# An Inverse Source Problem for Singular Parabolic Equations with Interior Degeneracy

Authors: Khalid Atifi; Idriss Boutaayamou; Hamed Ould Sidi; Jawad Salhi
Journal: Abstract and Applied Analysis
 (2018)

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## Abstract

The main purpose of this work is to study an inverse source problem for degenerate/singular parabolic equations with degeneracy and singularity occurring in the interior of the spatial domain. Using Carleman estimates, we prove a Lipschitz stability estimate for the source term provided that additional measurement data are given on a suitable interior subdomain. For the numerical solution, the reconstruction is formulated as a minimization problem using the output least squares approach with the Tikhonov regularization. The Fréchet differentiability of the Tikhonov functional and the Lipschitz continuity of the Fréchet gradient are proved. These properties allow us to apply gradient methods for numerical solution of the considered inverse source problem.


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## Body

## 1. Introduction

Inverse problems appear in a wide range of scientific applications, such as geophysics, biological and medical imaging, material and structure characterization, electrical, mechanical and civil engineering, and finances. The resolution of inverse problems consists of estimating the parameters of the observed system or structure from available data of solutions. The unknown quantities are diverse, according to the inverse problems and phenomena studied, but typical unknowns are spatially varying coefficients and source terms.In the present paper, we study the inverse problem of determining the source term in a degenerate heat equation perturbed by a singular potential from the theoretical analysis and numerical computation angles. More precisely, we consider the following problem:(1)ut-auxx-λbxu=ht,x,t,x∈Q,u0=u1=0,t∈0,T,u0,x=u0x,x∈0,1,where u0∈L2(0,1), T>0 fixed, and Q≔(0,T)×(0,1). Moreover, we assume that the constant λ satisfies suitable assumptions described below and the functions a and b degenerate at the same interior point x0 of the spatial domain (0,1) (for the precise assumptions we refer to Section 2). Let us recall that, in inverse source problems, the source term has to satisfy some condition; otherwise uniqueness may be false; see [1]. Let C0>0 be given and for t0∈(0,T) given, let T′≔(T+t0)/2. In [2, 3], the authors make the assumption that source terms h satisfy the condition(2)htt,x≤C0hT′,x,for  almost  allt,x∈Q.Therefore they define the set S(C0) of admissible source terms as (3)SC0≔h∈H10,T;L20,1:hsatisfies2.The Carleman estimate is a class of weighted energy estimates with a large parameter for a solution to a PDE and it is one of the major tools used in the study of unique continuation, observability, and controllability problems for various kinds of PDEs. The idea of using global Carleman estimates to solve inverse problems and prove Lipschitz stability results was first introduced by Puel and Yamamoto [4] in 1996 in the context of the wave equation, using a modification of the idea of [5]. Later on, it also has been applied to the standard heat equation by Imanuvilov and Yamamoto [2] in 1998. Their method is based on the use of global Carleman estimates for parabolic problems that were developed by Fursikov and Imanuvilov [6] and used to solve null controllability issues. The novelty of their work is that they not only solve the uniqueness question but they also provide unconditional Lipschitz stability result concerning the reconstruction of the source.In the last recent years an increasing interest has been devoted to (1) in the case when λ=0 and the degeneracy can occur at the boundary or in the interior of the space domain. For example, we recall the works [7–9], where the authors obtain results concerning Carleman estimates and null controllability.These results are complemented in [3, 10–12], where the authors obtain results concerning inverse problems for purely degenerate (i.e., λ=0) parabolic equations and parabolic coupled systems, addressing, in particular, issues such as uniqueness and stability. If λ≠0, the first results in this direction are obtained in [1] for the nondegenerate heat operator (i.e., a>0) with a singular potential. But, the study of numerical reconstruction questions are rarely taken into account; see [13, 14].Furthermore, in both theoretical and numerical aspects, very few results are known regarding the identification of coefficients in degenerate/singular parabolic equations, even though this class of operators occurs in interesting theoretical and applied problems. As far as we know, [15] is the unique published work on this subject; it concerns the reconstruction of the initial heat distribution in a degenerate/singular parabolic equation with degeneracy and singularity at the boundary of the domain.From the mathematical point of view and in connection with the work of Fragnelli and Mugnai (see [16]), we focus on identifying, on the basis of some observations, the source term, in a parabolic equation presenting both a degenerate diffusion coefficient and a singular potential with degeneracy and singularity inside the spatial domain.In particular, our results complement the ones of [1, 3] in the purely degenerate case and in the purely singular one, respectively. More precisely, we will follow the approach introduced in [2] for the treatment of uniformly parabolic problems which is based on the use of global Carleman estimates. For this purpose, we use and extend some recent Carleman estimates for degenerate/singular equations obtained by Fragnelli and Mugnai [16]. As a consequence, we prove a stability estimate of Lipschitz type in determining the term source using the following observations data: (4)auxx+λbuT′,·,andutt0,T×ω,where the subregion of measurements ω is a nonempty subinterval of (0,1) that is assumed to satisfy the following.Hypothesis 1.
The set of observationω⊂⊂(0,1) is such that (5)ω=ω1∪ω2, where ωi⊂(0,1), i=1,2 are intervals with ω1⊂⊂(0,x0), ω2⊂⊂(x0,1), and x0∉ω¯.For fixedT>T′>0, the main result of this paper can be stated as follows.Theorem 2.
LetC0>0 and suppose that Hypotheses 1 and 14 are satisfied. Then, there exists C=C(T,t0,x0,C0)>0 such that, for all h∈S(C0) and u0∈L2(0,1),(6)hL2Q2≤CauxxT′,·+λb·uT′,·L20,12+utL2t0,T×ω2.Remark 3.
(i)
It is worth noting that the result announced in Theorem2 is still valid also in the case in which the observation set ω is an interval containing the degeneracy point. Indeed, if x0∈ω one can always find two subintervals ω1⊂(0,x0), ω2=⊂(x0,1) such that (ω1∪ω2)⊂⊂ω∖{x0}. (ii)
If we restrict ourselves to the particular caseh∈{rg,g∈L2(0,1)} for some given function r∈C1([0,T]×[0,1]), positive at some time t=T′ and g is the unknown function that we want to recover, then uniqueness result can be shown as an immediate consequence of the Lipschitz stability result; see [17, Theorem 2.11].In fact, we will not only investigate the theoretical aspect of the inverse source problem due to our interest in mathematics, but also consider the numerical reconstruction of the source termh(t,x). To this end, we adopt the classical Tikhonov regularization to reformulate the inverse problem into a related optimization problem, for which we develop an iterative thresholding algorithm by using the corresponding adjoint system. In particular, we will focus on the determination of the unknown source term from the measured data at the final time. The resolution of this problem is standard and it is based on the gradient of the cost functional. More precisely, the most important issue in numerical solutions of inverse problems is the Lipschitz continuity of the Fréchet gradient. Indeed, in order to construct an effective minimization algorithm for an inverse problem, one needs to analyze the gradient of the considered cost functional. There is a vast literature on inverse problems for linear parabolic equations with final overdetermination. For example, we mention the pioneering work [18]. Compared to a standard parabolic equation, the main challenge here is the nonstandard degeneracy of the diffusion coefficient as well as the singularity of the potential of the partial differential equation (1).The rest of this article is organized as follows. In Section2, we recall the well-posedness of the problem (1). Then Section 3 is devoted to the proof of the main stability result of Lipschitz type. In Section 4, we reformulate our inverse source problem as a minimization problem with the Tikhonov regularization and provide a monotone iteration scheme based on a gradient method.Throughout the paper,C denotes a generic positive constant, which may vary from line to line.


## 2. Well-Posedness

The ways in whicha and b degenerate at x0 can be quite different, and for this reason, following [16], to establish our results, we give the following definitions and assumptions.Hypothesis 4 (double weakly degenerate case (WWD)).
There existsx0∈(0,1) such that a(x0)=b(x0)=0, a,b>0 in [0,1]∖{x0}, a,b∈C1([0,1]∖{x0}) and there exists K,L∈(0,1) such that x-x0a′≤Ka and x-x0b′≤Lb a.e. in [0,1].Hypothesis 5 (weakly strongly degenerate case (WSD)).
There existsx0∈(0,1) such that a(x0)=b(x0)=0, a,b>0 in [0,1]∖{x0}, a∈C1([0,1]∖{x0}), b∈C1([0,1]∖{x0})∩W1,∞(0,1),∃K∈(0,1),L∈[1,2) such that x-x0a′≤Ka and x-x0b′≤Lb a.e. in [0,1].Hypothesis 6 (strongly weakly degenerate case (SWD)).
There existsx0∈(0,1) such that a(x0)=b(x0)=0, a,b>0 in [0,1]∖{x0}, a∈C1([0,1]∖{x0})∩W1,∞(0,1), b∈C1([0,1]∖{x0}), ∃K∈[1,2), L∈(0,1) such that x-x0a′≤Ka and x-x0b′≤Lb a.e. in [0,1].Hypothesis 7 (double strongly degenerate case (SSD)).
There existsx0∈(0,1) such that a(x0)=b(x0)=0, a,b>0 in [0,1]∖{x0}, a,b∈C1([0,1]∖{x0})∩W1,∞(0,1), there exists K,L∈[1,2) such that x-x0a′≤Ka and x-x0b′≤Lb a.e. in [0,1].For the well-posedness of the problem (1), as in [16], we consider different classes of weighted Hilbert spaces, which are suitable to study the four different situations given above, namely, the (WWD), (WSD), (SWD), and (SSD) cases. Thus, we consider the Hilbert spaces (7)Ha10,1≔u∈W01,10,1:aux∈L20,1and (8)Ha,b10,1≔u∈Ha10,1:ub∈L20,1 endowed with the inner products (9)u,vHa1≔∫01au′v′dx+∫01uvdx, and (10)u,vHa,b1≔∫01au′v′dx+∫01uvdx+∫01uvbdx, respectively.In order to deal with the singularity ofb we need the following inequality proved in [16, Proposition 2.14].Lemma 8.
If one among Hypotheses4–6 holds with K+L≤2, then there exists a constant C>0 such that for all u∈Ha,b1(0,1) we have(11)∫01u2bxdx≤C∫01axu′2dx.In order to study well-posedness of problem (1) and in view of Lemma 8, we consider the space(12)H≔Ha,b10,1, where the Hardy-Poincaré inequality (11) holds.We underline that, from Lemma8, the standard norm ·H2 is equivalent to (13)·~2≔∫01au′2dx.From now on, we make the following assumptions ona, b, and λ.Hypothesis 9.
( 1 ) One among the Hypotheses 4, 5, or 6 holds true with K+L≤2 and we assume that(14)λ∈0,1C⋆.(2) Hypotheses 4, 5, 6, or 7 hold with λ<0.Using Lemma8, the next inequality is proved in [16, Proposition 2.18], which is crucial not only to obtain the well-posedness of problem (1), but also to prove that the inverse problem posed as weak solution minimization problem has a solution.Proposition 10.
Assume Hypothesis9. Then there exists a positive constant Λ∈(0,1] such that, for all u∈H, there holds(15)∫01au′2dx-λ∫01u2bdx≥Λ∫01au′2dx.Now, let us go back to problem (1), recalling the following definition.Definition 11.
Letu0∈L2(0,1) and h∈L2(Q). A function u is said to be a (weak) solution of (1) if (16)u∈C0,T;L20,1∩L20,T;Hand satisfies the following differential equation: (17)∫01uT,xφT,xdx-∫01u0xφ0,xdx-∬Qut,xφtt,xdxdt=-∬Qaxuxt,xφxt,xdxdt+λ∬Qut,xφt,xbdxdt+∬Qht,xφt,xdxdtfor all φ∈H1(0,T;L2(0,1))∩L2(0,T;H).Finally, we introduce the Hilbert space(18)Ha,b20,1≔u∈Ha10,1:au′∈H10,1andAu∈L20,1, where (19)Au≔au′′+λbu, with domain (20)DA≔Ha,b20,1.Remark 12.
Observe that ifu∈D(A), then u/b∈L2(0,1), so that u∈H and inequality (11) holds.Hence, the next result holds thanks to the theory of semigroups.Proposition 13.
The following assertions hold.
(i) The operator(A,D(A)) is the infinitesimal generator of a strongly continuous semigroup of contractions on L2(0,1). Moreover, the semigroup is analytic.
(ii) For allu0∈D(A) and h∈H1(0,T;L2(0,1)), problem (1) admits a unique strict solution belonging to the class (21)u∈C0,T;DA∩C10,T;L20,1. Moreover, if u0∈L2(0,1), then for all ε∈(0,T) there holds (22)u∈Cε,T;DA∩C1ε,T;L20,1.
(iii) For allu0∈L2(0,1) and for all h∈L2(0,T;L2(0,1)), problem (1) has a unique weak solution u∈C([0,T];L2(0,1))∩L2(0,T;H) such that for all ε∈(0,T) there holds (23)u∈L2ε,T;DA∩H1ε,T;L20,1. Moreover, if h∈H1(0,T;L2(0,1)) and ε∈(0,T), we have (24)u∈H1ε,T;DA∩H2ε,T;L20,1Proof.
The proof of statement(i) can be found in [16], whereas statements (ii) and (iii) are a consequence of (i) and [19, Proposition 3.3 and Proposition 3.8].


