{aligned} u(x,t) ={}& \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \bigl( \cosh \bigl((\mathcal {X} - x) \sqrt{i\xi} \bigr) \widehat{\phi}(\xi ) \bigr) e^{i\xi t} \,\mathrm {d} \xi \\ & -\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \biggl( \frac {\sinh ((\mathcal {X} - x) \sqrt{i\xi} )}{\sqrt{i\xi}} \widehat{ \psi}(\xi) \biggr) e^{i\xi t} \,\mathrm {d} \xi \\ & - \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \biggl( \int_{ x}^{\mathcal {X}} \frac{\sinh ((z-x) \sqrt{i\xi} )}{\sqrt {i\xi}} \widehat{\mathcal {W}} (u ) (z,\xi) \,\mathrm {d}z \biggr) e^{i\xi t} \,\mathrm {d} \xi. \end{aligned}$$

(2.6)

We say that u is a mild solution of problem (1.1) if u satisfies integral (2.6). We know that the three functions

$$\begin{aligned} \cosh \bigl((\mathcal {X}- x)\sqrt{i\xi} \bigr),\qquad \frac{\sinh ((\mathcal {X}- x)\sqrt{i\xi} )}{\sqrt{i\xi}},\qquad \frac{\sinh ((z- x)\sqrt{i\xi} )}{\sqrt{i\xi}} \end{aligned}$$

(2.7)

are unbounded as a function of the variable ξ. Consequently, small errors in high frequency components can blow up and completely destroy the solution for $0< x<z< \mathcal {X}$. A natural idea to stabilize the problem is to replace them by a bounded approximation. In a natural way, we can replace the terms in (2.7) by

$$\cosh^{\gamma(\delta)} \bigl((\mathcal {X}- x)\sqrt{i\xi} \bigr) ,\qquad \frac{\sinh^{\gamma(\delta)} ((\mathcal {X}- x)\sqrt{i\xi} )}{\sqrt{i\xi}}, \qquad\frac{\sinh^{\gamma(\delta)} ((z- x)\sqrt{i\xi} )}{\sqrt{i\xi}} $$

(respectively), with $\delta> 0$ is a small positive number representing the level of noise and the parameter $\gamma(\delta) >0$ is small (regularization parameter). We introduce the first regularized solution $U^{\delta}_{\gamma(\delta )}$ obtained by

$$\begin{aligned} &U^{\delta}_{\gamma(\delta)}( x,t) \\ &\quad= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \bigl( \cosh ^{\gamma(\delta)} \bigl(( \mathcal {X}- x)\sqrt{i\xi} \bigr) \widehat {\phi^{\delta}}(\xi) \bigr)e^{i\xi t} \,\mathrm {d} \xi \\ &\qquad- \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \biggl( \frac {\sinh^{\gamma(\delta)} ((\mathcal {X}- x)\sqrt{i\xi} )}{\sqrt{i\xi}} \widehat{ \psi^{\delta}}(\xi) \biggr)e^{i\xi t} \, \mathrm {d} \xi \\ &\qquad{}- \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \biggl( \int _{x}^{\mathcal{X}} \frac{\sinh^{\gamma(\delta)} ((z- x)\sqrt{i\xi } )}{\sqrt{i\xi}} \widehat{ \mathcal {W}} \bigl(U^{\delta}_{\gamma (\delta)} \bigr) (z,\xi) \,\mathrm {d}z \biggr) e^{i\xi t} \,\mathrm {d} \xi. \end{aligned}$$

(2.8)

Here $\cosh^{\gamma(\delta)} (y\sqrt{i\xi} )$, $\sinh ^{\gamma(\delta)} (y\sqrt{i\xi} )$ are defined for all $0 \leq y \leq\mathcal {X}$ and $\xi\in\mathbb {R}$ in the following:

$$\begin{aligned}& \cosh^{\gamma(\delta)} (y\sqrt{i\xi} ) = \cosh (y\sqrt{i\xi} ) + \frac{1 }{ 2} \bigl(\mathcal {F}_{\gamma(\delta)}^{\alpha }(\mathcal {X},\xi) - 1 \bigr) e^{\sqrt{i\xi }y} , \end{aligned}$$

(2.9)

$$\begin{aligned}& \sinh^{\gamma(\delta)} (y\sqrt{i\xi} ) =\sinh (y\sqrt{i\xi} )+ \frac{1}{ 2} \bigl( \mathcal {F}_{\gamma (\delta)}^{\alpha }(\mathcal {X},\xi) - 1 \bigr) e^{\sqrt{i\xi}y}, \end{aligned}$$

(2.10)

where

$$ \mathcal {F}_{\gamma(\delta)}^{\alpha }(\mathcal {X},\xi)= \bigl(1+\gamma (\delta) e^{ \sqrt{|\xi|/2} \mathcal {X} } \bigr)^{-\alpha } \quad\mbox{for } \alpha \in\mathbb {N}^{*}. $$

(2.11)

We introduce some notations and assumptions that are needed for our analysis.

Definition 2.1

(Gevrey space)

The Gevrey class of functions of order $\theta\geq0$ defined as

$$\begin{aligned} \text{``GEVREY''} := \biggl\{ w \in \mathscr {L}^{2}(\mathbb {R}) : \int_{-\infty }^{\infty} \exp \bigl(\sqrt{2 \vert \xi \vert }\theta \bigr) \bigl\vert \widehat {w}(\xi) \bigr\vert ^{2} \, \mathrm {d} \xi\leq\infty \biggr\} \end{aligned}$$

(2.12)

is equipped with the norm defined by

$$\begin{aligned} \Vert w \Vert _{G_{\theta}(\mathbb {R})} = \sqrt{ \int_{-\infty}^{\infty} \exp \bigl(\sqrt{2 \vert \xi \vert }\theta \bigr) \bigl\vert \widehat{w}(\xi) \bigr\vert ^{2} \, \mathrm {d} \xi} \leq\infty. \end{aligned}$$

(2.13)

Definition 2.2

For a Hilbert space $\mathbb {X}$, we denote by $\mathscr {L}^{p} (0,\mathcal {X};\mathbb {X} )$ (respectively, $\mathscr {C} ( [0,\mathcal {X} ];\mathbb {X} )$) the Banach spaces of measurable (respectively, continuous functions) functions $w:[0,\mathcal {X}]\to\mathbb {X}$ such that

$$\begin{aligned}& \Vert w \Vert _{ \mathscr {L}^{p} (0,\mathcal {X};\mathbb {X} )}= \biggl( \int_{0}^{\mathcal {X}} \bigl\Vert w (x ) \bigr\Vert _{\mathbb {X}}^{p}\,\mathrm {d}x \biggr)^{\frac{1}{p}}< \infty ,\quad1 \le p< \infty, \\ & \Vert w \Vert _{ \mathscr {L}^{\infty} (0,\mathcal {X};\mathbb {X} )}= \underset{0\le x \le\mathcal {X}}{\operatorname {ess\,sup}} \bigl\Vert w (x ) \bigr\Vert _{\mathbb {X}}< \infty,\quad p=\infty \\ & \Bigl(\text{respectively}, \Vert w \Vert _{\mathscr {C} ( [0,\mathcal {X} ];\mathbb {X} )}=\sup_{0\le x\le\mathcal {X}} \bigl\Vert w (x ) \bigr\Vert _{\mathbb {X}}< \infty\Bigr). \end{aligned}$$

We assume the following:

$(H_{1})$:

The data $\phi,\psi\in\mathscr {L}^{2}(\mathbb {R})$ are noisy and are represented by the observation data $\phi^{\delta},\psi^{\delta}\in\mathscr {L}^{2}(\mathbb {R})$ satisfying

$$\begin{aligned} \bigl\Vert \phi^{\delta}- \phi \bigr\Vert _{\mathscr {L}^{2} (\mathbb {R} )} \leq\delta, \qquad \bigl\Vert \psi^{\delta}- \psi \bigr\Vert _{\mathscr {L}^{2} (\mathbb {R} )} \leq\delta, \end{aligned}$$

(2.14)

here $\delta>0$ is a small positive number representing the level of noise.

$(H_{2})$:

The source functions $f: \mathbb {R} \to\mathbb {R}$ and $G: [0, \mathcal {X}] \times\mathbb {R}^{2} \to\mathbb {R}$ are globally Lipschitz-continuous, i.e., there exist the constants $K_{f},K_{G} \geq0$ such that

$$\begin{aligned}& \bigl\vert f(u) - f(v) \bigr\vert \leq K_{f} \vert u - v \vert , \end{aligned}$$

(2.15a)

$$\begin{aligned}& \bigl\vert G(x,t;u) - G(x,t;v) \bigr\vert \leq K_{G} \vert u - v \vert \end{aligned}$$

(2.15b)

for all $\mathbf{(}x,t) \in[0,\mathcal {X}] \times\mathbb {R}$, $u,v \in \mathbb {R}$.

$(H_{3})$:

There exist the constants $B_{f},B_{G} \geq0$ such that

$$\begin{aligned}& \bigl\Vert f(u) \bigr\Vert _{\mathscr {L}^{\infty}(0,\mathcal {X};\mathscr {L}^{2}(\mathbb {R}))} \leq B_{f}, \end{aligned}$$

(2.16a)

$$\begin{aligned}& \bigl\Vert G( x,\cdot;u) \bigr\Vert _{\mathscr {L}^{\infty}(0,\mathcal {X};\mathscr {L}^{2}(\mathbb {R}))} \leq B_{G} \end{aligned}$$

(2.16b)

for all $u \in\mathscr {C} ([0,\mathcal {X}];\mathscr {L}^{2}(\mathbb {R}) )$.

