\documentclass[reqno]{amsart} \usepackage{hyperref} \usepackage{graphicx} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2017 (2017), No. 135, pp. 1--17.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu} \thanks{\copyright 2017 Texas State University.} \vspace{8mm}} \begin{document} \title[\hfilneg EJDE-2017/135\hfil Data assimilation and null controllability] {Data assimilation and null controllability of degenerate/singular parabolic problems} \author[K. Atifi, E.-H. Essoufi \hfil EJDE-2017/135\hfilneg] {Khalid Atifi, El-Hassan Essoufi} \address{Khalid Atifi \newline Laboratoire de Math\'ematiques, Informatique et Sciences de l'ing\'enieur (MISI), Universit\'e Hassan 1, Settat 26000, Morocco} \email{k.atifi.uhp@gmail.com} \address{El-Hassan Essoufi \newline Laboratoire de Math\'ematiques, Informatique et Sciences de l'ing\'enieur (MISI), Universit\'e Hassan 1, Settat 26000, Morocco} \email{e.h.essoufi@gmail.com} \dedicatory{Communicated by Jerome A. Goldstein} \thanks{Submitted February 24, 2017. Published May 18, 2017.} \subjclass[2010]{15A29, 47A52, 93C20, 35K05, 35K65, 35K65, 93B05} \keywords{Data assimilation; null controllability; regularization; heat equation; \hfil\break\indent inverse problem; degenerate equations; optimization} \begin{abstract} In this article, we use the variational method in data assimilation to study numerically the null controllability of degenerate/singular parabolic problem \begin{gather*} \partial _{t}\psi - \partial_{x}(x^\alpha\partial _{x}\psi(x)) -\frac{\lambda }{x^{\beta }}\psi=f,\quad (x,t)\in ]0,1[\times]0,T[,\\ \psi(x,0)=\psi_0, \quad \psi\big|_{x=0}=\psi\big|_{x=1}=0. \end{gather*} To do this, we determine the source term $f$ with the aim of obtaining $\psi(\cdot ,T)=0$, for all $\psi_0 \in L^2(]0,1[)$. This problem can be formulated in a least-squares framework, which leads to a non-convex minimization problem that is solved using a regularization approach. Also we present some numerical experiments. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \allowdisplaybreaks \section{Introduction} In this article, we study an inverse problem of identifying the source term in degenerate/singular parabolic equation. This in the aim to study the null controllability, which has important applications in various areas of applied science and engineering. Controllability properties of degenerate/singular parabolic equations has been widely studied (see \cite{31,30,35,34,24}) using Carleman estimates. Our main contribution is to study numerically the null controllability of problem \eqref{AAA}, below, using the variational method in data assimilation. The problem can be stated as follows: Estimate the source term in the degenerate parabolic equation with singular potential \begin{equation}\label{AAA} \partial _{t}\psi - \partial_{x}(x^\alpha\partial _{x}\psi(x)) -\frac{\lambda }{x^{\beta }}\psi=f,\quad (x,t)\in\Omega\times]0,T[ \end{equation} where $\Omega=]0,1[$, $\alpha\in]0,1[$, $\beta\in ]0,2-\alpha[$, $\lambda\leq 0 $, and $f\in L^2( \Omega \times ] 0,T[)$. The mathematical model leads to a non-convex minimization problem \begin{equation}\label{BBB} \begin{gathered} \text{find $\hat{f}\in A_{ad}$ such that} \\ E(\hat{f})=\min_{f \in A_{ad}} E(f), \end{gathered} \end{equation} where the cost function $E$ is \begin{equation}\label{eqProblPositonprobJ} E(f)=\frac{1}{2}\| \psi (t=T)\| _{L^2(\Omega)}^2, \end{equation} subject to $\psi $ being the weak solution of the parabolic problem \eqref{AAA} with source term $f$. Problem \eqref{BBB} is ill-posed in the sense of Hadamard, some regularization technique is needed to guarantee numerical stability of the computational process, maybe with noisy input data. The problem thus consists in minimizing a functional of the form \begin{equation}\label{eqProblPositonprobJJJJJ} J(f)=\frac{1}{2}\| \psi (t=T)\| _{L^2(\Omega)}^2 + \frac{\varepsilon}{2}\| f\| _{ L^2( \Omega \times ] 0,T[)}^2. \end{equation} The last term in \eqref{eqProblPositonprobJJJJJ} stands for the so called Tikhonov-type regularization \cite{99,999}, $\varepsilon$ being a small regularizing coefficient that provides extra convexity to the functional $J$. First we prove that the functional $J$ is continuous, and G-derivable. Numerical experiments are presented later. \section{Problem statement and main result} Consider the problem \begin{equation}\label{eqProblPositonprob} \begin{gathered} \partial _{t}\psi +A(\psi)=f\\ \psi (0,t)=\psi (1,t)=0 \quad \forall t\in ]0,T[ \\ \psi (x,0)=\psi _0(x)\quad \forall x\in \Omega \end{gathered} \end{equation} where, $\Omega=]0,1[$, $f\in L^2( \Omega \times ] 0,T[)$, $\psi_0 \in