# Regularization techniques for estimating the source in a complete parabolic
equation in $\mathbb{R}^{n}$

GUILLERMO FEDERICO UMBRICHT Departamento de Matemática, Facultad de Ciencias
Empresariales, Universidad Austral Paraguay 1950, Rosario, Santa Fe, ARGENTINA
Consejo Nacional de Investigaciones Científicas y Técnicas (CONICET) Godoy
Cruz 2290, CABA, ARGENTINA DIANA RUBIO ITECA (UNSAM-CONICET), CEDEMA, ECyT,
Universidad Nacional de General San Martín 25 de mayo y Francia, San Martín,
Buenos Aires, ARGENTINA

Abstract: In this article, the problem of identifying the source term in
transport processes given by a complete parabolic equation is studied
mathematically from noisy measurements taken at an arbitrary fixed time. The
problem is solved analytically with Fourier techniques and it is shown that
this solution is not stable. Three single parameter families of regularization
operators are proposed to dealt with the instability of the solution. Each of
them is designed to compensate the factor that causes the instability of the
inverse operator. Moreover, a rule of choice for the regularization parameter
is included and a Hölder error bound type is obtained for each estimation.
Numerical examples of different characteristics are presented to demonstrate
the benefits of the proposed strategies. Key–Words: Transfer problems,
Parabolic equation, Inverse problems, Ill-posed problems, Regularization
operators. AMS Subject Classification: 35R30; 35R25; 47A52; 58J35; 65T50.

##  1 Introduction

Complete parabolic equations are used to model transport phenomena. These are
highly studied due to the number of direct applications they are related to in
different fields of physics, as they are used to analyze the transfer of mass,
energy, and information in different processes [1, 2]. Among the most
important applications, we can mention the heat transfer of a body immersed in
a moving fluid [3], heat transfer in biological tissue [4], and the
concentration of a certain compound in a flow of water [5].

On the other hand, the problem of source determination has been analyzed in
recent years in various areas of applied physics and has garnered considerable
attention in current research. It finds applications in fields such as heat
conduction [6, 7, 8, 9], crack identification [10], electromagnetic theory
[11, 12, 13], geophysical prospecting [14], contaminant detection [15], and
detection of tumor cells [16], among others.

Several methods are available in the literature for determining a source, and
among the most important tools are the potential logarithmic method [17], the
projective method [18], Green functions [19], limit element methods of dual
reciprocity [20], the method of dual reciprocity [21], and the method of
fundamental solution [22].

In the context of various transport processes or complete parabolic
differential equations, there is a limited number of articles published for
the general case [23, 24, 25], with the majority of available literature
focusing on the one-dimensional heat equation. Different methods, techniques,
and strategies have been employed to retrieve sources in this equation, as
seen in [26, 27, 28]. Many articles analyze specific cases with
simplifications or restrictions in mathematical models, source types, boundary
conditions, or selected domains, such as [26, 27, 28, 29, 30, 31, 32]. The
most commonly used methods in these cases include the limit element method
[26, 28], the fundamental solution method [27, 32], the Ritz-Galerkin method
[33], the finite difference method [34], the meshless method [35], the
conditional stability method [36], and the shooting method [29].

The problem of source determination is considered ill-posed in the sense of
Hadamard [37], as the solution does not depend continuously on the data.
Regularization methods [38, 39] play a crucial role in estimating unstable
solutions. Among the most commonly used are the iterative regularization
method [40, 41], the simplified Tikhonov regularization method [42, 43, 44,
45, 46], the modified regularization method [8, 25, 46, 47, 48, 49], Fourier
truncation [48], and the mollification method [50].

When it comes to source determination in parabolic equations, [51] focuses on
a convection-diffusion equation, while [7, 34, 47, 49, 52, 53, 54, 55, 56]
exclusively consider diffusion. More recently, the problem of finding the
source term in a complete parabolic equation has been addressed using the
quasi-reversibility method, as seen in [57]. Notably, this method extends its
applicability to solving the inverse source problem in nonlinear parabolic
equations [58].

The issue addressed in this article stems from modeling various transportation
problems and involves identifying the source term, which is independent of
time, in a complete parabolic evolutionary equation using measurements or
noisy data taken at an arbitrary fixed time. This problem is ill-posed, as
high-frequency components of arbitrarily small data errors can result in
disproportionately large errors in the solution. To tackle this challenge,
three uniparametric families of regularization operators are devised to
counteract the instability of the inverse operator. These regularization
operators create families of well-posed problems that approximate the
originally ill-posed problem. Additionally, a guideline is provided for
selecting the regularization parameter. The stability and convergence of each
method are analyzed, and an optimal H”older bound is derived for the error of
each estimate.

The three strategies presented here generalize the ideas used by other authors
for the case of the one-dimensional heat equation. Here, they are applied to a
complete parabolic equation where temperature measurements can be taken at any
instant. On the other hand, the proposal is framed within the theory of
operators, giving rise to applications in more general contexts and problems.
Any important application is the detection of contaminants in groundwater
layers, this is a problem that concerns all urbanized cities. Being able to
determine the focus of contamination from measurements in a particular place,
minimizes the costs used for the search [15, 59, 60]. Another application is
the estimation of the metabolic heat in a biological tissue [24] using the
Pennes model [4]. Any existing abnormality within the body can lead to
variations in temperature and heat flux at the surface. The presence of a
tumor produces local inflammation, increased metabolic activity, among other
symptoms. Due to this, the diseased cells act as a source of temperature that
produces abnormal thermal profiles in the skin and its measurements can be
used to identify, locate and characterize the sick cells. The two problems
mentioned above can be addressed with the tools that are explained in this
paper. There are many other applications in different disciplines, these two
are mentioned as an example to highlight the importance of the problem and the
proposed methods.

To illustrate the performance of the proposed regularizations and in order to
compare the different regularization operators introduced in this article;
numerical examples of different characteristics are included.

##  2 Source identification

In this section, the inverse problem to be studied is formally presented, an
analytical expression is given for the solution of the problem of interest,
and it is shown that the problem is ill-posed.

###  2.1 Presentation of the problem

Given $u:\mathbb{R}^{n}\times\mathbb{R}^{+}\longrightarrow\mathbb{R}$, we seek
to determine the source term in the following complete parabolic equation in
an unbounded domain

$\dfrac{\partial u}{\partial t}({\mathbf{x}},t)=\alpha^{2}\Delta
u(\mathbf{x},t)-\bm{\beta}\cdot\nabla(u(\mathbf{x},t))-\nu
u(\mathbf{x},t)+f(\mathbf{x}),\qquad\mathbf{x}\in\mathbb{R}^{n},\,\,\,t>0,$
(1)

where $\Delta$,$\nabla$ denote the Laplacian and Nabla operators,
respectively, and $``\cdot"$ represents the usual inner product in
$\mathbb{R}^{n}$. Without loss of generality, the null initial condition is
considered, i.e.,

$u(\mathbf{x},t)=0,\qquad\mathbf{x}\in\mathbb{R}^{n},\,\,\,t=0.$ (2)

The determination of the source term in Equation (1) with the condition (2),
is carried out using experimental or simulated noisy data, at an instant of
time $t_{0}$.

$u(\mathbf{x},t)=y_{\delta}(\mathbf{x}),\qquad\mathbf{x}\in\mathbb{R}^{n},\,\,\,t=t_{0}>0,$
(3)

where $y_{\delta}$ represents the noisy data or measurements and $\delta$ is
the noise in the data. It is also assumed that said noise is bounded, that is,

$||y(\mathbf{x})-y_{\delta}(\mathbf{x})||_{L^{2}(\mathbb{R}^{n})}\leq\delta,\qquad
0<\delta\leq\delta_{M},\\\ $ (4)

where $\delta_{M}\in\mathbb{R}^{+}$ is the maximum noise level tolerated. In
practice $\delta_{M}$ is obtained from measurement, instrumentation and
calibration errors.

###  2.2 Problem solution

The source estimation problem will be solved using the $n$-dimensional Fourier
transform. It is included here for completeness reasons.

######  Definition 1.

Let $g\in L^{2}(\mathbb{R}^{n})$, the $n$-dimensional Fourier transform [61]
is defined by

$\widehat{g}(\bm{\xi}):=\left(\dfrac{1}{\sqrt{2\pi}}\right)^{n}\displaystyle\int\limits_{\mathbb{R}^{n}}{e^{-i\bm{\xi}\cdot\mathbf{x}}\,g(\mathbf{x})\,d\mathbf{x}},\qquad\bm{\xi}\in\mathbb{R}^{n}.$

Let $\widehat{g}\in L^{2}(\mathbb{R}^{n})$, the Fourier antitransform is
defined by

$g(\mathbf{x}):=\left(\dfrac{1}{\sqrt{2\pi}}\right)^{n}\displaystyle\int\limits_{\mathbb{R}^{n}}{e^{i\bm{\xi}\cdot\mathbf{x}}\,\widehat{g}(\bm{\xi})\,d\bm{\xi}},\qquad\mathbf{x}\in\mathbb{R}^{n}.$

######  Proposition 1.

The $n$-dimensional Fourier transform is a linear integral operator that
satisfies the following property. Let $g\in L^{2}(\mathbb{R}^{n})$ hence:

$\widehat{\nabla
g}(\bm{\xi})=i\,\bm{\xi}\,\widehat{g}(\bm{\xi}),\qquad\widehat{\Delta
g}(\bm{\xi})=-\left\|\bm{\xi}\right\|^{2}\,\widehat{g}(\bm{\xi}),\qquad\bm{\xi}\in\mathbb{R}^{n}.$

From the use of the definition 1 and the proposition it is possible to find
the analytical solution of the problem of interest given by the equations
(1)-(4). This is given in the following result.