3 Error estimate in $\mathscr {L}^{p}(\omega,\mathcal {X};\mathscr {L}^{2}(\mathbb {R}))$, $0\leq\omega<\mathcal {X}$

First, we have the following lemmas which will be useful.

Lemma 3.1

For$\xi\in\mathbb{R}$, $\alpha \in\mathbb {N}^{*}$, we have the following inequalities:

$$\begin{aligned}& \mathrm{(a)}\quad \bigl\vert \mathcal {F}^{\alpha }_{\gamma(\delta)}(\mathcal {X},\xi) e^{\sqrt{i\xi} \chi} \bigr\vert \leq \bigl\vert \gamma(\delta) \bigr\vert ^{-\frac{\alpha \chi}{\mathcal {X}}} \quad \textit{for all } 0 \leq \chi\leq\mathcal {X}, \end{aligned}$$

(3.1a)

$$\begin{aligned}& \mathrm{(b)}\quad \bigl\vert \bigl(\mathcal {F}^{\alpha }_{\gamma(\delta)}(\mathcal {X}, \xi) -1 \bigr)e^{-\sqrt{i\xi} \chi} \bigr\vert \leq \bigl\vert \gamma (\delta) \bigr\vert ^{\frac{\alpha \chi}{\mathcal {X}}} \quad\textit{for all } 0 \leq\chi\leq\mathcal {X}, \end{aligned}$$

(3.1b)

$$\begin{aligned}& \mathrm{(c)}\quad \bigl\vert \cosh^{\gamma(\delta)} (\sqrt{i\xi}\chi ) \bigr\vert \le \bigl\vert \gamma(\delta) \bigr\vert ^{-\frac{\alpha \chi }{\mathcal {X}}} \quad\textit{for all } 0 \leq\chi\leq\mathcal {X}, \end{aligned}$$

(3.1c)

$$\begin{aligned}& \mathrm{(d)}\quad \bigl\vert \sinh^{\gamma(\delta)} (\sqrt{i\xi} \chi ) \bigr\vert \le \bigl\vert \gamma(\delta) \bigr\vert ^{-\frac{\alpha \chi }{\mathcal {X}}} \quad\textit{for all } 0 \leq\chi\leq\mathcal {X}. \end{aligned}$$

(3.1d)

Proof

(a) From (2.11) and Euler’s formula, we have

$$\begin{aligned} \bigl\vert \mathcal {F}_{\gamma(\delta)}^{\alpha }(\mathcal {X},\xi) e^{ \sqrt{i\xi} \chi} \bigr\vert &= \biggl\vert \frac{e^{ \sqrt{i\xi} \chi}}{ (1 + \gamma(\delta)e^{\sqrt{ \vert \xi \vert /2 }\mathcal {X}} )^{\alpha }} \biggr\vert = \frac{ \vert e^{ \sqrt{i\xi} \chi} \vert }{ (1 + \gamma(\delta) e^{\sqrt { \vert \xi \vert /2 }\mathcal {X}} )^{\alpha }} \leq\frac{e^{-\alpha \sqrt{ \vert \xi \vert /2} (\mathcal {X}-\chi)}}{ (e^{-\sqrt{ \vert \xi \vert /2}\mathcal {X} } + \gamma(\delta) )^{\alpha }} \\ &= \frac{e^{-\alpha \sqrt{ \vert \xi \vert /2} (\mathcal {X}-\chi)}}{ (e^{-\sqrt{ \vert \xi \vert /2}\mathcal {X} } + \gamma(\delta) )^{\frac{\alpha (\mathcal {X}-\chi)}{\mathcal {X}}} (e^{-\sqrt{ \vert \xi \vert /2}\mathcal {X} } + \gamma (\delta) )^{\frac{\alpha \chi}{\mathcal {X}}} } \\ & \leq\frac{1}{ (e^{-\sqrt{ \vert \xi \vert /2}\mathcal {X} } + \gamma (\delta) )^{\frac{\alpha \chi}{\mathcal {X}}}} \leq \bigl\vert \gamma (\delta) \bigr\vert ^{-\frac{\alpha \chi}{\mathcal {X}}} . \end{aligned}$$

(3.2)

(b) Similarly, from (2.11) and for $a< b$, we have $a^{\alpha }-b^{\alpha }\leq(a-b)^{\alpha }$, so

$$\begin{aligned} & \bigl\vert \bigl(\mathcal {F}^{\alpha }_{\gamma(\delta)}(\mathcal {X},\xi) -1 \bigr)e^{-\sqrt{i\xi}\chi} \bigr\vert \\ &\quad\leq \bigl\vert \gamma(\delta) \bigr\vert ^{\alpha }\frac{ \vert e^{-\sqrt{i\xi }\chi} \vert }{ (e^{-\sqrt{ \vert \xi \vert /2}\mathcal {X} } + \gamma(\delta) )^{\alpha }} \\ &\quad\leq \bigl\vert \gamma(\delta) \bigr\vert ^{\alpha }\frac{e^{-\alpha \sqrt{ \vert \xi \vert /2}\chi}}{ (e^{-\sqrt{ \vert \xi \vert /2}\mathcal {X}} + \gamma(\delta ) )^{\frac{\alpha \chi}{\mathcal {X}}} (e^{-\sqrt{ \vert \xi \vert /2}\mathcal {X} } + \gamma(\delta) )^{\frac{\alpha (\mathcal {X}-\chi)}{\mathcal {X}}}} \\ & \quad\leq\frac{ \vert \gamma(\delta) \vert ^{\alpha }}{ (e^{-\sqrt { \vert \xi \vert /2}\mathcal {X} } + \gamma(\delta) )^{\frac{\alpha (\mathcal {X}-\chi)}{\mathcal {X}}}} \leq \bigl\vert \gamma(\delta) \bigr\vert ^{\frac {\alpha \chi}{\mathcal {X}}}. \end{aligned}$$

(3.3)

(c) From (2.9) and (3.1a), we obtain

$$\begin{aligned} \bigl\vert \cosh^{\gamma(\delta)} (\sqrt{i\xi} \chi ) \bigr\vert &\le \frac{1}{2} \bigl\vert \mathcal {F}^{\alpha }_{\gamma(\delta)}(\mathcal {X},\xi) e^{\sqrt{i\xi} \chi} \bigr\vert + \frac{1}{2} \bigl\vert e^{-\sqrt {i\xi} \chi} \bigr\vert \\ &\le\frac{1}{2} \bigl\vert \gamma(\delta) \bigr\vert ^{-\frac{\alpha \chi }{\mathcal {X}}}+ \frac{1}{2} \\ &\le \bigl\vert \gamma(\delta) \bigr\vert ^{-\frac{\alpha \chi}{\mathcal {X}}}. \end{aligned}$$

From (2.9), an argument similar to the previous one implies (3.1d). □

Lemma 3.2

For$0 \le x \le\mathcal {X}$, $\xi\in\mathbb{R}$, we have

$$\begin{aligned}[b] &\widehat{u}(x,\xi) - \frac{\widehat{\partial_{x} u}( x,\xi)}{ \sqrt{i\xi}} \\ &\quad = e^{\sqrt{i\xi}(\mathcal {X}-x)} \biggl( \widehat{\phi}(\xi) - \frac{\widehat{\psi}(\xi)}{\sqrt{i\xi}} \biggr) - \int _{x}^{\mathcal{X}} \frac{e^{\sqrt{i\xi}(z- x)}}{\sqrt{i\xi}} \widehat { \mathcal {W}}(u) (z,\xi)\, \mathrm {d}z. \end{aligned} $$

(3.4)

Proof

Differentiating (2.6) with respect to x gives

$$\begin{aligned} -\frac{\widehat{\partial_{x}u}( x,\xi)}{\sqrt{i\xi}} &= \sinh \bigl((\mathcal {X}- x)\sqrt{i\xi} \bigr) \widehat{ \phi}(\xi) - \frac {\cosh ((\mathcal {X}- x)\sqrt{i\xi} )}{\sqrt{i\xi}} \widehat{\psi}(\xi) \\ &\quad- \int_{x}^{\mathcal{X}} \frac{\sinh ((z- x)\sqrt{i\xi} )}{\sqrt{i\xi}} \widehat{\mathcal {W}}(u) (z,\xi)\,\mathrm {d}z, \end{aligned}$$

(3.5)

and adding (3.5) to (2.6) completes the proof. □

Our result is in the next theorem.