L^2(\Omega)$, and $A$ is the operator defined as \begin{equation*} A(\psi)=-\partial_{x}(a(x) \partial _{x}\psi(x)) -\frac{\lambda }{x^{\beta }}\psi,\quad a(x)=x^\alpha \end{equation*} with $\alpha\in]0,1[$, $\beta\in ]0,2-\alpha[$, and $\lambda\leq 0$. The minimization problem with regularization associated to this problem is \begin{equation} \label{eqProblMin} \begin{gathered} \text{find $\hat{f}\in A_{ad}$ such that} \\ J(\hat{f})=\min_{f \in A_{ad}} J(f), \end{gathered} \end{equation} where the cost function $J$ is defined as \begin{equation}\label{eqProblPositonprobJJ} J(f)= \frac{1}{2}\| \psi (t=T)\| _{L^2(\Omega)}^2 +\frac{\varepsilon}{2}\| f\| _{L^2( \Omega \times ] 0,T[)}^2, \end{equation} subject to $\psi $ being the weak solution of the parabolic problem \eqref{eqProblPositonprob} with source term $f$, \begin{equation} A_{ad} =\{ u\in L^2(\Omega\times ]0,T[): \| u\|_{L^2(\Omega\times ]0,T[)} \leq r \}, \end{equation} where $r$ is a real strictly positive constant. We now specify some notation. Let us introduce the functional spaces (see \cite{31,4,33}) \begin{gather*} V=\{ u\in L^2(\Omega):u\text{ absolutely continuous on }[ 0,1] \}, \\ S=\{ u\in L^2(\Omega):\sqrt{a}u_{x}\in L^2(\Omega)\text{ and }u(0)=u(1)=0 \} ,\\ H_a^{1}(\Omega)=V \cap S, \\ H_a^2(\Omega)=\{ u\in H_a^{1}(\Omega): au_{x}\in H^{1}(\Omega)\}, \\ H^1_{\alpha,0}= \{ u\in H^1_{\alpha}: u(0)=u(1)=0 \}, \\ H^1_{\alpha}= \{ {u\in L^2(\Omega)\cap H^1_{\rm loc}(]0,1]): x^{\frac{\alpha}{2}}u_x\in L^2(\Omega)} \}, \end{gather*} with \begin{gather*} \| u\| ^2_{H^1_a(\Omega)}=\| u\| ^2_{L^2(\Omega)}+\| \sqrt{a} u_x\| ^2_{L^2(\Omega)},\\ \| u\|^2_{H^2_a(\Omega)}=\| u\| ^2_{H^1_a(\Omega)}+\| (au_x)_x\| ^2_{L^2(\Omega)},\\ \langle u,v \rangle_{H^1_{\alpha}}=\int_{\Omega} (uv+ x^{\alpha}u_xv_x)\,dx. \end{gather*} We recall that (see \cite{33}) $H^1_a$ is an Hilbert space and it is the closure of $C^\infty_c(0,1)$ for the norm $\|\cdot \|_{H^1_a}$. If $\frac{1}{\sqrt{a}}\in L^1(\Omega)$ then the following injections \begin{gather*} H^1_a(\Omega)\hookrightarrow L^2(\Omega), \\ H^2_a(\Omega)\hookrightarrow H^1_a(\Omega), \\ H^1(0,T;L^2(\Omega))\cap L^2(0,T;D(A)) \hookrightarrow L^2(0,T;H^1_a) \cap C(0,T;L^2(\Omega)) \end{gather*} are compact. The weak formulation of problem \eqref{eqProblPositonprob} is \begin{equation} \int_{\Omega}\partial_t \psi v \,dx +\int_{\Omega} \Big(a(x)\partial_x \psi \partial_x v-\frac{\lambda }{x^{\beta }} \psi v) \,dx =\int_{\Omega}fv \,dx, \quad \forall v \in H_0^1(\Omega). \end{equation} Let \begin{equation} B[\psi,v]=\int_{\Omega} \Big(a(x)\partial_x \psi \partial_x v-\frac{\lambda }{x^{\beta }}\psi v\Big) \,dx. \end{equation} We discuss the cases non-coercive and subcritical potential cases separately. \subsection*{Non-coercive case: $\lambda=0$} In this case the bilinear form $B$ becomes \begin{equation} B[\psi,v]=\int_{\Omega} (a(x)\partial_x \psi \partial_x v)\,dx. \end{equation} We have $a(x)=0$ at $x=0$, from where the bilinear form $B$ will be non-coercive. We recall the following theorem. \begin{theorem}[\cite{31,35,34}]\label{prop1} For all $f\in L^2 ( \Omega \times ]0,T[ ) $ and $\psi _0\in L^2(\Omega) $, there exists a unique weak solution to \eqref{eqProblPositonprob} such that \begin{equation*} \psi \in C( [ 0,T] ;L^2(\Omega)) \cap L^2(0,T;H_a^{1}) \end{equation*} and there is a constant $C_{T}$ such that for any solution of \eqref{eqProblPositonprob}, \[ \sup_{t\in [ 0,T] } \| \psi (t)\| _{L^2(\Omega)}^2 +\int_0^{T}\| \sqrt{a}\psi _{x}(t)\| _{L^2(\Omega)}^2dt \\ \leq C_{T}\Big( \| \psi _0\| _{L^2(\Omega)}^2 +\| f\| _{L^2( \Omega \times ]0,T[)}^2\Big)\,. \] Furthermore, if $\psi _0\in H_a^{1}(\Omega) $ then \[ \psi \in C( [ 0;T] ,H_a^{1}) \cap L^2( 0,T;H_a^2) \cap H^{1}( 0,T;L^2(\Omega)) \] and there is a constant $C_{T}$ such that \begin{align*} &\sup_{t\in [0,T]} \| \psi (t)\| _{H_a^{1}}^2 +\int_0^{T}( \| \psi _{t}\| _{L^2(\Omega)}^2 +\| ( a\psi _{x}) _{x}(t)\| _{L^2(\Omega)}^2) dt \\ &\leq C_{T}( \| \psi _0\| _{H_a^{1}}^2 +\| f\| _{L^2(\Omega \times ]0,T[)}^2). \end{align*} \end{theorem} The continuity of the functional $J$ is deduced from the continuity of the function $\varphi: f \to \psi$, where $\psi$ is the weak solution of \eqref{eqProblPositonprob} with source term $f$. \begin{theorem} \label{prop3} Let $\psi$ be the weak solution of \eqref{eqProblPositonprob}. In the non-coercive case, if $\psi _0\in H_a^{1}(\Omega)$, then the functon $\varphi : L^2(\Omega \times ]0,T[)\to C( [ 0,T] ;H_a^{1}(\Omega)) \cap L^2( 0,T;H_a^2(\Omega)) \\ \cap H^{1}( 0,T;L^2(\Omega))$, defined by \[ \varphi(f)= \psi \] is continuous. If $\psi _0\in L^2(\Omega)$, then $\varphi : L^2(\Omega \times ]0,T[)\to C( [ 0,T] ;L^2(\Omega)) \cap L^2(0,T;H_a^{1})$, $\varphi (f)= \psi$ is continuous. \end{theorem} The differentiability of the functional $J$ is deduced from the differentiability of the function $\varphi: f\to \psi$. \begin{theorem} \label{theodiffer2} Let $\psi$ be the weak solution of \eqref{eqProblPositonprob}. If $\psi _0\in H_a^{1}(\Omega)$, then the function $\varphi : L^2(\Omega \times ]0,T[)\to C( [ 0,T] ;H_a^{1}(\Omega)) \cap L^2( 0,T;H_a^2(\Omega)) \cap H^{1}( 0,T;L^2(\Omega))$, $\varphi(f)= \psi$ is G-derivable. If $\psi _0\in L^2(\Omega)$, then $\varphi : L^2(\Omega \times ]0,T[)\to C( [ 0,T] ;L^2(\Omega)) \cap L^2(0,T;H_a^{1})$, $\varphi(f)= \psi$ is G-derivable. \end{theorem} \subsection*{Sub-critical potential case: $\lambda\neq 0$} (see\cite{24,30}) In this case the bilinear form $B$ becomes \begin{equation} B[\psi,v]=\int_{\Omega} \Big( a(x)\partial_x \psi \partial_x v -\frac{\lambda }{x^{\beta }}\psi v ) \,dx. \end{equation} Since $a(x)=0$ at $x=0$ and $ \lim_{x \to 0} \frac{\lambda }{x^{\beta }} = +\infty$, the bilinear form $B$ is non-coercive and is non continuous at $x=0$. Consider the unbounded operator $(K,D(K))$ where \begin{equation} Ku=(x^{\alpha}u_x)_x+\frac{\lambda}{x^{\beta}}u, \end{equation} for $u$ in \begin{equation*} D(k)=[ {u\in H^1_{\alpha,0}\cap H^2_{\rm loc}(]0,1])\big| (x^{\alpha}u_x)_x +\frac{\lambda}{x^{\beta}}u\in L^2(\Omega)}] . \end{equation*} \begin{theorem}[\cite{4,24}] \label{lemme1} If $f = 0$, then for all $\psi_0\in L^2(\Omega)$, problem \eqref{eqProblPositonprob} has a unique weak solution \begin{equation} \psi\in C([0,T];L^2(\Omega))\cap C(]0,T];D(K))\cap C^1(]0,T];L^2(\Omega))\,. \end{equation} If $\psi _0\in D(K) $ then \begin{equation} \psi\in C([0,T];D(K))\cap C^1([0,T];L^2(\Omega))\,. \end{equation} If $f\in L^2(\Omega\times ]0,T[ )$ then for all $\psi_0\in L^2(\Omega)$, problem \eqref{eqProblPositonprob} has a unique solution \begin{equation} \psi\in C([0,T];L^2(\Omega)). \end{equation} \end{theorem} We have the following results. \begin{theorem} \label{prop4} Let $\psi$ be the weak solution of \eqref{eqProblPositonprob}. In the sub-critical potential case, the function $\varphi : L^2(\Omega\times ]0,T[ )\to C([0,T];L^2(\Omega))$, $\varphi(f)= \psi$ is continuous. \end{theorem} \begin{theorem} \label{theodiffer3} Let $\psi$ be the weak solution of \eqref{eqProblPositonprob}. Then $\varphi : L^2(\Omega\times ]0,T[ )\to C([0,T];L^2(\Omega))$, $\varphi(f)= \psi$ is G-derivable. \end{theorem} \section{Proof of main results.} \begin{proof}[Proof of Theorem \ref{prop3}] Let $\psi _0\in H_a^{1}(\Omega)$, and $\delta f$ a small variation such that $f+\delta f \in A_{ad}$. Consider $\delta\psi =\psi ^{\delta }-\psi $, with $\psi $ is the weak solution of \eqref{eqProblPositonprob} with source term $f$ and $\psi ^{\delta }$ is the weak solution of \eqref{eqProblPositonprob} with source term $f^{\delta }=f+\delta f$. Consequently, $\delta\psi $ is solution of the variational problem \begin{equation} \begin{gathered} \int_\Omega\partial _{t}\delta\psi v\ dx +\int_\Omega a(x) \partial _{x}\delta\psi(x)\partial_{x}v\ dx =\int_\Omega\delta fv dx\\ \delta\psi (0,t)=\delta\psi (1,t)=0\quad \forall t\in ] 0,T[ \\ \delta\psi (x,0)=0 \quad \forall x\in \Omega. \end{gathered} \end{equation} Hence, $\delta\psi$ is the weak solution of \eqref{eqProblPositonprob} with source term $\delta f$. We apply the estimate in theorem \ref{prop1}, to obtain a constant $C_{T}$ such that \begin{equation}\label{EqretCrit} \begin{aligned} &\sup_{t\in [0,T]} \| \delta\psi (t)\| _{H_a^{1}(\Omega)}^2 +\int_0^{T}( \| \partial_t \delta\psi \| _{L^2(\Omega)}^2 +\| \partial_x( a\partial_x \delta\psi)(t)\| _{L^2(\Omega)}^2) dt \\ &\leq C_{T} \| \delta f\| _{L^2(\Omega\times ]0,T [)}^2; \end{aligned} \end{equation} therefore, \begin{gather} \sup_{t\in [0,T]} \| \delta\psi (t)\| _{H_a^{1}(\Omega)}^2 \leq C_{T} \| \delta f\| _{L^2(\Omega\times ]0,T [)}^2, \\ \label{01} \| \delta\psi\|^2 _{C([0,T];H_a^{1}(\Omega))} \leq C_{T} \| \delta f\| _{L^2(\Omega\times ]0,T [)}^2. \end{gather} Then from \eqref{EqretCrit} we have \begin{gather*} \| \delta\psi (t)\| _{H_a^{1}(\Omega)}^2 +\int_0^{T}\|\partial_x ( a\partial_x \delta\psi )(t)\| _{L^2(\Omega)}^2 dt \leq C_{T} \| \delta f\| _{L^2(\Omega\times ]0,T [)}^2, \\ \int_0^T\| \delta\psi (t)\| _{H_a^{1}(\Omega)}^2dt+T\int_0^{T} \| \partial_x( a\partial_x \delta\psi )(t)\| _{L^2(\Omega)}^2 dt \leq TC_{T}\| \delta f\| _{L^2(\Omega\times ]0,T [)}^2, \\ \begin{aligned} &\inf (1,T)( \int_0^T\| \delta\psi (t)\| _{H_a^{1}(\Omega)}^2dt +\int_0^{T}\| \partial_x( a\partial_x\delta\psi ) (t)\| _{L^2(\Omega)}^2 dt) \\ &\leq TC_{T} \| \delta f\| _{L^2(\Omega\times ]0,T [)}^2, \end{aligned}\\ \begin{aligned} &\int_0^T\| \delta\psi (t)\| _{H_a^{1}(\Omega)}^2dt +\int_0^{T}\| \partial_x( a\partial_x\delta\psi ) (t)\| _{L^2(\Omega)}^2 dt \\ &\leq\frac{TC_{T}}{\inf (1,T)} \| \delta f\| _{L^2(\Omega\times ]0,T [)}^2. \end{aligned} \end{gather*} Hence, \begin{equation}\label{02} \| \delta\psi\|^2 _{L^2(0,T,H^2_a(\Omega))} \leq \frac{TC_{T}}{\inf (1,T)} \| \delta f\| _{L^2(\Omega\times ]0,T [)}^2. \end{equation} In addition, from \eqref{EqretCrit} we have \begin{gather} \| \delta\psi (t)\| _{H_a^{1}(\Omega)}^2 +\int_0^{T}\|\partial_t \delta\psi (t)\| _{L^2(\Omega)}^2 dt \leq C_{T} \| \delta f\| _{L^2(\Omega\times ]0,T [)}^2, \quad \forall t \in [0,T], \nonumber\\ \begin{aligned} &\| \delta\psi (t)\| _{L^2(\Omega)}^2 +\|\sqrt{a} \partial_x \delta\psi (t)\| _{L^2(\Omega)}^2 +\int_0^{T}\|\partial_t \delta\psi (t)\| _{L^2(\Omega)}^2 dt \nonumber\\ &\leq C_{T} \| \delta f\| _{L^2(\Omega\times ]0,T [)}^2, \quad \forall t \in [0,T], \end{aligned} \nonumber\\ \| \delta\psi (t)\| _{L^2(\Omega)}^2 +\int_0^{T}\|\partial_t \delta\psi (t)\| _{L^2(\Omega)}^2 dt \leq C_{T} \| \delta f\| _{L^2(\Omega\times ]0,T [)}^2, \quad \forall t \in [0,T], \nonumber\\ \int_0^T\| \delta\psi (t)\| _{L^2(\Omega)}^2dt+T\int_0^{T}\|\partial_t \delta\psi (t)\| _{L^2(\Omega)}^2 dt\leq TC_{T} \| \delta f\| _{L^2(\Omega\times ]0,T [)}^2, \nonumber\\ \label{03} \| \delta\psi\|^2_{H^1(0,T;L^2(\Omega))} \leq \frac{TC_{T}}{\inf (1,T)} \| \delta f\| _{L^2(\Omega\times ]0,T [)}^2. \end{gather} Inequalities \eqref{01}, \eqref{02} and \eqref{03} imply the continuity of the function \\ $\varphi : L^2(\Omega \times ]0,T[)\to C( [ 0,T] ;H_a^{1}(\Omega)) \cap L^2( 0,T;H_a^2(\Omega)) \cap H^{1}( 0,T;L^2(\Omega))$, $\varphi(f)=\psi$. In the same way we can prove that if $\psi _0\in L^2(\Omega)$, then the function $\varphi : L^2(\Omega \times ]0,T[)\to C( [ 0,T] ;L^2(\Omega)) \cap L^2(0,T;H_a^{1})$, $\varphi(f)=\psi$ is continuous. Hence, the cost $J$ is continuous. \end{proof} \begin{proof}[Proof of Theorem \ref{theodiffer2}] Let $\psi _0\in H_a^{1}(\Omega)$, and $\delta f$ a small variation such that $f+\delta f \in A_{ad}$, we define the function \begin{equation} \varphi'(f) : \delta f\in A_{ad} \to \delta \psi, \end{equation} where $\delta \psi$ is the solution of the variational problem \begin{equation} \begin{gathered} \int_{\Omega}\partial_t (\delta\psi) v \,dx +\int_{\Omega} a(x)\partial_x (\delta\psi) \partial_x v\,dx =\int_\Omega\delta f v dx\quad \forall v \in H^1_0(\Omega) \\ \delta\psi(0,t)=\delta \psi(1,t)=0\quad \forall t\in ]0,T[ \\ \delta \psi(x,0)= 0\quad \forall x\in \Omega \end{gathered} \end{equation} and we set \begin{equation} \phi(f)=\varphi(f+\delta f)-\varphi(f)-\varphi'(f)\delta f. \end{equation} We want to show that \begin{equation} \phi(f)=o(\delta f). \end{equation} We easily verify that the function $\phi$ is solution of following variational problem \begin{equation} \begin{gathered} \int_{\Omega}\partial_t \phi v \,dx +\int_{\Omega} a(x)\partial_x \phi \partial_x v \,dx =\int_\Omega(\delta f-(\delta f)^2)v dx\quad \forall v \in H^1_0(\Omega)\\ \phi(0,t)=\phi(1,t)=0\quad \forall t\in ]0,T[ \\ \phi(x,0)=0 \quad \forall x\in \Omega. \end{gathered} \end{equation} In the same way as in the proof of continuity, we deduce that \begin{gather} \| \phi \|^2_{C([0,T],H^1_a(\Omega))} \leq C_T\| \delta f-(\delta f)^2 \|^2_{L^2(\Omega\times ]0,T [)}, \\ \| \phi\| ^2_{L^2(0,T,H^2_a(\Omega))} \leq \frac{TC_{T}}{\inf (1,T)} \| \delta f-(\delta f)^2 \|^2_{L^2(\Omega\times ]0,T [)}, \\ \| \phi\|^2 _{H^1(0,T;L^2(\Omega))} \leq \frac{TC_{T}}{\inf (1,T)} \| \delta f-(\delta f)^2 \|^2_{L^2(\Omega\times ]0,T [)}. \end{gather} Therefore, the function $\varphi : L^2(\Omega \times ]0,T[)\to C( [ 0,T] ;H_a^{1}(\Omega)) \cap L^2( 0,T;H_a^2(\Omega)) \cap H^{1}( 0,T;L^2(\Omega))$ $\varphi(f)=\psi$ is G-derivable. In the same way we prove that if $\psi _0\in L^2(\Omega)$, then the function $\varphi : L^2(\Omega \times ]0,T[)\to C( [ 0,T] ;L^2(\Omega)) \cap L^2(0,T;H_a^{1})$, $\varphi(f)=\psi$ is G-derivable. Hence, we deduce the existence of the gradient of the functional $J$. \end{proof} \begin{proof}[Proof of Theorem \ref{prop4}] Let $\delta f$ be a small variation such that $f+\delta f \in A_{ad}$. Consider $\delta\psi =\psi ^{\delta }-\psi $, with $\psi $ a the weak solution of \eqref{eqProblPositonprob}, with source term $f$, and consider $\psi ^{\delta }$ the weak solution of \eqref{eqProblPositonprob} with source term $f^{\delta }=f+\delta f$. Consequently, $\delta\psi $ is the solution of variational problem \begin{equation}\label{ProbDpsicritical} \begin{gathered} \int_{\Omega}\partial_t \delta\psi v\,dx +\int_{\Omega}\Big(a(x)\partial_x \delta\psi \partial_x v -\frac{\lambda }{x^{\beta }}\delta\psi v\Big)dx =\int_{\Omega} \delta f v dx,\quad \forall v \in H_0^1(\Omega) \\ \delta\psi (0,t)=\delta\psi (1,t)=0\quad \forall t\in ] 0,T[ \\ \delta\psi (x,0)=0\quad \forall x\in \Omega. \end{gathered} \end{equation} Take $v=\delta\psi$, this gives \begin{equation} \int_{\Omega}\partial_t \delta\psi \delta\psi\,dx +\int_{\Omega}\Big(a(x)(\partial_x \delta\psi)^2-\frac{\lambda }{x^{\beta }} (\delta\psi)^2 \Big)dx =\int_\Omega \delta f\delta\psi\,dx, \end{equation} $\Omega$ is independent of $t$, which gives \begin{equation} \frac{1}{2}\frac{d}{dt} \int_{\Omega}(\delta\psi)^2dt +\int_{\Omega}\Big(a(x)(\partial_x \delta\psi)^2-\frac{\lambda }{x^{\beta }} (\delta\psi)^2 \Big)dx =\int_\Omega \delta f\delta\psi\,dx, \end{equation} recall that $\delta\psi(t=0)=0$, by integrating between $0$ and $t$ with $t\in [0,T] $ we obtain \begin{equation} \begin{aligned} &\frac{1}{2} \| \delta\psi(t)\|^2_{L^2(\Omega)} +\int_0^t\int_{\Omega}(a(x)(\partial_x \delta\psi)^2 -\frac{\lambda }{x^{\beta }}(\delta\psi)^2 )\,dx\,ds \\ &=\int_0^t\int_\Omega \delta f\delta\psi\,dx\,ds. \end{aligned} \end{equation} We have $2ab\leq a^2+b^2$, for all $(a,b)\in R$, therefore \begin{equation} \begin{aligned} &\frac{1}{2} \| \delta\psi(t)\|^2_{L^2(\Omega)} +\int_0^t\int_{\Omega}\Big(a(x)(\partial_x \delta\psi)^2 -\frac{\lambda }{x^{\beta }}(\delta\psi)^2 \Big)\,dx\,ds\\ & \leq\frac{1}{2}\int_0^t \| \delta f\|^2_{L^2(\Omega)}dt + \frac{1}{2}\int_0^t\| \delta\psi \|^2_{L^2(\Omega)}ds. \end{aligned} \end{equation} Then \begin{equation} \begin{aligned} &\frac{1}{2} \| \delta\psi(t)\|^2_{L^2(\Omega)} +\int_0^t\int_{\Omega}(a(x)(\partial_x \delta\psi)^2 -\frac{\lambda }{x^{\beta }}(\delta\psi)^2 )\,dx\,ds\\ &\leq\frac{1}{2} \| \delta f\|_{L^2(\Omega\times ]0,T [)}^2 + \frac{1}{2}\int_0^t\| \delta\psi \|^2_{L^2(\Omega)}ds. \end{aligned} \end{equation} Therefore \begin{equation} \| \delta\psi(t)\|^2_{L^2(\Omega)} \leq\| \| \delta f\|_{L^2(\Omega\times ]0,T [)}^2 + \int_0^t\| \delta\psi \|^2_{L^2(\Omega)}ds. \end{equation} Gronwall's Lemma gives \begin{gather*} \| \delta\psi(t)\|^2_{L^2(\Omega)} \leq \| \delta f\|_{L^2(\Omega\times ]0,T [)}^2 \exp(\int_0^t ds) \quad \forall t \in [0,T], \\ \| \delta\psi(t)\|^2_{L^2(\Omega)} \leq \exp(T)\| \delta f\|_{L^2(\Omega\times ]0,T [)}^2 \quad \forall t \in [0,T], \end{gather*} from where \begin{equation}\label{AQretourL2} \| \delta\psi\| _{C( [ 0;T] ,{L^2(\Omega)}) }^2\leq \exp(T) \| \delta f\|_{L^2(\Omega\times ]0,T [)}^2. \end{equation} Which implies the continuity of the function $\varphi : {L^2(\Omega\times ]0,T [)}\to C([0,T];L^2(\Omega))$, $\varphi(f)= \psi$. Hence, the cost $J$ is continuous. \end{proof} \begin{proof}[Proof of Theorem \ref{theodiffer3}] Let $\delta f$ be a small variation such that $f+\delta f \in A_{ad}$, we define the function \begin{equation} \varphi'(f) : \delta f\in A_{ad} \to \delta \psi, \end{equation} where $\delta \psi$ is the solution of the variational problem \begin{equation} \begin{gathered} \int_{\Omega}\partial_t (\delta\psi) v \,dx +\int_{\Omega} (a(x)\partial_x (\delta\psi) \partial_x v -\frac{\lambda }{x^{\beta }}\delta\psi v) \,dx =\int_{\Omega}\delta f v dx \quad \forall v \in H^1_0(\Omega) \\ \delta\psi(0,t)=\delta \psi(1,t)=0\quad \forall t\in ]0,T[ \\ \delta \psi(x,0)= 0\quad \forall x\in \Omega. \end{gathered} \end{equation} We set \begin{equation} \phi(f)=\varphi(f+\delta f)-\varphi(f)-\varphi'(f)\delta f. \end{equation} We want to show that \begin{equation} \phi(f)=o(\delta f). \end{equation} We easily verify that the function $\phi$ is the solution of variational problem \begin{equation} \begin{gathered} \int_{\Omega}\partial_t \phi v \,dx +\int_{\Omega} \big(a(x)\partial_x \phi \partial_x v-\frac{\lambda }{x^{\beta }} \phi v\Big) \,dx=\int_{\Omega}(\delta f-(\delta f)^2)v dx \quad \forall v \in H^1_0(\Omega)\\ \phi(0,t)=\phi(1,t)=0\quad \forall t\in ]0,T[ \\ \phi(x,0)=0 \quad \forall x\in \Omega. \end{gathered} \end{equation} In the same way as that used in the proof of continuity, we deduce \begin{equation} \| \phi \|_{C([0,T],L^2(\Omega))}^2 \leq \exp(T) \|\ \delta f-(\delta f)^2 \|_{L^2(\Omega\times ]0,T [)}^2. \end{equation} Hence, in all cases, the function $\varphi(f)=\psi$ is G-derivable and we deduce the existence of the gradient of the functional $J$. \end{proof} Now, we compute the gradient of $J$ using the adjoint state method. \section{Gradient of $J$} We define the G{\^a}teaux derivative of $\psi$ at $f$ in the