######  Theorem 1 (Solving the source identification problem).

For $\bm{\beta}\in\mathbb{R}^{n}$ and
$\alpha^{2},\nu,t_{0},\delta,\delta_{M}\in\mathbb{R}^{+}$ such that
$\delta<\delta_{M}$, are the functions
$u(\cdot,t),f(\cdot),y(\cdot),y_{\delta}(\cdot)\in L^{2}(\mathbb{R}^{n})$ with
$||y-y_{\delta}||_{L^{2}(\mathbb{R}^{n})}\leq\delta$ satisfying the parabolic
problem

$\begin{cases}\dfrac{\partial u}{\partial t}({\mathbf{x}},t)=\alpha^{2}\Delta
u(\mathbf{x},t)-\bm{\beta}\cdot\nabla(u(\mathbf{x},t))-\nu
u(\mathbf{x},t)+f(\mathbf{x}),\quad&\mathbf{x}\in\mathbb{R}^{n},\,\,\,t>0,\\\
u(\mathbf{x},t)=0,\quad&\mathbf{x}\in\mathbb{R}^{n},\,\,\,t=0,\\\
u(\mathbf{x},t)=y_{\delta}(\mathbf{x}),\quad&\mathbf{x}\in\mathbb{R}^{n},\,\,\,t=t_{0}>0.\end{cases}$
(5)

Then, the expression for the source is given by

$f_{\delta}(\mathbf{x})=\left(\dfrac{1}{\sqrt{2\pi}}\right)^{n}\displaystyle\int\limits_{\mathbb{R}^{n}}e^{i\bm{\xi}\cdot\mathbf{x}}\Lambda(\bm{\xi})\left[\left(\dfrac{1}{\sqrt{2\pi}}\right)^{n}\displaystyle\int\limits_{\mathbb{R}^{n}}e^{-i\bm{\xi}\cdot\mathbf{x}}y_{\delta}(\mathbf{x})d\mathbf{x}\right]d\bm{\xi},$

where

$\Lambda(\bm{\xi})=\dfrac{z(\bm{\xi})}{1-e^{-z(\bm{\xi})\,t_{0}}},$

with

$z(\bm{\xi})=\alpha^{2}\left\|\bm{\xi}\right\|^{2}+i\,\bm{\beta}\cdot\bm{\xi}+\nu\,\in\mathbb{C}.$

###### Proof.

For the proof of theorem 1 it is used the $n$-dimensional Fourier transform
given in 1 on the space variables of the system (5). Then the proposition 1 is
used to obtain,

$\begin{cases}\widehat{\dfrac{\partial u}{\partial
t}}({\bm{\xi}},t)=-(\alpha^{2}\left\|\bm{\xi}\right\|^{2}+i\bm{\beta}\cdot\bm{\xi}+\nu)\,\widehat{u}(\bm{\xi},t)+\widehat{f}(\bm{\xi}),\qquad&\bm{\xi}\in\mathbb{R}^{n},\,\,\,t>0,\\\
\widehat{u}(\bm{\xi},t)=0,\qquad&\bm{\xi}\in\mathbb{R}^{n},\,\,\,t=0,\\\
\widehat{u}(\bm{\xi},t)=\widehat{y_{\delta}}(\bm{\xi}),\qquad&\bm{\xi}\in\mathbb{R}^{n},\,\,\,t=t_{0}>0,\end{cases}$
(6)

where $\bm{\xi}$ is the $n$-dimensional Fourier variable. Equivalently, the
identification problem (5) can be rewritten fron (6) in frequency space as
follows

$\begin{cases}\widehat{\dfrac{\partial u}{\partial
t}}({\bm{\xi}},t)=-z(\bm{\xi})\,\widehat{u}(\bm{\xi},t)+\widehat{f}(\bm{\xi}),\qquad&\bm{\xi}\in\mathbb{R}^{n},\,\,\,t>0,\\\
\widehat{u}(\bm{\xi},t)=0,\qquad&\bm{\xi}\in\mathbb{R}^{n},\,\,\,t=0,\\\
\widehat{u}(\bm{\xi},t)=\widehat{y_{\delta}}(\bm{\xi}),\qquad&\bm{\xi}\in\mathbb{R}^{n},\,\,\,t=t_{0}>0,\end{cases}$
(7)

where $z(\bm{\xi})$ is given by the expression,

$z(\bm{\xi})=\alpha^{2}\left\|\bm{\xi}\right\|^{2}+i\,\bm{\beta}\cdot\bm{\xi}+\nu.$
(8)

The system (7) is represented by a first order non-homogeneous differential
equation with initial condition. The analytical solution to this equation is,

$\widehat{u}(\bm{\xi},t)=\dfrac{1-e^{-z(\bm{\xi})\,t}}{z(\bm{\xi})}\widehat{f}(\bm{\xi}).$
(9)

Because $\widehat{u}(\bm{\xi},t_{0})=\widehat{y_{\delta}}(\bm{\xi})$,
evaluating Equation (9) at $t=t_{0}$ produces a linear expression for the
source in frequency space,

$\widehat{f_{\delta}}(\bm{\xi})=\Lambda(\bm{\xi})\widehat{y_{\delta}}(\bm{\xi}),$
(10)

where

$\Lambda(\bm{\xi})=\dfrac{z(\bm{\xi})}{1-e^{-z(\bm{\xi})\,t_{0}}}.$ (11)

Equivalently, using the definition 1 (of the Fourier antitransform) results
from the Equation(10)

$f_{\delta}(\mathbf{x})=\left(\dfrac{1}{\sqrt{2\pi}}\right)^{n}\displaystyle\int\limits_{\mathbb{R}^{n}}e^{i\bm{\xi}\cdot\mathbf{x}}\Lambda(\bm{\xi})\left[\left(\dfrac{1}{\sqrt{2\pi}}\right)^{n}\displaystyle\int\limits_{\mathbb{R}^{n}}e^{-i\bm{\xi}\cdot\mathbf{x}}y_{\delta}(\mathbf{x})d\mathbf{x}\right]d\bm{\xi},$
(12)

which ends the proof. ∎

###  2.3 Ill-posed problem

The problem of identifying the source in a complete parabolic equation from
noisy measurements turns out to be a ill-posed problem in the sense of
Hadamard [37] since the solution does not depend continuously on the data.
This fact can be seen in the following result.

######  Theorem 2 (The problem is ill-posed).

Under the assumptions used so far, the identification problem (5) given in
Theorem 1. It is an ill-posed problem in Hadamard’s sense, since the solutions
do not vary continuously with the data.

###### Proof.

We denote
$\widehat{f_{\delta}}(\bm{\xi})=\Lambda(\bm{\xi})\widehat{y_{\delta}}(\bm{\xi})$.
It’s easy to see that

$\|\widehat{f}-\widehat{f_{\delta}}\|_{L^{2}(\mathbb{R}^{n})}=\|\Lambda(\bm{\xi})(\widehat{y}(\bm{\xi})-\widehat{y_{\delta}}(\bm{\xi}))\|_{L^{2}(\mathbb{R}^{n})}=\left\|\Lambda(\bm{\xi})(\widehat{y}(\bm{\xi})-\widehat{y_{\delta}}(\bm{\xi}))\right\|_{L^{2}(\mathbb{R}^{n})},$
(13)

on the other hand, using the Equations (8) and (11) it is found that,

$\displaystyle|\Lambda(\bm{\xi})|=\left|\dfrac{z(\bm{\xi})}{1-e^{-z(\bm{\xi})\,t_{0}}}\right|=\left|\dfrac{\alpha^{2}\left\|\bm{\xi}\right\|^{2}+i\,\bm{\beta}\cdot\bm{\xi}+\nu}{1-e^{-(\alpha^{2}\left\|\bm{\xi}\right\|^{2}+i\,\bm{\beta}\cdot\bm{\xi}+\nu)\,t_{0}}}\right|\geq\dfrac{|\alpha^{2}\left\|\bm{\xi}\right\|^{2}+i\,\bm{\beta}\cdot\bm{\xi}+\nu|}{1+e^{-(\alpha^{2}\left\|\bm{\xi}\right\|^{2}+\nu)\,t_{0}}},$
(14)

from (14) it is evident that $\Lambda(\bm{\xi})$ is not bounded, since it
tends to infinity as $\left\|\bm{\xi}\right\|\to\infty$. As can be seen in
(13) this fact amplifies the error of the measurements at high frequencies and
this can lead to a large estimation error
$\|\widehat{f}-\widehat{f_{\delta}}\|_{L^{2}(\mathbb{R}^{n})}$ even for very
small observation or measurement errors. In other words, the solution to the
identification problem (5) does not vary continuously with the data (see
[38]).

∎

##  3 Regularization operators

When an inverse problem is ill-posed, a regularization method is usually
applied to stabilize the solution. In this section, three regularization
operators are proposed for comparative purposes. The existence of the
regularization parameter leading to three convergent methods is proved, and
basic theoretical issues related to regularization operators are included.
Readers unfamiliar with this topic may find more information at[38, 39].

###  3.1 Regularization solutions

To stabilize the ill-posed problem, regularization operators will be used.

######  Definition 2.

Let $\mathbb{X}$ and $\mathbb{Y}$ be Hilbert spaces and
$T:\mathbb{Y}\longrightarrow\mathbb{X}$ a linear bounded operator. A
regularization strategy for $T$ is a family of linear bounded operators
satisfying

$R_{\mu}:\mathbb{Y}\longrightarrow\mathbb{X},\quad\mu>0,\quad/\lim_{\mu\to
0^{+}}R_{\mu}y=Ty,\quad\forall y\in\mathbb{Y}.$

For our particular case, the parametric families of linear operators are
defined by $R_{\mu}^{i}:L^{2}(\mathbb{R}^{n})\to L^{2}(\mathbb{R}^{n})$ with
$\mu\in\mathbb{R}^{+}$ and $i=1,2,3$; such that

$R_{\mu}^{1}\,\widehat{y}:=\dfrac{\Lambda(\bm{\xi})}{1+\mu^{2}\left\|\bm{\xi}\right\|^{2}}\,\widehat{y},\qquad
R_{\mu}^{2}\,\widehat{y}:=\dfrac{\Lambda(\bm{\xi})}{1+\mu^{2}\left\|\bm{\xi}\right\|^{4}}\,\widehat{y},\qquad
R_{\mu}^{3}\,\widehat{y}:=\dfrac{\Lambda(\bm{\xi})}{e^{\mu^{2}\,\left\|\bm{\xi}\right\|^{2}/4}}\,\widehat{y},$
(15)

where $\Lambda(\bm{\xi})$ is defined in (11), $R_{\mu}^{i}$ with $i=1,2,3$,
are regularization strategies for $\Lambda(\bm{\xi})$ and $\mu$ is the
regularization parameter.

######  Note 1.

Note that the denominators of the expressions (15) that define the linear
operators $R_{\mu}^{i}$ with $i=1,2,3$, were introduced in the solution only
for stabilization purposes.

######  Theorem 3 (Convergent regularization operators).

Consider the source identification problem (5). Let $u(\cdot,t),f(\cdot)\in
L^{2}(\mathbb{R}^{n})$ that satisfy the following differential equation with
initial condition

$\begin{cases}\dfrac{\partial u}{\partial t}({\mathbf{x}},t)=\alpha^{2}\Delta
u(\mathbf{x},t)-\bm{\beta}\cdot\nabla(u(\mathbf{x},t))-\nu
u(\mathbf{x},t)+f(\mathbf{x}),\quad&\mathbf{x}\in\mathbb{R}^{n},\,\,\,t>0,\\\
u(\mathbf{x},t)=0,\quad&\mathbf{x}\in\mathbb{R}^{n},\,\,\,t=0\\\ \end{cases}$
(16)

and let $\\{R_{\mu}^{i}\\}$ with $i=1,2,3$ be the families of operators
defined in (15). Then, for all $y(\mathbf{x})=u(\mathbf{x},t_{0})$ there
exists an a priori parameter choice rule for $\mu>0$ such that the pairs
$(R_{\mu}^{i},\mu)$ for $i=1,2,3$ are convergent regularization methods for
the identification problem (16).