Theorem 3.1

Let$\gamma(\delta) \in(0,1)$be such that

$$\begin{aligned} \textstyle\begin{cases} \lim_{\delta\to0^{+}} \gamma(\delta) =0, \\ \lim_{\delta\to0^{+}} \delta \vert \gamma(\delta) \vert ^{-\alpha } =0 &\textit{for } \alpha \in\mathbb {N}^{*}. \end{cases}\displaystyle \end{aligned}$$

(3.6)

Then the nonlinear integral equation (2.8) has a solution$U_{\gamma(\delta)}^{\delta}\in\mathscr {C} ([0, \mathcal {X}];\mathscr {L}^{2}(\mathbb {R}))$. Assume that problem (1.1) has a solutionusatisfying

$$\begin{aligned} \Vert u \Vert _{\mathscr {L}^{\infty}(0,\mathcal {X};G_{\mathcal {X}}(\mathbb {R}) )} + \frac{1}{\rho_{0}} \Vert \partial_{x} u \Vert _{\mathscr {L}^{\infty}(0,\mathcal {X};G_{\mathcal {X}}(\mathbb {R}) )} \leq P(u) \end{aligned}$$

(3.7)

for some known constant$P(u)>0$. Then (for all$\omega\in[0,\mathcal {X})$)

$$\begin{aligned} & \bigl\Vert U_{\gamma(\delta)}^{\delta} - u \bigr\Vert _{\mathscr {L}^{p} (\omega,\mathcal {X};\mathscr {L}^{2}(\mathbb {R}) )} \\ &\quad\le C(\alpha ,p,\mathcal {X},\rho_{0},\widetilde{K}) \bigl( \bigl\vert \gamma (\delta) \bigr\vert ^{-\alpha } \delta+ P(u) \bigr) \biggl( \frac{ \vert \gamma(\delta) \vert ^{\frac{p\alpha \omega}{\mathcal {X}}} - \vert \gamma(\delta) \vert ^{p\alpha }}{\log (\frac{1}{ \vert \gamma (\delta) \vert } ) } \biggr)^{\frac{1}{p}}, \end{aligned}$$

(3.8)

where the positive constant$C(\alpha ,p,\mathcal {X},\rho_{0},\widetilde {K})$is a positive constant independent ofωandδ.

Remark 3.1

From (3.6) we infer that the right-hand side of (3.8) tends to zero as ${\delta\to0^{+}}$. Let us choose a parameter regularization $\gamma(\delta)=\delta^{\kappa}$ ($\kappa \leq\frac{1}{\alpha }$), and then the error $\| U_{\gamma(\delta)}^{\delta} - u\|_{\mathscr {L}^{p} (\omega,\mathcal {X};\mathscr {L}^{2}(\mathbb {R}) )}$ is of the order $(\frac{\delta^{\frac{\kappa p \alpha \omega }{\mathcal {X}}} - \delta^{\kappa p \alpha } }{\log (\frac{1}{\delta ^{\kappa}} )} )^{\frac{1}{p}}$, which goes to zero as $\delta\to0^{+}$ for all $\omega\in[0,\mathcal {X})$.

Proof

The proof is divided into two parts.

1st part. Integral equation (2.8) has a unique solution$U_{\gamma(\delta)}^{\delta}\in\mathscr {C} ([0,\mathcal {X}];\mathscr {L}^{2}(\mathbb {R}) )$. We put

$$\begin{aligned}& \frac{1}{\overline{d}}=N= \biggl\lfloor \frac{ \vert \gamma(\delta ) \vert ^{-\alpha } B_{f}B_{G} }{\mho- \vert \gamma(\delta) \vert ^{-\alpha } ( \Vert \phi_{j}^{\delta} \Vert _{\mathscr {L}^{2}(\mathbb {R})} + \Vert \psi_{j}^{\delta} \Vert _{\mathscr {L}^{2}(\mathbb {R})} ) } \biggr\rfloor + \lfloor B_{G}K_{f}+B_{f}K_{G} \rfloor, \\& \quad\mho> \bigl\vert \gamma(\delta) \bigr\vert ^{-\alpha } \bigl( \bigl\Vert \phi_{j}^{\delta}\bigr\Vert _{\mathscr {L}^{2}(\mathbb {R})} + \bigl\Vert \psi_{j}^{\delta}\bigr\Vert _{\mathscr {L}^{2}(\mathbb {R})} \bigr), \end{aligned}$$

(3.9)

where $\lfloor y \rfloor$ is the integer part of the real number y. Let us define

$$\begin{aligned} \mathcal {X}_{j} &= \mathcal {X}-j\overline{d},\quad j=0,1,\ldots,N, \end{aligned}$$

and

$$\begin{aligned} \mathscr{M}_{j} ={}& \Bigl\{ \vartheta\in\mathscr {C} \bigl({[\mathcal {X}_{j+1},\mathcal {X}_{j}}]; \mathscr {L}^{2}(\mathbb {R}) \bigr) \big| \bigl( \vartheta(\mathcal {X}_{j},t), \partial_{x} \vartheta(\mathcal {X}_{j},t) \bigr)= \bigl(\phi_{j}^{\delta}(t), \psi_{j}^{\delta}(t) \bigr), \\ & \Vert \vartheta \Vert _{\oplus}= \sup_{x\in[\mathcal {X}_{j},\mathcal {X}_{j+1}]} \bigl\Vert \vartheta(x,\cdot) \bigr\Vert _{\mathscr {L}^{2}(\mathbb {R})} \leq\mho, x\in{[ \mathcal {X}_{j+1},\mathcal {X}_{j}]} \Bigr\} , \quad j=0,1,\ldots,N-1. \end{aligned}$$

We also define (for $\mathcal {X}_{j} -\overline{d} \leq x \leq z \leq \mathcal {X}_{j}$) $j=0,1,\ldots,N-1$

$$\begin{aligned} \mathscr{B}_{j}^{\gamma(\delta)}(\vartheta) (x,t) &:= { \frac {1}{\sqrt{2\pi}}} \int_{-\infty}^{\infty} \cosh^{\gamma(\delta)} \bigl((\mathcal {X}_{j}- x)\sqrt{i\xi} \bigr) \widehat{\phi_{j}^{\delta}}(\xi) e^{i\xi t} \,\mathrm {d} \xi \\ &\quad- {\frac{1}{\sqrt{2\pi}}} \int_{-\infty}^{\infty} \frac{\sinh ^{\gamma(\delta)} ((\mathcal {X}_{j}- x)\sqrt{i\xi} )}{\sqrt {i\xi}} \widehat{ \psi_{j}^{\delta}}(\xi) e^{i\xi t} \,\mathrm {d} \xi \\ &\quad- {\frac{1}{\sqrt{2\pi}}} \int_{-\infty}^{\infty} \biggl( \int _{x}^{\mathcal {X}_{j}}\frac{ \sinh^{\gamma(\delta)} ((z-x)\sqrt {i\xi} )}{\sqrt{i\xi}} \widehat{\mathcal {W}}(\vartheta) (z,\xi ) \,\mathrm {d}z \biggr) e^{i\xi t} \,\mathrm {d} \xi. \end{aligned}$$

For $\vartheta\in\mathscr{M}_{j}$ and $\mathcal {X}_{j} -\overline{d} \leq x \leq z \leq\mathcal {X}_{j}$, using Lemma 3.1, we obtain

$$\begin{aligned} \bigl\Vert \mathscr{B}_{j}^{\gamma(\delta)}(\vartheta) (\cdot,t) \bigr\Vert _{\mathscr {L}^{2}(\mathbb {R})} & \leq \bigl\vert \gamma(\delta) \bigr\vert ^{\frac{\alpha (x-\mathcal {X}_{j})}{\mathcal {X}_{j}}} \bigl\Vert \phi_{j}^{\delta}\bigr\Vert _{\mathscr {L}^{2}(\mathbb {R})} + \bigl\vert \gamma(\delta) \bigr\vert ^{\frac{\alpha (x-\mathcal {X}_{j})}{\mathcal {X}_{j}}} \bigl\Vert \psi_{j}^{\delta}\bigr\Vert _{\mathscr {L}^{2}(\mathbb {R})} \\ &\quad+ \int_{x}^{\mathcal {X}_{j}} \bigl\vert \gamma(\delta) \bigr\vert ^{\frac{\alpha (x-z)}{\mathcal {X}_{j}}} \bigl\Vert \mathcal {W} (z,\cdot;\vartheta ) \bigr\Vert _{\mathscr {L}^{2}(\mathbb {R})}. \end{aligned}$$

Multiplying by $|\gamma(\delta) |^{\frac{-\alpha x }{\mathcal {X}_{j}}}$ on both sides, we obtain

$$\begin{aligned} \bigl\vert \gamma(\delta) \bigr\vert ^{\frac{-\alpha x}{\mathcal {X}_{j}}} \bigl\Vert \mathscr{B}_{j}^{\gamma(\delta)}(\vartheta) (\cdot,t) \bigr\Vert _{\mathscr {L}^{2}(\mathbb {R})} &\leq \bigl\vert \gamma(\delta) \bigr\vert ^{-\alpha } \bigl( \bigl\Vert \phi _{j}^{\delta}\bigr\Vert _{\mathscr {L}^{2}(\mathbb {R})} + \bigl\Vert \psi_{j}^{\delta}\bigr\Vert _{\mathscr {L}^{2}(\mathbb {R})} \bigr) \\ &\quad+ \int_{x}^{\mathcal {X}_{j}} \bigl\vert \gamma(\delta) \bigr\vert ^{\frac{-\alpha z}{\mathcal {X}_{j}}} \bigl\Vert \mathcal {W} (z,\cdot;\vartheta ) \bigr\Vert _{\mathscr {L}^{2}(\mathbb {R})} \,\mathrm {d}z. \end{aligned}$$

(3.10)