## 3. Lipschitz Stability Result

In this section, we aim at obtaining a Lipschitz stability result on determining the source termh(x,t) in problem (1) in the spirit of the result by Imanuvilov and Yamamoto [2]. The key ingredient to obtain such a result is Carleman estimates. Here we use specific Carleman estimates for degenerate/singular parabolic equations (inspired by [16]). Thus, we first recall this fundamental tool in the following section before proving Theorem 2 in Section 3.2.


### 3.1. Carleman Estimate

The aim of this subsection is to prove a Carleman type inequality for solutions of problem (1). First of all, let us make precise the assumptions under which we consider problem (1).Hypothesis 14.
( 1 ) Hypothesis 9 holds. Moreover, if K>4/3, then there exists ϑ∈(0,K] that the following condition is satisfied.
( C ) The function x↦a/x-x0ϑ is nonincreasing on the left of x=x0 and nondecreasing on the right of x=x0.
( 2 ) Moreover, if λ<0, we require that(25)x-x0b′x≥0in0,1.As usual, the derivation of global Carleman estimates relies on the introduction of some suitable weight function of the form(26)φt,x≔θtψx,∀t,x∈t0,T×-1,1, where(27)θt≔1t-t0T-t4,andψx≔c1∫x0xy-x0a~ydy-c2,with c2>c~2⋆≔sup-1,1∫x0xy-x0/a~ydy, c1>0, and where a~ is defined by the following way:(28)a~x≔ax,x∈0,1,a-x,x∈-1,0.Observe that θ(t)→+∞ast→t0+,T-, and clearly (29)-c1c2≤ψx<0for  everyx∈-1,1. Eventually, we define as in [3] the second time weight function:(30)ηt≔T+t0-2t,∀t∈t0,T.Let us now turn to the following linear initial-boundary value problem:(31)∂u∂t-auxx-λbxu=ht,x,t,x∈t0,T×0,1,u0=u1=0,t∈t0,T,where h∈L2(t0,T,L2(0,1)). In the following, we denote (32)Qt0T≔t0,T×0,1andωt0T≔t0,T×ω.Now we are ready to state global Carleman estimates with boundary observation for system (31).Theorem 15.
Assume Hypothesis14. Then, there exist two positive constants C and s0 such that the solution u of (31) in L2ε,T;Ha,b2∩H1[ε,T],H satisfies, for all s≥s0,(33)∬Qt0Tsθaxux2+s3θ3x-x02axu2+sθ3/2ηψu2+1sθut2e2sφdxdt≤C∬Qt0Th2e2sφdxdt+sc1∫t0Tθax-x0ux2e2sφx=0x=1dt.Some part of estimate (33) is already proved in [16] and, even if we refer to [20] a few times, our proof is quite self-contained. In [16], the authors prove a Carleman inequality that estimates the integrals of sθa(x)ux2 and s3θ3((x-x0)2/ax)u2 (that were sufficient for control purposes). For inverse problems, these estimates are not sufficient and one also needs some additional estimate of u with a special weight and some estimate of the derivative term ut that we added here in the statement of Theorem 15. The proof is based on the methods developed in [3].Proof.
The proof of Theorem15 relies on the change of variables w=esφu with s>0. Then, from (31), we obtain(34)-awxx-sφtw-s2aφx2w-λbw︸Ls+w+wt+2saφxwx+saφxxw︸Ls-w=hesφ︸hs. Moreover, w(t0,x)=w(T,x)=0. This property allows us to apply the Carleman estimates established in [16, Lemma 3.8] to w with Qt0T in place of (0,T)×(0,1)(35)Ls+wQt0T2+Ls-wQt0T2+∬Qt0Ts3θ3x-x02axw2+sθaxwx2dxdt≤C∬Qt0Th2e2sφdxdt+sc1∫t0Tθax-x0ux2e2sφx=0x=1dt.The operators Ls+ and Ls- are not exactly the ones of [16]. However, one can prove that the Carleman estimates do not change. Using the previous estimate, we aim at proving estimate (33) that concerns the variable u.
Step 1. Estimate of ∬Qt0Tsθa(x)ux2e2sφdxdt and ∬Qt0Ts3θ3(x-x0)2/axu2e2sφdxdt. Replacing w by esφu, we immediately get from (35)(36)∬Qt0Ts3θ3x-x02axu2e2sφdxdt≤C∬Qt0Th2e2sφdxdt+sc1∫t0Tθax-x0ux2e2sφx=0x=1dt.Moreover, wx=sφxuesφ+uxesφ. Therefore,(37)∬Qt0Tsθaux2e2sφdxdt≤2∬Qt0Tsθawx2dxdt+2∬Qt0Ts3θ3x-x02axu2e2sφdxdt.In conclusion, thanks to (35)-(37), we get(38)∬Qt0Tsθaxux2+s3θ3x-x02axu2e2sφdxdt≤C∬Qt0Th2e2sφdxdt+sc1∫t0Tθax-x0ux2e2sφx=0x=1dt.Step 2. Estimate ∬Qt0Tsθ3/2ηψu2e2sφdxdt. Observe that, since η≤T-t0 and ψ≤c1c2, one has(39)∬Qt0Tsθ3/2ηψw2dxdt≤C∬Qt0Tsθ3/2w2dxdt.To estimate the integral on the right-hand side of (39), we follow the technique of [20, Lemma 4.1]. Using the Young inequality, we find (40)∬Qt0Tsθ3/2w2dxdt=s∬Qt0Tθa1/3x-x02/3w23/4θ3x-x02aw21/4dxdt≤s34∬Qt0Tθa1/3x-x02/3w2dxdt+s14∬Qt0Tθ3x-x02aw2dxdt. Now, if K≤4/3, we consider the function p(x)=x-x04/3. Obviously, there exists q∈(1,4/3) such that the function p(x)/x-x0q is nonincreasing on the left of x0 and nondecreasing on the right of x0. Then, we can apply the Hardy-Poincaré inequality given in [20, Proposition 1.1], obtaining (41)∫01a1/3x-x02/3w2dx≤maxx∈0,1a1/3x∫011x-x02/3w2dx=maxx∈0,1⁡a1/3x∫01pxx-x02w2dx≤maxx∈0,1⁡a1/3xCHP∫01pxwx2dx=maxx∈0,1⁡a1/3xCHP∫01ax-x04/3awx2dx≤maxx∈0,1⁡a1/3xCHPC1∫01awx2dx, where CHP is the Hardy-Poincaré constant and C1=max(x04/3/a(0),(1-x0)4/3/a(1)).
In the previous inequality, we have used the property that the mapx↦x-x0γ/a(x) is nonincreasing on the left of x0 and nondecreasing on the right of x0 for all γ≥K; see [20, Lemma 2.1]. If K>4/3, we can consider the function p(x)=(a(x)x-x04)1/3. Then we have (42)a1/3x-x02/3=pxx-x02,px=axx-x02ax2/3≤C2ax,where (43)C2≔max⁡x02a02/3,1-x02a12/3.
Moreover, using the condition(C) given in Hypothesis 14, one has that the function p(x)/x-x0q, with q≔(4+ϑ)/3∈(1,2), is nonincreasing on the left of x0 and nondecreasing on the right of x0. The Hardy-Poincaré inequality given in [20, Proposition 1.1] implies (44)∫01a1/3x-x02/3w2dx=∫01px-x02w2dx≤CHP∫01pwx2dx≤CHPC2∫01awx2dx.Thus, in every case, for s large enough there holds(45)∬Qt0Tsθ3/2ηψw2dxdt≤C∬Qt0Tsθawx2+s3θ3x-x02aw2dxdt.Hence, (46)∬Qt0Tsθ3/2ηψw2dxdt≤C∬Qt0Th2e2sφdxdt+sc1∫t0Tθax-x0ux2e2sφx=0x=1dt. Finally, coming back to u, we get(47)∬Qt0Tsθ3/2ηψu2e2sφdxdt≤C∬Qt0Th2e2sφdxdt+sc1∫t0Tθax-x0ux2e2sφx=0x=1dt.Step 3. Estimate of ∬Qt0T(1/sθ)ut2e2sφdxdt. First of all, coming back to the definition of Ls-w, we have(48)1sθLs-w=1sθwt+2c1sθx-x0wx+c1sθw. Note that 1/θ is bounded on [t0,T] and x-x02/a(x)≤max⁡x02/a0,(1-x0)2/a1 (see [20, Lemma 2.1]). Therefore, there exists C>0 such that(49)∬Qt0T1sθwt2dxdt≤CLs-wQt0T2+∬Qt0Tsθx-x02aawx2dxdt+∬Qt0Tsθw2dxdt≤CLs-wQt0T2+∬Qt0Tsθawx2dxdt+∬Qt0Tsθw2dxdt.We then estimate ∬Qt0Tsθw2dxdt thanks to Hardy inequality, as we have done in the previous step in (45). In this way, we find (50)∬Qt0Tsθw2dxdt=s∬Qt0Tθa1/3x-x02/3w23/4θx-x02aw21/4dxdt≤s34∬Qt0Tθa1/3x-x02/3w2dxdt+s4∬Qt0Tθx-x02aw2dxdt≤C∫∫Qt0Tsθawx2+s3θ3x-x02aw2dxdt.Finally, using (49) and (35), one has(51)∬Qt0T1sθwt2dxdt≤C∬Qt0Th2e2sφdxdt+sc1∫t0Tθax-x0ux2e2sφx=0x=1dt.Now, in order to obtain the estimate of ut we have to use the estimate of (47). From the definition of w, we have wt=utesφ+sφtw. Hence (52)∬Qt0T1sθut2e2sφdxdt≤2∬Qt0T1sθwt2dxdt+∬Qt0Ts2φt2sθw2dxdt. The second term in the above right-hand side is estimated as follows: (53)∬Qt0Ts2φt2sθw2dxdt=16∬Qt0Tsθ3/2η2ψ2w2dxdt≤16T-t02c1c2∬Qt0Tsθ3/2ηψw2dxdt. Hence using (47) and (51) we conclude that(54)∬Qt0T1sθut2e2sφdxdt≤C∬Qt0Th2e2sφdxdt+sc1∫t0Tθax-x0ux2e2sφx=0x=1dt.Conclusion. We immediately deduce the expected Carleman estimate (33) from (38), (47), and (54).