###### Proof.

The operators $\\{R_{\mu}^{i}\\}$ with $i=1,2,3$ defined in (15) are
continuous for all $\bm{\xi}\in\mathbb{R}^{n}$ and are bounded, since it is
evident that,

$\begin{split}0\leq\lim_{\left\|\bm{\xi}\right\|\to\infty}R_{\mu}^{i}<\infty,\qquad
i=1,2,3.\end{split}$

Therefore, for each parameter $\mu>0$, the operators $R_{\mu}^{i}$ with
$i=1,2,3$ are linear, continuous and it has to be

$\lim_{\mu\to 0^{+}}R_{\mu}^{i}\,\widehat{y}=\Lambda\,\widehat{y},\qquad
i=1,2,3,$

where $\widehat{y}\in L^{2}(\mathbb{R}^{n})$, then $R_{\mu}^{i}$ with
$i=1,2,3$ are regularization strategies for $\Lambda$. Therefore, according to
[38], there exist _a priori_ parameter choice rules $\mu$ such that
$(R_{\mu}^{i},\mu)$ with $i=1,2,3$ are convergent regularization methods to
solve (10). ∎

The regularized solution of the inverse source identification problem, in
frequency space, is given by

$\widehat{f}_{\delta,\mu}^{i}(\bm{\xi})=R_{\mu}^{i}\,\widehat{y}_{\delta}(\bm{\xi}),\qquad
i=1,2,3.$ (17)

Therefore, from the equation (17) an approximate expression for the function
$f$ is obtained, solution of the problem given in (16). This is,

$f_{\delta,\mu}^{i}(\mathbf{x})=\left(\dfrac{1}{\sqrt{2\pi}}\right)^{n}\displaystyle\int\limits_{\mathbb{R}^{n}}e^{i\bm{\xi}\cdot\mathbf{x}}R_{\mu}^{i}\,\widehat{y}_{\delta}(\bm{\xi})\,d\bm{\xi},\qquad
i=1,2,3,$

equivalently

$f_{\delta,\mu}^{i}(\mathbf{x})=\left(\dfrac{1}{\sqrt{2\pi}}\right)^{n}\displaystyle\int\limits_{\mathbb{R}^{n}}e^{i\bm{\xi}\cdot\mathbf{x}}R_{\mu}^{i}\,\left[\left(\dfrac{1}{\sqrt{2\pi}}\right)^{n}\displaystyle\int\limits_{\mathbb{R}^{n}}e^{-i\bm{\xi}\cdot\mathbf{x}}\,y_{\delta}(\mathbf{x})d\mathbf{x}\right]\,d\bm{\xi},\qquad
i=1,2,3.$ (18)

Using in the equation (18), the definitions of the regularization operators
$(R_{\mu}^{i},\mu)$ with $i=1,2,3$ given by the equations (15), three analytic
expressions for the source estimate are obtained (one for each regularization
strategy). These are,

$f_{\delta,\mu}^{1}(\mathbf{x})=\left(\dfrac{1}{\sqrt{2\pi}}\right)^{n}\displaystyle\int\limits_{\mathbb{R}^{n}}e^{i\bm{\xi}\cdot\mathbf{x}}\dfrac{\Lambda(\bm{\xi})}{1+\mu^{2}\left\|\bm{\xi}\right\|^{2}}\,\left[\left(\dfrac{1}{\sqrt{2\pi}}\right)^{n}\displaystyle\int\limits_{\mathbb{R}^{n}}e^{-i\bm{\xi}\cdot\mathbf{x}}\,y_{\delta}(\mathbf{x})d\mathbf{x}\right]\,d\bm{\xi},$
$f_{\delta,\mu}^{2}(\mathbf{x})=\left(\dfrac{1}{\sqrt{2\pi}}\right)^{n}\displaystyle\int\limits_{\mathbb{R}^{n}}e^{i\bm{\xi}\cdot\mathbf{x}}\dfrac{\Lambda(\bm{\xi})}{1+\mu^{2}\left\|\bm{\xi}\right\|^{4}}\,\left[\left(\dfrac{1}{\sqrt{2\pi}}\right)^{n}\displaystyle\int\limits_{\mathbb{R}^{n}}e^{-i\bm{\xi}\cdot\mathbf{x}}\,y_{\delta}(\mathbf{x})d\mathbf{x}\right]\,d\bm{\xi}$

and

$f_{\delta,\mu}^{3}(\mathbf{x})=\left(\dfrac{1}{\sqrt{2\pi}}\right)^{n}\displaystyle\int\limits_{\mathbb{R}^{n}}e^{i\bm{\xi}\cdot\mathbf{x}}\dfrac{\Lambda(\bm{\xi})}{e^{\mu^{2}\,\left\|\bm{\xi}\right\|^{2}/4}}\,\left[\left(\dfrac{1}{\sqrt{2\pi}}\right)^{n}\displaystyle\int\limits_{\mathbb{R}^{n}}e^{-i\bm{\xi}\cdot\mathbf{x}}\,y_{\delta}(\mathbf{x})d\mathbf{x}\right]\,d\bm{\xi}.$

###  3.2 Error Analysis

In this section, a bound will be obtained for the error committed by each
estimation.

####  3.2.1 Auxiliary results

In order to analyze the behavior of the proposed regularizations, some results
are first introduced that will be used later to obtain an error bound between
the source $f(\mathbf{x})$ and its estimates $f_{\delta,\mu}^{i}(\mathbf{x}).$

######  Note 2.

Some of the auxiliary results may seem trivial to the reader. However, and in
order to carry out a complete and self-contained work, a brief demonstration
is presented for each one of them.

######  Lemma 1.

Let $\omega\in\mathbb{C}$ with $Re(\omega)>0$ then
$\displaystyle\left|\dfrac{1}{1-e^{-\omega}}\right|\leq\dfrac{1}{1-e^{-Re(\omega)}}.$

###### Proof.

Euler’s formula is used for complex numbers and it is found that,

$\displaystyle{\left|1-e^{-\omega}\right|^{2}=\left(1-e^{-Re(\omega)}\right)^{2}+2e^{-Re(\omega)}\left(1-cos\left(Im(\omega)\right)\right)}\geq\left(1-e^{-Re(\omega)}\right)^{2},$

therefore,

$\left|1-e^{-\omega}\right|\geq
1-e^{-Re(\omega)}\Longrightarrow\left|\dfrac{1}{1-e^{-\omega}}\right|\leq\dfrac{1}{1-e^{-Re(\omega)}}.$

∎

######  Lemma 2.

The function $f:\mathbb{R}^{+}\to\mathbb{R}$ defined by $\displaystyle
f(x):=\begin{cases}\dfrac{x}{1-e^{-x}},&0<x<1,\\\ \dfrac{1}{1-e^{-x}},&1\leq
x,\end{cases}$ satisfies $f(x)<2$.

###### Proof.

First, consider $f$ at $(0,1)$. Differentiating, in this case, we have

$f^{\prime}(x)=\left(\dfrac{x}{1-e^{-x}}\right)^{\prime}=\dfrac{1+e^{-x}(-1+x)}{(1-e^{-x})^{2}}>0,$

the function $f$ is increasing in $(0,1)$. Then
$\dfrac{x}{1-e^{-x}}\leq\dfrac{1}{1-e^{-1}}.$

On the other hand, for $x\geq 1$, we have

$f^{\prime}(x)=\left(\dfrac{1}{1-e^{-x}}\right)^{\prime}=\dfrac{-e^{-x}}{(1-e^{-x})^{2}}<0,$

the function $f$ is decreasing $\forall x\geq 1$. Then
$\dfrac{x}{1-e^{-x}}\leq\dfrac{1}{1-e^{-1}}.$

So, $f(x)\leq\dfrac{1}{1-e^{-1}}<2$, ∎

######  Lemma 3.

Let $x\in\mathbb{R}-\left\\{0\right\\}$. If $0<\mu<1$ we have
$\dfrac{x^{2}}{(1-e^{-x^{2}})e^{(\mu^{2}x^{2})/4}}<\dfrac{4}{\mu^{2}}.$

###### Proof.

Using the lemma 2 the proof is immediate since,

$0\leq
x^{2}<1\Longrightarrow\dfrac{x^{2}}{(1-e^{-x^{2}})e^{(\mu^{2}x^{2})/4}}<\dfrac{2}{e^{(\mu^{2}x^{2})/4}}<2<\frac{2}{\mu^{2}}<\frac{4}{\mu^{2}},$
$x^{2}\geq
1\Longrightarrow\dfrac{x^{2}}{(1-e^{-x^{2}})e^{(\mu^{2}x^{2})/4}}<\dfrac{2x^{2}}{e^{(\mu^{2}x^{2})/4}}<\dfrac{2}{e^{\mu^{2}/4}}<\frac{4}{\mu^{2}},$

∎

######  Lemma 4.

Let $x\in\mathbb{R}-\left\\{0\right\\}$. So $\dfrac{1-e^{-x^{2}}}{x^{2}}<1$.

###### Proof.

Let $g:\mathbb{R}^{+}\to\mathbb{R}$ be defined by
$g(x):=\dfrac{1-e^{-x^{2}}}{x^{2}}$. Differentiating we get,

$g^{\prime}(x)=\dfrac{-2x\left[1-(x^{2}+1)e^{-x^{2}}\right]}{x^{2}}<0,\quad\forall
x>0.$

Then, the function $g$ is strictly decreasing and also,

$\lim_{x\to 0^{+}}g(x)=1,\quad\lim_{x\to+\infty}g(x)=0.$

Then $g(x)<\lim_{x\to 0^{+}}g(x)=1,\quad\forall x>0.$ ∎

######  Lemma 5.

Let $a,b,x\in\mathbb{R}^{+}$. So
$\dfrac{x}{ax^{2}+b}\leq\dfrac{1}{2\sqrt{ab}}$.

###### Proof.

Let $h:\mathbb{R}^{+}\to\mathbb{R}$ be defined by $h(x):=\dfrac{x}{ax^{2}+b}$.
Differentiating we get,

$h^{\prime}(x)=\dfrac{b-ax^{2}}{ax^{2}+b},\quad\forall x>0.$

Then the function $h$ has its only critical point at
$x_{0}=\sqrt{\dfrac{b}{a}}$. Noting that $f(0)=0$ and
$\lim_{x\to+\infty}h(x)=0$, it follows that $x_{0}$ is the absolute maximum of
$h$. So,

$h(x)=\dfrac{x}{ax^{2}+b}\leq h(x_{0})=\dfrac{1}{2\sqrt{ab}},\quad\forall
x>0,$

∎

######  Lemma 6.

Let $\rho\in\mathbb{R}$. If $0<\mu<1$ we have
$\dfrac{|\rho|}{1+\rho^{2}\mu^{2}}<\dfrac{1}{2\mu^{2}}.$ Furthermore, for
$\alpha^{2},\nu>0$ the following inequality is valid
$\dfrac{\alpha^{2}\rho^{2}+\nu}{1+\rho^{2}\mu^{2}}<\dfrac{1}{\mu^{2}}\left(\nu+\alpha^{2}\right).$

###### Proof.

For all $\rho\in\mathbb{R}$ holds

$0\leq\displaystyle(1-|\rho|\mu)^{2}=1-2|\rho|\mu+|\rho|^{2}\mu^{2}\Longrightarrow
1+|\rho|^{2}\mu^{2}\geq 2|\rho|\mu.$

In addition, since $\mu>0$ the first inequality is obtained,
$\dfrac{|\rho|}{1+\rho^{2}\mu^{2}}\leq\dfrac{1}{2\mu}<\dfrac{1}{2\mu^{2}}.$

On the other hand, let $\displaystyle
k(\rho):=\frac{\alpha^{2}\rho^{2}+\nu}{1+\rho^{2}\mu^{2}}$ then
$k^{\prime}(\rho)=\dfrac{2\rho(\alpha^{2}-\nu\mu^{2})}{(1+\rho^{2}\mu^{2})^{2}}$
and $k$ has a single critical point $\rho=0$. Three cases are considered:

  * •

$\alpha^{2}=\nu\mu^{2}$: we have $k(\rho)=\nu$, constant
$\forall\rho\in\mathbb{R}$.