Using $(H_{3})$, the second term in (3.10) becomes

$$\begin{aligned} & \int_{x}^{\mathcal {X}_{j}} \bigl\vert \gamma(\delta) \bigr\vert ^{\frac{-\alpha z}{\mathcal {X}_{j}}} \bigl\Vert \mathcal {W} (z,\cdot;\vartheta ) \bigr\Vert _{\mathscr {L}^{2}(\mathbb {R})}\,\mathrm {d} z \\ &\quad= \int _{x}^{\mathcal {X}_{j}} \bigl\vert \gamma(\delta) \bigr\vert ^{\frac{-\alpha z}{\mathcal {X}_{j}}} \bigl\Vert f (\vartheta ) \bigr\Vert _{\mathscr {L}^{2}(\mathbb {R})} \bigl\Vert G (z,\cdot;\vartheta ) \bigr\Vert _{\mathscr {L}^{2}(\mathbb {R})} \,\mathrm {d} z \\ &\quad\leq(\mathcal {X}_{j}-x) \bigl\vert \gamma(\delta) \bigr\vert ^{-\alpha } B_{f}B_{G}. \end{aligned}$$

(3.11)

From (3.10), (3.11), we have for $\mathcal {X}_{j} - \overline{d} \leq x \leq z \leq\mathcal {X}_{j}$

$$\begin{aligned} \bigl\Vert \mathscr{B}_{j}^{\gamma(\delta)} (\vartheta) \bigr\Vert _{\oplus} &\leq \bigl\vert \gamma(\delta) \bigr\vert ^{-\alpha } \bigl( \bigl\Vert \phi_{j}^{\delta}\bigr\Vert _{\mathscr {L}^{2}(\mathbb {R})} + \bigl\Vert \psi _{j}^{\delta}\bigr\Vert _{\mathscr {L}^{2}(\mathbb {R})} + (\mathcal {X}_{j}-x) B_{f}B_{G} \bigr). \end{aligned}$$

From (3.9), we have

$$\begin{aligned} \frac{1}{\overline{d}} \geq\frac{ \vert \gamma(\delta) \vert ^{-\alpha } B_{f}B_{G} }{\mho- \vert \gamma(\delta) \vert ^{-\alpha } ( \Vert \phi_{j}^{\delta} \Vert _{\mathscr {L}^{2}(\mathbb {R})} + \Vert \psi_{j}^{\delta} \Vert _{\mathscr {L}^{2}(\mathbb {R})} ) }. \end{aligned}$$

(3.12)

This implies that

$$\begin{aligned} \overline{d}< \frac{\mho- \vert \gamma(\delta) \vert ^{-\alpha } ( \Vert \phi_{j}^{\delta} \Vert _{\mathscr {L}^{2}(\mathbb {R})} + \Vert \psi_{j}^{\delta} \Vert _{\mathscr {L}^{2}(\mathbb {R})} ) }{ \vert \gamma(\delta) \vert ^{-\alpha } B_{f}B_{G}}, \end{aligned}$$

(3.13)

so $\Vert \mathscr{B}_{j}^{\gamma(\delta)} (\vartheta) \Vert _{\oplus} \leq\mho$, i.e., $\mathscr{B}_{j}^{\gamma(\delta)} (\mathscr{M}_{j} ) \subset\mathscr{M}_{j}$.

Now, for $\vartheta_{1}, \vartheta_{2} \in\mathscr{M}_{j}$, we invoke Lemma 3.1 to deduce that, for $\mathcal {X}_{j} -\overline{d} \leq x \leq z \leq\mathcal {X}_{j}$,

$$\begin{aligned} & \bigl\Vert \mathscr{B}_{j}^{\gamma(\delta)}(\vartheta_{1}) ( x,\cdot) - \mathscr{B}_{j}^{\gamma(\delta)}(\vartheta_{2}) (x,\cdot) \bigr\Vert _{\mathscr {L}^{2}(\mathbb {R})} \\ & \quad\leq \int_{x}^{\mathcal {X}_{j}} \bigl\vert \gamma(\delta) \bigr\vert ^{\frac{\alpha (x-z)}{\mathcal {X}_{j}}} \bigl\Vert \mathcal {W} (z,\cdot ;\vartheta_{1} ) - \mathcal {W} (z,\cdot;\vartheta_{2} ) \bigr\Vert _{\mathscr {L}^{2}(\mathbb {R})}\, \mathrm {d}z. \end{aligned}$$

(3.14)

From hypotheses $(H_{2})$ and $(H_{3})$, we infer that

$$\begin{aligned} & \bigl\Vert \mathcal {W} ( x,\cdot;\vartheta_{1} ) - \mathcal {W} ( x, \cdot;\vartheta_{2} ) \bigr\Vert _{\mathscr {L}^{2}(\mathbb {R})} \\ &\quad = \bigl\Vert f (\vartheta_{1} )G ( x,\cdot;\vartheta_{1} ) - f (\vartheta_{2} ) G ( x,\cdot;\vartheta_{2} ) \bigr\Vert _{\mathscr {L}^{2}(\mathbb {R})} \\ &\quad\leq \bigl\Vert G ( x,\cdot;\vartheta_{1} ) \bigr\Vert _{\mathscr {L}^{2}(\mathbb {R})} \bigl\Vert f ( \vartheta_{1} ) - f (\vartheta _{2} ) \bigr\Vert _{\mathscr {L}^{2}(\mathbb {R})} \\ &\quad\quad+ \bigl\Vert f (\vartheta_{2} ) \bigr\Vert _{\mathscr {L}^{2}(\mathbb {R})} \bigl\Vert G ( x,\cdot;\vartheta_{1} ) - G ( x,\cdot ; \vartheta_{2} ) \bigr\Vert _{\mathscr {L}^{2}(\mathbb {R})} \\ &\quad\leq (B_{G}K_{f} + B_{f}K_{G} ) \bigl\Vert \vartheta_{1}( x,\cdot) - \vartheta_{2} (x, \cdot) \bigr\Vert _{\mathscr {L}^{2}(\mathbb {R})}. \end{aligned}$$

(3.15)

We put (3.15) into (3.14) and multiply both sides of the above result by $|\gamma(\delta) |^{\frac{-\alpha x}{\mathcal {X}_{j}}}$ to get

$$\begin{aligned} & \bigl\vert \gamma(\delta) \bigr\vert ^{\frac{-\alpha x}{\mathcal {X}_{j}}} \bigl\Vert \mathscr{B}_{j}^{\gamma(\delta)}(\vartheta_{1}) ( x,\cdot) - \mathscr{B}_{j}^{\gamma(\delta)}(\vartheta_{2}) (x,\cdot) \bigr\Vert _{\mathscr {L}^{2}(\mathbb {R})} \\ &\quad\leq \int_{x}^{\mathcal {X}_{j}} \bigl\vert \gamma(\delta) \bigr\vert ^{\frac{-\alpha z}{\mathcal {X}_{j}}} (B_{G}K_{f}+B_{f}K_{G} ) \bigl\Vert \vartheta _{1}(z,\cdot) - \vartheta_{2}(z, \cdot) \bigr\Vert _{\mathscr {L}^{2}(\mathbb {R})} \,\mathrm {d}z \\ &\quad\leq(\mathcal {X}_{j}-x) (B_{G}K_{f}+B_{f}K_{G} ) \Vert \vartheta_{1} - \vartheta_{2} \Vert _{\oplus}, \end{aligned}$$

so

$$\begin{aligned} \bigl\Vert \mathscr{B}_{j}^{\gamma(\delta)}(\vartheta_{1}) - \mathscr {B}_{j}^{\gamma(\delta)}(\vartheta_{2}) \bigr\Vert _{\oplus} &\leq\overline{d} (B_{G}K_{f}+B_{f}K_{G} ) \Vert \vartheta_{1} - \vartheta_{2} \Vert _{\oplus}. \end{aligned}$$

(3.16)

From (3.9), we have

$$\frac{1}{\overline{d}} \geq B_{G}K_{f}+B_{f}K_{G}, $$

which implies that

$$\begin{aligned} \overline{d} \leq\frac{1}{B_{G}K_{f} + B_{f}K_{G}}. \end{aligned}$$

(3.17)