### 3.2. Proof of Lipschitz Stability

The object of this subsection is to prove Theorem2 which recovers a source term h from the measured final data and the partial knowledge of ut over the subdomain ω⊂(0,1). In proving these kinds of stability estimates, the global Carleman estimate obtained in Theorem 15 will play a crucial part along with certain energy estimates.In order to obtain our main result, we need to define the following weights function associated with nondegenerate Carleman estimates in a general interval(A,B) which are suited to our purpose. For x∈[A,B],(55)Φt,x=θtΨx,Ψx=eρσx-e2ρσ∞,where ρ>0 and σ a C2([A,B]) function such that σ(x)>0 in (A,B), σ(A)=σ(B)=0 and σx(x)≠0 in [A,B]∖ω~, ω~ is an open subset of ω.Now, choose the constantc1 in (27) so that(56)c1≥e2ρσ∞-1c2-c~2⋆.Thus, one can show that weight functions satisfy the following properties which are needed in the sequel.Lemma 16.
For(t,x)∈[0,T]×[0,1], we have(57)φt,x≤Φt,xandφt,-x≤Φt,x.Proof.
First, let us setc2⋆≔sup[0,1]∫x0x(y-x0/a~y)dy.
(i) φ ≤ Φ. For all x∈[0,1], we have (58)-c1c2≤ψx≤c1c2⋆-c2,and1-e2ρσ∞≤Ψx≤eρσ∞-e2ρσ∞. Thus, to show that ψ≤Ψ, it suffices to have (59)c1c2⋆-c2≤1-e2ρσ∞. This means c1≥(e2ρσ∞-1)/(c2-c2⋆) and the conclusion follows immediately.
(ii) φ ( t , - x ) ≤ Φ ( t , x ). In a similar manner, using (56), one has max{ψ(-1),ψ(0)}≤Ψ(x) and the thesis follows.Proof of Theorem2.
Letu0∈L2(0,1), h∈S(C0), and t0∈(0,T). Observe that, according to Proposition 13(iii), the solution u of the problem (1) belongs to (60)H1t0,T;DA∩H2t0,T;L20,1.It follows that the solution u of (1) satisfies sufficient regularity properties to proceed to the following computations. Let z≔ut where u satisfies (1). Then z∈L2([t0,T];D(A))∩H1([t0,T];L2(0,1)) and satisfies(61)zt-azxx-λbxz=htt,x,t,x∈Q,z0=z1=0,t∈0,T,z0,x=au0xx+λbxu0+h0,x,x∈0,1.
Let us recall that our goal is to provide an estimate ofhL2(Q). For this purpose, we divide the proof into four steps and we recall that h∈H1(t0,T,L2(0,1)).
Step 1 (Carleman estimates with locally distributed observation). Such estimates are obtained by studying some auxiliary problems, introduced by means of a suitable cut-off argument. First, by the assumption on the observation set, we can fix two subintervals ω1≔(α1,β1)⊂⊂(0,x0), ω2≔(α2,β2)⊂⊂(x0,1) and four points γi,γi′∈(αi,βi), i=1,2, with γi<γi′. Then, fix β~2∈(α2,γ2) and consider a smooth function η:[0,1]→[0,1] such that(62)ηx=1,x∈γ2,1,0,x∈0,β~2. Let us now introduce Z=ηz where z is the solution of (61). Hence, fixed α~2∈(α2,β~2); Z is a solution of the nondegenerate nonsingular parabolic equation (63)Zt-aZxx-λbZ=H,t,x∈t0,T×α~2,1,Zt,α~2=Zt,1=0,t∈t0,T, with H≔ηht-(aηxz)x-aηxzx. Observe that, by the assumption on a and the fact that ηx is supported in ω˘≔(β~2,γ2)⊂⊂ω2, we have H∈L2((t0,T)×(α~2,1)).
Thus, we can apply the analogue of [6, Lemma 1.2] for Z in (α~2,1), obtaining that there exist two positive constants C and s0 (s0 sufficiently large), such that Z satisfies, for all s≥s0, (64)∫t0T∫α~211sθZt2+sθZx2+s3θ3Z2e2sΦdxdt≤C∫t0T∫α~21H2e2sΦdxdt+∬ωt0Ts3θ3Z2e2sΦdxdt. Using again the fact that ηx is supported in ω˘ and the boundedness of a (far away from x0 in the weakly degenerate case and since a∈W1,∞(0,1) in the strongly degenerate case), we have, by the Caccioppoli inequality for the nondegenerate case, (65)∫t0T∫α~21H2e2sΦdxdt≤C∫t0T∫α~21η2ht2+axηxzx+axηxzx2e2sΦdxdt≤C∫t0T∫α~21η2ht2e2sΦdxdt+∬ω˘t0z2+zx2e2sΦdxdt≤C∫t0T∫α~21η2ht2e2sΦdxdt+∬ω˘t0ht2+s2θ2z2e2sΦdxdt. Hence, we can choose s0 so large that for all s≥s0 and for a positive constant C, the following estimate holds:(66)∫t0T∫α~211sθZt2+sθZx2+s3θ3Z2e2sΦdxdt≤C∫t0T∫α~21η2ht2e2sΦdxdt+∬ω˘t0ht2+s3θ3z2e2sΦdxdt.Furthermore, using the fact that sθ3/2ηψZ2≤Cs3θ3Z2, by (66) one has(67)∫t0T∫α~21sθ3/2ηψZ2e2sΦdxdt≤C∫t0T∫α~21η2ht2e2sΦdxdt+∬ω˘t0ht2+s3θ3z2e2sΦdxdt.The estimates (66)-(67) lead to(68)∫t0T∫α~211sθZt2+sθ3/2ηψZ2+sθZx2+s3θ3Z2e2sΦdxdt≤C∫t0T∫α~21η2ht2e2sΦdxdt+∬ω˘t0ht2+s3θ3z2e2sΦdxdt.Now, since x0∉(α~2,1), by Lemma 16 we can prove that there exists a positive constant k such that for every (t,x)∈[0,T]×[α~2,1](69)axe2sφt,x≤ke2sΦandx-x02axe2sφt,x≤ke2sΦ.Then, using (69), (68) becomes (70)∫t0T∫α~211sθZt2+sθ3/2ηψZ2+sθaZx2+s3θ3x-x02aZ2e2sφdxdt≤k∫t0T∫α~211sθZt2+sθ3/2ηψZ2+sθZx2+s3θ3Z2e2sΦdxdt≤kC∫t0T∫α~21η2ht2e2sΦdxdt+∫∫ω˘t0ht2+s3θ3z2e2sΦdxdt. Hence, using the definition of Z, it results(71)∫t0T∫γ211sθzt2+sθ3/2ηψz2+sθazx2+s3θ3x-x02az2e2sφdxdt=∫t0T∫γ211sθZt2+sθ3/2ηψZ2+sθaZx2+s3θ3x-x02aZ2e2sφdxdt≤∫t0T∫α~211sθZt2+sθ3/2ηψZ2+sθaZx2+s3θ3x-x02aZ2e2sφdxdt≤C∬Qt0Tht2e2sΦdxdt+∬ωt0Tht2+s3θ3z2e2sΦdxdt≤C∬Qt0Tht2e2sΦdxdt+∬ωt0Ts3θ3z2e2sΦdxdt,for a positive constant C and for s large enough.
To complete the proof, it is sufficient to prove a similar inequality on the interval[0,γ2]. To this aim, we follow a reflection procedure. Consider the function (72)wt,x≔zt,x,x∈0,1,-zt,-x,x∈-1,0. Therefore, w solves the problem(73)wt-a~wxx-λb~w+=G,t,x∈0,T×-1,1,wt,-1=wt,1=0,t∈0,T,being (74)b~x≔bx,x∈0,1,b-x,x∈-1,0,G=htx,x∈0,1,ht-x,x∈-1,0. Now, consider a smooth function τ:[-1,1]→[0,1] such that(75)τx=1,x∈-γ1,γ2,0,x∈-1,-γ1′∪γ2′,1, and define the function W=τw, where w is the solution of (73). Then W solves (76)Wt-a~Wxx-λb~W=τG-a~τxwx-a~τxwx≔F,t,x∈0,T×-β1,1,Wt,-β1=Wt,1=0,t∈0,T.Observe that Wx(t,-β1)=Wx(t,1)=0 and, by the assumption on a and the fact that τx is supported in [-γ1′,-γ1]∪[γ2,γ2′], F∈L2((0,T)×I), where I≔(-β1,1). Thus, we can apply the analogue of Theorem 15 (which still holds true, since a~ belongs to W1,1(-1,1) in the weakly degenerate case and to W1,∞(-1,1) in the strongly degenerate one) on (-β1,1) in place of (0,1), obtaining that there exist two positive constants C and s0 (s0 sufficiently large), such that W satisfies, for all s≥s0, (77)∫t0T∫-β11sθa~Wx2+s3θ3x-x02a~W2+sθ3/2ηψW2+1sθWt2e2sφdxdt≤C∫t0T∫-β11F2e2sφdxdt. Using again the fact that τx is supported in [-γ1′,-γ1]∪γ2,γ2′ and the boundedness of a~ and a~′ (recall that, using the assumption on a, a~ is C1 far away from x0 and 0 in the weakly degenerate case and it is W1,∞(-1,1) in the strongly degenerate one), it follows that(78)
∫
t
0
T
∫
-
β
1
1
s
θ
a
~
W
x
2
+
s
3
θ
3
x
-
x
0
2
a
~
W
2
+
s
θ
3
/
2
η
ψ
W
2
+
1
s
θ
W
t
2
e
2
s
φ
d
x
d
t
≤
C
∫
t
0
T
∫
-
β
1
1
τ
2
G
2
e
2
s
φ
d
x
d
t
+
∫
t
0
T
∫
-
γ
1
′
-
γ
1
​
+
∫
γ
2
γ
2
′
​
w
2
+
w
x
2
e
2
s
φ
d
x
d
t
.Now, by the definitions of w, we note that (79)∫t0T∫-γ1′-γ1w2+wx2e2sφxdxdt=∫t0T∫-γ1′-γ1z2-x+zx2-xe2sφxdxdt. Using a change of variable, by (57), one has (80)∫t0T∫-γ1′-γ1z2-x+zx2-xe2sφxdxdt=∫t0T∫γ1γ1′z2x+zx2xe2sφ-xdxdt≤∫t0T∫γ1γ1′z2x+zx2xe2sΦxdxdt. Consequently,(81)∫t0T∫-γ1′-γ1w2+wx2e2sφxdxdt≤∫t0T∫γ1γ1′z2+zx2e2sΦxdxdt.Going back to (78), by (81), and using the fact that φ≤Φ, we obtain(82)
∫
t
0
T
∫
-
β
1
1
s
θ
a
~
W
x
2
+
s
3
θ
3
x
-
x
0
2
a
~
W
2
+
s
θ
3
/
2
η
ψ
W
2
+
1
s
θ
W
t
2
e
2
s
φ
d
x
d
t
≤
C
∫
t
0
T
∫
-
β
1
1
τ
2
G
2
e
2
s
Φ
d
x
d
t
+
∫
t
0
T
∫
γ
1
γ
1
′
​
+
∫
γ
2
γ
2
′
​
z
2
+
z
x
2
e
2
s
Φ
d
x
d
t
.Thus, applying the Caccioppoli inequality and recalling that G is defined through ht, one can find C>0 such that (83)
∫
t
0
T
∫
-
β
1
1
s
θ
a
~
W
x
2
+
s
3
θ
3
x
-
x
0
2
a
~
W
2
+
s
θ
3
/
2
η
ψ
W
2
+
1
s
θ
W
t
2
e
2
s
φ
d
x
d
t
≤
C
∫
t
0
T
∫
-
β
1
1
τ
2
G
2
e
2
s
Φ
d
x
d
t
+
∫
t
0
T
∫
γ
1
γ
1
′
​
+
∫
γ
2
γ
2
′
​
z
2
e
2
s
Φ
d
x
d
t
+
∫
t
0
T
∫
ω
1
​
+
∫
ω
2
​
h
t
2
+
s
2
θ
2
z
2
e
2
s
Φ
d
x
d
t
≤
C
∬
Q
t
0
T
h
t
2
e
2
s
Φ
d
x
d
t
+
∫
t
0
T
∫
ω
1
​
+
∫
ω
2
​
h
t
2
+
s
2
θ
2
z
2
e
2
s
Φ
d
x
d
t
≤
C
∬
Q
t
0
T
h
t
2
e
2
s
Φ
d
x
d
t
+
∬
ω
t
0
T
s
2
θ
2
z
2
e
2
s
Φ
d
x
d
t
. Hence, using the definitions of w and W, there exists a constant C>0 such that(84)∫t0T∫0γ2sθazx2+s3θ3x-x02az2+sθ3/2ηψz2+1sθzt2e2sφdxdt=∫t0T∫0γ2sθa~wx2+s3θ3x-x02a~w2+sθ3/2ηψw2+1sθwt2e2sφdxdt=∫t0T∫0γ2sθa~Wx2+s3θ3x-x02a~W2+sθ3/2ηψW2+1sθWt2e2sφdxdt≤∫t0T∫-β11sθa~Wx2+s3θ3x-x02a~W2+sθ3/2ηψW2+1sθWt2e2sφdxdt≤C∬Qt0Tht2e2sΦdxdt+∬ωt0Ts2θ2z2e2sΦdxdt.Finally adding up (71) and (84), we conclude that(85)I0≔∬Qt0Tsθazx2+s3θ3x-x02az2+sθ3/2ηψz2+1sθzt2e2sφdxdt≤C∬Qt0Tht2e2sΦdxdt+∬ωt0Ts3θ3z2e2sΦdxdt︸I1,for all s≥s0 and for a positive constant C.
Step 2 (estimate from above of I1). In this step, our purpose is to show that there exists some constant C>0 such that(86)I1≤C1s∫01h2T′,xe2sΦT′,xdx+utL2ωt0T2.Let us recall that, by the definition of Ψ, there exists some constant p<0 such that Ψ(x)≤p for all x∈[0,1]. Therefore we have (87)s3θ3e2sΦ≤s3θ3e2spθ→0ast→t0orT.As a consequence, setting C=supR{y↦y3e2spy}>0, we have(88)∬ωt0Ts3θ3z2e2sΦdxdt≤CzL2ωt0T2=utL2ωt0T2.In order to complete the proof of (86), it remains to prove the following lemma.
Lemma  17.There exists a constantC>0such that(89)∬Qt0Tht2e2sΦdxdt≤Cs∫01h2T′,xe2sΦT′,xdx.
We omit the proof of Lemma 17 which is classical and we refer the reader to [2]. Using (88) and (89), we obtain (86).
Step 3 (estimate from below of I0). The purpose of the step is to provide the following estimate: there exists a constant C>0 such that(90)∫01z2T′,xe2sφT′,xdx≤CI0.Since limt→t0z2(t,x)e2sφ(t,x)=0, for a.e. x∈(0,1), we can write(91)∫01z2T′,xe2sφT′,xdx=∫t0T′∂∂t∫01z2t,xe2sφt,xdxdt=∫t0T′∫012zzt+2sφtz2e2sφt,xdxdt.Using Young’s inequality, one has(92)∫t0T′∫012zzte2sφdxdt=∫t0T′∫012sθzesφ1sθztesφdxdt≤∬Qt0Tsθz2e2sφ+1sθzt2e2sφdxdt≤∬Qt0Tsθz2e2sφdxdt+I0.Moreover, by the Hardy-Poincaré inequality, and using the same computation as in the proof of (45), we get(93)∫t0T′∫012zzte2sφdxdt≤C∬Qt0Tsθaxzx2+s3θ3x-x02axz2e2sφdxdt+I0≤CI0.Next, using the fact that φt≤Cηψθ3/2, we have(94)∫t0T′∫01sφtz2e2sφdxdt≤C∬Qt0Tsθ3/2ηψz2e2sφdxdt≤CI0.Eventually, (91) associated with (92) and (93) gives (90).
Step 4 (conclusion). Using (90), (85), and next (86), there exists some constant C>0 such that(95)∫01z2T′,xe2sφT′,xdxdt≤C1s∫01h2T′,xe2sΦT′,xdx+utL2ωt0T2.On the other hand, let us recall that z satisfies (96)zT′,x=utT′,x=auxxT′,x+λbuT′,x+hT′,xfor  a.e.x∈0,1. Therefore,(97)∫01h2T′,xe2sφT′,xdx≤CauxxT′,x+λbuT′,xL20,12+∫01z2T′,xe2sφT′,xdx.Applying (95) to estimate the term ∫01z2(T′,x)e2sφ(T′,x)dx, we obtain (98)∫01h2T′,xe2sφT′,xdx≤CauxxT′,x+λbuT′,xL20,12+utL2ωt0T2+1s∫01h2T′,xe2sΦT′,xdx. Choosing s large enough such that C/s=1/2, we get (99)∫01h2T′,xe2sφT′,xdx≤CauxxT′,x+λbuT′,xL20,12+utL2ωt0T2. On the other hand, s being now fixed, there exists some constant γ>0 such that (100)e2sφT′,x≥γ>0. So we can write (101)γ∫01h2T′,xdx≤CauxxT′,x+λbuT′,xL20,12+utL2ωt0T2. Moreover, in view of (2),(102)ht,x≤hT′,x+∫T′thts,xds≤ChT′,x. Hence, (103)hL2Q2≤C∬Qh2T′,xdxdt=CT∫01h2T′,xdx≤CauxxT′,x+λbuT′,xL20,12+utL2ωt0T2, which concludes the proof.


