A quasi-reversibility method for solving nonhomogeneous sideways heat equation

Yu Qiao; Xiangtuan Xiong

doi:10.1515/math-2024-0049

Article Open Access

A quasi-reversibility method for solving nonhomogeneous sideways heat equation

Yu Qiao and Xiangtuan Xiong

Published/Copyright: September 16, 2024

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$Open Mathematics$

From the journal Open Mathematics Volume 22 Issue 1

Abstract

We consider solving a severely ill-posed problem for determining the surface temperature and heat flux distribution of the nonhomogeneous sideways heat equation by a quasi-reversibility regularization method. An analytical solution is deduced based on the Fourier transform, and then, the logarithmic-type error estimate for the regularized solution is achieved. Finally, three numerical examples, including smooth and non-smooth functions, validate the feasibility and effectiveness of the proposed approach.

Keywords: ill-posed problem; nonhomogeneous sideways heat equation; a quasi-reversibility method; error estimate

MSC 2010: 35R25; 35R30; 65J20; 65M30

1 Introduction

The sideways heat equation (SHE), which is a Cauchy problem for the heat equation, has attracted lots of researchers in the last few decades. It is a model of a problem where one wants to determine the temperature or heat flux on both sides of a thick wall, but where one side is inaccessible to measurements. Under these circumstances, the missing information is compensated for by additional observations made through measurement on the complementary accessible parts of the boundary of the body. SHE has many applications, such as the accurate determination of the temperature spike in a cannon at firing [1], continuous data monitoring [2], and re-entry vehicles [3].

The homogeneous SHE is formulated as follows:

(1.1) u t − u x x = 0 , x > 0 , t > 0 , u ( x , 0 ) = 0 , x ≥ 0 , u ( 1 , t ) = h ( t ) , t ≥ 0 , u ( x , t ) ∣ x → ∞ bounded .

The goal here is to seek temperature distribution for 0 ≤ x < 1 from the measurement data h ( t ) . In general, this problem is ill-posed in the sense of Hadamard [4], i.e., any small errors in the observation data will cause large deviations in the estimate quantities. As a consequence, its solution does not meet the general requirement of existence, uniqueness, and stability under small perturbation to the input data. Then, regularization techniques [5] are required to make it a well-posed problem, e.g., Tikhonov method [6], spectral method [7], quasi-reversibility method [8–11], quasi-boundary value method and iteration method [12], wavelet method [13–16], multi-resolution method [17], and wavelet-Galerkin method [18,19]. Besides, it has been shown that difference schemes have a regularizing effect [20], so Eldén and Guo et al. [21–23] employed the finite difference method to deal with this ill-posed problem.

Still in the homogeneous situation, Xiong et al. addressed the difference regularization method [24] and the Fourier method [25] to discuss the non-standard inverse heat conduction problem

(1.2) u t + u x = u x x , x > 0 , t > 0 , u ( x , 0 ) = 0 , x ≥ 0 , u ( 1 , t ) = φ ( t ) , t ≥ 0 , u ( x , t ) ∣ x → ∞ bounded .

This article acquired stable estimates of temperature and flux distribution and used numerical example to show that the computational effect of their methods is satisfactory. Furthermore, Xiong and Fu [26] investigated a more general problem

(1.3) u t ( x , t ) = a ( x ) u x x + b ( x ) u x + c ( x ) , x > 0 , t > 0 , u ( x , 0 ) = 0 , x ≥ 0 , u ( 1 , t ) = ψ ( t ) , t ≥ 0 , u ( x , t ) ∣ x → ∞ bounded .

In this work, the authors adopted the wavelet dual least-squares method to determine surface temperature and heat flux and proved the error estimates between the approximate solution and the exact one are order optimal. Subsequent to [26], Liu and Deng et al. [27–29] used the method of iteration to compute the approximate solution of (1.3) and showed that their solution matched the exact solution well. Furthermore, there are also some articles (see, e.g., [30–34] and references therein) that study the sideways problems in two-dimensional and higher-dimensional space.

However, severe ill-posedness and the existence of the nonhomogeneous term f ( x , t ) make the use of a priori bound E more troublesome. Therefore, it is much more difficult to solve SHE in the nonhomogeneous case than in the homogeneous case. In particular, the determination of heat flux is scarcely discussed. To the knowledge of the author, there are still very few results on the Cauchy problem of the heat conduction equation in the nonhomogeneous case. Liu and Chang [35] solved a three-dimensional nonhomogeneous SHE in a cuboid by a Fourier sine series method, and the analysis of the regularization parameter and the stability of the solution were worked out. This study points out that their method is quite accurate but does not give an error estimate of the regularized solution. Luan and Khanh [36] investigated the two-dimensional inhomogeneous heat equation in the presence of a general source term and proposed a kernel regularization scheme to recover the temperature and thermal flux distribution from the measured data. However, an error estimate at the boundary is not given. In the present study, we are devoted to extending the work [10] and [11] to a more general case, i.e., the following nonhomogeneous system:

(1.4) u t − u x x = f ( x , t ) , x > 0 , t > 0 , u ( x , 0 ) = 0 , x ≥ 0 , u ( 1 , t ) = g ( t ) , t ≥ 0 , u ( x , t ) ∣ x → ∞ bounded ,

where the function f ( x , t ) is the heat source density. There are many ways to solve partial differential equations [37,38]. Here, we aim at obtaining an analytical solution to (1.4) via Fourier transform techniques. Due to the ill-posedness of the problem, a fourth-order modified method is derived for the stable reconstruction of the solution. Moreover, under some appropriate a priori assumptions, the logarithmic-type error estimate for the temperature and flux regularized solutions in the whole domain is obtained, which proves the usefulness of the quasi-reversibility regularization method. The quasi-reversibility method was first introduced by Lattès and Lions [39], and since then, the idea of this method has been successfully used for solving various types of ill-posed problems. Throughout this article, we assume that f ˆ ∈ L 2 ( 0 , 1 ; L 2 ( R ) ) satisfies

(1.5) ∫ 0 1 ∣ f ˆ ( s , ξ ) ∣ 2 d s < e − 3 ∣ ξ ∣ , ∀ ξ ∈ R .

2 Mathematical analysis and quasi-reversibility method

In order to apply the Fourier transform, we extend all the functions in this article to the whole line − ∞ < t < ∞ by making them zero outside the original domain, and our theoretical analysis will be performed in L 2 ( R ) . Let g ˆ denote the Fourier transform of g ( t ) defined by

g ˆ ( ξ ) ≔ 1 2 π ∫ − ∞ ∞ g ( t ) e − i ξ t d t .

We also define the following norms:

‖ u ( 0 , ⋅ ) ‖ p ≔ ∫ − ∞ ∞ ( 1 + ξ 2 ) p ∣ u ˆ ( 0 , ξ ) ∣ 2 d ξ 1 2 ,

‖ f ‖ L 2 ( 0 , 1 ; H p ( R ) ) ≔ ∫ 0 1 ‖ f ( x , ⋅ ) ‖ p 2 d x 1 2 .

When p = 0 , ‖ ⋅ ‖ p = ‖ ⋅ ‖ denotes the L 2 ( R ) norm.

Applying the Fourier transform with respect to t to both sides of (1.4), we obtain the following second-order ordinary differential equation in the frequency space:

(2.1) u ˆ x x ( x , ξ ) − i ξ u ˆ ( x , ξ ) = − f ˆ ( x , ξ ) , ξ ∈ R , u ˆ ( 1 , ξ ) = g ˆ ( ξ ) , ξ ∈ R , u ˆ ( x , ξ ) ∣ x → ∞ bounded , ξ ∈ R .

The solution of (2.1) is given by

(2.2) u ˆ ( x , ξ ) = e θ ( ξ ) ( 1 − x ) g ˆ ( ξ ) + ∫ x 1 f ˆ ( s , ξ ) sinh ( θ ( ξ ) ( s − x ) ) θ ( ξ ) d s , 0 ≤ x < 1 ,

and equivalently,

(2.3) u ( x , t ) = 1 2 π ∫ − ∞ ∞ e θ ( ξ ) ( 1 − x ) g ˆ ( ξ ) + ∫ x 1 f ˆ ( s , ξ ) sinh ( θ ( ξ ) ( s − x ) ) θ ( ξ ) d s e i ξ t d ξ , 0 ≤ x < 1 ,

where

(2.4) θ ( ξ ) ≔ i ξ = ∣ ξ ∣ 2 ( 1 + σ i ) , σ = sign ( ξ ) , ∀ ξ ∈ R .