Hence we have $\Vert \mathscr{B}_{j}^{\gamma(\delta)}(\vartheta_{1}) - \mathscr {B}_{j}^{\gamma(\delta)}(\vartheta_{2}) \Vert _{\oplus} \leq\varrho \Vert \vartheta_{1} - \vartheta_{2} \Vert _{\oplus}$, $\varrho\in (0,1)$, so $\mathscr{B}_{j}^{\gamma(\delta)}$ is a contraction on $\mathscr{M}_{j}$. The Banach fixed point principle guarantees that there exists unique $\vartheta\in\mathscr{M}_{j}$ such that

$$\mathscr{B}_{j}^{\gamma(\delta)}(\vartheta) (x ,t)=\vartheta(x ,t). $$

In fact, for $j=0$, we put $\phi_{0}^{\delta}(t) = \phi^{\delta}(t)$ and $\psi_{0}^{\delta}(t) = \psi^{\delta}(t)$. Using the results just obtained, we can find $U_{\gamma(\delta),0}^{\delta}\in\mathscr {C} ([\mathcal {X}_{1},\mathcal {X}];\mathscr {L}^{2} (\mathbb {R} ) )$ such that $\mathscr{B}_{0} (U_{\gamma(\delta ),0}^{\delta}) = U_{\gamma(\delta),0}^{\delta}$. Assume that we can find $U_{\gamma(\delta),j-1}^{\delta}\in\mathscr {C} ([\mathcal {X}_{j},\mathcal {X}_{j-1}];\mathscr {L}^{2} (\mathbb {R} ) )$ such that $\mathscr{B}_{j-1} (U_{\gamma(\delta ),j-1}^{\delta}) = U_{\gamma(\delta),j-1}^{\delta}$. We now prove that we can extend this solution to the interval $[\mathcal {X}_{j+1},\mathcal {X}_{j}]$. Indeed, put $\phi^{\delta}_{j}(t) = U_{\gamma (\delta),j}^{\delta}( \mathcal {X}_{j},t)$ and $\psi^{\delta}_{j}(t) = \partial_{x} U^{\delta}(\mathcal {X}_{j},t)$. We can find $U_{\gamma(\delta ),j}^{\delta}\in\mathscr {C} ([\mathcal {X}_{j+1},\mathcal {X}_{j}];\mathscr {L}^{2} (\mathbb {R} ) )$ such that $\mathscr{B}_{j} U_{\gamma(\delta),j}^{\delta}= U_{\gamma(\delta ),j}^{\delta}$. So we can extend the solution $U_{\gamma(\delta )}^{\delta}$ on $[\mathcal {X}_{j+1},\mathcal {X}_{j}]$ by putting $U^{\delta}_{\gamma(\delta)}( x ,t)=U^{\delta}_{\gamma(\delta),j}(x,t)$ for $\mathcal {X}_{j+1} \leq x \leq\mathcal {X}_{j}$. By induction, we can complete the proof of this step.

2nd part. Error estimate$\Vert U_{\gamma(\delta )}^{\delta}- u \Vert _{\mathscr {L}^{p} (\omega,\mathcal {X};\mathscr {L}^{2}(\mathbb {R}) )}$. Note

$$\begin{aligned} & \bigl\Vert U^{\delta}_{\gamma(\delta)}-u \bigr\Vert _{\mathscr {L}^{p} (\omega,\mathcal {X};\mathscr {L}^{2}(\mathbb {R}) )} \\ &\quad\le \bigl\Vert U^{\delta}_{\gamma(\delta)}-V_{\gamma(\delta)} \bigr\Vert _{\mathscr {L}^{p} (\omega,\mathcal {X};\mathscr {L}^{2}(\mathbb {R}) )} + \Vert V_{\gamma(\delta)} - u \Vert _{\mathscr {L}^{p} (\omega ,\mathcal {X};\mathscr {L}^{2}(\mathbb {R}) )}, \end{aligned}$$

(3.18)

where $V_{\gamma(\delta)}( x,t)$ is defined by

$$\begin{aligned} V_{\gamma(\delta)}(x,t) &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \bigl( \cosh ^{\gamma(\delta)} \bigl(( \mathcal {X}- x)\sqrt{i\xi} \bigr) \widehat {\phi}(\xi) \bigr) e^{i\xi t} \, \mathrm {d} \xi \\ &\quad- \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \biggl( \frac {\sinh^{\gamma(\delta)} ((\mathcal {X}- x)\sqrt{i\xi} )}{\sqrt{i\xi}} \widehat{ \psi}(\xi) \biggr) e^{i\xi t} \,\mathrm {d} \xi \\ &\quad- \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \biggl( \int_{ x}^{\mathcal{X}}\frac{ \sinh^{\gamma(\delta)} ((z- x)\sqrt{i\xi } )}{\sqrt{i\xi}} \widehat{\mathcal {W}} (V_{\gamma(\delta )} ) (z,\xi) \,\mathrm {d}z \biggr) e^{i\xi t} \,\mathrm {d} \xi. \end{aligned}$$

(3.19)

From (2.6) and Lemma 3.2, we have

$$\begin{aligned} \widehat{u}(x,\xi) & = \cosh^{\gamma(\delta)} \bigl((\mathcal {X}- x)\sqrt{i\xi} \bigr) \widehat{\phi}(\xi) - \frac{\sinh^{\gamma (\delta)} ((\mathcal {X}- x)\sqrt{i\xi} )}{\sqrt{i\xi}} \widehat{\psi}(\xi) \\ &\quad- \int_{ x}^{\mathcal{X}}\frac{ \sinh^{\gamma(\delta)} ((z- x)\sqrt{i\xi} )}{\sqrt{i\xi}} \widehat{\mathcal {W}} (u ) (z,\xi) \,\mathrm {d}z + \frac{1}{2} \bigl(1-\mathcal {F}^{\alpha }_{\gamma (\delta)}(\mathcal {X},\xi) \bigr) \\ &\quad\times \biggl(e^{\sqrt{i\xi}(\mathcal {X}- x)} \biggl(\widehat{\phi }(\xi)+\frac{\widehat{\psi}(\xi)}{\sqrt{i\xi}} \biggr) - \int _{x}^{\mathcal{X}} \frac{e^{\sqrt{i\xi}(z- x)}}{\sqrt{i\xi}}\widehat { \mathcal {W}}(u) (z,\xi) \,\mathrm {d}z \biggr) \\ &= \cosh^{\gamma(\delta)} \bigl((\mathcal {X}- x)\sqrt{i\xi} \bigr) \widehat{\phi}( \xi) - \frac{\sinh^{\gamma(\delta)} ((\mathcal {X}- x)\sqrt{i\xi} )}{\sqrt{i\xi}} \widehat{\psi }(\xi) \\ & \quad- \int_{ x}^{\mathcal{X}}\frac{ \sinh^{\gamma(\delta)} ((z- x)\sqrt{i\xi} )}{\sqrt{i\xi}} \widehat{\mathcal {W}} (u ) (z,\xi) \,\mathrm {d}z \\ &\quad+ \frac{1}{2} \bigl(1-\mathcal {F}^{\alpha }_{\gamma(\delta)}(\mathcal {X},\xi) \bigr) \biggl[\widehat{u}(x,\xi) - \frac{\widehat{\partial _{x} u}(x,\xi)}{\sqrt{i\xi}} \biggr]. \end{aligned}$$

(3.20)

Note

$$\begin{aligned} & \bigl\Vert V_{\gamma(\delta)}( x,\cdot)-u( x,\cdot) \bigr\Vert _{\mathscr {L}^{2}(\mathbb {R})}^{2} \\ &\quad= \bigl\Vert \widehat{V_{\gamma(\delta)}}( x,\cdot)-\widehat{u}( x,\cdot) \bigr\Vert _{\mathscr {L}^{2}(\mathbb {R})}^{2} \\ &\quad \le2 \int_{-\infty}^{\infty} \int_{x}^{\mathcal{X}} \biggl\vert \frac {\sinh^{\gamma(\delta)} ((z- x)\sqrt{i\xi} )}{\sqrt {i\xi}} \biggr\vert ^{2} \bigl\vert \widehat{\mathcal {W}}(V_{\gamma(\delta )} ) (z,\xi)-\widehat{\mathcal {W}}(u ) (z,\xi) \bigr\vert ^{2} \,\mathrm {d}z \,\mathrm {d} \xi \\ &\qquad+ 2 \int_{-\infty}^{\infty} \biggl\vert \frac{1}{2} \bigl( \mathcal {F}^{\alpha }_{{\gamma(\delta)}} (\mathcal {X},\xi )-1 \bigr)e^{-\sqrt{i\xi} x} \biggr\vert ^{2} \biggl\vert e^{\sqrt{i\xi} x} \widehat {u}(x,\xi) - e^{\sqrt{i\xi} x}\frac{\widehat{\partial_{x}u}( x,\xi )}{\sqrt{i\xi}} \biggr\vert ^{2} \,\mathrm {d} \xi \\ &\quad\le2 \int_{-\infty}^{\infty} \biggl( \int_{x}^{\mathcal{X}} \biggl\vert \frac{\sinh^{\gamma(\delta)} ((z- x)\sqrt{i\xi} )}{\sqrt{i\xi}} \biggr\vert ^{2} \,\mathrm {d}z \int_{x}^{\mathcal{X}} \bigl\vert \widehat{\mathcal {W}}(V_{\gamma(\delta)} ) (z,\xi)-\widehat {\mathcal {W}}(u ) (z,\xi) \bigr\vert ^{2} \,\mathrm {d}z \biggr) \,\mathrm {d} \xi \\ &\qquad+ 4 \int_{-\infty}^{\infty} \biggl\vert \frac{1}{2} \bigl( \mathcal {F}^{\alpha }_{\gamma(\delta)} (\mathcal {X},\xi )-1 \bigr)e^{-\sqrt {i\xi}x} \biggr\vert ^{2} \biggl( \bigl\vert e^{\sqrt{i\xi} x} \widehat{u}( x,\xi) \bigr\vert ^{2} + \biggl\vert e^{\sqrt{i\xi} x} \frac{\widehat{\partial _{x}u}( x,\xi)}{\sqrt{i\xi}} \biggr\vert ^{2} \biggr) \,\mathrm {d} \xi. \end{aligned}$$