## 3.1. Carleman Estimate

The aim of this subsection is to prove a Carleman type inequality for solutions of problem (1). First of all, let us make precise the assumptions under which we consider problem (1).Hypothesis 14.
( 1 ) Hypothesis 9 holds. Moreover, if K>4/3, then there exists ϑ∈(0,K] that the following condition is satisfied.
( C ) The function x↦a/x-x0ϑ is nonincreasing on the left of x=x0 and nondecreasing on the right of x=x0.
( 2 ) Moreover, if λ<0, we require that(25)x-x0b′x≥0in0,1.As usual, the derivation of global Carleman estimates relies on the introduction of some suitable weight function of the form(26)φt,x≔θtψx,∀t,x∈t0,T×-1,1, where(27)θt≔1t-t0T-t4,andψx≔c1∫x0xy-x0a~ydy-c2,with c2>c~2⋆≔sup-1,1∫x0xy-x0/a~ydy, c1>0, and where a~ is defined by the following way:(28)a~x≔ax,x∈0,1,a-x,x∈-1,0.Observe that θ(t)→+∞ast→t0+,T-, and clearly (29)-c1c2≤ψx<0for  everyx∈-1,1. Eventually, we define as in [3] the second time weight function:(30)ηt≔T+t0-2t,∀t∈t0,T.Let us now turn to the following linear initial-boundary value problem:(31)∂u∂t-auxx-λbxu=ht,x,t,x∈t0,T×0,1,u0=u1=0,t∈t0,T,where h∈L2(t0,T,L2(0,1)). In the following, we denote (32)Qt0T≔t0,T×0,1andωt0T≔t0,T×ω.Now we are ready to state global Carleman estimates with boundary observation for system (31).Theorem 15.
Assume Hypothesis14. Then, there exist two positive constants C and s0 such that the solution u of (31) in L2ε,T;Ha,b2∩H1[ε,T],H satisfies, for all s≥s0,(33)∬Qt0Tsθaxux2+s3θ3x-x02axu2+sθ3/2ηψu2+1sθut2e2sφdxdt≤C∬Qt0Th2e2sφdxdt+sc1∫t0Tθax-x0ux2e2sφx=0x=1dt.Some part of estimate (33) is already proved in [16] and, even if we refer to [20] a few times, our proof is quite self-contained. In [16], the authors prove a Carleman inequality that estimates the integrals of sθa(x)ux2 and s3θ3((x-x0)2/ax)u2 (that were sufficient for control purposes). For inverse problems, these estimates are not sufficient and one also needs some additional estimate of u with a special weight and some estimate of the derivative term ut that we added here in the statement of Theorem 15. The proof is based on the methods developed in [3].Proof.
The proof of Theorem15 relies on the change of variables w=esφu with s>0. Then, from (31), we obtain(34)-awxx-sφtw-s2aφx2w-λbw︸Ls+w+wt+2saφxwx+saφxxw︸Ls-w=hesφ︸hs. Moreover, w(t0,x)=w(T,x)=0. This property allows us to apply the Carleman estimates established in [16, Lemma 3.8] to w with Qt0T in place of (0,T)×(0,1)(35)Ls+wQt0T2+Ls-wQt0T2+∬Qt0Ts3θ3x-x02axw2+sθaxwx2dxdt≤C∬Qt0Th2e2sφdxdt+sc1∫t0Tθax-x0ux2e2sφx=0x=1dt.The operators Ls+ and Ls- are not exactly the ones of [16]. However, one can prove that the Carleman estimates do not change. Using the previous estimate, we aim at proving estimate (33) that concerns the variable u.
Step 1. Estimate of ∬Qt0Tsθa(x)ux2e2sφdxdt and ∬Qt0Ts3θ3(x-x0)2/axu2e2sφdxdt. Replacing w by esφu, we immediately get from (35)(36)∬Qt0Ts3θ3x-x02axu2e2sφdxdt≤C∬Qt0Th2e2sφdxdt+sc1∫t0Tθax-x0ux2e2sφx=0x=1dt.Moreover, wx=sφxuesφ+uxesφ. Therefore,(37)∬Qt0Tsθaux2e2sφdxdt≤2∬Qt0Tsθawx2dxdt+2∬Qt0Ts3θ3x-x02axu2e2sφdxdt.In conclusion, thanks to (35)-(37), we get(38)∬Qt0Tsθaxux2+s3θ3x-x02axu2e2sφdxdt≤C∬Qt0Th2e2sφdxdt+sc1∫t0Tθax-x0ux2e2sφx=0x=1dt.Step 2. Estimate ∬Qt0Tsθ3/2ηψu2e2sφdxdt. Observe that, since η≤T-t0 and ψ≤c1c2, one has(39)∬Qt0Tsθ3/2ηψw2dxdt≤C∬Qt0Tsθ3/2w2dxdt.To estimate the integral on the right-hand side of (39), we follow the technique of [20, Lemma 4.1]. Using the Young inequality, we find (40)∬Qt0Tsθ3/2w2dxdt=s∬Qt0Tθa1/3x-x02/3w23/4θ3x-x02aw21/4dxdt≤s34∬Qt0Tθa1/3x-x02/3w2dxdt+s14∬Qt0Tθ3x-x02aw2dxdt. Now, if K≤4/3, we consider the function p(x)=x-x04/3. Obviously, there exists q∈(1,4/3) such that the function p(x)/x-x0q is nonincreasing on the left of x0 and nondecreasing on the right of x0. Then, we can apply the Hardy-Poincaré inequality given in [20, Proposition 1.1], obtaining (41)∫01a1/3x-x02/3w2dx≤maxx∈0,1a1/3x∫011x-x02/3w2dx=maxx∈0,1⁡a1/3x∫01pxx-x02w2dx≤maxx∈0,1⁡a1/3xCHP∫01pxwx2dx=maxx∈0,1⁡a1/3xCHP∫01ax-x04/3awx2dx≤maxx∈0,1⁡a1/3xCHPC1∫01awx2dx, where CHP is the Hardy-Poincaré constant and C1=max(x04/3/a(0),(1-x0)4/3/a(1)).
In the previous inequality, we have used the property that the mapx↦x-x0γ/a(x) is nonincreasing on the left of x0 and nondecreasing on the right of x0 for all γ≥K; see [20, Lemma 2.1]. If K>4/3, we can consider the function p(x)=(a(x)x-x04)1/3. Then we have (42)a1/3x-x02/3=pxx-x02,px=axx-x02ax2/3≤C2ax,where (43)C2≔max⁡x02a02/3,1-x02a12/3.
Moreover, using the condition(C) given in Hypothesis 14, one has that the function p(x)/x-x0q, with q≔(4+ϑ)/3∈(1,2), is nonincreasing on the left of x0 and nondecreasing on the right of x0. The Hardy-Poincaré inequality given in [20, Proposition 1.1] implies (44)∫01a1/3x-x02/3w2dx=∫01px-x02w2dx≤CHP∫01pwx2dx≤CHPC2∫01awx2dx.Thus, in every case, for s large enough there holds(45)∬Qt0Tsθ3/2ηψw2dxdt≤C∬Qt0Tsθawx2+s3θ3x-x02aw2dxdt.Hence, (46)∬Qt0Tsθ3/2ηψw2dxdt≤C∬Qt0Th2e2sφdxdt+sc1∫t0Tθax-x0ux2e2sφx=0x=1dt. Finally, coming back to u, we get(47)∬Qt0Tsθ3/2ηψu2e2sφdxdt≤C∬Qt0Th2e2sφdxdt+sc1∫t0Tθax-x0ux2e2sφx=0x=1dt.Step 3. Estimate of ∬Qt0T(1/sθ)ut2e2sφdxdt. First of all, coming back to the definition of Ls-w, we have(48)1sθLs-w=1sθwt+2c1sθx-x0wx+c1sθw. Note that 1/θ is bounded on [t0,T] and x-x02/a(x)≤max⁡x02/a0,(1-x0)2/a1 (see [20, Lemma 2.1]). Therefore, there exists C>0 such that(49)∬Qt0T1sθwt2dxdt≤CLs-wQt0T2+∬Qt0Tsθx-x02aawx2dxdt+∬Qt0Tsθw2dxdt≤CLs-wQt0T2+∬Qt0Tsθawx2dxdt+∬Qt0Tsθw2dxdt.We then estimate ∬Qt0Tsθw2dxdt thanks to Hardy inequality, as we have done in the previous step in (45). In this way, we find (50)∬Qt0Tsθw2dxdt=s∬Qt0Tθa1/3x-x02/3w23/4θx-x02aw21/4dxdt≤s34∬Qt0Tθa1/3x-x02/3w2dxdt+s4∬Qt0Tθx-x02aw2dxdt≤C∫∫Qt0Tsθawx2+s3θ3x-x02aw2dxdt.Finally, using (49) and (35), one has(51)∬Qt0T1sθwt2dxdt≤C∬Qt0Th2e2sφdxdt+sc1∫t0Tθax-x0ux2e2sφx=0x=1dt.Now, in order to obtain the estimate of ut we have to use the estimate of (47). From the definition of w, we have wt=utesφ+sφtw. Hence (52)∬Qt0T1sθut2e2sφdxdt≤2∬Qt0T1sθwt2dxdt+∬Qt0Ts2φt2sθw2dxdt. The second term in the above right-hand side is estimated as follows: (53)∬Qt0Ts2φt2sθw2dxdt=16∬Qt0Tsθ3/2η2ψ2w2dxdt≤16T-t02c1c2∬Qt0Tsθ3/2ηψw2dxdt. Hence using (47) and (51) we conclude that(54)∬Qt0T1sθut2e2sφdxdt≤C∬Qt0Th2e2sφdxdt+sc1∫t0Tθax-x0ux2e2sφx=0x=1dt.Conclusion. We immediately deduce the expected Carleman estimate (33) from (38), (47), and (54).