Remark 2.1

When f = 0 , (2.2) and (2.3) give the following solution:

u ˆ ( x , ξ ) = e θ ( ξ ) ( 1 − x ) g ˆ ( ξ ) , 0 ≤ x < 1 , u ( x , t ) = 1 2 π ∫ − ∞ ∞ e θ ( ξ ) ( 1 − x ) g ˆ ( ξ ) e i ξ t d ξ , 0 ≤ x < 1 ,

which is the solution of problem (1.1) [40]. Thus, the solution of nonhomogeneous problem is consistent with its homogeneous system.

Remark 2.2

In (2.2), since sinh ( θ ) θ is an exponentially increasing function, in order to obtain a valid estimate, we need to assume that f ˆ is a rapidly decreasing function. Condition (1.5) is therefore reasonable and will be used frequently afterward.

Note that the real part of θ ( ξ ) is an increasing positive function of ξ . Therefore, the terms e θ ( ξ ) ( 1 − x ) and sinh ( θ ( ξ ) ( s − x ) ) θ ( ξ ) are unbounded as ∣ ξ ∣ → ∞ . This means small errors in the data can blow up and completely destroy the solution for 0 ≤ x < 1 . Thus, the Cauchy problem of the nonhomogeneous SHE is severely ill-posed. Next, we apply a quasi-reversibility regularization method to construct an approximate solution for (1.4). Here, the quasi-reversibility method is to find the solution of the following problem:

(2.5) u t α , δ = u x x α , δ − α 2 u x x t t α , δ + f δ ( x , t ) , x > 0 , t > 0 , u α , δ ( x , 0 ) = 0 , x ≥ 0 , u α , δ ( 1 , t ) = g δ ( t ) , t ≥ 0 , u α , δ ( x , t ) ∣ x → ∞ bounded ,

where α > 0 is a regularization parameter and the measured data ( g δ , f δ ) ∈ L 2 ( R ) × L 2 ( ( 0 , 1 ) ; L 2 ( R ) ) satisfy

(2.6) max { ‖ g δ − g ‖ , ‖ f δ − f ‖ L 2 ( 0 , 1 ; L 2 ( R ) ) } ≤ δ ,

where δ > 0 is called an error level. Similarly, applying the Fourier transform with respect to t to both sides of (2.5), the formal solution of problem (2.5) in the frequency space can be obtained as follows:

(2.7) u ˆ α , δ ( x , ξ ) = e θ ˜ ( ξ ) ( 1 − x ) g ˆ δ ( ξ ) + ∫ x 1 f ˆ δ ( s , ξ ) sinh ( θ ˜ ( ξ ) ( s − x ) ) θ ˜ ( ξ ) d s , 0 ≤ x < 1 ,

where

(2.8) θ ˜ ( ξ ) ≔ θ ( ξ ) θ 1 and θ 1 = 1 1 + α 2 ξ 2 ≤ 1 .

Lemma 2.1

[8] For arbitrary β ≥ 0 , z 1 , z 2 ∈ C , the following inequalities hold:

(2.9) 1 − e − β ≤ β ,

(2.10) ∣ 1 − e ± β i ∣ ≤ β ,

(2.11) ∣ sinh ( z 1 ) − sinh ( z 2 ) ∣ ≤ ∣ e z 1 − e z 2 ∣ .

Lemma 2.2

[11] The function

h j ( ∣ ξ ∣ ) = e − ∣ ξ ∣ 2 x ∣ ξ ∣ j 2 ,

obtains the maximum h max j = j 2 x e j at ∣ ξ ∣ = j 2 x , here j = 4 , 5 , 6 .

Lemma 2.3

[36] For arbitrary z ∈ C , x ∈ ( 0 , 1 ] , and η ∈ [ 0 , x ) , we have

(2.12) sinh ( ( x − η ) z ) z ≤ ( x − η ) e ( x − η ) ∣ z ∣ ,

(2.13) ∣ cosh ( x z ) ∣ ≤ e x ℜ ( z ) ≤ e x ∣ z ∣ ,

where ℜ ( z ) denotes the real parts of z.

3 Determination of temperature structure and error estimate

So as to acquire a more sharp convergence, we use the following a priori bound on the exact solution:

(3.1) ‖ u ( 0 , ⋅ ) ‖ ≤ E ,

and the regularization parameter α is chosen as

(3.2) α = 1 2 ln E δ 2 .

Lemma 3.1

If conditions (1.5) and (3.1) hold, let A ( ξ ) = u ˆ ( 0 , ξ ) − ∫ 0 1 f ˆ sinh ( θ s ) θ d s , then

‖ A ( ξ ) ‖ ≤ E + M 1 ,

where M 1 is a constant.

Proof

Using successively the triangle inequality, (3.1), the Cauchy-Schwarz integral inequality, (2.12) and (1.5), we obtain

‖ A ( ξ ) ‖ ≤ ‖ u ˆ ( 0 , ξ ) ‖ + ∫ 0 1 f ˆ sinh ( θ s ) θ d s ≤ E + ∫ − ∞ ∞ ∫ 0 1 sinh ( θ s ) θ 2 d s ∫ 0 1 ∣ f ˆ ∣ 2 d s d ξ 1 2 ≤ E + ∫ − ∞ ∞ ∫ 0 1 ∣ s e s ∣ θ ∣ ∣ 2 d s ∫ 0 1 ∣ f ˆ ∣ 2 d s d ξ 1 2 ≤ E + ∫ − ∞ ∞ 1 2 ∣ ξ ∣ e − ∣ ξ ∣ d ξ 1 2 .

It is easy to know that the generalized integral on the right-hand side of the last inequality converges, here we introduce the notation

M 1 ≔ ∫ − ∞ ∞ 1 2 ∣ ξ ∣ e − ∣ ξ ∣ d ξ 1 2 .

Therefore,

‖ A ( ξ ) ‖ ≤ E + M 1 ,

where M 1 is a constant.□

Theorem 3.1

Let u ˆ ( x , ξ ) given by (2.2) be the exact solution of problem (1.4) in the frequency space, u ˆ α , δ ( x , ξ ) given by (2.7) be the exact solution of problem (2.5) in the frequency space. Conditions (1.5), (2.6), and (3.1) hold, and the regularization parameter α is chosen by (3.2). Then, for a fixed x ∈ ( 0 , 1 ) , we have

‖ u α , δ ( x , ⋅ ) − u ( x , ⋅ ) ‖ ≤ 1 2 ln E δ + 1 E 1 − x δ x + C 1 ln E δ 4 ( E + M 1 ) + C 2 ln E δ 4 ,

where C 1 and C 2 are some constants that only depend on x and M 1 is a constant.

Proof

By the triangle inequality, we have

(3.3) ‖ u α , δ ( x , ⋅ ) − u ( x , ⋅ ) ‖ ≤ ‖ u α , δ ( x , ⋅ ) − u α , 0 ( x , ⋅ ) ‖ ︸ ℐ 1 + ‖ u α , 0 ( x , ⋅ ) − u ( x , ⋅ ) ‖ ︸ ℐ 2 ,

where u α , 0 ( x , t ) is the exact solution of problem (2.5) in the frequency space with noise-free data. Next, we divide the argument into two steps.

Step 1. Estimate the term ℐ 1 in (3.3). It follows immediately from Parseval’s equality and the triangle inequality that

ℐ 1 = ‖ u ˆ α , δ ( x , ⋅ ) − u ˆ α , 0 ( x , ⋅ ) ‖ ≤ ‖ e θ ˜ ( 1 − x ) ( g ˆ δ − g ˆ ) ‖ ︸ ℐ 1 ¯ + ∫ x 1 sinh ( θ ˜ ( s − x ) ) θ ˜ ( f ˆ δ − f ˆ ) d s ︸ ℐ 2 ¯ .

By (2.6), (3.2), and note that

(3.4) ∣ θ ˜ ∣ = i ξ 1 + α 2 ξ 2 ≤ 1 2 α ,

we have

ℐ 1 ¯ ≤ δ sup ξ ∈ R ∣ e θ ˜ ( 1 − x ) ∣ ≤ δ e ∣ θ ˜ ∣ ( 1 − x ) ≤ δ e 1 2 α ( 1 − x ) = E 1 − x δ x .