From Lemma 3.1, we get

$$\begin{aligned} & \bigl\Vert V_{\gamma(\delta)}(x ,\cdot)-u( x,\cdot) \bigr\Vert _{\mathscr {L}^{2}(\mathbb {R})}^{2} \\ &\quad\le2 \int_{-\infty}^{\infty} \frac{\mathcal {X} }{\rho_{0}} \bigl\vert \gamma( \delta) \bigr\vert ^{\frac{2\alpha (x-z)}{\mathcal {X}}} \int _{x}^{\mathcal{X}} \bigl\vert \widehat{\mathcal {W}}( V_{\gamma(\delta)} ) (z,\xi)-\widehat{\mathcal {W}}(u ) (z,\xi) \bigr\vert ^{2} \,\mathrm {d}z \, \mathrm {d} \xi \\ &\qquad+ 4 \int_{-\infty}^{\infty} \bigl\vert \gamma(\delta) \bigr\vert ^{\frac {2\alpha x}{\mathcal {X}}} \biggl(e^{\sqrt{2 \vert \xi \vert }\mathcal {X}} \bigl\vert \widehat {u}( x,\xi) \bigr\vert ^{2} + \frac{e^{\sqrt{2 \vert \xi \vert }\mathcal {X}}}{\rho _{0}} \bigl\vert \widehat{ \partial_{x}u}( x,\xi) \bigr\vert ^{2} \biggr) \,\mathrm {d} \xi \\ &\quad\le2 \int_{x}^{\mathcal{X}} \frac{\mathcal {X}}{\rho_{0}} \bigl\vert \gamma (\delta) \bigr\vert ^{\frac{2\alpha (x-z)}{\mathcal {X}}} \bigl\Vert \widehat {\mathcal {W}}( V_{\gamma(\alpha )} ) (z,\cdot)-\widehat{\mathcal {W}}(u ) (z,\cdot) \bigr\Vert _{\mathscr {L}^{2}(\mathbb {R})}^{2} \,\mathrm {d}z \\ &\qquad+ 4 \bigl\vert \gamma(\delta) \bigr\vert ^{\frac{2\alpha x}{\mathcal {X}}} \biggl( \bigl\Vert \widehat{u}( x,\cdot) \bigr\Vert _{ G_{\mathcal{X}}(\mathbb {R})}^{2} + \frac{1}{\rho_{0}} \bigl\Vert \widehat{\partial_{x}u}(x,\cdot) \bigr\Vert _{G_{\mathcal{X}}(\mathbb {R})}^{2} \biggr). \end{aligned}$$

Hence, we get

$$\begin{aligned} & \bigl\Vert V_{\gamma(\delta)}(x ,\cdot)-u( x,\cdot) \bigr\Vert _{\mathscr {L}^{2}(\mathbb {R})}^{2} \\&\quad\le2 \int_{x}^{\mathcal{X}} \frac{\mathcal {X} }{\rho_{0}} \bigl\vert \gamma(\delta) \bigr\vert ^{\frac{2\alpha (x-z)}{\mathcal {X}}} \bigl\Vert \mathcal {W}(z,\cdot; V_{\gamma(\delta)} ) - \mathcal {W} (z,\cdot;u ) \bigr\Vert _{\mathscr {L}^{2}(\mathbb {R})}^{2} \,\mathrm {d}z \\ &\qquad+4 \bigl\vert \gamma(\delta) \bigr\vert ^{\frac{2\alpha x}{\mathcal {X}}} \biggl( \Vert u \Vert _{\mathscr {L}^{\infty}(0,\mathcal {X};G_{\mathcal{X}}(\mathbb {R}) )}^{2} + \frac{1}{\rho_{0}} \Vert \partial_{x} u \Vert _{\mathscr {L}^{\infty}(0,\mathcal {X}; G_{\mathcal{X}}(\mathbb {R}) )}^{2} \biggr) \\ &\quad\le2 \int_{x}^{\mathcal{X}} \frac{\mathcal {X}}{\rho_{0}} \bigl\vert \gamma (\delta) \bigr\vert ^{\frac{2\alpha (x-z)}{\mathcal {X}}} \bigl\Vert \mathcal {W} ( z, \cdot;V_{\gamma(\delta)} ) - \mathcal {W} (z,\cdot;u ) \bigr\Vert _{\mathscr {L}^{2}(\mathbb {R})}^{2} \,\mathrm {d}z + 4 \bigl\vert \gamma (\delta) \bigr\vert ^{\frac{2\alpha x}{\mathcal {X}}} P^{2}(u). \end{aligned}$$

Moreover, like in (3.15), we obtain

$$\begin{aligned} \bigl\Vert \mathcal {W} ( x,\cdot,V_{\gamma(\delta)} ) - \mathcal {W} (x,\cdot,u ) \bigr\Vert _{\mathscr {L}^{2}(\mathbb {R})} &\le (K_{f} B_{G} + K_{G} B_{f} ) \bigl\Vert V_{\gamma(\delta)}( x,\cdot) - u( x,\cdot) \bigr\Vert _{\mathscr {L}^{2}(\mathbb {R})} \\ &\leq\widetilde{K} \bigl\Vert V_{\gamma(\delta)}(x,\cdot) - u( x,\cdot) \bigr\Vert _{\mathscr {L}^{2}(\mathbb {R})}, \end{aligned}$$

where $\widetilde{K}:=K_{f} B_{G} + K_{G} B_{f}$.

Note

$$\begin{aligned} & \bigl\Vert V_{\gamma(\delta)}( x,\cdot) - u( x,\cdot) \bigr\Vert _{\mathscr {L}^{2}(\mathbb {R})}^{2} \\ &\quad\le4 \bigl\vert \gamma(\delta) \bigr\vert ^{\frac{2\alpha x}{\mathcal {X}}} P^{2}(u) + \frac{2\mathcal {X} }{\rho_{0}} \widetilde{K} \int_{x}^{\mathcal{X}} \bigl\vert \gamma(\delta) \bigr\vert ^{\frac{2\alpha (x-z)}{\mathcal {X}}} \bigl\Vert V_{\gamma(\delta)}(z,\cdot) - u(z,\cdot) \bigr\Vert _{\mathscr {L}^{2}(\mathbb {R})}^{2} \,\mathrm {d}z. \end{aligned}$$

Thus, we have the following inequality:

$$\begin{aligned} & \bigl\vert \gamma(\delta) \bigr\vert ^{-\frac{2\alpha x}{\mathcal {X}}} \bigl\Vert V_{\gamma(\delta)}(x,\cdot) - u( x,\cdot) \bigr\Vert _{\mathscr {L}^{2}(\mathbb {R})}^{2} \\ &\quad\le4 P^{2}(u) + \frac{2\mathcal {X}}{\rho_{0}} \widetilde{K} \int _{x}^{\mathcal{X}} \bigl\vert \gamma(\delta) \bigr\vert ^{-\frac{2\alpha z}{\mathcal {X}}} \bigl\Vert V_{\gamma(\delta)}(z,\cdot) - u(z,\cdot) \bigr\Vert _{\mathscr {L}^{2}(\mathbb {R})}^{2} \,\mathrm {d}z. \end{aligned}$$

From Gronwall’s inequality, we conclude that

$$\bigl\vert \gamma(\delta) \bigr\vert ^{-\frac{2\alpha x}{\mathcal {X}}} \bigl\Vert V_{\gamma(\delta)}(x,\cdot) - u( x,\cdot) \bigr\Vert _{\mathscr {L}^{2}(\mathbb {R})}^{2} \le4 P^{2}(u) \exp \biggl( \frac{2\mathcal {X}(\mathcal {X}- x)}{\rho_{0}} \widetilde{K} \biggr). $$

Hence, we deduce that

$$\begin{aligned} & \Vert V_{\gamma(\delta)} - u \Vert _{\mathscr {L}^{p} (\omega ,\mathcal {X};\mathscr {L}^{2}(\mathbb {R}) )} \\ &\quad\le2 \frac{\mathcal {X}}{p\alpha } P(u) \exp \biggl(\frac{\mathcal {X}^{2}\widetilde{K} }{\rho_{0}} \biggr) \biggl( \frac{ \vert \gamma (\delta) \vert ^{\frac{p\alpha \omega}{\mathcal {X}}} - \vert \gamma (\delta) \vert ^{p\alpha }}{\log (\frac{1}{ \vert \gamma(\delta ) \vert } ) } \biggr)^{\frac{1}{p}}. \end{aligned}$$

(3.21)