direction $h\in L^2( \Omega \times ] 0,T[)$, by \begin{equation} \hat{\psi}=\lim_{s \to 0} \frac{\psi(f+s h)-\psi(f) }{s }, \end{equation} $\psi(f+s h)$ is the weak solution of \eqref{eqProblPositonprob} with source term $f+s h$, and $\psi(f) $ is the weak solution of \eqref{eqProblPositonprob} with source term $f$. We compute the G{\^a}teaux (directional) derivative of \eqref{eqProblPositonprob} at $f$ in some direction $h\in L^2( \Omega \times ] 0,T[)$, and we get the so-called tangent linear model: \begin{equation} \begin{gathered} \partial _{t}\hat{\psi}+A\hat{\psi}=h \\ \hat{\psi}(0,t)=\hat{\psi}(1,t)=0\quad \forall t\in ]0,T[ \\ \hat{\psi}(x,0)=0\quad \forall x\in \Omega. \end{gathered} \end{equation} We introduce the adjoint variable $P$, and we integrate, \begin{gather} \int_0^{1}\int_0^{T}\partial _{t}\hat{\psi}P \,dt\,dx +\int_0^{1}\int_0^{T}A \hat{\psi}P\,dx =\int_0^{1}\int_0^{T}hP\,dt\,dx, \\ \int_0^{1}\Big( [ \hat{\psi}P] _0^{T}-\int_0^{T}\hat{\psi} \partial _{t}P\,dt\Big) \,dx +\int_0^{T}\langle A\hat{\psi} ,P\rangle_{L^2(\Omega)}dt =\langle h ,P\rangle_{L^2(\Omega\times ]0,T [)}, \\ \label{EqI1} \begin{aligned} &\int_0^{1}[ \hat{\psi}(T)P(T)-\hat{\psi}(0)P(0)]dx -\int_0^{T}\langle \hat{\psi},\partial _{t}P\rangle_{L^2(\Omega)} dt +\int_0^{T}\langle A\hat{\psi},P\rangle_{L^2(\Omega)}dt \\ & =\langle h ,P\rangle_{L^2(\Omega\times ]0,T [)}. \end{aligned} \end{gather} Let us take $P(x=0)=P(x=1)=0$, then we may write $\langle \hat{\psi} ,AP\rangle_{L^2(\Omega)} =\langle A\hat{\psi},P\rangle_{L^2(\Omega)}$. With $P(T)=0$ we may now rewrite \eqref{EqI1} as \begin{equation*} \int_0^{T}\langle \hat{\psi},\partial _{t}P-AP\rangle_{L^2(\Omega)} dt=-\langle h ,P\rangle_{L^2(\Omega\times ]0,T [)} \end{equation*} this gives \begin{equation}\label{EqJEquation00} \begin{gathered} \int_0^{T}\langle \hat{\psi},\partial _{t}P-AP\rangle_{L^2(\Omega)} dt =-\langle h ,P\rangle_{L^2(\Omega\times ]0,T [)} \\ P(x=0)=P(x=1)=0,\quad P(T)=0. \end{gathered} \end{equation} The discretization in time of \eqref{EqJEquation00}, using the Rectangular integration method, gives \begin{equation}\label{EqJEquation} \begin{gathered} \sum_{j=0}^{M+1}\langle \hat{\psi}(t_j),\partial _{t}P(t_j) -AP(t_j)\rangle_{L^2(\Omega)} \Delta t=\langle -P,h \rangle_{L^2(\Omega\times ]0,T [)}\\ P(x=0)=P(x=1)=0 ,\quad P(T)=0. \end{gathered} \end{equation} With \begin{equation*} t_j=j\Delta t, \quad j\in \{0,1,2,\dots,M+1\}, \end{equation*} where $\Delta t$ is the step in time and $T=(M+1)\Delta t$. The G{\^a}teaux derivative of $J$ at $f$ in the direction $h\in L^2(\Omega)$ is given by \begin{equation*} \hat{J}(h)=\lim_{s \to 0} \frac{J(f+s h)-J(f)}{s }. \end{equation*} After some computations, we arrive at \begin{equation}\label{EqJ3} \hat{J}(h)=\langle \psi(T),\hat{\psi}(T)\rangle _{L^2(\Omega)} +\langle\varepsilon f,h\rangle _{L^2(\Omega\times ]0,T [)}. \end{equation} The adjoint model is \begin{equation}\label{ProbAdjointPremProb} \begin{gathered} \partial _{t}P(T)-AP(T)=\frac{1}{\Delta t}\psi(T), \quad \partial _{t}P(t_j)-AP(t_j)=0 \quad \forall t_j\neq T\\ P(x=0)=P(x=1)=0\quad \forall t_j\in ] 0;T[ \\ P(T)=0. \end{gathered} \end{equation} From equations \eqref{EqJEquation}, \eqref{EqJ3} and \eqref{ProbAdjointPremProb}, the gradient of $J$ is given by \begin{equation} \frac{\partial J}{\partial f}=-P+\varepsilon f. \end{equation} Problem \eqref{ProbAdjointPremProb} is retrograde, we make the change of variable $t\longleftrightarrow T-t$. \section{Discretized problem} \noindent\textbf{Step 1. Full discretization.} Discrete approximations of these problems need to be made for numerical implementation. To resolve the Direct problem and adjoint problem, we use the Method $\protect\theta $-schema in time. This method is unconditionally stable for $1 >\theta \geq \frac{1}{2}$. Let $h$ be the step in space and $\Delta t$ the step in time. Let \begin{gather*} x_{i}=ih, \quad i\in \{ 0,1,2,\dots, N+1\}, \\ c(x_{i})=a(x_{i})+\gamma, \\ t_j=j\Delta t, \quad j\in \{0,1,2,\dots, M+1\}, \\ f_{i}^{j}=f(x_{i},t_j). \end{gather*} We put \begin{gather*} \psi _{i}^{j}=\psi (x_{i},t_j), \\ da(x_{i})=\frac{c(x_{i+1})-c(x_{i})}{h}, \\ b(x)=-\frac{\lambda }{x^{\beta }}. \end{gather*} Therefore \begin{equation} \partial _{t}\psi +A\psi =f \end{equation} is approximated by \begin{align*} &-\frac{\theta \Delta t}{h^2}c(x_{i})\psi _{i-1}^{j+1} +\Big( 1+\frac{2\theta \Delta t}{h^2}c(x_{i}) +da(x_{i})\frac{\theta \Delta t}{h}+b(x_{i})\theta\Delta t\Big) \psi _{i}^{j+1} \\ &-\big(\frac{\theta \Delta t}{h^2}c(x_{i}) +da(x_{i})\frac{\theta \Delta t}{h}\big)\psi _{i+1}^{j+1}\\ &=\frac{( 1-\theta ) \Delta t}{h^2}c(x_{i})\psi_{i-1}^{j} +\Big( 