Using the Cauchy-Schwarz integral inequality, (2.12), (3.4), and (2.6) yields

ℐ 2 ¯ ≤ ∫ − ∞ ∞ ∫ x 1 sinh ( θ ˜ ( s − x ) ) θ ˜ 2 d s ∫ x 1 ∣ f ˆ δ − f ˆ ∣ 2 d ξ 1 2 ≤ ∫ − ∞ ∞ ∫ x 1 e 2 ( s − x ) 1 2 α d s ∫ 0 1 ∣ f ˆ δ − f ˆ ∣ 2 d s d ξ 1 2 ≤ 2 α 2 e 2 ( 1 − x ) 1 2 α 1 2 δ = 1 2 ln E δ E 1 − x δ x .

Hence,

(3.5) ℐ 1 ≤ 1 2 ln E δ + 1 E 1 − x δ x .

Step 2. Estimate the term ℐ 2 in (3.3). Again, in view of Parseval’s identity and the triangle inequality,

ℐ 2 = ‖ u ˆ α , 0 ( x , ⋅ ) − u ˆ ( x , ⋅ ) ‖ ≤ ∥ ( e θ ˜ ( 1 − x ) − e θ ( 1 − x ) ) g ˆ ∥ ︸ ℐ 1 ˜ + ∫ x 1 sinh ( θ ˜ ( s − x ) ) θ ˜ − sinh ( θ ( s − x ) ) θ f ˆ d s ︸ ℐ 2 ˜ .

We start by estimating the first term above. Let

A ˜ 1 ( ξ ) = ∣ e θ ˜ ( 1 − x ) − θ − e − θ x ∣ ,

by (2.2), we obtain

ℐ 1 ˜ = ( e θ ˜ ( 1 − x ) − θ − e − θ x ) u ˆ ( 0 , ξ ) − ∫ 0 1 f ˆ sinh ( θ s ) θ d s ≤ sup ξ ∈ R A ˜ 1 ( ξ ) ‖ A ( ξ ) ‖ .

Next, we estimate A ˜ 1 ( ξ ) , let

(3.6) γ = θ − θ ˜ = γ 1 ( 1 + σ i ) , γ 1 = ∣ ξ ∣ 2 1 − 1 1 + α 2 ξ 2 ≥ 0 ,

by the triangle inequality, (2.9), and (2.10), we obtain

A ˜ 1 ( ξ ) = ∣ e − θ x ∣ ∣ 1 − e − γ ( 1 − x ) ∣ = e − ∣ ξ ∣ 2 x ( ∣ 1 − e − γ 1 ( 1 − x ) ∣ + ∣ e − γ 1 ( 1 − x ) − e − γ 1 ( 1 − x ) ( 1 + i σ ) ∣ ) ≤ e − ∣ ξ ∣ 2 x ( γ 1 ( 1 − x ) + e − γ 1 ( 1 − x ) ∣ 1 − e − γ 1 ( 1 − x ) i σ ∣ ) ≤ 2 γ 1 ( 1 − x ) e − ∣ ξ ∣ 2 x .

Note that

(3.7) γ 1 ≤ ∣ ξ ∣ 2 ( 1 + α 2 ξ 2 − 1 ) ≤ 1 2 2 α 2 ∣ ξ ∣ 5 2 ,

by Lemma 2.2 and (3.2), we have

A ˜ 1 ( ξ ) ≤ 1 − x 2 α 2 5 2 x e 5 = C 1 ln E δ 4 ,

where

(3.8) C 1 = 1 − x 4 2 5 2 x e 5 .

Combining with Lemma 3.1, we obtain

(3.9) ℐ 1 ˜ ≤ C 1 ln E δ 4 ( E + M 1 ) .

Let

A ˜ 2 ( ξ ) = ∣ e θ ˜ ( s − x ) − e θ ( s − x ) ∣ , A ˜ 3 ( ξ ) = sinh ( θ ˜ ( s − x ) ) θ ˜ − sinh ( θ ( s − x ) ) θ .

To estimate ℐ 2 ˜ , we first estimate A ˜ 2 ( ξ ) and A ˜ 3 ( ξ ) . Using (3.6), the triangle inequality, (2.9), (2.10), and (3.7) in order, for ∀ s ∈ ( x , 1 ) , we have

(3.10) A ˜ 2 ( ξ ) = ∣ e ( s − x ) θ ∣ ∣ 1 − e − ( s − x ) γ ∣ ≤ e ( s − x ) ∣ ξ ∣ 2 ( ∣ 1 − e − ( s − x ) γ 1 σ i ∣ + ∣ e − ( s − x ) γ 1 σ i − e − ( s − x ) γ 1 ( 1 + σ i ) ∣ ) ≤ 2 e ( s − x ) ∣ ξ ∣ 2 ( s − x ) γ 1 ≤ 2 2 e ∣ ξ ∣ 2 ( s − x ) ∣ ξ ∣ 5 2 α 2 .

Using successively the triangle inequality, (2.12), (2.11), note that 1 − θ 1 ≤ 1 2 α 2 ∣ ξ ∣ 2 and ∣ θ ˜ ∣ ≤ ∣ θ ∣ , then for ∀ s ∈ ( x , 1 ) , we obtain

A ˜ 3 ( ξ ) = sinh ( θ ˜ ( s − x ) ) − θ 1 sinh ( θ ( s − x ) ) θ ˜ ≤ ( 1 − θ 1 ) sinh ( θ ˜ ( s − x ) ) θ ˜ + sinh ( θ ˜ ( s − x ) ) − sinh ( θ ( s − x ) ) θ ≤ 1 2 α 2 ∣ ξ ∣ 2 ( s − x ) e ∣ ξ ∣ ( s − x ) + A ˜ 2 ( ξ ) ∣ ξ ∣ ≤ 2 ( s − x ) e ∣ ξ ∣ ∣ ξ ∣ 2 α 2 .

Now, we estimate ℐ 2 ˜ . Using the Cauchy-Schwarz integral inequality yields

ℐ 2 ˜ ≤ ∫ − ∞ ∞ ∫ x 1 ∣ f ˆ ∣ 2 d s ∫ x 1 sinh ( θ ˜ ( s − x ) ) θ ˜ − sinh ( θ ( s − x ) ) θ 2 d s d ξ 1 2 ≤ ∫ − ∞ ∞ ∫ x 1 ( 2 ( s − x ) e ∣ ξ ∣ ∣ ξ ∣ 2 α 2 ) 2 d s e − 3 ∣ ξ ∣ d ξ 1 2 ≤ 6 3 ( 1 − x ) 3 2 α 2 ∫ − ∞ ∞ ∣ ξ ∣ 4 e − ∣ ξ ∣ d ξ 1 2 .

Similarly, since the generalized integral on the right-hand side of the last inequality converges, we introduce the notation

M 2 ≔ ∫ − ∞ ∞ ∣ ξ ∣ 4 e − ∣ ξ ∣ d ξ 1 2 .

By (3.2), we obtain

(3.11) ℐ 2 ˜ ≤ C 2 ln E δ 4 ,

where

(3.12) C 2 = ( 1 − x ) 3 2 2 6 M 2 .

Substituting (3.9) and (3.11) into ℐ 2 , we have

(3.13) ℐ 2 ≤ C 1 ln E δ 4 ( E + M 1 ) + C 2 ln E δ 4 ,

where C 1 and C 2 are given by (3.8) and (3.12), respectively. Theorem 3.1 follows by combining (3.3), (3.5), and (3.13).□

Remark 3.1

In this article, we consider that both the data g ( t ) and the source term f ( x , t ) are given approximately. From the proof of Theorem 3.1, it follows that the total error ‖ u α , δ ( x , ⋅ ) − u ( x , ⋅ ) ‖ can be split into the noise propagation error ℐ 1 and the approximation error ℐ 2 . Clearly, as α tends to 0, ℐ 1 tends to infinity, yet ℐ 2 tends to 0. Therefore, we need to choose α = α ( δ ) dependent on δ to keep the total error as small as possible.

Note that the asymptotics of the regularized solution become progressively lower as x → 0 . To obtain the continuous dependence of the solution at x = 0 , we have to introduce stronger assumptions

(3.14) ‖ u ( 0 , ⋅ ) ‖ p ≤ E , p > 0 ,

(3.15) ( 1 + ξ 2 ) p ∫ 0 1 ∣ f ˆ ( s , ξ ) ∣ 2 d s < e − 3 ∣ ξ ∣ , ∀ ξ ∈ R ,

and the regularization parameter α is chosen as

(3.16) α = 1 2 ln E δ ln E δ − 2 p 2 .