Next, we estimate $\Vert U^{\delta}_{\gamma(\delta)}-V_{\gamma (\delta)} \Vert _{\mathscr {L}^{p} (\omega,\mathcal {X};\mathscr {L}^{2}(\mathbb {R}) )}$. Using the basic inequality $(a + b + c)^{2} \leq3(a^{2} + b^{2} + c^{2})$ and Hölder’s inequality, we obtain

$$\begin{aligned} & \bigl\Vert U^{\delta}_{\gamma(\delta)}( x,\cdot)-V_{\gamma(\delta )}(x, \cdot) \bigr\Vert _{\mathscr {L}^{2}(\mathbb {R})}^{2}\\&\quad = \bigl\Vert \widehat {U^{\delta}_{\gamma(\delta)}}( x,\cdot)-\widehat{V_{\gamma(\delta )}}( x,\cdot) \bigr\Vert _{\mathscr {L}^{2}(\mathbb {R})}^{2} \\ &\quad\le3 \int_{-\infty}^{\infty} \bigl\vert \cosh^{\gamma(\delta)} \bigl((\mathcal {X}- x)\sqrt{i\xi} \bigr) \bigr\vert ^{2} \bigl\vert \widehat{\phi ^{\delta}}(\xi)-\widehat{\phi}(\xi) \bigr\vert ^{2} \,\mathrm {d} \xi \\ &\qquad+ 3 \int_{-\infty}^{\infty} \biggl\vert \frac{\sinh^{\gamma(\delta)} ((\mathcal {X}- x)\sqrt{i\xi} )}{\sqrt{i\xi}} \biggr\vert ^{2} \bigl\vert \widehat{\psi^{\delta}}(\xi) - \widehat{ \psi}(\xi) \bigr\vert ^{2} \,\mathrm {d} \xi \\ &\qquad+ 3 \int_{-\infty}^{\infty} \int_{x}^{\mathcal{X}} \biggl\vert \frac{ \sinh ^{\gamma(\delta)} ((z-x)\sqrt{i\xi} )}{\sqrt{i\xi }} \biggr\vert ^{2} \,\mathrm {d} z\\&\qquad\times \int_{x}^{\mathcal{X}} \bigl\vert \widehat {\mathcal {W}} \bigl(U^{\delta}_{\gamma(\delta)} \bigr) (z,\xi) - \widehat {\mathcal {W}} (V_{\gamma(\delta)} ) (z,\xi) \bigr\vert ^{2} \,\mathrm {d}z \,\mathrm {d} \xi. \end{aligned}$$

Similar calculations as in (3.15) yield

$$\begin{aligned} & \| \mathcal {W} \bigl(x,\cdot,U^{\delta}_{\gamma(\delta)} \bigr) - \mathcal {W} ( x,\cdot,V_{\gamma(\delta)} \|_{\mathscr {L}^{2}(\mathbb {R})} \leq\widetilde{K} \bigl\Vert U^{\delta}_{\gamma(\delta)}( x,\cdot) - V_{\gamma(\delta )}( x,\cdot) \bigr\Vert _{\mathscr {L}^{2}(\mathbb {R})}. \end{aligned}$$

From Lemma 3.1 and using the Lipschitzian property of $\mathcal {W}$, we get the following inequality:

$$\begin{aligned} & \bigl\Vert U^{\delta}_{\gamma(\delta)}( x,\cdot)-V_{\gamma(\delta )}(x, \cdot) \bigr\Vert _{\mathscr {L}^{2}(\mathbb {R})}^{2} \\ &\quad\le3 \bigl\vert \gamma(\delta) \bigr\vert ^{\frac{2\alpha (x-\mathcal {X})}{\mathcal {X}}} \bigl\Vert \phi^{\delta}- \phi \bigr\Vert ^{2}_{\mathscr {L}^{2}(\mathbb {R})} + 3 \bigl\vert \gamma(\delta) \bigr\vert ^{\frac{2\alpha (x-\mathcal {X})}{\mathcal {X}}} \bigl\Vert \psi^{\delta}- \psi \bigr\Vert _{\mathscr {L}^{2}(\mathbb {R})}^{2} \\ &\qquad+ 3 (\mathcal {X}- x)\widetilde{K} \int_{x}^{\mathcal{X}} \bigl\vert \gamma (\delta) \bigr\vert ^{\frac{2\alpha (x-z)}{\mathcal {X}}} \bigl\Vert U^{\delta}_{\gamma(\delta)}(z,\cdot) - V_{\gamma(\delta )}(z,\cdot) \bigr\Vert ^{2}_{\mathscr {L}^{2}(\mathbb {R})} \,\mathrm {d}z \\ & \quad\le6 \bigl\vert \gamma(\delta) \bigr\vert ^{\frac{2\alpha (x-\mathcal {X})}{\mathcal {X}}} \delta^{2} +3 \mathcal {X} \widetilde{K} \int_{x}^{\mathcal{X}} \bigl\vert \gamma(\delta) \bigr\vert ^{\frac{2\alpha (x-z)}{\mathcal {X}}} \bigl\Vert U^{\delta}_{\gamma(\delta)}(z,\cdot) - V_{\gamma(\delta )}(z,\cdot) \bigr\Vert ^{2}_{\mathscr {L}^{2}(\mathbb {R})} \,\mathrm {d}z. \end{aligned}$$

This implies that

$$\begin{aligned} & \bigl\vert \gamma(\delta) \bigr\vert ^{-\frac{2\alpha x}{\mathcal {X}}} \bigl\Vert U^{\delta}_{\gamma(\delta)}( x,\cdot)-V_{\gamma(\delta)}( x,\cdot ) \bigr\Vert _{\mathscr {L}^{2}(\mathbb {R})}^{2} \\ &\quad\le 6 \bigl\vert \gamma(\delta) \bigr\vert ^{-2\alpha } \delta^{2} +3 \mathcal {X} \widetilde{K} \int_{x}^{\mathcal{X}} \bigl\vert \gamma(\delta) \bigr\vert ^{-\frac{2\alpha z}{\mathcal {X}}} \bigl\Vert U^{\delta}_{\gamma(\delta )}(z, \cdot)-V_{\gamma(\delta)}(z,\cdot) \bigr\Vert _{\mathscr {L}^{2}(\mathbb {R})}^{2}\,\mathrm {d}z. \end{aligned}$$

(3.22)

Applying Grönwall’s inequality, we have

$$\begin{aligned} \bigl\vert \gamma(\delta) \bigr\vert ^{-\frac{2\alpha x}{\mathcal {X}}} \bigl\Vert U^{\delta}_{\gamma(\delta)}( x,\cdot)-V_{\gamma(\delta)}(x,\cdot ) \bigr\Vert _{\mathscr {L}^{2}(\mathbb {R})}^{2} \le6 \bigl\vert \gamma(\delta) \bigr\vert ^{-2\alpha }\delta^{2} \exp \bigl\{ 3\mathcal {X}(\mathcal {X}- x) \widetilde {K} \bigr\} . \end{aligned}$$

Thus

$$\begin{aligned} & \bigl\Vert U^{\delta}_{\gamma(\delta)}-V_{\gamma(\delta)} \bigr\Vert _{\mathscr {L}^{p} (\omega,\mathcal {X};\mathscr {L}^{2}(\mathbb {R}) )} \\ &\quad\le\sqrt{6}\frac{\mathcal {X}}{p\alpha } \bigl\vert \gamma(\delta) \bigr\vert ^{-\alpha }\delta\exp \biggl\{ \frac{3}{2}\mathcal {X}^{2} \widetilde {K} \biggr\} \biggl(\frac{ \vert \gamma(\delta) \vert ^{\frac{p\alpha \omega}{\mathcal {X}}} - \vert \gamma(\delta) \vert ^{p\alpha }}{\log (\frac{1}{\gamma(\delta)} ) } \biggr)^{\frac{1}{p}}. \end{aligned}$$

(3.23)

Combining (3.18), (3.21), and (3.23), we obtain

$$\begin{aligned} & \bigl\Vert U^{\delta}_{\gamma(\delta)}-u \bigr\Vert _{\mathscr {L}^{p} (\omega,\mathcal {X};\mathscr {L}^{2}(\mathbb {R}) )} \\ &\quad\le \bigl\Vert U^{\delta}_{\gamma(\delta)}-V_{\gamma(\delta)} \bigr\Vert _{\mathscr {L}^{p} (\omega,\mathcal {X};\mathscr {L}^{2}(\mathbb {R}) )} + \Vert V_{\gamma(\delta)} - u \Vert _{\mathscr {L}^{p} (\omega ,\mathcal {X};\mathscr {L}^{2}(\mathbb {R}) )} \\ &\quad \leq C(\alpha ,\mathcal {X},\rho_{0},\widetilde{K}) \bigl( \bigl\vert \gamma (\delta) \bigr\vert ^{-\alpha }\delta+ P(u) \bigr) \biggl( \frac{ \vert \gamma(\delta) \vert ^{\frac{p\alpha \omega}{\mathcal {X}}} - \vert \gamma(\delta) \vert ^{p\alpha }}{\log (\frac{1}{\gamma(\delta )} ) } \biggr)^{\frac{1}{p}}, \end{aligned}$$

where

$$C(\alpha ,p,\mathcal {X},\rho_{0},\widetilde{K}) = \max \biggl\{ 2 \frac {\mathcal {X}}{p\alpha } \exp \biggl\{ \frac{\mathcal {X}^{2} }{\rho_{0}} \widetilde{K} \biggr\} ; \sqrt{6} \frac{\mathcal {X}}{p\alpha } \exp \biggl\{ \frac{3}{2}\mathcal {X}^{2} \widetilde{K} \biggr\} \biggr\} . $$

The proof of the theorem is completed. □

4 Numerical test

In this section, we show a numerical simulation for the following inverse sideways heat equation by the filter method:

$$\begin{aligned} \textstyle\begin{cases} \frac{\partial u}{\partial t}=\frac{\partial^{2} u}{\partial x^{2}} + f(u)G( x,t;u),& (x,t) \in(0,\mathcal {X}) \times\mathbb {R}, \\ u( x,t)|_{t\to\pm\infty} = 0, & (x,t) \in(0,\mathcal {X}) \times \mathbb {R},\\ u(\mathcal {X},t)=\phi(t),& t \in\mathbb {R}, \\ \frac{\partial u}{\partial x}(\mathcal {X},t)=\psi(t),& t \in\mathbb {R}. \end{cases}\displaystyle \end{aligned}$$