## 3.2. Proof of Lipschitz Stability

The object of this subsection is to prove Theorem2 which recovers a source term h from the measured final data and the partial knowledge of ut over the subdomain ω⊂(0,1). In proving these kinds of stability estimates, the global Carleman estimate obtained in Theorem 15 will play a crucial part along with certain energy estimates.In order to obtain our main result, we need to define the following weights function associated with nondegenerate Carleman estimates in a general interval(A,B) which are suited to our purpose. For x∈[A,B],(55)Φt,x=θtΨx,Ψx=eρσx-e2ρσ∞,where ρ>0 and σ a C2([A,B]) function such that σ(x)>0 in (A,B), σ(A)=σ(B)=0 and σx(x)≠0 in [A,B]∖ω~, ω~ is an open subset of ω.Now, choose the constantc1 in (27) so that(56)c1≥e2ρσ∞-1c2-c~2⋆.Thus, one can show that weight functions satisfy the following properties which are needed in the sequel.Lemma 16.
For(t,x)∈[0,T]×[0,1], we have(57)φt,x≤Φt,xandφt,-x≤Φt,x.Proof.
First, let us setc2⋆≔sup[0,1]∫x0x(y-x0/a~y)dy.
(i) φ ≤ Φ. For all x∈[0,1], we have (58)-c1c2≤ψx≤c1c2⋆-c2,and1-e2ρσ∞≤Ψx≤eρσ∞-e2ρσ∞. Thus, to show that ψ≤Ψ, it suffices to have (59)c1c2⋆-c2≤1-e2ρσ∞. This means c1≥(e2ρσ∞-1)/(c2-c2⋆) and the conclusion follows immediately.
(ii) φ ( t , - x ) ≤ Φ ( t , x ). In a similar manner, using (56), one has max{ψ(-1),ψ(0)}≤Ψ(x) and the thesis follows.Proof of Theorem2.
Letu0∈L2(0,1), h∈S(C0), and t0∈(0,T). Observe that, according to Proposition 13(iii), the solution u of the problem (1) belongs to (60)H1t0,T;DA∩H2t0,T;L20,1.It follows that the solution u of (1) satisfies sufficient regularity properties to proceed to the following computations. Let z≔ut where u satisfies (1). Then z∈L2([t0,T];D(A))∩H1([t0,T];L2(0,1)) and satisfies(61)zt-azxx-λbxz=htt,x,t,x∈Q,z0=z1=0,t∈0,T,z0,x=au0xx+λbxu0+h0,x,x∈0,1.
Let us recall that our goal is to provide an estimate ofhL2(Q). For this purpose, we divide the proof into four steps and we recall that h∈H1(t0,T,L2(0,1)).
Step 1 (Carleman estimates with locally distributed observation). Such estimates are obtained by studying some auxiliary problems, introduced by means of a suitable cut-off argument. First, by the assumption on the observation set, we can fix two subintervals ω1≔(α1,β1)⊂⊂(0,x0), ω2≔(α2,β2)⊂⊂(x0,1) and four points γi,γi′∈(αi,βi), i=1,2, with γi<γi′. Then, fix β~2∈(α2,γ2) and consider a smooth function η:[0,1]→[0,1] such that(62)ηx=1,x∈γ2,1,0,x∈0,β~2. Let us now introduce Z=ηz where z is the solution of (61). Hence, fixed α~2∈(α2,β~2); Z is a solution of the nondegenerate nonsingular parabolic equation (63)Zt-aZxx-λbZ=H,t,x∈t0,T×α~2,1,Zt,α~2=Zt,1=0,t∈t0,T, with H≔ηht-(aηxz)x-aηxzx. Observe that, by the assumption on a and the fact that ηx is supported in ω˘≔(β~2,γ2)⊂⊂ω2, we have H∈L2((t0,T)×(α~2,1)).
Thus, we can apply the analogue of [6, Lemma 1.2] for Z in (α~2,1), obtaining that there exist two positive constants C and s0 (s0 sufficiently large), such that Z satisfies, for all s≥s0, (64)∫t0T∫α~211sθZt2+sθZx2+s3θ3Z2e2sΦdxdt≤C∫t0T∫α~21H2e2sΦdxdt+∬ωt0Ts3θ3Z2e2sΦdxdt. Using again the fact that ηx is supported in ω˘ and the boundedness of a (far away from x0 in the weakly degenerate case and since a∈W1,∞(0,1) in the strongly degenerate case), we have, by the Caccioppoli inequality for the nondegenerate case, (65)∫t0T∫α~21H2e2sΦdxdt≤C∫t0T∫α~21η2ht2+axηxzx+axηxzx2e2sΦdxdt≤C∫t0T∫α~21η2ht2e2sΦdxdt+∬ω˘t0z2+zx2e2sΦdxdt≤C∫t0T∫α~21η2ht2e2sΦdxdt+∬ω˘t0ht2+s2θ2z2e2sΦdxdt. Hence, we can choose s0 so large that for all s≥s0 and for a positive constant C, the following estimate holds:(66)∫t0T∫α~211sθZt2+sθZx2+s3θ3Z2e2sΦdxdt≤C∫t0T∫α~21η2ht2e2sΦdxdt+∬ω˘t0ht2+s3θ3z2e2sΦdxdt.Furthermore, using the fact that sθ3/2ηψZ2≤Cs3θ3Z2, by (66) one has(67)∫t0T∫α~21sθ3/2ηψZ2e2sΦdxdt≤C∫t0T∫α~21η2ht2e2sΦdxdt+∬ω˘t0ht2+s3θ3z2e2sΦdxdt.The estimates (66)-(67) lead to(68)∫t0T∫α~211sθZt2+sθ3/2ηψZ2+sθZx2+s3θ3Z2e2sΦdxdt≤C∫t0T∫α~21η2ht2e2sΦdxdt+∬ω˘t0ht2+s3θ3z2e2sΦdxdt.Now, since x0∉(α~2,1), by Lemma 16 we can prove that there exists a positive constant k such that for every (t,x)∈[0,T]×[α~2,1](69)axe2sφt,x≤ke2sΦandx-x02axe2sφt,x≤ke2sΦ.Then, using (69), (68) becomes (70)∫t0T∫α~211sθZt2+sθ3/2ηψZ2+sθaZx2+s3θ3x-x02aZ2e2sφdxdt≤k∫t0T∫α~211sθZt2+sθ3/2ηψZ2+sθZx2+s3θ3Z2e2sΦdxdt≤kC∫t0T∫α~21η2ht2e2sΦdxdt+∫∫ω˘t0ht2+s3θ3z2e2sΦdxdt. Hence, using the definition of Z, it results(71)∫t0T∫γ211sθzt2+sθ3/2ηψz2+sθazx2+s3θ3x-x02az2e2sφdxdt=∫t0T∫γ211sθZt2+sθ3/2ηψZ2+sθaZx2+s3θ3x-x02aZ2e2sφdxdt≤∫t0T∫α~211sθZt2+sθ3/2ηψZ2+sθaZx2+s3θ3x-x02aZ2e2sφdxdt≤C∬Qt0Tht2e2sΦdxdt+∬ωt0Tht2+s3θ3z2e2sΦdxdt≤C∬Qt0Tht2e2sΦdxdt+∬ωt0Ts3θ3z2e2sΦdxdt,for a positive constant C and for s large enough.