Lemma 3.2

Let conditions (3.14) and (3.15) hold, A ˜ ( ξ ) = ( 1 + ξ 2 ) p 2 u ˆ ( 0 , ξ ) − ∫ 0 1 f ˆ sinh ( θ s ) θ d s , then

‖ A ˜ ( ξ ) ‖ ≤ E + M 1 ,

where M 1 is a constant.

The proof is similar to the proof of Lemma 3.1, and we omit it here.

Theorem 3.2

Let u ˆ ( x , ξ ) given by (2.2) be the exact solution of problem (1.4) in the frequency space and u ˆ α , δ ( x , ξ ) given by (2.7) be the exact solution of problem (2.5) in the frequency space. Conditions (2.6), (3.14), (3.15) hold, and the regularization parameter α is chosen by (3.16). Then, for p > 0 , we have

‖ u α , δ ( 0 , ⋅ ) − u ( 0 , ⋅ ) ‖ ≤ E ln E δ − 2 p 1 2 ln E δ ln E δ − 2 p + 1 + ε 1 ( E + M 1 ) + ε 2 M 3 ,

where M 1 and M 3 are some constants, and ε 1 ≕ max 2 α 4 5 p , 2 2 α 2 , ε 2 ≕ max 2 α 4 5 p , 2 α 2 .

Proof

By the triangle inequality, we have

(3.17) ‖ u α , δ ( 0 , ⋅ ) − u ( 0 , ⋅ ) ‖ ≤ ‖ u α , δ ( 0 , ⋅ ) − u α , 0 ( 0 , ⋅ ) ‖ ︸ ℐ 3 + ‖ u α , 0 ( 0 , ⋅ ) − u ( 0 , ⋅ ) ‖ ︸ ℐ 4 .

Next, we divide the argument into two steps.

Step 1. Estimate the term ℐ 3 in (3.17). Taking a similar procedure of the estimate of ℐ 1 , and by (3.16), we obtain

(3.18) ℐ 3 ≤ ‖ e θ ˜ ( g ˆ δ − g ˆ ) ‖ + ∫ 0 1 sinh ( θ ˜ s ) θ ˜ ( f ˆ δ − f ˆ ) d s ≤ δ e 1 2 α + δ 2 α 2 e 2 2 α 1 2 = E ln E δ − 2 p 1 2 ln E δ ln E δ − 2 p 1 2 + 1 .

Step 2. Estimate the term ℐ 4 in (3.17).

(3.19) ℐ 4 ≤ ‖ ( e θ ˜ − e θ ) g ˆ ‖ ︸ ℐ 3 ˜ + ∫ 0 1 sinh ( θ ˜ s ) θ ˜ − sinh ( θ s ) θ f ˆ d s ︸ ℐ 4 ˜ .

We first estimate ℐ 3 ˜ in (3.19). Let

A ˜ 4 ( ξ ) = ∣ e θ ˜ − θ − 1 ∣ ( 1 + ξ 2 ) − p 2 .

By (2.2), we obtain

ℐ 3 ˜ = ( e θ ˜ − θ − 1 ) u ˆ ( 0 , ξ ) − ∫ 0 1 sinh ( θ s ) θ f ˆ d s ≤ sup ξ ∈ R A ˜ 4 ( ξ ) ‖ A ˜ ( ξ ) ‖ .

Let ξ 0 = α − 4 5 , we discuss in the following three situations to estimate A ˜ 4 ( ξ ) .

Case 1: ∣ ξ ∣ ≥ ξ 0 . By the triangle inequality, (3.6), we have

A ˜ 4 ( ξ ) ≤ ( ∣ e − γ ∣ + 1 ) ( 1 + ξ 2 ) − p 2 ≤ 2 ∣ ξ ∣ − p ≤ 2 α 4 5 p .

Case 2: 1 ≤ ∣ ξ ∣ < ξ 0 . Note that A ˜ 1 ( ξ ) ≤ 2 γ 1 when x = 0 , combine with (3.7), we obtain

A ˜ 4 ( ξ ) ≤ 2 γ 1 ( 1 + ξ 2 ) − p 2 ≤ 1 2 α 2 ∣ ξ ∣ 5 2 − p .

If 0 < p ≤ 5 2 , we have

A ˜ 4 ( ξ ) ≤ 1 2 α 2 ( α − 4 5 ) 5 2 − p ≤ 2 α 4 5 p .

If p > 5 2 , we obtain

A ˜ 4 ( ξ ) ≤ 2 2 α 2 .

Case 3: ∣ ξ ∣ < 1 .

A ˜ 4 ( ξ ) ≤ 2 γ 1 ( 1 + ξ 2 ) − p 2 ≤ 1 2 α 2 ∣ ξ ∣ 5 2 ≤ 2 2 α 2 .

In conclusion,

(3.20) A ˜ 4 ( ξ ) ≤ max 2 α 4 5 p , 2 2 α 2 ≕ ε 1 .

By Lemma 3.2, we obtain

(3.21) ℐ 3 ˜ ≤ ε 1 ( E + M 1 ) .

Now, we estimate ℐ 4 ˜ in (3.19). Let

A ˜ 5 ( ξ ) = sinh ( θ ˜ s ) θ ˜ − sinh ( θ s ) θ ( 1 + ξ 2 ) − p 2 e − ∣ ξ ∣ .

Similarly, we discuss A ˜ 5 ( ξ ) in three cases.

Case 1: ∣ ξ ∣ ≥ ξ 0 . By the triangle inequality, (2.12), and note that ∣ θ ˜ ∣ ≤ ∣ θ ∣ , then for ∀ s ∈ ( x , 1 ) , we have

A ˜ 5 ( ξ ) ≤ sinh ( θ ˜ s ) θ ˜ + sinh ( θ s ) θ ( 1 + ξ 2 ) − p 2 e − ∣ ξ ∣ ≤ 2 e ∣ θ ∣ ∣ ξ ∣ − p e − ∣ ξ ∣ ≤ 2 α 4 5 p .

Case 2: 1 ≤ ∣ ξ ∣ < ξ 0 . Note that A ˜ 3 ( ξ ) ≤ 2 s e ∣ ξ ∣ ∣ ξ ∣ 2 α 2 when x = 0 , then for ∀ s ∈ ( x , 1 ) , we obtain

A ˜ 5 ( ξ ) ≤ 2 ∣ ξ ∣ 2 α 2 ( 1 + ξ 2 ) − p 2 ≤ 2 α 2 ∣ ξ ∣ 2 − p .

If 0 < p ≤ 2 , note that α ∈ ( 0 , 1 ) , we have

A ˜ 5 ( ξ ) ≤ 2 α 2 ( α − 4 5 ) 2 − p ≤ 2 α 4 5 p + 2 5 ≤ 2 α 4 5 p .

If p > 2 , we obtain

A ˜ 5 ( ξ ) ≤ 2 α 2 .

Case 3: ∣ ξ ∣ < 1 .

A ˜ 5 ( ξ ) ≤ 2 ∣ ξ ∣ 2 α 2 ( 1 + ξ 2 ) − p 2 ≤ 2 α 2 ∣ ξ ∣ 2 ≤ 2 α 2 .

In summary,

(3.22) A ˜ 5 ( ξ ) ≤ max 2 α 4 5 p , 2 α 2 ≕ ε 2 .

Hence, for ∀ s ∈ ( x , 1 ) , we have

ℐ 4 ˜ ≤ ∫ − ∞ ∞ ∫ 0 1 ∣ f ˆ ∣ 2 d s ∫ 0 1 sinh ( θ ˜ s ) θ ˜ − sinh ( θ s ) θ 2 d s d ξ 1 2 ≤ ∫ − ∞ ∞ ∫ 0 1 sinh ( θ ˜ s ) θ ˜ − sinh ( θ s ) θ e − ∣ ξ ∣ ( 1 + ξ 2 ) − p 2 2 d s e − ∣ ξ ∣ d ξ 1 2 ≤ ε 2 ∫ − ∞ ∞ e − ∣ ξ ∣ d ξ 1 2 .