  * •

$\alpha^{2}<\nu\mu^{2}$: then the function $k$ reaches its global maximum
value $\nu$ at $\rho=0$.

  * •

$\alpha^{2}>\nu\mu^{2}$: $k$ is an even and increasing function for $\rho>$0
with $\displaystyle\lim_{\rho\to\pm\infty}k(\rho)=\dfrac{\alpha^{2}}{\mu^{2}}$
we have $k(\rho)\leq\displaystyle\dfrac{\alpha^{2}}{\mu^{2}}$.

Therefore,

$\dfrac{\alpha^{2}\rho^{2}+\nu}{1+\rho^{2}\mu^{2}}\leq\max\left\\{\nu,\dfrac{\alpha^{2}}{\mu^{2}}\right\\}\leq\max\left\\{\dfrac{\nu}{\mu^{2}},\dfrac{\alpha^{2}}{\mu^{2}}\right\\}<\dfrac{1}{\mu^{2}}\left(\nu+\alpha^{2}\right).$

∎

######  Lemma 7.

Let $\rho\in\mathbb{R}^{+}$. If $0<\mu<1$ we have
$\dfrac{\rho}{1+\rho^{4}\mu^{2}}<\dfrac{1}{\mu^{2}}.$ Also, for
$\,\alpha^{2},\nu>0$ the following inequality is valid
$\dfrac{\alpha^{2}\rho^{2}+\nu}{1+\rho^{4}\mu^{2}}<\dfrac{1}{\mu^{2}}\left(\nu+\alpha^{2}\right).$

###### Proof.

For all $\rho\geq 1$ the lemma 6 is used and the proof is immediate, since

$\dfrac{\rho}{1+\rho^{4}\mu^{2}}\leq\dfrac{\rho}{1+\rho^{2}\mu^{2}}<\dfrac{1}{2\,\mu^{2}}<\dfrac{1}{\mu^{2}},\qquad\dfrac{\alpha^{2}\rho^{2}+\nu}{1+\rho^{4}\mu^{2}}\leq\dfrac{\alpha^{2}\rho^{2}+\nu}{1+\rho^{2}\mu^{2}}<\dfrac{1}{\mu^{2}}\left(\nu+\alpha^{2}\right).$

On the other hand, for the case $0<\rho<1$, we have,

$\dfrac{\rho}{1+\rho^{4}\mu^{2}}<\dfrac{1}{1+\rho^{4}\mu^{2}}<1<\dfrac{1}{\mu^{2}},\qquad\dfrac{\alpha^{2}\rho^{2}+\nu}{1+\rho^{4}\mu^{2}}<\dfrac{\alpha^{2}+\nu}{1+\rho^{4}\mu^{2}}<\alpha^{2}+\nu<\dfrac{1}{\mu^{2}}(\alpha^{2}+\nu).$

∎

######  Lemma 8.

Let $\bm{\beta}\in\mathbb{R}^{n}$; $\alpha^{2},\nu,t_{0}>0$; $0<\mu<1$ and
$R_{\mu}^{i}$ with $i=1,2,3$ given by (15). So

$\left|R_{\mu}^{i}(\bm{\xi})\right|<\dfrac{M_{i}}{\mu^{2}},\quad i=1,2,3,$

where

$M_{1}=\max\left\\{2\,\nu+2\,\alpha^{2}+\sqrt{n}\left\|\bm{\beta}\right\|_{\infty};\dfrac{2}{t_{0}}+\dfrac{\sqrt{n}\left\|\bm{\beta}\right\|_{\infty}}{\nu\,t_{0}}\right\\},$
(19)
$M_{2}=\max\left\\{2\,\nu+2\,\alpha^{2}+2\,\sqrt{n}\left\|\bm{\beta}\right\|_{\infty};\dfrac{2}{t_{0}}+\dfrac{2\,\sqrt{n}\left\|\bm{\beta}\right\|_{\infty}}{\nu\,t_{0}}\right\\},$
(20)
$M_{3}=\max\left\\{(\alpha^{2}+\nu)\left[8+\frac{4\,\sqrt{n}\left\|\bm{\beta}\right\|_{\infty}}{\sqrt{\alpha^{2}\,\nu}}\right];\dfrac{8}{t_{0}}+\frac{4\,\sqrt{n}\left\|\bm{\beta}\right\|_{\infty}}{t_{0}\sqrt{\alpha^{2}\,\nu}}\right\\}.$
(21)

###### Proof.

Since the three regularization operators given in (15) contain the expression
$\Lambda(\bm{\xi})$ defined in (11), we begin by bounding this operator. The
lemma 1 applied to the absolute value of (11) is used and we immediately
obtain,

$\left|\Lambda(\bm{\xi})\right|=\left|\dfrac{z(\bm{\xi})}{1-e^{-z(\bm{\xi})\,t_{0}}}\right|=\left|\dfrac{\alpha^{2}\left\|\bm{\xi}\right\|^{2}+i\,\bm{\beta}\cdot\bm{\xi}+\nu}{1-e^{-(\alpha^{2}\left\|\bm{\xi}\right\|^{2}+i\,\bm{\beta}\cdot\bm{\xi}+\nu)\,t_{0}}}\right|\leq\left|\dfrac{\alpha^{2}\left\|\bm{\xi}\right\|^{2}+i\,\bm{\beta}\cdot\bm{\xi}+\nu}{1-e^{-(\alpha^{2}\left\|\bm{\xi}\right\|^{2}+\nu)\,t_{0}}}\right|.$
(22)

_If $(\alpha^{2}\left\|\bm{\xi}\right\|^{2}+\nu)\,t_{0}\geq 1$ _ : From the
inequality (22), using the triangular inequality, the lemma 2 and the lemma 6,
we have,

$\begin{split}\phantom{space}\left|R_{\mu}^{1}(\bm{\xi})\right|&=\left|\dfrac{\Lambda(\bm{\xi})}{1+\left\|\bm{\xi}\right\|^{2}\mu^{2}}\right|\leq\dfrac{|\alpha^{2}\,\left\|\bm{\xi}\right\|^{2}+\nu+\bm{\beta}\cdot\bm{\xi}\,i|}{(1-e^{-(\alpha^{2}\left\|\bm{\xi}\right\|^{2}+\nu)\,t_{0}})(1+\left\|\bm{\xi}\right\|^{2}\mu^{2})}\\\
&<2\left(\dfrac{\alpha^{2}\left\|\bm{\xi}\right\|^{2}+\nu}{1+\left\|\bm{\xi}\right\|^{2}\mu^{2}}+\dfrac{\left\|\bm{\beta}\right\|\left\|\xi\right\|}{1+\left\|\bm{\xi}\right\|^{2}\mu^{2}}\right)<2\left(\dfrac{1}{\mu^{2}}\left(\nu+\alpha^{2}\right)+\dfrac{\sqrt{n}\left\|\bm{\beta}\right\|_{\infty}}{2\mu^{2}}\right)\\\
&=\dfrac{1}{\mu^{2}}\left(2\nu+2\alpha^{2}+\sqrt{n}\left\|\bm{\beta}\right\|_{\infty}\right).\end{split}$
(23)

Similarly, from the inequality (22), using the triangular inequality, the
lemma 2 and the lemma 7, we obtain,

$\begin{split}\phantom{space}\left|R_{\mu}^{2}(\bm{\xi})\right|&=\left|\dfrac{\Lambda(\bm{\xi})}{1+\left\|\bm{\xi}\right\|^{4}\mu^{2}}\right|\leq\dfrac{|\alpha^{2}\,\left\|\bm{\xi}\right\|^{2}+\nu+\bm{\beta}\cdot\bm{\xi}\,i|}{(1-e^{-(\alpha^{2}\left\|\bm{\xi}\right\|^{2}+\nu)\,t_{0}})(1+\left\|\bm{\xi}\right\|^{4}\mu^{2})}\\\
&<2\left(\dfrac{\alpha^{2}\left\|\bm{\xi}\right\|^{2}+\nu}{1+\left\|\bm{\xi}\right\|^{4}\mu^{2}}+\dfrac{\left\|\bm{\beta}\right\|\left\|\xi\right\|}{1+\left\|\bm{\xi}\right\|^{4}\mu^{2}}\right)<2\left(\dfrac{1}{\mu^{2}}\left(\nu+\alpha^{2}\right)+\dfrac{\sqrt{n}\left\|\bm{\beta}\right\|_{\infty}}{\mu^{2}}\right)\\\
&=\dfrac{2}{\mu^{2}}\left(\nu+\alpha^{2}+\sqrt{n}\left\|\bm{\beta}\right\|_{\infty}\right).\end{split}$
(24)

Finally, from the inequality (22), using the triangular inequality, the lemmas
2, 3, 4 and 5 hold that,

$\begin{split}\phantom{space}\left|R_{\mu}^{3}(\bm{\xi})\right|&=\left|\dfrac{\Lambda(\bm{\xi})}{e^{(\left\|\bm{\xi}\right\|^{2}\mu^{2})/4}}\right|\leq\dfrac{|\alpha^{2}\,\left\|\bm{\xi}\right\|^{2}+\nu+\bm{\beta}\cdot\bm{\xi}\,i|}{(1-e^{-(\alpha^{2}\left\|\bm{\xi}\right\|^{2}+\nu)\,t_{0}})(e^{(\left\|\bm{\xi}\right\|^{2}\mu^{2})/4})}\\\
&<2\left(\dfrac{\alpha^{2}\left\|\bm{\xi}\right\|^{2}+\nu+\left\|\bm{\beta}\right\|\left\|\xi\right\|}{e^{(\left\|\bm{\xi}\right\|^{2}\mu^{2})/4}}\right)\dfrac{\left\|\bm{\xi}\right\|^{2}}{1-e^{-\left\|\bm{\xi}\right\|^{2}}}\dfrac{1-e^{-\left\|\bm{\xi}\right\|^{2}}}{\left\|\bm{\xi}\right\|^{2}}\\\
&<\dfrac{8}{\mu^{2}}(\alpha^{2}\left\|\bm{\xi}\right\|^{2}+\nu)\left(1+\frac{\left\|\bm{\beta}\right\|\left\|\xi\right\|}{\alpha^{2}\left\|\bm{\xi}\right\|^{2}+\nu}\right)\dfrac{1-e^{-\left\|\bm{\xi}\right\|^{2}}}{\left\|\bm{\xi}\right\|^{2}}\\\
&=\dfrac{8}{\mu^{2}}\left(\dfrac{\alpha^{2}\left\|\bm{\xi}\right\|^{2}(1-e^{-\left\|\bm{\xi}\right\|^{2}})}{\left\|\bm{\xi}\right\|^{2}}+\nu\,\dfrac{1-e^{-\left\|\bm{\xi}\right\|^{2}}}{\left\|\bm{\xi}\right\|^{2}}\right)\left(1+\frac{\left\|\bm{\beta}\right\|\left\|\xi\right\|}{\alpha^{2}\left\|\bm{\xi}\right\|^{2}+\nu}\right)\\\
&<\dfrac{8}{\mu^{2}}(\alpha^{2}+\nu)\left[1+\frac{\sqrt{n}\left\|\bm{\beta}\right\|_{\infty}}{2\sqrt{\alpha^{2}\,\nu}}\right].\end{split}$
(25)

_If $(\alpha^{2}\left\|\bm{\xi}\right\|^{2}+\nu)\,t_{0}\in(0,1)$ _ : From the
inequality (22), using the triangular inequality, the lemma 2 and the lemma 6,
we have,