To complete the proof, it is sufficient to prove a similar inequality on the interval[0,γ2]. To this aim, we follow a reflection procedure. Consider the function (72)wt,x≔zt,x,x∈0,1,-zt,-x,x∈-1,0. Therefore, w solves the problem(73)wt-a~wxx-λb~w+=G,t,x∈0,T×-1,1,wt,-1=wt,1=0,t∈0,T,being (74)b~x≔bx,x∈0,1,b-x,x∈-1,0,G=htx,x∈0,1,ht-x,x∈-1,0. Now, consider a smooth function τ:[-1,1]→[0,1] such that(75)τx=1,x∈-γ1,γ2,0,x∈-1,-γ1′∪γ2′,1, and define the function W=τw, where w is the solution of (73). Then W solves (76)Wt-a~Wxx-λb~W=τG-a~τxwx-a~τxwx≔F,t,x∈0,T×-β1,1,Wt,-β1=Wt,1=0,t∈0,T.Observe that Wx(t,-β1)=Wx(t,1)=0 and, by the assumption on a and the fact that τx is supported in [-γ1′,-γ1]∪[γ2,γ2′], F∈L2((0,T)×I), where I≔(-β1,1). Thus, we can apply the analogue of Theorem 15 (which still holds true, since a~ belongs to W1,1(-1,1) in the weakly degenerate case and to W1,∞(-1,1) in the strongly degenerate one) on (-β1,1) in place of (0,1), obtaining that there exist two positive constants C and s0 (s0 sufficiently large), such that W satisfies, for all s≥s0, (77)∫t0T∫-β11sθa~Wx2+s3θ3x-x02a~W2+sθ3/2ηψW2+1sθWt2e2sφdxdt≤C∫t0T∫-β11F2e2sφdxdt. Using again the fact that τx is supported in [-γ1′,-γ1]∪γ2,γ2′ and the boundedness of a~ and a~′ (recall that, using the assumption on a, a~ is C1 far away from x0 and 0 in the weakly degenerate case and it is W1,∞(-1,1) in the strongly degenerate one), it follows that(78)
∫
t
0
T
∫
-
β
1
1
s
θ
a
~
W
x
2
+
s
3
θ
3
x
-
x
0
2
a
~
W
2
+
s
θ
3
/
2
η
ψ
W
2
+
1
s
θ
W
t
2
e
2
s
φ
d
x
d
t
≤
C
∫
t
0
T
∫
-
β
1
1
τ
2
G
2
e
2
s
φ
d
x
d
t
+
∫
t
0
T
∫
-
γ
1
′
-
γ
1
​
+
∫
γ
2
γ
2
′
​
w
2
+
w
x
2
e
2
s
φ
d
x
d
t
.Now, by the definitions of w, we note that (79)∫t0T∫-γ1′-γ1w2+wx2e2sφxdxdt=∫t0T∫-γ1′-γ1z2-x+zx2-xe2sφxdxdt. Using a change of variable, by (57), one has (80)∫t0T∫-γ1′-γ1z2-x+zx2-xe2sφxdxdt=∫t0T∫γ1γ1′z2x+zx2xe2sφ-xdxdt≤∫t0T∫γ1γ1′z2x+zx2xe2sΦxdxdt. Consequently,(81)∫t0T∫-γ1′-γ1w2+wx2e2sφxdxdt≤∫t0T∫γ1γ1′z2+zx2e2sΦxdxdt.Going back to (78), by (81), and using the fact that φ≤Φ, we obtain(82)
∫
t
0
T
∫
-
β
1
1
s
θ
a
~
W
x
2
+
s
3
θ
3
x
-
x
0
2
a
~
W
2
+
s
θ
3
/
2
η
ψ
W
2
+
1
s
θ
W
t
2
e
2
s
φ
d
x
d
t
≤
C
∫
t
0
T
∫
-
β
1
1
τ
2
G
2
e
2
s
Φ
d
x
d
t
+
∫
t
0
T
∫
γ
1
γ
1
′
​
+
∫
γ
2
γ
2
′
​
z
2
+
z
x
2
e
2
s
Φ
d
x
d
t
.Thus, applying the Caccioppoli inequality and recalling that G is defined through ht, one can find C>0 such that (83)
∫
t
0
T
∫
-
β
1
1
s
θ
a
~
W
x
2
+
s
3
θ
3
x
-
x
0
2
a
~
W
2
+
s
θ
3
/
2
η
ψ
W
2
+
1
s
θ
W
t
2
e
2
s
φ
d
x
d
t
≤
C
∫
t
0
T
∫
-
β
1
1
τ
2
G
2
e
2
s
Φ
d
x
d
t
+
∫
t
0
T
∫
γ
1
γ
1
′
​
+
∫
γ
2
γ
2
′
​
z
2
e
2
s
Φ
d
x
d
t
+
∫
t
0
T
∫
ω
1
​
+
∫
ω
2
​
h
t
2
+
s
2
θ
2
z
2
e
2
s
Φ
d
x
d
t
≤
C
∬
Q
t
0
T
h
t
2
e
2
s
Φ
d
x
d
t
+
∫
t
0
T
∫
ω
1
​
+
∫
ω
2
​
h
t
2
+
s
2
θ
2
z
2
e
2
s
Φ
d
x
d
t
≤
C
∬
Q
t
0
T
h
t
2
e
2
s
Φ
d
x
d
t
+
∬
ω
t
0
T
s
2
θ
2
z
2
e
2
s
Φ
d
x
d
t
. Hence, using the definitions of w and W, there exists a constant C>0 such that(84)∫t0T∫0γ2sθazx2+s3θ3x-x02az2+sθ3/2ηψz2+1sθzt2e2sφdxdt=∫t0T∫0γ2sθa~wx2+s3θ3x-x02a~w2+sθ3/2ηψw2+1sθwt2e2sφdxdt=∫t0T∫0γ2sθa~Wx2+s3θ3x-x02a~W2+sθ3/2ηψW2+1sθWt2e2sφdxdt≤∫t0T∫-β11sθa~Wx2+s3θ3x-x02a~W2+sθ3/2ηψW2+1sθWt2e2sφdxdt≤C∬Qt0Tht2e2sΦdxdt+∬ωt0Ts2θ2z2e2sΦdxdt.Finally adding up (71) and (84), we conclude that(85)I0≔∬Qt0Tsθazx2+s3θ3x-x02az2+sθ3/2ηψz2+1sθzt2e2sφdxdt≤C∬Qt0Tht2e2sΦdxdt+∬ωt0Ts3θ3z2e2sΦdxdt︸I1,for all s≥s0 and for a positive constant C.
Step 2 (estimate from above of I1). In this step, our purpose is to show that there exists some constant C>0 such that(86)I1≤C1s∫01h2T′,xe2sΦT′,xdx+utL2ωt0T2.Let us recall that, by the definition of Ψ, there exists some constant p<0 such that Ψ(x)≤p for all x∈[0,1]. Therefore we have (87)s3θ3e2sΦ≤s3θ3e2spθ→0ast→t0orT.As a consequence, setting C=supR{y↦y3e2spy}>0, we have(88)∬ωt0Ts3θ3z2e2sΦdxdt≤CzL2ωt0T2=utL2ωt0T2.In order to complete the proof of (86), it remains to prove the following lemma.
Lemma  17.There exists a constantC>0such that(89)∬Qt0Tht2e2sΦdxdt≤Cs∫01h2T′,xe2sΦT′,xdx.
We omit the proof of Lemma 17 which is classical and we refer the reader to [2]. Using (88) and (89), we obtain (86).
Step 3 (estimate from below of I0). The purpose of the step is to provide the following estimate: there exists a constant C>0 such that(90)∫01z2T′,xe2sφT′,xdx≤CI0.Since limt→t0z2(t,x)e2sφ(t,x)=0, for a.e. x∈(0,1), we can write(91)∫01z2T′,xe2sφT′,xdx=∫t0T′∂∂t∫01z2t,xe2sφt,xdxdt=∫t0T′∫012zzt+2sφtz2e2sφt,xdxdt.Using Young’s inequality, one has(92)∫t0T′∫012zzte2sφdxdt=∫t0T′∫012sθzesφ1sθztesφdxdt≤∬Qt0Tsθz2e2sφ+1sθzt2e2sφdxdt≤∬Qt0Tsθz2e2sφdxdt+I0.Moreover, by the Hardy-Poincaré inequality, and using the same computation as in the proof of (45), we get(93)∫t0T′∫012zzte2sφdxdt≤C∬Qt0Tsθaxzx2+s3θ3x-x02axz2e2sφdxdt+I0≤CI0.Next, using the fact that φt≤Cηψθ3/2, we have(94)∫t0T′∫01sφtz2e2sφdxdt≤C∬Qt0Tsθ3/2ηψz2e2sφdxdt≤CI0.Eventually, (91) associated with (92) and (93) gives (90).
Step 4 (conclusion). Using (90), (85), and next (86), there exists some constant C>0 such that(95)∫01z2T′,xe2sφT′,xdxdt≤C1s∫01h2T′,xe2sΦT′,xdx+utL2ωt0T2.On the other hand, let us recall that z satisfies (96)zT′,x=utT′,x=auxxT′,x+λbuT′,x+hT′,xfor  a.e.x∈0,1. Therefore,(97)∫01h2T′,xe2sφT′,xdx≤CauxxT′,x+λbuT′,xL20,12+∫01z2T′,xe2sφT′,xdx.Applying (95) to estimate the term ∫01z2(T′,x)e2sφ(T′,x)dx, we obtain (98)∫01h2T′,xe2sφT′,xdx≤CauxxT′,x+λbuT′,xL20,12+utL2ωt0T2+1s∫01h2T′,xe2sΦT′,xdx. Choosing s large enough such that C/s=1/2, we get (99)∫01h2T′,xe2sφT′,xdx≤CauxxT′,x+λbuT′,xL20,12+utL2ωt0T2. On the other hand, s being now fixed, there exists some constant γ>0 such that (100)e2sφT′,x≥γ>0. So we can write (101)γ∫01h2T′,xdx≤CauxxT′,x+λbuT′,xL20,12+utL2ωt0T2. Moreover, in view of (2),(102)ht,x≤hT′,x+∫T′thts,xds≤ChT′,x. Hence, (103)hL2Q2≤C∬Qh2T′,xdxdt=CT∫01h2T′,xdx≤CauxxT′,x+λbuT′,xL20,12+utL2ωt0T2, which concludes the proof.