Likewise, since the generalized integral on the right-hand side of the last inequality converges, we introduce the notation

M 3 ≔ ∫ − ∞ ∞ e − ∣ ξ ∣ d ξ 1 2 .

Therefore,

(3.23) ℐ 4 ˜ ≤ ε 2 M 3 .

Substituting (3.21) and (3.23) into (3.19), we have

(3.24) ℐ 4 ≤ ε 1 ( E + M 1 ) + ε 2 M 3 ,

where ε 1 and ε 2 are given by (3.20) and (3.22). We arrive at the final conclusion by (3.17), (3.18), and (3.24).□

Remark 3.2

Since the regularization parameter α → 0 ( δ → 0 ), we can easily find that, for p > 0 , ε 1 , ε 2 → 0 ( δ → 0 ). Therefore,

lim δ → 0 ‖ u α , δ ( 0 , ⋅ ) − u ( 0 , ⋅ ) ‖ = 0 , p > 0 .

Remark 3.3

Theorems 3.1 and 3.2 are the main results of this section, which are not provided in other articles.

4 Determination of flux structure and error estimate

In this section, we seek heat flux structure u x ( x , t ) for 0 ≤ x < 1 from the measurement data g ( t ) . By (2.1) and (2.13), we obtain

(4.1) u ˆ x ( x , ξ ) = − θ e θ ( 1 − x ) g ˆ ( ξ ) − ∫ x 1 f ˆ ( s , ξ ) cosh ( θ ( s − x ) ) d s , 0 ≤ x < 1 ,

and

(4.2) u ˆ x α , δ ( x , ξ ) = − θ ˜ e θ ˜ ( 1 − x ) g ˆ δ ( ξ ) − ∫ x 1 f ˆ δ ( s , ξ ) cosh ( θ ˜ ( s − x ) ) d s , 0 ≤ x < 1 .

Theorem 4.1

‖ u x α , δ ( x , ⋅ ) − u x ( x , ⋅ ) ‖ ≤ ln E δ + 1 2 ln E δ E 1 − x δ x + C 3 ln E δ 4 ( E + M 1 ) + C 4 ln E δ 4 ,

where C 3 and C 4 are some constants that only depend on x and M 1 is a constant.

Proof

By the triangle inequality, we have

(4.3) ‖ u x α , δ ( x , ⋅ ) − u x ( x , ⋅ ) ‖ ≤ ‖ u x α , δ ( x , ⋅ ) − u x α , 0 ( x , ⋅ ) ‖ ︸ J 1 + ‖ u x α , 0 ( x , ⋅ ) − u x ( x , ⋅ ) ‖ ︸ J 2 .

Next, we divide the argument into two steps.

Step 1. Estimate the term J 1 in (4.3). Using Parseval’s equality and the triangle inequality, we have

J 1 = ‖ u ˆ x α , δ ( x , ⋅ ) − u ˆ x α , 0 ( x , ⋅ ) ‖ ≤ ‖ θ ˜ e θ ˜ ( 1 − x ) ( g ˆ − g ˆ δ ) ‖ ︸ J 1 ¯ + ∫ x 1 ( f ˆ − f ˆ δ ) cosh ( θ ˜ ( s − x ) ) d s ︸ J 2 ¯ .

By (2.6), (3.4), and (3.2), we obtain

(4.4) J 1 ¯ ≤ δ sup ξ ∈ R ∣ θ ˜ e θ ˜ ( 1 − x ) ∣ ≤ δ ∣ θ ˜ ∣ e ∣ θ ˜ ∣ ( 1 − x ) ≤ δ 1 2 α e 1 2 α ( 1 − x ) = E 1 − x δ x ln E δ .

Using the Cauchy-Schwarz integral inequality, (2.13), and (3.2) yields

(4.5) J 2 ¯ ≤ ∫ − ∞ ∞ ∫ x 1 ∣ cosh ( θ ˜ ( s − x ) ∣ 2 ) d s ∫ x 1 ∣ f ˆ − f ˆ δ ∣ 2 d s d ξ 1 2 ≤ ∫ − ∞ ∞ ∫ x 1 ∣ e ∣ θ ˜ ∣ ( s − x ) ∣ 2 d s ∫ 0 1 ∣ f ˆ − f ˆ δ ∣ 2 d s d ξ 1 2 ≤ 2 α 2 e 2 2 α ( 1 − x ) 1 2 ‖ f ˆ − f ˆ δ ‖ L 2 ( 0 , 1 ; L 2 ( R ) ) ≤ 1 2 ln E δ E 1 − x δ x .

Thus,

(4.6) J 1 ≤ ln E δ + 1 2 ln E δ E 1 − x δ x .

Step 2. Estimate the term J 2 in (4.3). Again, using Parseval’s identity and the triangle inequality, we have

J 2 = ‖ u ˆ x α , 0 ( x , ⋅ ) − u ˆ x ( x , ⋅ ) ‖ ≤ ‖ ( θ e θ ( 1 − x ) − θ ˜ e θ ˜ ( 1 − x ) ) g ˆ ‖ ︸ J 1 ˜ + ∫ x 1 f ˆ [ cosh ( θ ( s − x ) ) − cosh ( θ ˜ ( s − x ) ) ] d s ︸ J 2 ˜ .

Let

B ˜ 1 ( ξ ) = ∣ θ e − θ x − θ ˜ e θ ˜ ( 1 − x ) − θ ∣ .

By (2.2), we obtain

J 1 ˜ = ( θ e − θ x − θ ˜ e θ ˜ ( 1 − x ) − θ ) u ˆ ( 0 , ξ ) − ∫ 0 1 f ˆ sinh ( θ s ) θ d s ≤ sup ξ ∈ R B ˜ 1 ( ξ ) ‖ A ( ξ ) ‖ .

Next, we estimate B ˜ 1 ( ξ ) . By (3.6), (2.9), (2.10), and (3.7), for ∀ x ∈ ( 0 , 1 ) , we have

B ˜ 1 ( ξ ) = ∣ e − θ x ∣ ∣ θ − θ ˜ e − γ ( 1 − x ) ∣ ≤ e − ∣ ξ ∣ 2 x ( ∣ θ − θ e − i γ 1 σ ( 1 − x ) ∣ + ∣ θ e − i γ 1 σ ( 1 − x ) − θ ˜ e − i γ 1 σ ( 1 − x ) ∣ + ∣ θ ˜ e − i γ 1 σ ( 1 − x ) − θ ˜ e − ( 1 + σ i ) γ 1 ( 1 − x ) ∣ ) ≤ e − ∣ ξ ∣ 2 x ( ∣ θ ∣ γ 1 ( 1 − x ) ∣ + 2 γ 1 + ∣ θ ∣ γ 1 ( 1 − x ) ) ≤ 2 ∣ ξ ∣ 2 + 1 2 α 2 ∣ ξ ∣ 5 2 e − ∣ ξ ∣ 2 x .

Note that h max 5 < h max 6 for x ∈ ( 0 , 1 ) , we attain

B ˜ 1 ( ξ ) ≤ 2 2 6 2 x e 6 + 1 2 5 2 x e 5 α 2 ≤ 2 6 2 x e 6 α 2 = 2 4 6 2 x e 6 1 ln E δ 4 .

By Lemma 3.1, we obtain

(4.7) J 1 ˜ ≤ C 3 ln E δ 4 ( E + M 1 ) ,

(4.8) C 3 = 2 4 6 2 x e 6 .

Let

B ˜ 2 ( ξ ) = ∣ cosh ( θ ˜ ( s − x ) ) − cosh ( θ ( s − x ) ) ∣ .

To estimate J 2 ˜ , we first estimate B ˜ 2 ( ξ ) :

B ˜ 2 ( ξ ) = ( e ( s − x ) θ − e ( s − x ) θ ˜ ) ( 1 − e − ( s − x ) ( θ ˜ + θ ) ) 2 ≤ A ˜ 2 ( ξ ) B ˜ 3 ( ξ ) ,

where A ˜ 2 ( ξ ) = ∣ e ( s − x ) θ − e ( s − x ) θ ˜ ∣ , B ˜ 3 ( ξ ) = ∣ 1 − e − ( s − x ) ( θ ˜ + θ ) ∣ .