$\begin{split}\phantom{space}\left|R_{\mu}^{1}(\bm{\xi})\right|&=\left|\dfrac{\Lambda(\bm{\xi})}{1+\left\|\bm{\xi}\right\|^{2}\mu^{2}}\right|\leq\dfrac{|\alpha^{2}\,\left\|\bm{\xi}\right\|^{2}+\nu+\bm{\beta}\cdot\bm{\xi}\,i|}{(1-e^{-(\alpha^{2}\left\|\bm{\xi}\right\|^{2}+\nu)\,t_{0}})(1+\left\|\bm{\xi}\right\|^{2}\mu^{2})}\\\
&=\dfrac{(\alpha^{2}\,\left\|\bm{\xi}\right\|^{2}+\nu)\,t_{0}}{1-e^{-(\alpha^{2}\left\|\bm{\xi}\right\|^{2}+\nu)\,t_{0}}}\left(\dfrac{1}{(1+\left\|\bm{\xi}\right\|^{2}\mu^{2})\,t_{0}}+\dfrac{\left|\bm{\beta}\cdot\bm{\xi}\,i\right|}{(1+\left\|\bm{\xi}\right\|^{2}\mu^{2})(\alpha^{2}\,\left\|\bm{\xi}\right\|^{2}+\nu)\,t_{0}}\right)\\\
&<2\left(\dfrac{1}{t_{0}}+\dfrac{\sqrt{n}\left\|\bm{\beta}\right\|_{\infty}}{2\,\mu^{2}\,\nu\,t_{0}}\right)<\dfrac{2}{\mu^{2}}\left(\dfrac{1}{t_{0}}+\dfrac{\sqrt{n}\left\|\bm{\beta}\right\|_{\infty}}{2\,\nu\,t_{0}}\right).\end{split}$
(26)

Similarly, from the inequality (22), using the triangular inequality, the
lemma 2 and the lemma 7, we obtain,

$\begin{split}\phantom{space}\left|R_{\mu}^{2}(\bm{\xi})\right|&=\left|\dfrac{\Lambda(\bm{\xi})}{1+\left\|\bm{\xi}\right\|^{4}\mu^{2}}\right|\leq\dfrac{|\alpha^{2}\,\left\|\bm{\xi}\right\|^{2}+\nu+\bm{\beta}\cdot\bm{\xi}\,i|}{(1-e^{-(\alpha^{2}\left\|\bm{\xi}\right\|^{2}+\nu)\,t_{0}})(1+\left\|\bm{\xi}\right\|^{4}\mu^{2})}\\\
&=\dfrac{(\alpha^{2}\,\left\|\bm{\xi}\right\|^{2}+\nu)\,t_{0}}{1-e^{-(\alpha^{2}\left\|\bm{\xi}\right\|^{2}+\nu)\,t_{0}}}\left(\dfrac{1}{(1+\left\|\bm{\xi}\right\|^{4}\mu^{2})\,t_{0}}+\dfrac{\left|\bm{\beta}\cdot\bm{\xi}\,i\right|}{(1+\left\|\bm{\xi}\right\|^{4}\mu^{2})(\alpha^{2}\,\left\|\bm{\xi}\right\|^{2}+\nu)\,t_{0}}\right)\\\
&<2\left(\dfrac{1}{t_{0}}+\dfrac{\sqrt{n}\left\|\bm{\beta}\right\|_{\infty}}{\mu^{2}\,\nu\,t_{0}}\right)<\dfrac{2}{\mu^{2}}\left(\dfrac{1}{t_{0}}+\dfrac{\sqrt{n}\left\|\bm{\beta}\right\|_{\infty}}{\nu\,t_{0}}\right).\end{split}$
(27)

Finally, from the inequality (22), using the triangular inequality, the lemmas
2, 3, 4 and 5 hold that,

$\begin{split}\phantom{space}\left|R_{\mu}^{3}(\bm{\xi})\right|&=\left|\dfrac{\Lambda(\bm{\xi})}{e^{(\left\|\bm{\xi}\right\|^{2}\mu^{2})/4}}\right|\leq\dfrac{|\alpha^{2}\,\left\|\bm{\xi}\right\|^{2}+\nu+\bm{\beta}\cdot\bm{\xi}\,i|\,(\alpha^{2}\,\left\|\bm{\xi}\right\|^{2}+\nu)\,t_{0}}{(1-e^{-(\alpha^{2}\left\|\bm{\xi}\right\|^{2}+\nu)\,t_{0}})\,(e^{(\left\|\bm{\xi}\right\|^{2}\mu^{2})/4})\,(\alpha^{2}\,\left\|\bm{\xi}\right\|^{2}+\nu)\,t_{0}}\\\
&<2\left(\dfrac{|\alpha^{2}\,\left\|\bm{\xi}\right\|^{2}+\nu+\bm{\beta}\cdot\bm{\xi}\,i|}{(e^{(\left\|\bm{\xi}\right\|^{2}\mu^{2})/4}\,(\alpha^{2}\,\left\|\bm{\xi}\right\|^{2}+\nu)\,t_{0}}\right)\dfrac{\left\|\bm{\xi}\right\|^{2}}{1-e^{-\left\|\bm{\xi}\right\|^{2}}}\dfrac{1-e^{-\left\|\bm{\xi}\right\|^{2}}}{\left\|\bm{\xi}\right\|^{2}}\\\
&<\dfrac{8}{\mu^{2}}\left(\dfrac{1}{t_{0}}+\frac{\left\|\bm{\beta}\right\|\left\|\xi\right\|}{(\alpha^{2}\left\|\bm{\xi}\right\|^{2}+\nu)\,t_{0}}\right)\dfrac{1-e^{-\left\|\bm{\xi}\right\|^{2}}}{\left\|\bm{\xi}\right\|^{2}}<\dfrac{8}{\mu^{2}}\left[\dfrac{1}{t_{0}}+\frac{\sqrt{n}\left\|\bm{\beta}\right\|_{\infty}}{2\,t_{0}\sqrt{\alpha^{2}\,\nu}}\right].\end{split}$
(28)

To finish the proof of the lemma 8, all that remains is to combine the
expressions (23)-(28). ∎

######  Lemma 9.

Let $\bm{\xi}\in\mathbb{R}^{n}$ and $0<\mu<1$ then the following inequality
holds

$\sup_{\left\|\bm{\xi}\right\|\in\mathbb{R}}\left|(1+\left\|\bm{\xi}\right\|^{2})^{-p/2}\left(1-\dfrac{R_{\mu}^{i}(\bm{\xi})}{\Lambda(\bm{\xi})}\right)\right|\leq\max\left\\{\mu^{p},\mu^{2}\right\\},\quad
i=1,2,3,$ (29)

where $\Lambda(\bm{\xi})$ and $R_{\mu}^{i}(\bm{\xi})$ are given by (11) and
(15) respectively.

###### Proof.

Let

$\Omega_{i}(\bm{\xi}):=(1+\left\|\bm{\xi}\right\|^{2})^{-p/2}\left(1-\dfrac{R_{\mu}^{i}(\bm{\xi})}{\Lambda(\bm{\xi})}\right),\quad
i=1,2,3.$

Three cases are considered for the proof.

Case 1
$(\left\|\bm{\xi}\right\|\geq\left\|\bm{\xi}_{0}\right\|:=\frac{1}{\mu})$. So

$\Omega_{i}(\bm{\xi})\leq(1+\left\|\bm{\xi}\right\|^{2})^{-p/2}\leq\left\|\bm{\xi}\right\|^{-p}\leq\left\|\bm{\xi}_{0}\right\|^{-p}=\mu^{p},\quad
i=1,2,3.$ (30)

Case 2 $(\left\|\bm{\xi}\right\|<1)$. So

$\Omega_{1}(\bm{\xi})=\dfrac{\left\|\bm{\xi}\right\|^{2}\mu^{2}}{1+\left\|\bm{\xi}\right\|^{2}\mu^{2}}(1+\left\|\bm{\xi}\right\|^{2})^{-p/2}\leq\left\|\bm{\xi}\right\|^{2}\mu^{2}(1+\left\|\bm{\xi}\right\|^{2})^{-p/2}\leq\mu^{2},$
(31)
$\Omega_{2}(\bm{\xi})=\frac{\left\|\bm{\xi}\right\|^{4}\mu^{2}}{1+\left\|\bm{\xi}\right\|^{4}\mu^{2}}(1+\left\|\bm{\xi}\right\|^{2})^{-p/2}\leq\left\|\bm{\xi}\right\|^{4}\mu^{2}(1+\left\|\bm{\xi}\right\|^{2})^{-p/2}\leq\mu^{2},$
(32)
$\Omega_{3}(\bm{\xi})=\left(1-e^{\frac{\left\|\bm{\xi}\right\|^{2}\mu^{2}}{4}}\right)(1+\left\|\bm{\xi}\right\|^{2})^{-p/2}\leq
1-e^{\frac{\left\|\bm{\xi}\right\|^{2}\mu^{2}}{4}}\leq\frac{\left\|\bm{\xi}\right\|^{2}\mu^{2}}{4}\leq\mu^{2}.$
(33)

Case 3
$(1\leq\left\|\bm{\xi}\right\|<\left\|\bm{\xi}_{0}\right\|:=\frac{1}{\mu})$.
So

$\Omega_{1}(\bm{\xi})=\dfrac{\left\|\bm{\xi}\right\|^{2}\mu^{2}}{1+\left\|\bm{\xi}\right\|^{2}\mu^{2}}(1+\left\|\bm{\xi}\right\|^{2})^{-p/2}\leq\dfrac{\left\|\bm{\xi}\right\|^{2-p}\,\mu^{2}}{1+\left\|\bm{\xi}\right\|^{2}\mu^{2}}\leq\left\|\bm{\xi}\right\|^{2-p}\,\mu^{2},$
(34)
$\Omega_{2}(\bm{\xi})=\dfrac{\left\|\bm{\xi}\right\|^{4}\mu^{2}}{1+\left\|\bm{\xi}\right\|^{4}\mu^{2}}(1+\left\|\bm{\xi}\right\|^{2})^{-p/2}\leq\dfrac{\left\|\bm{\xi}\right\|^{2-p}\,\mu^{2}\,\left\|\bm{\xi}\right\|^{2}}{1+\left\|\bm{\xi}\right\|^{4}\mu^{2}}\leq\dfrac{\left\|\bm{\xi}\right\|^{2-p}}{\left\|\bm{\xi}\right\|^{4}\,\mu^{2}}\leq\left\|\bm{\xi}\right\|^{2-p}\,\mu^{2},$
(35)
$\Omega_{3}(\bm{\xi})=\left(1-e^{\frac{\left\|\bm{\xi}\right\|^{2}\mu^{2}}{4}}\right)(1+\left\|\bm{\xi}\right\|^{2})^{-p/2}\leq\left\|\bm{\xi}\right\|^{-p}\dfrac{\left\|\bm{\xi}\right\|^{2}\mu^{2}}{4}\leq\left\|\bm{\xi}\right\|^{2-p}\,\mu^{2}.$
(36)

If $0<p\leq 2,$ you have

$\Omega_{i}(\bm{\xi})\leq\left\|\bm{\xi}\right\|^{2-p}\,\mu^{2}\leq\left\|\bm{\xi}_{0}\right\|^{2-p}\,\mu^{2}=\mu^{p},\quad
i=1,2,3.$ (37)

If $p>2,$ follow that

$\Omega_{i}(\bm{\xi})\leq\left\|\bm{\xi}\right\|^{2-p}\,\mu^{2}\leq\mu^{2},\quad
i=1,2,3.$ (38)

The expressions given by (30)-(38) are combined to obtain (29). This concludes
the proof. ∎

######  Remark 1.

Notice that

$\lim_{\mu\to 0^{+}}\dfrac{R_{\mu}^{i}(\bm{\xi})}{\Lambda(\bm{\xi})}=1,\qquad
i=1,2,3.$

####  3.2.2 Analytical bound of error

Now it is possible to obtain a bound for the error estimate.