## 4. Numerical Approach

In this section, we develop an algorithm for numerically reconstructing the unknown source term from the measured final data.


### 4.1. Solvability of the Inverse Problem and Gradient Formula

As the theoretical stability is guaranteed by Theorem2, in this subsection we study the inverse source problem from the numerical viewpoint. To this end, let us define our inverse problem which we use in computations.Inverse Source Problem (ISP). Let u be the solution to (1). Determine the source term h(t,x) from the measured data at the final time u(T,.).Problem(ISP) will be defined as the inverse source problem associated with the singular parabolic equation (1). Accordingly, for a given function h∈U, from some class of admissible source functions U, which will be defined below, the singular parabolic problem (1) will be referred to as a direct problem. A solution of the direct problem (1), corresponding to the function h∈U, will be defined as u(t,x;h).Remark 18.
It should be mentioned that we do not need the supplement distributed measurements to obtain the numerical solution of the inverse problem.Numerically, we treat ISP by interpreting its solution as a minimizer of the following least squares functional with the Tikhonov regularization:(104)minh∈U⁡Jh,Jh=12uT,x;h-u~xL20,12+ε2h-h0L2Q2,where u~∈L2(0,1) is the observation data with noise, h0 is the initial guess for h, ε>0 stands for the regularization parameter, and U is the set of admissible unknown sources defined in the following way:(105)U≔h∈H10,T;L20,1:hH10,T;L20,1≤r,r>0.Evidently, the set U is a bounded, closed, and convex subset of H1(0,T;L2(0,1)).We are now going to show the existence of minimizers to problem (104). To do so, we need the following lemma.Lemma 19.
Assume Hypothesis9. Let u be the weak solution of (1) corresponding to a given source term h. Then, the function F:H1(0,T;L2(0,1))→C([0,T];L2(0,1))∩L2(0,T;H) defined as F(h)≔u is continuous.Proof.
First, takeu0∈D(A). Then, let the source term h be perturbed by a small amount δh such that h+δh∈U. Consider δu=uδ-u, where uδ is the weak solution of (1) with source term hδ≔h+δh. Then δu∈C1([0,T];L2(0,1))∩C(0,T;D(A)) satisfies the following sensitivity problem:(106)∂tδu-∂xa∂xδu-λbδu=δht,x,t,x∈Q,δu0=δu1=0,t∈0,T,δu0,x=0,x∈0,1.Let v(t,x) be a smooth function. From (106) and by the Gauss Green identity [16, Lemma 2.21], we have (107)∫01∂tδuvdx+∫01a∂xδuvx-λbδuvdx=∫01δhvdx. We take δu as a mutual test function for v to deduce (108)12ddt∫01δu2dx+∫01a∂xδu2-λbδu2dx=∫01δhδudx. Then, using Lemma 8, by the Cauchy-Schwarz inequality we obtain(109)12ddtδutL20,12+Λ∫01a∂xδut2dx≤12δutL20,12+12δhtL20,12,for every t≤T, from which (110)ddtδutL20,12≤δutL20,12+δhtL20,12. Applying Gronwall’s inequality, we obtain(111)δutL20,12≤eTδu0L20,12+δhL2Q2=eTδhL2Q2,for every t≤T. From (109) and (111), we immediately get(112)∫0TaδuxtL20,12dt≤CTδhL2Q2,for every t≤T and some universal constant CT>0. Thus, by (111) and (112), we obtain (113)supt∈0,T⁡δutL20,12+∫0TδutH2dt≤CTδhL2Q2, from which it follows that (114)δuC0,T;L20,12+δuL20,T;H2≤CδhH10,T;L20,12, if u0∈D(A). Since D(A) is dense in L2(0,1), the same inequality holds if u0∈L2(0,1). This completes the proof Lemma 19.An immediate consequence of Lemma19 is the following result.Proposition 20.
Assume Hypothesis9. Then, the functional J is continuous on U and there exists a minimizer h⋆∈U of J(h), i.e., (115)Jh⋆=minh∈U⁡Jh.Here we will propose an iterative method to solve the nonlinear optimization problem (104), and thus the information of the derivatives of the objective functional plays a significant and important role. Our approach for computing the Fréchet derivatives is based on adjoint state method. This used method is also called the variational adjoint method or the adjoint problem approach [18, 21, 22]. A distinct advantage of using such a method is due to the fact that it permits to reduce the computational costs as well as its relatively simple numerical implementation.The following proposition characterizes the derivative of the cost functional (104).Proposition 21.
The Tikhonov functionalJ is Fréchet differentiable and its derivative at each h∈U is given by(116)J′h=v+εh-h0,where v∈C([0,T];L2(0,1)) is the mild solution of the following adjoint equation:(117)vt+avxx+λbxv=0,t,x∈Q,v0=v1=0,t∈0,T,vT,x=uT-u~,x∈0,1.For the proof of the above result, we shall use the following lemma which derives an integral relationship relating the changeδh in the term source to the change of the output δu(T) through the solution of an adjoint problem.Lemma 22.
Leth,hδ be given elements. Denote by u and uδ the solutions of the direct problem (1) corresponding to the given admissible term sources h,hδ∈U. Then the following integral identity holds:(118)∫01δuTqxdx=∫0T∫01δhvdxdt,where δu(T)=uδ(T)-u(T), δh=hδ-h, and v(t,x)≔v(t,x;q) solves the adjoint problem(119)vt+avxx+λbxv=0,t,x∈Q,v0=v1=0,t∈0,T,vT,x=qx,x∈0,1,with an arbitrary input q(x)∈L2(0,1).Proof of Proposition21.
Letu and uδ be the solutions of the direct problem (1) corresponding the source terms h,hδ=h+δh∈U. By calculating the increment δJ(h)≔J(hδ-J(h)) of the cost functional, we deduce that(120)δJh=∫01uT-u~δuTdx+12δuTL20,12+ε∫0T∫01h-h0δhdxdt+ε2δhL2Q2,for all h,hδ∈U, where δu=uδ-u.
Choosing the arbitrary inputq∈L2(0,1) in the adjoint problem (119) as q(x)≔(u(T)-u~), we deduce from the integral relationship (118) that(121)∫01uT-u~δuTdx=∫0T∫01δhvdxdt,Using the integral equality (121) in the increment formula (120), we deduce that(122)δJh=∫0T∫01δhvdxdt+ε∫0T∫01h-h0δhdxdt+12δuTL20,12+ε2δhL2Q2.To obtain the Fréchet differential of δJ(h), by definition, we need to show that (123)δJh=v+εh-h0,δhL2Q+OδhL2Q2. To this aim, observe that if we substitute t by T in (111) it follows that (124)δuTL20,12≤eTδhL2Q2. i.e., the second right-hand side integral of (122) is of the order O(δhL2(Q)2). So, since the last two integrals in (122) are of the order O(δhL2(Q)2), we deduce that the cost functional J(h) is Frćhet-differentiable, with Fréchet differential: (125)J′h=v+εh-h0,∀h∈U.Proof of Lemma22.
The functionδu(t)=uδ(t)-u(t) solves the initial-boundary value problem(126)∂tδu=∂xa∂xδu+λbδu+δh,t,x∈Q,δu0=δu1=0,t∈0,T,δu0,x=0,x∈0,1.Multiply both sides of the parabolic equation (126) by v, integrate on Q, and then apply the integration by parts formula on both sides multiple times. Then, taking into account the initial and boundary conditions in problem (126), we obtain the following identity:(127)∫01δuTvTdx-∫0T∫01δuvtdxdt=∫0T∫01∂xa∂xδu+λbδu+δhvdxdt=∫0T∫01avxx+λbvδudxdt+∫0T∫01δhvdxdt.Since v(T)=q(x), thanks to (119) and (127), we arrive at (128)∫01δuTqxdx=∫01δuTvTdx=∫0T∫01vt+avxx+λbvδudxdt+∫0T∫01δhvdxdt=∫0T∫01δhvdxdt.This ends the proof.