(4.1)

The tests are performed using software MATLAB R2014b (version 64-bit). Problem (4.1) is considered for $(x,t) \in(0,1) \times(-\infty ,+\infty)$ and the functions are

$$\begin{aligned} &f = u, \qquad G = 1-2t, \end{aligned}$$

(4.2)

$$\begin{aligned} &\phi= \exp\bigl(-t^{2}\bigr)\sin(1), \qquad\psi= \exp \bigl(-t^{2}\bigr)\cos(1). \end{aligned}$$

(4.3)

Based on the Fourier transform, for $\mathscr{F} \in\mathscr {L}^{2}(\mathbb {R})$, we have

$$\widehat{\mathscr{F}}(\xi)= \frac{1}{\sqrt{2\pi}} \int_{-\infty }^{\infty} \mathscr{F}(t) e^{-i\xi t} \, \mathrm {d} t, \quad\xi\in \mathbb {R}. $$

The exact solution of problem (4.1)–(4.3) is given by

$$\begin{aligned} u(x,t) & = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \bigl( \cosh \bigl((\mathcal {X}- x) \sqrt{i\xi} \bigr) \widehat{\phi}(\xi) \bigr) e^{i\xi t} \,\mathrm {d} \xi \\ &\quad-\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \biggl( \frac {\sinh ((\mathcal {X} - x) \sqrt{i\xi} )}{\sqrt{i\xi}} \widehat{ \psi}(\xi) \biggr) e^{i\xi t} \,\mathrm {d} \xi \\ & \quad- \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \biggl( \int_{ x}^{\mathcal {X}} \frac{\sinh ((z-x) \sqrt{i\xi} )}{\sqrt {i\xi}} \widehat{\mathcal {W}} (u ) (z,\xi) \,\mathrm {d}z \biggr) e^{i\xi t} \,\mathrm {d} \xi. \end{aligned}$$

(4.4)

The data $\phi,\psi\in\mathscr {L}^{2}(\mathbb {R})$ are noisy and are represented by the observation data $\phi^{\delta},\psi^{\delta}\in\mathscr {L}^{2}(\mathbb {R})$ satisfying

$$\begin{aligned} \bigl\Vert \phi^{\delta}- \phi \bigr\Vert _{\mathscr {L}^{2} (\mathbb {R} )} \leq\delta, \qquad \bigl\Vert \psi^{\delta}- \psi \bigr\Vert _{\mathscr {L}^{2} (\mathbb {R} )} \leq\delta, \end{aligned}$$

(4.5)

here $\delta>0$ is a small positive number representing the level of noise ($\delta\to0^{+}$).

We recall the regularized solution $U^{\delta}_{\gamma(\delta)}$ obtained by

$$\begin{aligned} &U^{\delta}_{\gamma(\delta)}(x,t) \\ &\quad= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \bigl( \cosh ^{\gamma(\delta)} \bigl(( \mathcal {X}- x)\sqrt{i\xi} \bigr) \widehat {\phi^{\delta}}(\xi) \bigr)e^{i\xi t} \,\mathrm {d} \xi \\ &\qquad- \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \biggl( \frac {\sinh^{\gamma(\delta)} ((\mathcal {X}- x)\sqrt{i\xi} )}{\sqrt{i\xi}} \widehat{ \psi^{\delta}}(\xi) \biggr)e^{i\xi t} \, \mathrm {d} \xi \\ &\qquad- \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \biggl( \int _{x}^{\mathcal{X}} \frac{\sinh^{\gamma(\delta)} ((z-x)\sqrt{i\xi} )}{\sqrt{i\xi}} \widehat{ \mathcal {W}} \bigl(U^{\delta}_{\gamma (\delta)} \bigr) (z,\xi) \,\mathrm {d}z \biggr) e^{i\xi t} \,\mathrm {d} \xi. \end{aligned}$$

(4.6)

Here $\cosh^{\gamma(\delta)} (y\sqrt{i\xi} )$, $\sinh ^{\gamma(\delta)} (y\sqrt{i\xi} )$ are defined for all $0 \leq y \leq\mathcal {X}$ and $\xi\in\mathbb {R}$ in the following:

$$\begin{aligned}& \cosh^{\gamma(\delta)} (y\sqrt{i\xi} ) = \cosh (y\sqrt{i\xi} ) + \frac{1 }{ 2} \bigl(\mathcal {F}_{\gamma(\delta)}^{\alpha }(\mathcal {X},\xi) - 1 \bigr) e^{\sqrt{i\xi }y} , \end{aligned}$$

(4.7)

$$\begin{aligned}& \sinh^{\gamma(\delta)} (y\sqrt{i\xi} ) =\sinh (y\sqrt{i\xi} )+ \frac{1}{ 2} \bigl( \mathcal {F}_{\gamma (\delta)}^{\alpha }(\mathcal {X},\xi) - 1 \bigr) e^{\sqrt{i\xi}y}, \end{aligned}$$

(4.8)

where

$$ \mathcal {F}_{\gamma(\delta)}^{\alpha }(\mathcal {X},\xi)= \bigl(1+\gamma (\delta) e^{ \sqrt{|\xi|/2} \mathcal {X} } \bigr)^{-\alpha } \quad\mbox{for } \alpha \in\mathbb {N}^{*}. $$

(4.9)

Next, we consider the problem of computing the Fourier transform as follows:

Let $m, n \in\mathbb{R}$, $m< n$ and assume that

$$\mathscr{F}(t) = \textstyle\begin{cases} \mathscr{F}(t) & \text{if $m < t < n$}, \\ 0 & \text{otherwise}. \end{cases} $$

Put $h_{t} = \frac{n-m}{N_{t}}$, $t_{i} = ih_{t} + n$, $i = \overline {1,N_{t}}$, respectively. Noting that $\xi_{k} = \frac{2\pi (k-\frac{N_{t}}{2} )}{n-m}$, $k = \overline{1,N_{t}}$, we obtain

$$\begin{aligned} \widehat{\mathscr {F}}(\xi_{k})= \frac{1}{\sqrt{2\pi}} \int_{m}^{n} \mathscr{F}(t) e^{-i\xi_{k} t} \, \mathrm {d} t = \frac{1}{\sqrt{2\pi}} \sum_{i=1}^{N_{t}} \mathscr{F}(t_{i}) e^{-i\xi_{k} t_{i}} h_{t}. \end{aligned}$$

(4.10)

We set up a uniform Cartesian grid $(x,t) \in(0,1) \times(T_{1},T_{2})$, which can be generated as follows:

$$\begin{aligned}& x_{p} = p \Delta x ,\quad p = \overline{0,P_{x}}, \Delta x = \frac {1}{P_{x}}, \\& t_{q} = q \Delta t + T_{1} ,\quad q = \overline{0,Q_{t}}, \Delta t = \frac{T_{2} - T_{1}}{Q_{t}}. \end{aligned}$$

To illustrate the theoretical results, we consider this example with some conditions given by

$$\begin{gathered} P_{x} = 200 ,\qquad Q_{t} = N_{t} = 200,\qquad m = -100,\qquad n = 100,\\ \alpha= 2,\quad\quad T_{1} = -10,\qquad T_{2} = 10, \\ \gamma(\delta) = \delta^{1/3},\quad \delta\in \bigl\{ 10^{-2}, 10^{-3}, 10^{-4} \bigr\} .\end{gathered} $$

The error between the exact and regularized solutions is evaluated by

$$\begin{aligned} \text{Err} = \sqrt{\frac{1}{Q_{t}+1} \sum _{q=0}^{Q_{t}} \bigl[ U^{\delta}_{\gamma(\delta)}( \cdot,t_{q}) - u(\cdot,t_{q}) \bigr]^{2}}. \end{aligned}$$

(4.11)

Figures 1(a), 2(a), 3(a) show the graphs of the exact and regularized solutions at $x \in\{0.3, 0.5, 0.8\}$ for $\delta\in \{10^{-2}, 10^{-3}, 10^{-4} \}$. The errors between u and $U^{\delta}_{\gamma (\delta)}$ at $x \in\{0.3, 0.5, 0.8\}$ for various amounts of noise $\delta\in \{10^{-2}, 10^{-3}, 10^{-4} \}$ are shown in Table 1. For convenience of comparison, we give the contour graphs between the exact and regularized solutions (see Figs. 1(b), 2(b), 3(b)).

Figure 1

The solutions and errors at $\delta= 0.01$ with $(x,t) \in\{ 0.3\} \times(-10,10)$

Figure 2

The solutions and errors at $\delta= 0.001$ with $(x,t) \in \{0.3\} \times(-10,10)$

Figure 3

The solutions and errors at $\delta= 0.0001$ with $(x,t) \in \{0.3\} \times(-10,10)$

Table 1 The error between the exact and regularized solutions at $\delta\in\{0.01, 0.001, 0.0001\}$ and $x \in\{ 0.3, 0.5, 0.8\}$

From them, we observe that the errors at $\delta= 0.001$ are greater than those at $\delta= 0.0001$ and smaller than those at $\delta= 0.01$. Furthermore, with the smaller errors of input data, the results obtained are more accurate, which verifies the theoretical results.