Using (2.4), (2.8), (2.9), and (2.10), we obtain

B ˜ 3 ( ξ ) = ∣ 1 − e − ( s − x ) ( 1 + θ 1 ) ∣ ξ ∣ 2 ( 1 + σ i ) ∣ ≤ 1 − e − ( s − x ) ( 1 + θ 1 ) ∣ ξ ∣ 2 σ i + e − ( s − x ) ( 1 + θ 1 ) ∣ ξ ∣ 2 σ i − e − ( s − x ) ( 1 + θ 1 ) ∣ ξ ∣ 2 ( 1 + σ i ) ≤ 2 2 ( s − x ) ∣ ξ ∣ .

Hence,

(4.9) B ˜ 2 ( ξ ) ≤ 2 ( s − x ) 2 α 2 e ∣ ξ ∣ 2 ∣ ξ ∣ 3 .

Then, using the Cauchy-Schwarz integral inequality, (1.5), and (4.9) yields

J 2 ˜ ≤ ∫ − ∞ ∞ ∫ x 1 ∣ f ˆ ∣ 2 d s ∫ x 1 ∣ cosh ( θ ( s − x ) ) − cosh ( θ ˜ ( s − x ) ) ∣ 2 d s d ξ 1 2 ≤ ∫ − ∞ ∞ ∫ 0 1 ∣ f ˆ ∣ 2 d s ∫ x 1 2 ( s − x ) 2 α 2 e ∣ ξ ∣ 2 ∣ ξ ∣ 3 2 d s d ξ 1 2 ≤ 2 5 ( 1 − x ) 5 2 α 2 ∫ − ∞ ∞ ∣ ξ ∣ 6 e 2 ∣ ξ ∣ − 3 ∣ ξ ∣ d ξ 1 2 .

Introducing the notation,

M 3 = ∫ − ∞ ∞ ∣ ξ ∣ 6 e 2 ∣ ξ ∣ − 3 ∣ ξ ∣ d ξ 1 2 .

Therefore,

(4.10) J 2 ˜ ≤ C 4 ln E δ 4 ,

where

(4.11) C 4 = 5 10 ( 1 − x ) 5 2 M 3 .

Combining (4.7) and (4.10), we have

(4.12) J 2 ≤ C 3 ln E δ 4 ( E + M 1 ) + C 4 ln E δ 4 ,

where C 3 and C 4 are given by (4.8) and (4.11), respectively. By substituting (4.6) and (4.12) into (4.3), we arrive at the final conclusion.□

Remark 4.1

Note that the stability estimate in Theorem 4.1 only solves our problem for 0 < x < 1 and does not give any useful information on the continuous dependence of the solution at x = 0 . The following Theorem 4.2 will use assumptions (3.14) and (3.15) to give the convergence of this boundary.

Theorem 4.2

Let u ˆ ( x , ξ ) given by (2.2) be the exact solution of problem (1.4) in the frequency space, u ˆ α , δ ( x , ξ ) given by (2.7) be the exact solution of problem (2.5) in the frequency space. Conditions (2.6), (3.14), and (3.15) hold, and the regularization parameter α is chosen by (3.16). Then, for p > 1 2 , we have

‖ u x α , δ ( 0 , ⋅ ) − u x ( 0 , ⋅ ) ‖ ≤ E ln E δ − 2 p ln E δ ln E δ − 2 p + 1 2 ln E δ ln E δ − 2 p + ε 3 ( E + M 1 ) + ε 4 M 4 ,

where M 1 and M 4 are some constants, ε 3 ≔ max 2 ( α 2 3 ) p − 1 2 , 2 2 α 5 3 + α 2 and ε 4 ≔ max 2 α 2 3 p , α 2 .

Proof

By the triangle inequality, we have

(4.13) ‖ u x α , δ ( 0 , ⋅ ) − u x ( 0 , ⋅ ) ‖ ≤ ‖ u x α , δ ( 0 , ⋅ ) − u x α , 0 ( 0 , ⋅ ) ‖ ︸ J 3 + ‖ u x α , 0 ( 0 , ⋅ ) − u x ( 0 , ⋅ ) ‖ ︸ J 4 .

Next, we divide the argument into two steps.

Step 1. Estimate the term J 3 in (4.13). Taking a similar procedure of the estimate of J 1 , we obtain

(4.14) J 3 ≤ ‖ θ ˜ e θ ˜ ( g ˆ − g ˆ δ ) ‖ + ∫ 0 1 ( f ˆ − f ˆ δ ) cosh ( θ ˜ s ) d s ≤ δ 1 2 α e 1 2 α + δ 2 α 2 e 2 2 α 1 2 = E ln E δ − 2 p ln E δ ln E δ − 2 p + 1 2 ln E δ ln E δ − 2 p .

Step 2. Estimate the term J 4 in (4.13). From the triangle inequality, we obtain

J 4 ≤ ‖ ( θ e θ − θ ˜ e θ ˜ ) g ˆ ‖ ︸ J 3 ˜ + ∫ 0 1 f ˆ [ cosh ( θ ˜ s ) − cosh ( θ s ) ] d s ︸ J 4 ˜ .

Let

B ˜ 4 ( ξ ) = ∣ θ − θ ˜ e θ ˜ − θ ∣ ( 1 + ξ 2 ) − p 2 .

By (2.2), we obtain

J 3 ˜ = ( θ − θ ˜ e θ ˜ − θ ) u ˆ ( 0 , ξ ) − ∫ 0 1 f ˆ sinh ( θ s ) θ d s ≤ sup ξ ∈ R B ˜ 4 ( ξ ) ‖ A ˜ ( ξ ) ‖ .

Let ξ 1 = α − 2 3 , we discuss in the following three situations to estimate B ˜ 4 ( ξ ) .

Case 1: ∣ ξ ∣ ≥ ξ 1 . By the triangle inequality, (3.6), we have

B ˜ 4 ( ξ ) ≤ ( ∣ θ ∣ + ∣ θ ˜ ∣ ∣ e − γ ∣ ) ( 1 + ξ 2 ) − p 2 ≤ ∣ θ ∣ ( 1 + e − γ ) ∣ ξ ∣ − p ≤ 2 ∣ ξ ∣ 1 2 − p ≤ 2 ( α 2 3 ) p − 1 2 , p > 1 2 .

Case 2: 1 ≤ ∣ ξ ∣ < ξ 1 . Note that B ˜ 4 ( ξ ) ≤ 2 ∣ ξ ∣ + 1 2 α 2 ∣ ξ ∣ 5 2 when x = 0 , we obtain

B ˜ 4 ( ξ ) ≤ 2 ∣ ξ ∣ + 1 2 α 2 ∣ ξ ∣ 5 2 − p ≤ 1 2 α 2 2 ∣ ξ ∣ 3 − p + ∣ ξ ∣ 5 2 − p .

If 1 2 < p ≤ 5 2 , note that α ∈ ( 0 , 1 ) , we have

B ˜ 4 ( ξ ) ≤ 1 2 α 2 2 ( α − 2 3 ) 3 − p + ( α − 2 3 ) 5 2 − p ≤ 2 2 α 2 3 p + α 2 3 p + 1 3 ≤ 2 α 2 3 ( p − 1 2 ) .

If 5 2 < p ≤ 3 , note that 3 − p ∈ [ 0 , 1 2 ) , we obtain

B ˜ 4 ( ξ ) ≤ 1 2 α 2 2 ( α − 2 3 ) 3 − p + 1 ≤ 1 2 α 2 2 ( α − 2 3 ) 1 2 + 1 ≤ 2 2 ( α 5 3 + α 2 ) .

If p > 3 , note that α ∈ ( 0 , 1 ) , we obtain

B ˜ 4 ( ξ ) ≤ 1 2 α 2 ( 2 + 1 ) ≤ 2 2 ( α 5 3 + α 2 ) .

Case 3: ∣ ξ ∣ < 1 .

B ˜ 4 ( ξ ) ≤ 1 2 α 2 ( 2 ∣ ξ ∣ + 1 ) ∣ ξ ∣ 5 2 ≤ 1 2 α 2 ( 2 + 1 ) ≤ 2 2 ( α 5 3 + α 2 ) .

In conclusion,

(4.15) B ˜ 4 ( ξ ) ≤ max 2 ( α 2 3 ) p − 1 2 , 2 2 ( α 5 3 + α 2 ) ≕ ε 3 , p > 1 2 .

By Lemma 3.2, we obtain

(4.16) J 3 ˜ ≤ ε 3 ( E + M 1 ) , p > 1 2 .