######  Definition 3.

The norm in Sobolev space $H^{p}(\mathbb{R}^{n})$ for $p>0$ is defined as

$\|f\|_{H^{p}(\mathbb{R}^{n})}:=\left(\,\int\limits_{\mathbb{R}^{n}}{|\widehat{f}|^{2}\left(1+\left\|\bm{\xi}\right\|^{2}\right)^{p}\,d\bm{\xi}}\right)^{1/2}.$
(39)

By means of the definition (39), a parameter choice rule is given for the
proposed regularizations and an analytical expression is found for the error
bound of each estimation. This can be seen in the following result.

######  Theorem 4 (Analytical bound for the estimation error).

Consider the inverse problem of determining the source $f(\mathbf{x})$ in (5).
Let $f_{\delta,\mu}^{i}(\mathbf{x})$ with $i=1,2,3$ be the regularization
solutions given in (18). It is assumed that there exists $C\in\mathbb{R}^{+}$
bounding the norm of $f$ in $H^{p}(\mathbb{R}^{n})$ for some $0<p<\infty$
(39). if chosen

$\mu^{2}=\left(\dfrac{\delta}{\delta_{M}}\right)^{2/p+2}<1,$ (40)

then there exist constants $K_{i}\,(i=1,2,3)$ independent of $\delta$ such
that

$\|f-f_{\delta,\mu}^{i}\|_{L^{2}(\mathbb{R}^{n})}\leq
K_{i}\,\max\left\\{\left(\dfrac{\delta}{\delta_{M}}\right)^{2/p+2},\left(\dfrac{\delta}{\delta_{M}}\right)^{p/p+2}\right\\},\qquad
i=1,2,3.$

###### Proof.

$\left\|\widehat{f}(\bm{\xi})-R_{\mu}^{i}(\bm{\xi})\,\widehat{y}(\bm{\xi})\right\|_{L^{2}(\mathbb{R}^{n})}=\left\|\widehat{f}(\bm{\xi})\left(1-\frac{R_{\mu}^{i}(\bm{\xi})}{\Lambda(\bm{\xi})}\right)\frac{(1+\left\|\bm{\xi}\right\|^{2})^{p/2}}{(1+\left\|\bm{\xi}\right\|^{2})^{p/2}}\right\|_{L^{2}(\mathbb{R}^{n})},$

rearranging

$\left\|\widehat{f}(\bm{\xi})-R_{\mu}^{i}(\bm{\xi})\,\widehat{y}(\bm{\xi})\right\|_{L^{2}(\mathbb{R}^{n})}\leq\sup_{\left\|\bm{\xi}\right\|\in\mathbb{R}}\left|(1+\left\|\bm{\xi}\right\|^{2})^{-p/2}\left(1-\frac{R_{\mu}^{i}(\bm{\xi})}{\Lambda(\bm{\xi})}\right)\right|\left\|\widehat{f}(\bm{\xi})(1+\left\|\bm{\xi}\right\|^{2})^{p/2}\right\|_{L^{2}(\mathbb{R}^{n})}.$

We use the definition of the norm in the Sobolev space
$(H^{p}(\mathbb{R}^{n}))$ given by the expression (39),

$\left\|\widehat{f}(\bm{\xi})-R_{\mu}^{i}(\bm{\xi})\,\widehat{y}(\bm{\xi})\right\|_{L^{2}(\mathbb{R}^{n})}\leq\sup_{\left\|\bm{\xi}\right\|\in\mathbb{R}}\left|(1+\left\|\bm{\xi}\right\|^{2})^{-p/2}\left(1-\frac{R_{\mu}^{i}(\bm{\xi})}{\Lambda(\bm{\xi})}\right)\right|\left\|f(\bm{\xi})\right\|_{H^{p}(\mathbb{R}^{n})}.$
(41)

By the triangle inequality we have,

$\left\|\widehat{f}-\widehat{f_{\delta,\mu}^{i}}\right\|_{L^{2}(\mathbb{R}^{n})}\leq\left\|\widehat{f}-R_{\mu}^{i}(\bm{\xi})\,\widehat{y}(\bm{\xi})\right\|_{L^{2}(\mathbb{R}^{n})}+\left\|R_{\mu}^{i}(\bm{\xi})\,\widehat{y}(\bm{\xi})-\widehat{f_{\delta,\mu}^{i}}\right\|_{L^{2}(\mathbb{R}^{n})}.$
(42)

From the inequalities (41)-(42) together with the definition of the
regularized source (17) in frequency space, we obtain,

$\left\|\widehat{f}-\widehat{f_{\delta,\mu}^{i}}\right\|_{L^{2}(\mathbb{R}^{n})}\leq\sup_{\left\|\bm{\xi}\right\|\in\mathbb{R}}\left|(1+\left\|\bm{\xi}\right\|^{2})^{-p/2}\left(1-\frac{R_{\mu}^{i}(\bm{\xi})}{\Lambda(\bm{\xi})}\right)\right|\left\|f\right\|_{H^{p}(\mathbb{R}^{n})}+\sup_{\left\|\bm{\xi}\right\|\in\mathbb{R}}\left|R_{\mu}^{i}(\bm{\xi})\right|\left\|\widehat{y}-\widehat{y}_{\delta}\right\|_{L^{2}(\mathbb{R}^{n})},$
(43)

Parseval’s identity [61] is used in (43), the fact that $C$ bounds $f$ in norm
$H^{p}(\mathbb{R}^{n})$ and the assumption that the error in the data is
bounded
$(\left\|y-y_{\delta}\right\|_{L^{2}(\mathbb{R}^{n})}=\left\|\widehat{y}-\widehat{y}_{\delta}\right\|_{L^{2}(\mathbb{R}^{n})}\leq\delta)$,
then

$\left\|f-f_{\delta,\mu}^{i}\right\|_{L^{2}(\mathbb{R}^{n})}=\left\|\widehat{f}-\widehat{f_{\delta,\mu}^{i}}\right\|_{L^{2}(\mathbb{R}^{n})}\leq
C\,\sup_{\left\|\bm{\xi}\right\|\in\mathbb{R}}\left|(1+\left\|\bm{\xi}\right\|^{2})^{-p/2}\left(1-\frac{R_{\mu}^{i}(\bm{\xi})}{\Lambda(\bm{\xi})}\right)\right|+\delta\,\sup_{\left\|\bm{\xi}\right\|\in\mathbb{R}}\left|R_{\mu}^{i}(\bm{\xi})\right|.$
(44)

By virtue of the lemmas 8, 9, the expression (44) can be rewritten as,

$\left\|f-f_{\delta,\mu}^{i}\right\|_{L^{2}(\mathbb{R}^{n})}\leq
C\,\max\left\\{\mu^{2},\mu^{p}\right\\}+\frac{\delta}{\mu^{2}}\,M_{i},$ (45)

where $M_{i}$ with $i=1,2,3$ is given by the equations (19), (20) and (21). It
is used in the equation (45)
$\mu^{2}=\left(\dfrac{\delta}{\delta_{M}}\right)^{2/p+2}$ (40) and it is
obtained,

$\left\|f-f_{\delta,\mu}^{i}\right\|_{L^{2}(\mathbb{R}^{n})}\leq
C\,\max\left\\{\left(\dfrac{\delta}{\delta_{M}}\right)^{2/p+2},\left(\dfrac{\delta}{\delta_{M}}\right)^{p/p+2}\right\\}+\delta_{M}\,M_{i}\,\left(\dfrac{\delta}{\delta_{M}}\right)^{p/p+2},$

equivalently

$\left\|f-f_{\delta,\mu}^{i}\right\|_{L^{2}(\mathbb{R}^{n})}\leq
K_{i}\,\max\left\\{\left(\dfrac{\delta}{\delta_{M}}\right)^{2/p+2},\left(\dfrac{\delta}{\delta_{M}}\right)^{p/p+2}\right\\},$
(46)

where $K_{i}=C+\delta_{M}\,M_{i}$ with $i=1,2,3$. This concludes the proof. ∎

######  Remark 2.

Note that the bound (46) satisfies

$\|f-f_{\delta,\mu}^{i}\|_{L^{2}(\mathbb{R}^{n})}\longrightarrow
0\,\,\,\text{if}\,\,\,\delta\longrightarrow 0,$

which means that the estimate ($f_{\delta,\mu}^{i}$) converges to the source
function ($f$) when the noise in the data ($\delta$) tends to $0$.

######  Remark 3.

Note that the case $p=\infty$ is not included in the hypotheses of the 4
theorem, the reason is that in this case,

$\|f-f_{\delta,\mu}^{i}\|_{L^{2}(\mathbb{R}^{n})}=\left\|\widehat{f}-\widehat{f_{\delta,\mu}^{i}}\right\|_{L^{2}(\mathbb{R}^{n})}\leq
K_{i}<C\nrightarrow 0.$

######  Remark 4.

Note that the bound obtained for the regularization error (46) is of Hölder
type and only depends on the smoothness of the source and the parameters of
the mathematical model.

######  Note 3.

A case of special interest in the bibliography is for $p=2$. In this the
expression (46) is reduced to:

$\left\|f-f_{\delta,\mu}^{i}\right\|_{L^{2}(\mathbb{R}^{n})}\leq
K_{i}\,\sqrt{\dfrac{\delta}{\delta_{M}}}.$

###  3.3 Numerical examples

We consider concrete examples of the estimation of the source $f$ in
$\mathbb{R}^{n}$ for $n=1,2,3$. For each of the examples discussed in this
section, different values are chosen for the parameters of the source
identification problem $(\alpha^{2},\bm{\beta},\nu$, $t_{0})$. In addition, to
simulate the noise in the data, a set of values of standard deviation
$\epsilon$. The space is discretized into a uniform $n$-dimensional mesh and a
data set $\\{y_{\delta_{1}},...,y_{\delta_{N}}\\}$ is obtained from the
evaluation of the solution $u(\mathbf{x},t)$ at a fixed time instant given
$t_{0}$ and adding noise, that is,

$y_{\delta_{i}}=y(\mathbf{x}_{i})+\eta_{i},\quad
i=1,...,N,\,\,\mathbf{x}_{i}\in{\cal G},$

where ${\cal G}$ is a uniform discretization on $\mathbb{R}^{n}$ chosen and
$\eta_{i},i=1,...,N$ are realizations of the normally distributed random
variable ${\eta}$ with mean 0 and deviation $\epsilon$. By denoting
$y_{i}=y(\mathbf{x}_{i}),i=1,..,N$ and taking into account the noise level
$\delta$ satisfying (4), the error

$y-y_{\delta}=(y_{1}-y_{\delta_{1}},...,y_{N}-y_{\delta_{N}})=(\eta_{1},...,\eta_{N})$

It is numerically calculated using the Simpson integration method. This
calculation allows obtaining an approximate value of the error $y-y_{\delta}$
that directly depends of the noise. It can be seen that the noise level
$\delta$ is a function of the standard deviation $\epsilon$, that is,
$\delta=\delta(\epsilon)$.