1-\frac{( 1-\theta ) \Delta t}{h}da(x_{i})-\frac{2( 1-\theta ) \Delta t}{h^2}c(x_{i}) \\ &\quad -( 1-\theta )b(x_{i})\Delta t\Big) \psi _{i}^{j} +\Big( \frac{( 1-\theta ) \Delta t}{h}da(x_{i}) +\frac{( 1-\theta ) \Delta t}{h^2}c(x_{i})\big) \psi_{i+1}^{j}\\ &\quad +\Delta t[( 1-\theta ) f_{i}^{j}+\theta f_{i}^{j+1}]. \end{align*} Let us define \begin{gather*} g_1(x_{i})=-\frac{\theta \Delta t}{h^2}c(x_{i}), \\ g_2(x_{i})=1+\frac{2\theta \Delta t}{h^2}c(x_{i})+da(x_{i}) \frac{\theta \Delta t}{h}+b(x_{i})\theta \Delta t, \\ g_3(x_{i})=-\frac{\theta \Delta t}{h^2}c(x_{i})-da(x_{i}) \frac{\theta \Delta t}{h}, \\ k_1(x_{i})=\frac{( 1-\theta ) \Delta t}{h^2}c(x_{i}), \\ k_2(x_{i})=1-\frac{( 1-\theta ) \Delta t}{h}da(x_{i}) -\frac{2(1-\theta ) \Delta t}{h^2}c(x_{i})-( 1-\theta )b(x_{i})\Delta t, \\ k_3(x_{i})=\frac{( 1-\theta ) \Delta t}{h}da(x_{i}) +\frac{(1-\theta ) \Delta t}{h^2}c(x_{i}). \end{gather*} Let $\psi ^{j}=( \psi _{i}^{j}) _{i\in \{ 1,2,\dots,N\} }$, finally we obtain \begin{equation}\label{SystemDiscri} \begin{gathered} D\psi ^{j+1}=B\psi ^{j}+V^{j}\text{ with }j\in \{ 1,2,\dots, M\}\\ \psi ^{0}=( f(ih)) _{i\in \{ 1,2,\dots, N\} }, \end{gathered} \end{equation} where \begin{equation*} D=\begin{bmatrix} g_2(x_1) & g_3(x_1) & 0 & & & & & 0 \\ g_1(x_2) & g_2(x_2) & g_3(x_2) & 0 & & & & \\ 0 & g_1(x_3) & g_2(x_3) & g_3(x_3) & 0 & & & \\ & 0 & g_1(x_{4}) & g_2(x_{4}) & g_3(x_{4}) & 0 & & \\ & & 0 & . & . & . & 0 & \\ & & & . & . & . & . & 0 \\ & & & & 0 & g_1(x_{N-1}) & g_2(x_{N-1}) & g_3(x_{N-1}) \\ 0 & & & & & 0 & g_1(x_{N}) & g_2(x_{N})% \end{bmatrix} \end{equation*} \begin{equation*} B=\begin{bmatrix} k_2(x_1) & k_3(x_1) & 0 & & & & & 0 \\ k_1(x_2) & k_2(x_2) & k_3(x_2) & 0 & & & & \\ 0 & k_1(x_3) & k_2(x_3) & k_3(x_3) & 0 & & & \\ & 0 & k_1(x_{4}) & k_2(x_{4}) & k_3(x_{4}) & 0 & & \\ & & 0 & . & . & . & 0 & \\ & & & . & . & . & . & 0 \\ & & & & 0 & k_1(x_{N-1}) & k_2(x_{N-1}) & k_3(x_{N-1}) \\ 0 & & & & & 0 & k_1(x_{N}) & k_2(x_{N})% \end{bmatrix}% \end{equation*} \begin{equation*} V^{j}=\begin{bmatrix} \Delta t[( 1-\theta ) f(x_1,t_j)+\theta f(x_1,t_j+\Delta t)]\\ \Delta t[( 1-\theta ) f(x_2,t_j)+\theta f(x_2,t_j+\Delta t)] \\ \vdots \\ \Delta t [( 1-\theta ) f(x_{N-1},t_j)+\theta f(x_{N-1},t_j+\Delta t)] \\ \Delta t[( 1-\theta ) f(x_{N},t_j)+\theta f(x_{N},t_j+\Delta t)] \end{bmatrix} \end{equation*} \smallskip \noindent\textbf{Step 2.} Discretization of the functional \begin{equation} J(u)=\frac{\varepsilon}{2}\int_0^1(u(x))^2 dx +\frac{1}{2}\int_0^1(\psi(x,T))^2 dx. \end{equation} We recall that the Simpson methods for calculate an integral is \begin{equation*} \int_a^b f(x) \, dx\simeq\frac{h}{2} \Big[ f(x_0)+2 \sum_{i=1}^{\frac{N+1}{2}-1}f(x_{2i}) +4 \sum_{i=1}^{\frac{N+1}{2}}f(x_{2i+1})+f(x_{N+1}) \Big] \end{equation*} with $x_0=a$, $x_{N+1}=b$, $x_i=a+ih$, $i \in [ 1,\dots,N+1]$. Let \begin{gather*} \phi(x)=(u(x))^2 \quad \forall x\in \Omega, \\ \varphi(x)=(\psi(x,T))^2 \quad \forall x\in \Omega\,. \end{gather*} We have \begin{gather*} \int_0^1 \phi(x) \, dx\simeq\frac{h}{2} \Big[ \phi(0)+2 \sum_{i=1}^{\frac{{N+1}}{2}-1}\phi(x_{2i}) +4 \sum_{i=1}^{\frac{{N+1}}{2}}\phi(x_{2i+1})+\phi(1) \Big], \\ \int_0^1 \varphi(x) \ ,dx\simeq\frac{h}{2} \Big[ \varphi(0)+2 \sum_{i=1}^{\frac{{N+1}}{2}-1}\varphi(x_{2i}) +4 \sum_{i=1}^{\frac{{N+1}}{2}}\varphi(x_{2i+1})+\varphi(1) \Big]. \end{gather*} Therefore, \begin{align*} J(u)&\simeq\frac{\varepsilon h}{4} \Big[ \phi(0)+2 \sum_{i=1}^{\frac{{N+1}}{2}-1}\phi(x_{2i}) +4 \sum_{i=1}^{\frac{{N+1}}{2}}\phi(x_{2i+1})+\phi(1) \Big]\\ &\quad +\frac{h}{4}[ \varphi(0)+2 \sum_{i=1}^{\frac{{N+1}}{2}-1} \varphi(x_{2i})+4 \sum_{i=1}^{\frac{{N+1}}{2}}\varphi(x_{2i+1})+\varphi(1) ]. \end{align*} The main steps for descent method at each iteration are: \begin{itemize} \item Calculate $\psi^{k}$ solution of \eqref{eqProblPositonprob} with source term $f^k$ \item Calculate $P^{k}$ solution of the adjoint problem \item Calculate the descent direction $d_{k}=-\nabla J(f^k)$ \item Find $ t_{k}=\underset{t>0}{argmin}$ $J(f^k+td_{k})$ \item Update the variable $f^{k+1}=f^k+t_{k}d_{k}$. \end{itemize} The algorithm ends when $ \big| J(f)\big|<\mu$, where $\mu$ is a given small precision. The value $t_{k}$ is chosen by the inaccurate linear search by the Armijo-Goldstein Rule as follows: Let $\alpha_{i}, \beta \in [0,1[$ and $\alpha>0$ if $J(f^{k}+\alpha_i d_{k})\leq J(f^{k})+\beta \alpha_{i} d^{T}_{k}d_{k}$, $t_{k}=\alpha_i$ and stop. if not, $\alpha_i = \alpha \alpha_i$. \section{Numerical experiments} We did all tests on a PC with the following configurations: Intel Core i3 CPU 2.27GHz; RAM 4GB (2.93 usable). For all tests, we take number of points in space $N=100$, number of points in time $M=100$, and initial state the function $\psi_0=\frac{x(x-1)}{T}$. In the figures below, $\psi_0$ is drawn red and the rebuilt function $\psi$ in blue. \subsection*{Noncoercive case} Let $\alpha =\frac{1}{2}$ and $\lambda=0$. Figure \ref{fig1} shows results without regularization. Figures \ref{fig2} and \ref{fig3} show results with regularization. \begin{figure}[ht] \begin{center} \includegraphics[width =0.5\textwidth]{fig1} %{Sans_regular_t_final.png} \end{center} \caption{Final temperature without regularization. It shows that we cannot have $\psi(T)\simeq0$.