Let

B ˜ 5 ( ξ ) = cosh ( θ ˜ s ) − cosh ( θ s ) ( 1 + ξ 2 ) − p 2 e − ∣ ξ ∣ 2 .

To estimate J 4 ˜ , we first estimate B ˜ 5 ( ξ ) . Similarly, divided into three cases.

Case 1: ∣ ξ ∣ ≥ ξ 1 . By the triangle inequality and (2.11), we have

B ˜ 5 ( ξ ) ≤ ( ∣ e θ s ∣ + ∣ e θ ˜ s ∣ ) ( 1 + ξ 2 ) − p 2 e − ∣ ξ ∣ 2 ≤ ( e ∣ ξ ∣ 2 + e ∣ ξ ∣ 2 θ 1 ) ∣ ξ ∣ − p e − ∣ ξ ∣ 2 ≤ 2 ∣ ξ ∣ − p ≤ 2 α 2 3 p .

Case 2: 1 ≤ ∣ ξ ∣ < ξ 1 . By (4.9), for ∀ s ∈ ( x , 1 ) , we obtain

B ˜ 5 ( ξ ) ≤ ∣ ξ ∣ 3 α 2 ( 1 + ξ 2 ) − p 2 ≤ ∣ ξ ∣ 3 − p α 2 .

If 0 < p ≤ 3 , we have

A ˜ 4 ( ξ ) ≤ ( α − 2 3 ) 3 − p α 2 ≤ 2 α 2 3 p .

If p > 3 , note that ∣ ξ ∣ ≥ 1 , we obtain

A ˜ 4 ( ξ ) ≤ α 2 .

Case 3: ∣ ξ ∣ < 1 .

B ˜ 5 ( ξ ) ≤ ∣ ξ ∣ 3 α 2 ( 1 + ξ 2 ) − p 2 ≤ ∣ ξ ∣ 3 α 2 ≤ α 2 .

In conclusion,

(4.17) B ˜ 5 ( ξ ) ≤ max 2 α 2 3 p , α 2 ≕ ε 4 , p > 1 2 .

We are now in a position to estimate J 4 ˜ .

J 4 ˜ ≤ ∫ − ∞ ∞ ∫ 0 1 ∣ f ˆ ∣ 2 d s ∫ 0 1 cosh ( θ s ) − cosh ( θ ˜ s ) 2 d s d ξ 1 2 ≤ ∫ − ∞ ∞ ( 1 + ξ 2 ) p ∫ 0 1 ∣ f ˆ ∣ 2 d s ∫ 0 1 ( 1 + ξ 2 ) − p 2 ∣ cosh ( θ s ) − cosh ( θ ˜ s ) ∣ 2 d s d ξ 1 2 ≤ ∫ − ∞ ∞ e 2 ∣ ξ ∣ − 3 ∣ ξ ∣ ∫ 0 1 ( 1 + ξ 2 ) − p 2 e − ∣ ξ ∣ 2 ∣ cosh ( θ s ) − cosh ( θ ˜ s ) ∣ 2 d s d ξ 1 2 ≤ ε 4 ∫ − ∞ ∞ e 2 ∣ ξ ∣ − 3 ∣ ξ ∣ d ξ 1 2 .

Introducing the notation,

M 4 = ∫ − ∞ ∞ e 2 ∣ ξ ∣ − 3 ∣ ξ ∣ d ξ 1 2 .

Then,

(4.18) J 4 ˜ ≤ ε 4 M 4 .

Substituting (4.16) and (4.18) into J 4 , we have

(4.19) J 4 ≤ ε 3 ( E + M 1 ) + ε 4 M 4 ,

where ε 3 and ε 4 are given by (4.15) and (4.17). Theorem 4.2 now follows from (4.13), (4.14), and (4.19).□

Remark 4.2

Since the regularization parameter α → 0 ( δ → 0 ), we can easily find that, for p > 1 2 , ε 3 , ε 4 → 0 ( δ → 0 ). Therefore,

lim δ → 0 ‖ u x α , δ ( 0 , ⋅ ) − u x ( 0 , ⋅ ) ‖ = 0 , p > 1 2 .

Remark 4.3

If conditions (3.1), (3.14), and (3.15) hold, and the regularization parameter α is chosen by (3.16) in Theorems 3.1 and 4.1, then we can also obtain convergence results.

Remark 4.4

Theorems 4.1 and 4.2 are the main results of this section, which are not provided in other articles.

Remark 4.5

The proposed method can be easily extended to the two-dimensional case, i.e.,

(4.20) u t − u x x − u y y = f ( x , y , t ) , x > 0 , y > 0 , t > 0 , u ( 1 , y , t ) = g ( y , t ) , y ≥ 0 , t ≥ 0 , u ( x , y , 0 ) = 0 , x ≥ 0 , y ≥ 0 , u ( x , 0 , t ) = 0 , x ≥ 0 , t ≥ 0 , u y ( x , 0 , t ) = 0 , x ≥ 0 , t ≥ 0 , u ( x , y , t ) ∣ x → ∞ bounded .

In fact, we can obtain error estimates for temperature and flux over the entire region by solving the following problem:

(4.21) u t α , δ = u x x α , δ + u y y α , δ − α 2 ( u x x y y α , δ + u x x t t α , δ ) = f δ ( x , y , t ) , x > 0 , y > 0 , t > 0 , u α , δ ( 1 , y , t ) = g δ ( y , t ) , y ≥ 0 , t ≥ 0 , u α , δ ( x , y , 0 ) = 0 , x ≥ 0 , y ≥ 0 , u α , δ ( x , 0 , t ) = 0 , x ≥ 0 , t ≥ 0 , u y α , δ ( x , 0 , t ) = 0 , x ≥ 0 , t ≥ 0 , u α , δ ( x , y , t ) ∣ x → ∞ bounded .

5 Application

In this section, we present several examples to illustrate the performance of the proposed method. In particular, for the convenience of testing, we take f ( x , t ) = 0 as an example for numerical experiments. First, in order to construct the data g ( t ) , we consider the direct problem

(5.1) u t − u x x = 0 , x > 0 , t > 0 , u ( x , 0 ) = 0 , x ≥ 0 , u ( 0 , t ) = h ( t ) , t ≥ 0 , u ( x , t ) ∣ x → ∞ bounded ,

which is well posed and may be solved by the method presented in [41]. The solution of (5.1) at x = 1 is

(5.2) g ( t ) ≔ u ( 1 , t ) = 1 2 π ∫ − ∞ ∞ e i ξ t e − i ξ h ˆ ( ξ ) d ξ .

In the numerical implementation, we give the data h ( t ) , sample on the equidistant grid and perform the discrete Fourier transform. Then, according to (5.2), we obtain the data g ( t ) by inverse discrete Fourier transform. The noise data g δ is generated by

g δ = g + ε ⋅ rand n ( size ( g ) ) ,

where ε denotes the error level, and the function rand n ( ⋅ ) generates arrays of random numbers whose elements are normally distributed with a mean of 0 and a variance of σ 2 = 1 . The total noise δ can be measured in the sense of root mean square error according to

(5.3) δ = ‖ g δ − g ‖ = 1 M ∑ i = 1 M ( g δ ( t i ) − g ( t i ) ) 2 .

The approximate solutions of temperature and flux are calculated by (2.7) and (4.2), respectively, and the regularization parameter α is selected by (3.16). To show the accuracy of numerical solutions, we compute the approximate relative errors in l 2 norm, denoted by

(5.4) e r ( u , ε ) = ‖ u α , δ ( x , t ) − u ( x , t ) ‖ l 2 ‖ u ( x , t ) ‖ l 2

and

(5.5) e r ( u x , ε ) = ‖ u x α , δ ( x , t ) − u x ( x , t ) ‖ l 2 ‖ u x ( x , t ) ‖ l 2 .

Application 5.1

Consider a smooth data function h ( t ) = t sin ( − 4 t ) .

In this example, we take p = 2 7 . Figure 1 gives the numerical results without using any regularization method for the error level ε = 0.03 . It is easy to see from Figure 1 that the approximate solutions without using any regularization method are unacceptable. Since problem (1.4) is ill-posed, we must apply some regularization methods to it. Figures 2 and 3 show the approximate results of temperature and flux at different locations. We can observe that the smaller the error level, the better the approximation. To further illustrate the proposed method, we compare the quasi-reversibility method with the Fourier regularization method [42] in Table 1. e r 1 ( u , ε ) and e r 1 ( u x , ε ) denote the relative error of the Fourier method, and e r 2 ( u , ε ) and e r 2 ( u x , ε ) denote the relative error of the quasi-reversibility method. As can be seen from Table 1, the proposed approach generally performs well in the current context.