In practice, the maximum tolerance value for the error in the data
$\delta_{M}$ given in (4) is obtained from the calibration, instrumentation
and measurement errors of the instruments used to perform the measurements.
Here, to make sure that the regularization parameter is less than 1, it is
chosen as the maximum value of $\delta$ plus one unit. That is to say,

$\delta_{M}=1+\max\\{\delta_{1},...,\delta_{N}\\}.$

Then $\\{\widehat{y}_{\delta_{1}},...,\widehat{y}_{\delta_{N}}\\}$ is
calculated using the FFT transform (Fast Fourier Transform) [62] and get the
regularization solutions $f^{i}_{\delta,\mu}$ with $i=1,2,3$ given in (18)
using the anti-fast Fourier transform [7, 62], where the regularization
parameter $\mu$ is chosen according to (40), that is to say,

$\mu^{2}=\displaystyle\left(\frac{\delta}{\delta_{M}}\right)^{2/p+2}.$

For each of the examples considered, the results of the estimated sources
without regularization and those obtained after using the regularization
methods presented in this article are graphed. An error table is also included
for each example. These tables consider the relative errors made, in the
estimation, by each one of the regularization operators used to address the
ill-posed problem. The disturbances used for the construction of the tables
are
$\\{\epsilon_{1};...;\epsilon_{5}\\}=\\{10^{-1},10^{-2},10^{-3};10^{-4};10^{-5}\\}$,
these were chosen based on the values of the solution.

######  Note 4.

The reader may notice that in some examples the $p$ considered for the
recovery does not correspond to the space $H^{p}(\mathbb{R}^{n})$ to which the
source belongs, since $f$ is not in that Hilbert space. These examples were
included to show that recovery is reasonable even in those cases.

####  3.3.1 Examples 1D

In this subsection, three examples of estimation of one-dimensional sources
with different characteristics are discussed.

######  Example 1.

For this example, the following parameters are considered $\alpha^{2}=2\times
10^{-5}$; $\bm{\beta}=1\times 10^{-5}$; $\nu=$1; $N=1001$ and $t_{0}=5$. Also,
the perturbation values used are $\epsilon\in\\{0.2;0.15;0.1;$0.05}. Finally,
the source to estimate in this case is:

$f(x)=\begin{cases}-1,\qquad&-10\leq x<-5,\\\ 1,\qquad&-5\leq x<0,\\\
-1,\qquad&0\leq x<5,\\\ 1,\qquad&5\leq x\leq 10,\\\ 0,\qquad&\text{in another
case}.\end{cases}$ (47)

The source (47) retrieved in example 1 is of interest in signal theory. This
type of function turns out to be a common example in [48, 50] source retrieval
problems. This is mainly due to the fact that as it is a discontinuous
function, the Gibbs [7, 62] phenomenon can appear. This phenomenon indicates
that when a Fourier series function is developed and it is not continuous in
the considered region, it is possible that there is not a good precision in
the neighborhoods of the discontinuities.

As can be seen in the graphs of figure 1, the regularization operators used
smooth the inverse operator making the recovery stable. As expected, the
estimation improves for smaller values of perturbation $\epsilon$. In
addition, no significant differences between the different operators are
noticeable at first glance, although the operator $R^{3}_{\mu}$ seems to have
a better performance in the neighborhoods of discontinuity points.

Figure 1: Example 1: Non-regularized font with $t_{0}=5$ (top-left);
Regularized sources with $t_{0}=5$ and $p=1$, using $R^{1}_{\mu}$ (top-right),
$R^{2}_{\mu}$ (bottom-left), $R^{3}_{\mu}$ (bottom-right) for different noise
levels.

The table 1 includes the relative estimation errors that also show the
goodness of the methods used. As noted, for this example, the operator
$R^{3}_{\mu}$ performs better. For example, for the case $\epsilon=10^{-3}$,
the relative error of the estimate of the regularized source with
$R^{1}_{\mu}$ is $2.74\%$, with $R^{2}_{\mu}$ is $8.08\%$ and with
$R^{3}_{\mu}$ it is $1.15\%$.

Relative errors
---
| $\epsilon$ | $\|f-f^{1}_{\delta,\mu}\|/\left\|f\right\|$ | $\|f-f^{2}_{\delta,\mu}\|/\left\|f\right\|$ | $\|f-f^{3}_{\delta,\mu}\|/\left\|f\right\|$
---|---|---|---
$10^{-1}$ | 0.1310 | 0.1472 | 0.1137
$10^{-2}$ | 0.0888 | 0.1217 | 0.0706
$10^{-3}$ | 0.0549 | 0.0996 | 0.0365
$10^{-4}$ | 0.0274 | 0.0808 | 0.0115
$10^{-5}$ | 0.0090 | 0.0635 | 0.0026
Table 1: Example 1: Relative estimation errors for $t_{0}=1$ and $p=1.$

######  Example 2.

For this example, the following parameters are considered $\alpha^{2}=1$;
$\bm{\beta}=0$; $\nu=$0; $N=1001$ and $t_{0}=0.1$. Furthermore, the
perturbation values used are $\epsilon\in\\{0.01;0.007;0.003;$0.001}. Finally,
the source to estimate in this case is:

$f(x)=\begin{cases}\left(-\frac{x^{3}}{4}+\frac{3x}{2}\right)e^{\frac{-x^{2}}{4}},\qquad&-10\leq
x\leq 10,\\\ 0,\qquad&\text{in another case}.\end{cases}$ (48)

The source (48) recovered in example 2 is of interest in heat transfer
problems. See for example, [53]. This type of function turns out to be a
common example in source recovery problems in heat transfer processes [48,
50]. The example 2 is interesting, since this type of sources have already
been recovered by means of different approaches in simpler problems where it
is used, instead of the full parabolic equation, the heat equation where only
diffusion is taken into account [32, 34, 35].

Table 2 includes the relative estimation errors that also show the goodness of
the methods used. As noted, for this example, the operator $R^{2}_{\mu}$
performs better. For example, for the case $\epsilon=10^{-3}$, the relative
error of the estimate of the regularized source with $R^{1}_{\mu}$ is
$5.89\%$, with $R^{2}_{\mu}$ is $4.82\%$ and with $R^{3}_{\mu}$ it is
$6.48\%$.

As it can be seen in the graphs of figure 2, the regularization operators
used, again, smooth the inverse operator making the recovery stable. In this
case, the stabilization effect is much more noticeable to the naked eye
(because smaller values of disturbances were considered); due to this and for
comparative purposes, a smaller scale was taken for the sources retrieved with
the regularization operators. Once again it can be seen that the estimation
improves for smaller values of disturbance $\epsilon$. Also, no significant
differences between the different operators are noticeable to the naked eye,
although the operator $R^{2}_{\mu}$ seems to have, in this example, a better
overall performance.

Relative errors
---
| $\epsilon$ | $\|f-f^{1}_{\delta,\mu}\|/\left\|f\right\|$ | $\|f-f^{2}_{\delta,\mu}\|/\left\|f\right\|$ | $\|f-f^{3}_{\delta,\mu}\|/\left\|f\right\|$
---|---|---|---
$10^{-1}$ | 0.8387 | 0.6170 | 0.9686
$10^{-2}$ | 0.1746 | 0.1074 | 0.1822
$10^{-3}$ | 0.0589 | 0.0482 | 0.0648
$10^{-4}$ | 0.0474 | 0.0471 | 0.0489
$10^{-5}$ | 0.0365 | 0.0321 | 0.0389
Table 2: Example 2: Relative estimation errors for $t_{0}=1$ and $p=1.$

Figure 2: Ejemplo 2: Non-regularized font with $t_{0}=0.1$ (top-left);
Regularized sources with $t_{0}=0.1$ and $p=2$, using $R^{1}_{\mu}$ (top-
right), $R^{2}_{\mu}$ (bottom-left), $R^{3}_{\mu}$ (bottom-right) for
different noise levels.

######  Example 3.

For this example, the following parameters are considered $\alpha^{2}=2$;
$\bm{\beta}=0$; $\nu=1$; $N=1001$ and $t_{0}=0.2$. Furthermore, the
perturbation values used are $\epsilon\in\\{0.004;0.003;0.002;$0.0001}.
Finally, the source to estimate in this case is:

$f(x)=\begin{cases}x+1,\qquad&-1\leq x<0,\\\ -x+1,\qquad&0\leq x\leq 1,\\\
0,\qquad&\text{in another case}.\end{cases}$ (49)

The source (49) that is recovered in the example 3 is of interest in various
problems with different characteristics due to the particularities that this
function presents. Although it is a continuous function, it has a point where
it is not differentiable. Recovery is often difficult at these points. Because
of this, this type of function also turns out to be a common example in source
recovery problems, both in signal theory and in heat transfer problems [48,
50].

Table 3 includes the relative estimation errors that again show the benefits
of using the methods presented in this article. As noted, for this example,
the $R^{2}_{\mu}$ operator performs better. For example, for the case
$\epsilon=10^{-3}$, the relative error of the estimate of the regularized
source with $R^{1}_{\mu}$ is $5.35\%$, with $R^{2}_{\mu}$ is $2.28\%$ and with
$R^{3}_{\mu}$ it is $3.73\%$.

Relative errors
---
| $\epsilon$ | $\|f-f^{1}_{\delta,\mu}\|/\left\|f\right\|$ | $\|f-f^{2}_{\delta,\mu}\|/\left\|f\right\|$ | $\|f-f^{3}_{\delta,\mu}\|/\left\|f\right\|$
---|---|---|---
$10^{-1}$ | 0.9157 | 0.3027 | 0.6716
$10^{-2}$ | 0.1607 | 0.0353 | 0.1772
$10^{-3}$ | 0.0535 | 0.0228 | 0.0373
$10^{-4}$ | 0.0478 | 0.0221 | 0.0309
$10^{-5}$ | 0.0239 | 0.0187 | 0.0277
Table 3: Example 3: Relative estimation errors for $t_{0}=1$ and $p=1.$

As can be seen in the graphs of figure3, the regularization operators used,
again, smooth the inverse operator making the recovery stable. In this case
the stabilization effect, too, is more noticeable to the naked eye than in the
example 1. This is because small values of perturbations were considered in
order to be able to compare the regularization solutions with the non-
regularized ones (which, again, presented many fluctuations). Due to this
reason and for comparative purposes, a smaller scale was taken for the sources
retrieved with the regularization operators. Once again it can be seen that
the estimation improves for smaller values of disturbance $\epsilon$.
Furthermore, no significant differences between the different operators are
noticeable to the naked eye, although the operator $R^{2}_{\mu}$ clearly
provides a better estimate.

Figure 3: Ejemplo 3: Non-regularized font with $t_{0}=0.2$ (top-left);
Regularized sources with $t_{0}=0.2$ and $p=4$, using $R^{1}_{\mu}$ (top-
right), $R^{2}_{\mu}$ (bottom-left), $R^{3}_{\mu}$ (bottom-right) for
different noise levels.