} \label{fig1} \end{figure} \begin{figure}[ht] \begin{center} \includegraphics[width=0.4\textwidth]{fig2a} % alpha_1_2_lamda_0_t_10.png \includegraphics[width=0.4\textwidth]{fig2b} % alpha_1_2_lamda_0_t_20.png \end{center} \caption{Temperature at $t=t_{10}$ (left), and at $t=t_{20}$ (right).} \label{fig2} \end{figure} \begin{figure}[ht] \begin{center} \includegraphics[width=0.4\textwidth]{fig3a} % alpha_1_2_lamda_0_t_50.png \includegraphics[width=0.4\textwidth]{fig3b} % alpha_1_2_lamda_0_t_100.png \end{center} \caption{Temperature at $t=t_{50}$ which is nearly $0$ (left). Final temperature showing that $\psi(T)\simeq 0$ (right).} \label{fig3} \end{figure} \begin{figure}[ht] \begin{center} \includegraphics[width=0.4\textwidth]{fig4a} % alpha_1_2_lamda_GraphJ.png \includegraphics[width=0.4\textwidth]{fig4b} % alpha_1_2_lamda_Graphgradient.png \end{center} \caption{Graph of $J$ (left). Norm of gradient (right).} \label{fig4} \end{figure} Next we have tests for $\alpha\geq 2$ and $\lambda=0$. Using the Carleman estimates, in \cite{3301} we prove that problem \eqref{eqProblPositonprob} is non-null controllable. In this tests we confim numerically this result; see Figures \ref{fig5} and \ref{fig6}. \begin{figure}[ht] \begin{center} \includegraphics[width=0.4\textwidth]{fig5a} % NonControl_alpha_2_lamda_0_t_10.png \includegraphics[width=0.4\textwidth]{fig5b} % NonControl_alpha_2_lamda_0_t_100.png \end{center} \caption{Temperature at $t=t_{10}$ with $\alpha= 2$ (left). Final temperature with $\alpha=2$ which shows the non-null controllability of \eqref{eqProblPositonprob} (right).} \label{fig5} \end{figure} \begin{figure}[ht] \begin{center} \includegraphics[width=0.4\textwidth]{fig6a} % NonControl_alpha_4_lamda_0_t_10.png \includegraphics[width=0.4\textwidth]{fig6b} % NonControl_alpha_4_lamda_0_t_100.png \end{center} \caption{Temperature at $t=t_{10}$ with $\alpha= 4$ (left). Final temperature with $\alpha= 4$ which shows the non-null controllability of \eqref{eqProblPositonprob} (right).} \label{fig6} \end{figure} \subsection{Sub-critical potential case} Let $\alpha=\frac{1}{2}$, $\lambda=-\frac{(1-\alpha)^2}{4}$, and $ \beta=\frac{2-\alpha}{2}$. Figure \ref{fig7} shows test without regularization. Figures \ref{fig8} and \ref{fig9} have regularization. \begin{figure}[ht] \begin{center} \includegraphics[width=0.5\textwidth]{fig7} % subc_sansregulari.png \end{center} \caption{Final temperature without regularization which shows that we cannot have $\psi(T)\simeq0$.} \label{fig7} \end{figure} \begin{figure}[ht] \begin{center} \includegraphics[width=0.4\textwidth]{fig8a} % alpha_1_2_subcriti_t_10.png \includegraphics[width=0.4\textwidth]{fig8b} % alpha_1_2_subcriti_t_20.png \end{center} \caption{Temperature at $t=t_{10}$ (left), and at $t=t_{20}$ (right).} \label{fig8} \end{figure} \begin{figure}[ht] \begin{center} \includegraphics[width=0.4\textwidth]{fig9a} % alpha_1_2_subcriti_t_50.png \includegraphics[width=0.4\textwidth]{fig9b} % alpha_1_2_subcriti_t_100.png \end{center} \caption{Temperature at $t=t_{50}$ which is nearly $0$ (left). Final temperature showing that $\psi(T)\simeq0$ (right).} \label{fig9} \end{figure} \begin{figure}[ht] \begin{center} \includegraphics[width=0.4\textwidth]{fig10a} % alpha_subc_1_2_graphJ.png \includegraphics[width=0.4\textwidth]{fig10b} % alpha_subcr_1_2_graphgradientJ.png \end{center} \caption{Graph of $J$ (left). Norm of gradient (right).} \label{fig10} \end{figure} \subsection*{Conclusion} This article presents a regularization method for determining the source term. This is done with the aim of studying numerically the null controllability of degenerate/singular parabolic problems. \begin{thebibliography}{10} \bibitem{31} F. Alabau-Boussouira, P. Cannarsa, G. Fragnelli; \emph{Carleman estimates for degenerate parabolic operators with applications to null controllability}, J. evol. equ., 6 (2006), 161--204. \bibitem{11222} C. Alves, A. L. Silvestre, T. Takahashi, M. 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