$Figure 1 Un-regularized solutions and exact solutions with ε = 0.03 \varepsilon =0.03 for Application 5.1: (a) temperature and (b) heat flux.$

Figure 1

Un-regularized solutions and exact solutions with ε = 0.03 for Application 5.1: (a) temperature and (b) heat flux.

$Figure 2 Numerical results of temperature for Application 5.1: (a) x = 0 x=0 , (b) x = 0.3 x=0.3 , (c) x = 0.6 x=0.6 , and (d) x = 0.9 x=0.9 .$

Figure 2

Numerical results of temperature for Application 5.1: (a) x = 0 , (b) x = 0.3 , (c) x = 0.6 , and (d) x = 0.9 .

$Figure 3 Numerical results of heat flux for Application 5.1: (a) x = 0 x=0 , (b) x = 0.3 x=0.3 , (c) x = 0.6 x=0.6 , and (d) x = 0.9 x=0.9 .$

Figure 3

Numerical results of heat flux for Application 5.1: (a) x = 0 , (b) x = 0.3 , (c) x = 0.6 , and (d) x = 0.9 .

Table 1

Comparison of numerical results between the Fourier Method ( e r 1 ) and the quasi-reversibility method ( e r 2 ) in Application 5.1 with ε = 0.01

x	0.05	0.2	0.35	0.5	0.65	0.875	0.95
e r 1 ( u , 0.01 )	0.2274	0.1821	0.1125	0.0798	0.0449	0.0282	0.0176
e r 2 ( u , 0.01 )	0.1603	0.1244	0.0779	0.0613	0.0442	0.0298	0.0218
e r 1 ( u x , 0.01 )	0.7835	0.5424	0.3073	0.2006	0.1181	0.0949	0.0548
e r 2 ( u x , 0.01 )	0.3906	0.3063	0.2280	0.1652	0.1070	0.0839	0.0584

Application 5.2

Choose a non-smooth but continuous data function

h ( t ) = t , 0 ≤ t ≤ 1 2 , 1 − t , 1 2 < t ≤ 1 .

Figures 4 and 5 and Table 2 give the numerical results of Application 5.2 with p = 3 7 . From Table 2, it can be found that the closer x is to 1, the better the numerical accuracy, which is reasonable in view of the character of the proposed problem. On the other hand, since Application 5.2 is a non-smooth function, its numerical results are slightly worse than those of Application 5.1. The numerical results are acceptable because of the ill-posedness of the problem. Figure 6 illustrates a plot of the error in temperature and heat flux as a function of p , from which it can be seen that for large p , the results are not necessarily good.

$Figure 4 Numerical results of temperature for Application 5.2: (a) x = 0.4 x=0.4 and (b) x = 0.8 x=0.8 .$

Figure 4

Numerical results of temperature for Application 5.2: (a) x = 0.4 and (b) x = 0.8 .

$Figure 5 Numerical results of heat flux for Application 5.2: (a) x = 0.4 x=0.4 and (b) x = 0.8 x=0.8 .$

Figure 5

Numerical results of heat flux for Application 5.2: (a) x = 0.4 and (b) x = 0.8 .

Table 2

Numerical results of Application 5.2 for different x with p = 3 7

x	0	0.15	0.3	0.45	0.6	0.75	0.9
e r ( u , 0.003 )	0.1008	0.0609	0.0422	0.0345	0.0223	0.0184	0.0142
e r ( u , 0.001 )	0.0479	0.0349	0.0211	0.0154	0.0100	0.0069	0.0044
e r ( u x , 0.003 )	0.2884	0.2162	0.1744	0.1705	0.1201	0.1151	0.1076
e r ( u x , 0.001 )	0.1670	0.1435	0.0997	0.0908	0.0726	0.0695	0.0555

$Figure 6 Numerical results for Application 5.2 with ε = 0.001 \varepsilon =0.001 at x = 0.8 x=0.8 : (a) temperature u ( x , t ) u\left(x,t) and (b) heat flux u x ( x , t ) {u}_{x}\left(x,t) .$

Figure 6

Numerical results for Application 5.2 with ε = 0.001 at x = 0.8 : (a) temperature u ( x , t ) and (b) heat flux u x ( x , t ) .

Application 5.3

Let the exact data function be h ( y , t ) = 4 cos ( 2 π y ) cos ( 2 π t ) .

The numerical results of Application 5.3 are presented in Figures 7 and 8, from which it is easy to see that the recovery of temperature is slightly better than the recovery of flux at the same location. On the other hand, it can be observed that the approximation of this example is not as good as that of Applications 5.1 and 5.2, due to the fact that problem (4.20) has more variables and is more ill-posed. In summary, the proposed quasi-reversibility method can be used to deal with similar problems in the two-dimensional case.

$Figure 7 Numerical results of temperature at x = 0.8 x=0.8 for Application 5.3: (a) exact u ( x , y , t ) u\left(x,y,t) and (b) numerical u δ ( x , y , t ) {u}^{\delta }\left(x,y,t) .$

Figure 7

Numerical results of temperature at x = 0.8 for Application 5.3: (a) exact u ( x , y , t ) and (b) numerical u δ ( x , y , t ) .

$Figure 8 Numerical results of heat flux at x = 0.8 x=0.8 for Application 5.3: (a) exact u x ( x , y , t ) {u}_{x}\left(x,y,t) and (b) numerical u x δ ( x , y , t ) {u}_{x}^{\delta }\left(x,y,t) .$

Figure 8

Numerical results of heat flux at x = 0.8 for Application 5.3: (a) exact u x ( x , y , t ) and (b) numerical u x δ ( x , y , t ) .

6 Conclusion

SHE is very important in some engineering contexts and industrial applications, so many researchers have been involved in this topic in recent decades. For the homogeneous setting, theoretical concepts and computational implementation are plentifully discussed and developed well. However, there are very few theoretical stability and convergence estimates for the nonhomogeneous system.

In this article, we use a quasi-reversibility method to recover the surface temperature and heat flux of the nonhomogeneous SHE. The convergence estimates of regularized solutions are obtained under some appropriate a prior conditions. Numerical verification of the efficiency and accuracy of the quasi-reversibility method is performed by solving some numerical examples. Furthermore, we point out that the proposed method can be extended to the two-dimensional case. In fact, our main contribution in this work is focused on the generalization of the standard inverse problems for heat equations to inverse problems for nonhomogeneous heat equations involving the source term f ( x , t ) . These mathematical problems have become an important tool in modeling many real-life problems. It is worth noting that the estimates obtained are sufficient to prove the results, but most of them are quite rough and can be improved. Here, the source term f ( x , t ) is a function of x and t . However, the source function may also be related to u , i.e., f = f ( x , t , u ) . Obviously, in this case, some new and strong techniques [3,43] are needed to estimate the errors. Moreover, the homogeneous initial condition, u ( x , 0 ) = 0 , is not important. A problem with non-zero initial data, u ( x , 0 ) = ϕ ( x ) , can be reduced to (1.4) by using the linearity of the differential equation. And the modified method can easily be extended to variable coefficient problems, which requires further exploration.

Acknowledgement

The authors thank the referees for their valuable comments and suggestions, which improved the presentation of this manuscript.

Funding information: The research was supported by the Natural Science Foundation of China (No. 11661072) and the Natural Science Foundation of Northwest Normal University (No. NWNU-LKQN-17-5).
Author contributions: Yu Qiao and Xiangtuan Xiong wrote jointly the manuscript and prepared all figures (tables). All authors read and approved the final manuscript.
Conflict of interest: The authors state no conflict of interest.
Data availability statement: Data sharing is not applicable to this article as no datasets were generated or analyzed during this study.

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Received: 2023-09-25

Revised: 2024-03-30

Accepted: 2024-07-25

Published Online: 2024-09-16

This work is licensed under the Creative Commons Attribution 4.0 International License.

Articles in the same Issue

https://doi.org/10.1515/math-2024-0049

Keywords for this article

ill-posed problem; nonhomogeneous sideways heat equation; a quasi-reversibility method; error estimate

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