####  3.3.2 Examples 2D

Here are two examples of two-dimensional estimation with different
characteristics. In order to visualize more clearly the improvements obtained
with the regularization operators, not only the recovered sources will be
graphed, but also contour lines will be made. These schemes will allow direct
comparisons to be made on the sources obtained with the different
regularization strategies.

######  Example 4.

For this example, the following parameters are considered $\alpha^{2}=0.2$;
$\bm{\beta}=(0,0)$; $\nu=0.99$; $N=1001\times 1001$; $t_{0}=1$ and
$\epsilon=0.025$. Finally, the source to estimate in this case is:

$f(x,y)=\begin{cases}\cos\left(\dfrac{x}{20}\right)\,\cos\left(\dfrac{y}{20}\right),\qquad&-40\leq
x,y\leq 40,\\\ 0,\qquad&\text{in another case}.\end{cases}$ (50)

The source (50) that is recovered in the example 4 is of interest in various
problems with different characteristics due to the particularities that this
function presents. Its graph is a surface, smooth, continuous and
differentiable.

As can be seen in the graphs of figure 4, the regularization operators used,
again, smooth the inverse operator making the recovery stable. No significant
differences between the different operators are noticeable with the naked eye,
although if the level curves are visualized and compared, the operator
$R^{2}_{\mu}$ provides a better estimate.

Table 4 includes the relative estimation errors that once again show the good
performance obtained with the methods introduced in this article. As noted,
for this example, the operator $R^{2}_{\mu}$ performs better overall. For
example, for the case $\epsilon=10^{-2}$, the relative error of the estimate
of the source regularized with $R^{1}_{\mu}$ is $1.18\%$, with $R^{2}_{\mu}$
is $0.67\%$ and with $R^{3}_{\mu}$ it is $1.41\%$.

Relative errors
---
| $\epsilon$ | $\|f-f^{1}_{\delta,\mu}\|/\left\|f\right\|$ | $\|f-f^{2}_{\delta,\mu}\|/\left\|f\right\|$ | $\|f-f^{3}_{\delta,\mu}\|/\left\|f\right\|$
---|---|---|---
$10^{-1}$ | 0.0846 | 0.0583 | 0.0989
$10^{-2}$ | 0.0118 | 0.0067 | 0.0141
$10^{-3}$ | 0.0029 | 0.0011 | 0.0045
$10^{-4}$ | 0.0008 | 0.0004 | 0.0013
$10^{-5}$ | 0.0004 | 0.0003 | 0.0004
Table 4: Example 4: Relative estimation errors for $t_{0}=1$ and $p=1.$

######  Example 5.

For this example, the following parameters are considered $\alpha^{2}=1$;
$\bm{\beta}=(0,0)$; $\nu=1$; $N=1001\times 1001$; $t_{0}=0.4$ and
$\epsilon=0.05$. Finally, the source to estimate in this case is:

$f(x,y)=\begin{cases}10+x-y,\qquad&-10\leq x\leq 0,\qquad 0\leq y\leq 10+x,\\\
10+x+y,\qquad&-10\leq x\leq 0,\qquad-10-x\leq y\leq 0,\\\ 10-x-y,\qquad&0\leq
x\leq 10,\qquad 0\leq y\leq 10-x,\\\ 10-x+y,\qquad&0\leq x\leq
10,\qquad-10+x\leq y\leq 0,\\\ 0,\qquad&\text{in another case}.\end{cases}$
(51)

Figure 4: Example 4: Fonts and contour lines. Not regularized with $t_{0}=1$
(First row); regularized with $t_{0}=1$ and $p=1$, using $R^{1}_{\mu}$ (second
row), $R^{2}_{\mu}$ (third row), $R^{3}_{\mu}$ (fourth row) for
$\epsilon=0.025$.

Figure 5: Example 5: Fonts and contour lines. Not regularized with $t_{0}=0.4$
(First row); regularized with $t_{0}=0.4$ and $p=0.6$, using $R^{1}_{\mu}$
(second row), $R^{2}_{\mu}$ (third row), $R^{3}_{\mu}$ (fourth row) for
$\epsilon=0.05$.

The source (51) retrieved in example 5 is of interest because it is a non-
differentiable surface. This type of surfaces can be approximated badly in
environments close to the points of non-differentiability.

As can be seen in the graphs of figure 5, once again, the regularization
operators used, smooth the inverse operator making the recovery stable. No
significant differences between the different operators are noticeable at
first glance, although if the level curves are visualized and compared, once
again, the operator $R^{2}_{\mu}$ provides a better estimate, more effectively
stabilizing the solution of the ill-posed problem.

Table 5 includes the relative estimation errors that once again show the good
performance obtained with the methods introduced in this paper. As observed,
for this example, the operator $R^{2}_{\mu}$ presents a better general
performance, since it achieves a better stabilization of the solution. For
example, for the case $\epsilon=10^{-2}$, the relative error of the estimate
of the regularized source with $R^{1}_{\mu}$ is $1.21\%$, with $R^{2}_{\mu}$
is $0.38\%$ and with $R^{3}_{\mu}$ it is $1.65\%$.

Relative errors
---
| $\epsilon$ | $\|f-f^{1}_{\delta,\mu}\|/\left\|f\right\|$ | $\|f-f^{2}_{\delta,\mu}\|/\left\|f\right\|$ | $\|f-f^{3}_{\delta,\mu}\|/\left\|f\right\|$
---|---|---|---
$10^{-1}$ | 0.0442 | 0.0193 | 0.0491
$10^{-2}$ | 0.0121 | 0.0038 | 0.0165
$10^{-3}$ | 0.0056 | 0.0025 | 0.0096
$10^{-4}$ | 0.0043 | 0.0013 | 0.0079
$10^{-5}$ | 0.0027 | 0.0009 | 0.0054
Table 5: Example 5: Relative estimation errors for $t_{0}=1$ and $p=1.$

####  3.3.3 Examples 3D

An example of three-dimensional estimation is now addressed. Since in this
case the source is a scalar function with domain in $R^{3}$, 4 dimensions are
required to be able to plot it, because of this, only the cuts with the
coordinate axes will be plotted. That is, the graphs of the retrieved sources
will be analyzed when one of the variables is null. In addition, to visualize
more clearly the improvements obtained with the different regularization
operators introduced and studied in this article, contour diagrams will be
made for each of the cuts considered and for each regularization.

######  Example 6.

For this example, the following parameters are considered $\alpha^{2}=0.4$;
$\bm{\beta}=(1,-0.5,-0.5)$; $\nu=0.997$; $N=129\times 129\times 129$;
$t_{0}=3$ and $\epsilon=0.035$. Finally, the source to estimate in this case
is:

$f(x,y,z)=\begin{cases}\sin\left(\dfrac{x+y+z}{20}\right),\qquad&-2\pi\leq
x,y,z\leq 2\pi,\\\ 0,\qquad&\text{in another case}.\end{cases}$ (52)

The source (52) that is recovered in the example 6 is of interest in various
problems with different characteristics due to the particularities that this
function presents. It is a function, smooth, continuous and differentiable;
moreover, it is symmetric with respect to the three coordinates.

Relative errors
---
| $\epsilon$ | $\|f-f^{1}_{\delta,\mu}\|/\left\|f\right\|$ | $\|f-f^{2}_{\delta,\mu}\|/\left\|f\right\|$ | $\|f-f^{3}_{\delta,\mu}\|/\left\|f\right\|$
---|---|---|---
$10^{-1}$ | 0.0740 | 0.0365 | 0.0904
$10^{-2}$ | 0.0115 | 0.0037 | 0.0127
$10^{-3}$ | 0.0089 | 0.0007 | 0.0093
$10^{-4}$ | 0.0017 | 0.0006 | 0.0089
$10^{-5}$ | 0.0014 | 0.0005 | 0.0053
Table 6: Example 6: Relative estimation errors for $t_{0}=1$ and $p=1.$

Figure 6: Example 6: Sources and contour lines with $(x=0)$. Not regularized
with $t_{0}=3$ (First row); regularized with $t_{0}=3$ and $p=3$, using
$R^{1}_{\mu}$ (second row), $R^{2}_{\mu}$ (third row), $R^{3}_{\mu}$ (fourth
row) for $\epsilon=0.035$.

Figure 7: Example 6: Sources and contour lines with $(y=0)$. Not regularized
with $t_{0}=3$ (First row); regularized with $t_{0}=3$ and $p=3$, using
$R^{1}_{\mu}$ (second row), $R^{2}_{\mu}$ (third row), $R^{3}_{\mu}$ (fourth
row) for $\epsilon=0.035$.

Figure 8: Example 6: Sources and contour lines with $(z=0)$. Not regularized
with $t_{0}=3$ (First row); regularized with $t_{0}=3$ and $p=3$, using
$R^{1}_{\mu}$ (second row), $R^{2}_{\mu}$ (third row), $R^{3}_{\mu}$ (fourth
row) for $\epsilon=0.035$.

Table 6 includes the relative estimation errors that once again show the good
performance obtained with the methods introduced in this article. As noted, in
this example, the $R^{2}_{\mu}$ operator performs better overall. For example,
for the case $\epsilon=10^{-2}$, the relative error of the estimate of the
regularized source with $R^{1}_{\mu}$ is $1.15\%$, with $R^{2}_{\mu}$ is
$0.37\%$ and with $R^{3}_{\mu}$ it is $9.04\%$.

As can be seen in the graphs of the figures 6, 7, 8 the regularization
operators used, again, smooth the inverse operator making the recovery in each
cut with the coordinate axes stable. In principle, notable differences between
the recovery of each cut are not appreciated; since they are all estimated
with the same order of fluctuation. On the other hand, no significant
differences between the approximations with the different regularization
operators are noticeable at first glance. If the level curves are carefully
observed, for each cut with the coordinate axes, it is seen that, once again,
the operator $R^{2}_{\mu}$ yields better results.

###  3.4 Discussion of results

The numerical examples carried out show that the three regularization
operators, introduced in this article, are useful to solve the problem of lack
of stability in the source estimation solution. Any of the three operators
studied smooth out the inverse operator making the recovery stable. This fact
is independent of the general characteristics of the function to be
determined, since the results are good, even for discontinuous and non-
differentiable sources.

However, in general terms the operator $R^{2}_{\mu}$ yielded better results
and allowed estimating the source with the lowest relative error, for the
different disturbances considered. Particularly the operator $R^{3}_{\mu}$
offered a better performance when the source to be estimated is a
discontinuous function.

##  4 Conclusions

This paper deals with the inverse problem of identifying the source in a
complete parabolic equation from noisy measurements. An analytical solution to
the estimation problem is given and shown that the problem is ill-posed since
said solution is not stable. In order to address this drawback, three families
of regularization operators specifically designed to compensate for the
instability factor in the inverse operator are defined. Besides, a choice rule
is included for the regularization parameters that is based on the level of
data noise and the smoothness of the source to be identified. It is shown that
for the proposed parameter choice rule the methods are stable. A bound is
obtained for the estimation errors that turn out to be optimal since they are
of Hölder type.

Several numerical examples of recovering sources that belong to different
spaces of Hilbert are included. It is observed that in all of them a good
performance of the adopted regularization approaches is obtained. On the other
hand, the sources recovered by means of the inverse operator are compared with
those obtained using the regularization operators and it is concluded that the
proposed regularization methods offer greater precision in the estimation of
the sources considered.

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