### 4.2. Lipschitz Continuity of the Gradient and the Monotone Iteration Scheme

As emphasized in [23, Chapter 3], any gradient method for the minimization problem (104) requires an estimation of the iteration parameter tn>0 in the iteration process(129)hn+1=hn-tnJ′hn,n=0,1,2,…,where h0∈U is a given initial iteration. The choice of tn defines different gradient methods, although in many situation the estimation of this parameter is a difficult problem. However, in the case of Lipschitz continuity of the gradient J′(h) the parameter tn can be estimated via the Lipschitz constant as follows (see [18]):(130)0<δ0≤tn≤2L+2δ1,where δ0,δ1>0 are arbitrary parameters.Now we will prove the Lipschitz continuity of the Fréchet gradient.Theorem 23.
Leth,δh∈U. Then the Fréchet gradient (116) of the Tikhonov functional is Lipschitz continuous, i.e.,(131)J′hδ-J′hL2Q≤LδhL2Q,∀h,δh∈U,where the Lipschitz constant L>0 is defined via the parameters T and ε as follows:(132)L=2TeT+ε2.Proof.
Let the functionsv(t,x;h) and v(t,x;hδ) be the solutions to the adjoint problem (119) with q(x)=u(T,x;h)-u~ and q(x)=u(T,x;hδ)-u~, respectively. We have(133)J′hδ-J′hL2Q2=∫0T∫01δv+εδh2dxdt≤2∫0T∫01δv2dxdt+2ε2δhL2Q2,where δv=δv(t,x;h)≔v(t,x;hδ)-v(t,x;h) is the solution of the backward parabolic problem(134)∂tδv+∂xa∂xδv+λbδv=0,t,x∈Q,δv0=δv1=0,t∈0,T,δvT,x=δuT,x;h,x∈0,1.Multiplying (134) by δv, integrating over (0,1), and using the boundary conditions, we get (135)12ddt∫01δvt,x;h2dx=∫01a∂xδvt,x;h2-λbδvt,x;h2dx.Using again (15), we have (136)12ddt∫01δvt,x;h2dx≥Λ∫01a∂xδvt,x;h2dx.Let (137)Ft≔∫01δvt,x;h2dx.Since F′(t)≥0, F is increasing on (0,T], and thus(138)Ft≤FT,∀t∈0,T.By (138), we obtain that (139)∫01δvt,x;h2dx≤∫01δvT,x;h2dx=∫01δuT,x;h2dx. Finally, by (111) it follows that(140)∫01δvt,x;h2dx≤eTδhL2Q2.Combining (133) and (140), we get the inequality (131).An important application of this theorem is the following lemma (see [23, Lemmas 3.4.4 and 3.4.5]).Lemma 24.
Let the Fréchet gradient of the Tikhonov functional (104) defined on the set of admissible source terms U be Lipschitz continuous with the Lipschitz constant L>0. Denote by {hn}⊂U the sequence of iterations obtained by the Landweber iteration algorithm (129). If the relaxation parameter tn∈(0,2/L), then the following statements hold: (i)
{J(hn)} is a monotone decreasing sequence;(ii)
limn→∞J′(hn)L2(Q)=0;(iii)
for the rate of convergence of the sequence{J(hn)}, the following estimate holds: (141)0≤Jhn-J⋆≤2Ldn-1,d>0,n=1,2,3,…,whereJ⋆≔infh∈U⁡J(h).


## 4.1. Solvability of the Inverse Problem and Gradient Formula

As the theoretical stability is guaranteed by Theorem2, in this subsection we study the inverse source problem from the numerical viewpoint. To this end, let us define our inverse problem which we use in computations.Inverse Source Problem (ISP). Let u be the solution to (1). Determine the source term h(t,x) from the measured data at the final time u(T,.).Problem(ISP) will be defined as the inverse source problem associated with the singular parabolic equation (1). Accordingly, for a given function h∈U, from some class of admissible source functions U, which will be defined below, the singular parabolic problem (1) will be referred to as a direct problem. A solution of the direct problem (1), corresponding to the function h∈U, will be defined as u(t,x;h).Remark 18.
It should be mentioned that we do not need the supplement distributed measurements to obtain the numerical solution of the inverse problem.Numerically, we treat ISP by interpreting its solution as a minimizer of the following least squares functional with the Tikhonov regularization:(104)minh∈U⁡Jh,Jh=12uT,x;h-u~xL20,12+ε2h-h0L2Q2,where u~∈L2(0,1) is the observation data with noise, h0 is the initial guess for h, ε>0 stands for the regularization parameter, and U is the set of admissible unknown sources defined in the following way:(105)U≔h∈H10,T;L20,1:hH10,T;L20,1≤r,r>0.Evidently, the set U is a bounded, closed, and convex subset of H1(0,T;L2(0,1)).We are now going to show the existence of minimizers to problem (104). To do so, we need the following lemma.Lemma 19.
Assume Hypothesis9. Let u be the weak solution of (1) corresponding to a given source term h. Then, the function F:H1(0,T;L2(0,1))→C([0,T];L2(0,1))∩L2(0,T;H) defined as F(h)≔u is continuous.Proof.
First, takeu0∈D(A). Then, let the source term h be perturbed by a small amount δh such that h+δh∈U. Consider δu=uδ-u, where uδ is the weak solution of (1) with source term hδ≔h+δh. Then δu∈C1([0,T];L2(0,1))∩C(0,T;D(A)) satisfies the following sensitivity problem:(106)∂tδu-∂xa∂xδu-λbδu=δht,x,t,x∈Q,δu0=δu1=0,t∈0,T,δu0,x=0,x∈0,1.Let v(t,x) be a smooth function. From (106) and by the Gauss Green identity [16, Lemma 2.21], we have (107)∫01∂tδuvdx+∫01a∂xδuvx-λbδuvdx=∫01δhvdx. We take δu as a mutual test function for v to deduce (108)12ddt∫01δu2dx+∫01a∂xδu2-λbδu2dx=∫01δhδudx. Then, using Lemma 8, by the Cauchy-Schwarz inequality we obtain(109)12ddtδutL20,12+Λ∫01a∂xδut2dx≤12δutL20,12+12δhtL20,12,for every t≤T, from which (110)ddtδutL20,12≤δutL20,12+δhtL20,12. Applying Gronwall’s inequality, we obtain(111)δutL20,12≤eTδu0L20,12+δhL2Q2=eTδhL2Q2,for every t≤T. From (109) and (111), we immediately get(112)∫0TaδuxtL20,12dt≤CTδhL2Q2,for every t≤T and some universal constant CT>0. Thus, by (111) and (112), we obtain (113)supt∈0,T⁡δutL20,12+∫0TδutH2dt≤CTδhL2Q2, from which it follows that (114)δuC0,T;L20,12+δuL20,T;H2≤CδhH10,T;L20,12, if u0∈D(A). Since D(A) is dense in L2(0,1), the same inequality holds if u0∈L2(0,1). This completes the proof Lemma 19.An immediate consequence of Lemma19 is the following result.Proposition 20.
Assume Hypothesis9. Then, the functional J is continuous on U and there exists a minimizer h⋆∈U of J(h), i.e., (115)Jh⋆=minh∈U⁡Jh.Here we will propose an iterative method to solve the nonlinear optimization problem (104), and thus the information of the derivatives of the objective functional plays a significant and important role. Our approach for computing the Fréchet derivatives is based on adjoint state method. This used method is also called the variational adjoint method or the adjoint problem approach [18, 21, 22]. A distinct advantage of using such a method is due to the fact that it permits to reduce the computational costs as well as its relatively simple numerical implementation.The following proposition characterizes the derivative of the cost functional (104).Proposition 21.
The Tikhonov functionalJ is Fréchet differentiable and its derivative at each h∈U is given by(116)J′h=v+εh-h0,where v∈C([0,T];L2(0,1)) is the mild solution of the following adjoint equation:(117)vt+avxx+λbxv=0,t,x∈Q,v0=v1=0,t∈0,T,vT,x=uT-u~,x∈0,1.For the proof of the above result, we shall use the following lemma which derives an integral relationship relating the changeδh in the term source to the change of the output δu(T) through the solution of an adjoint problem.Lemma 22.
Leth,hδ be given elements. Denote by u and uδ the solutions of the direct problem (1) corresponding to the given admissible term sources h,hδ∈U. Then the following integral identity holds:(118)∫01δuTqxdx=∫0T∫01δhvdxdt,where δu(T)=uδ(T)-u(T), δh=hδ-h, and v(t,x)≔v(t,x;q) solves the adjoint problem(119)vt+avxx+λbxv=0,t,x∈Q,v0=v1=0,t∈0,T,vT,x=qx,x∈0,1,with an arbitrary input q(x)∈L2(0,1).Proof of Proposition21.
Letu and uδ be the solutions of the direct problem (1) corresponding the source terms h,hδ=h+δh∈U. By calculating the increment δJ(h)≔J(hδ-J(h)) of the cost functional, we deduce that(120)δJh=∫01uT-u~δuTdx+12δuTL20,12+ε∫0T∫01h-h0δhdxdt+ε2δhL2Q2,for all h,hδ∈U, where δu=uδ-u.
Choosing the arbitrary inputq∈L2(0,1) in the adjoint problem (119) as q(x)≔(u(T)-u~), we deduce from the integral relationship (118) that(121)∫01uT-u~δuTdx=∫0T∫01δhvdxdt,Using the integral equality (121) in the increment formula (120), we deduce that(122)δJh=∫0T∫01δhvdxdt+ε∫0T∫01h-h0δhdxdt+12δuTL20,12+ε2δhL2Q2.To obtain the Fréchet differential of δJ(h), by definition, we need to show that (123)δJh=v+εh-h0,δhL2Q+OδhL2Q2. To this aim, observe that if we substitute t by T in (111) it follows that (124)δuTL20,12≤eTδhL2Q2. i.e., the second right-hand side integral of (122) is of the order O(δhL2(Q)2). So, since the last two integrals in (122) are of the order O(δhL2(Q)2), we deduce that the cost functional J(h) is Frćhet-differentiable, with Fréchet differential: (125)J′h=v+εh-h0,∀h∈U.Proof of Lemma22.
The functionδu(t)=uδ(t)-u(t) solves the initial-boundary value problem(126)∂tδu=∂xa∂xδu+λbδu+δh,t,x∈Q,δu0=δu1=0,t∈0,T,δu0,x=0,x∈0,1.Multiply both sides of the parabolic equation (126) by v, integrate on Q, and then apply the integration by parts formula on both sides multiple times. Then, taking into account the initial and boundary conditions in problem (126), we obtain the following identity:(127)∫01δuTvTdx-∫0T∫01δuvtdxdt=∫0T∫01∂xa∂xδu+λbδu+δhvdxdt=∫0T∫01avxx+λbvδudxdt+∫0T∫01δhvdxdt.Since v(T)=q(x), thanks to (119) and (127), we arrive at (128)∫01δuTqxdx=∫01δuTvTdx=∫0T∫01vt+avxx+λbvδudxdt+∫0T∫01δhvdxdt=∫0T∫01δhvdxdt.This ends the proof.


## 4.2. Lipschitz Continuity of the Gradient and the Monotone Iteration Scheme

As emphasized in [23, Chapter 3], any gradient method for the minimization problem (104) requires an estimation of the iteration parameter tn>0 in the iteration process(129)hn+1=hn-tnJ′hn,n=0,1,2,…,where h0∈U is a given initial iteration. The choice of tn defines different gradient methods, although in many situation the estimation of this parameter is a difficult problem. However, in the case of Lipschitz continuity of the gradient J′(h) the parameter tn can be estimated via the Lipschitz constant as follows (see [18]):(130)0<δ0≤tn≤2L+2δ1,where δ0,δ1>0 are arbitrary parameters.Now we will prove the Lipschitz continuity of the Fréchet gradient.Theorem 23.
Leth,δh∈U. Then the Fréchet gradient (116) of the Tikhonov functional is Lipschitz continuous, i.e.,(131)J′hδ-J′hL2Q≤LδhL2Q,∀h,δh∈U,where the Lipschitz constant L>0 is defined via the parameters T and ε as follows:(132)L=2TeT+ε2.Proof.
Let the functionsv(t,x;h) and v(t,x;hδ) be the solutions to the adjoint problem (119) with q(x)=u(T,x;h)-u~ and q(x)=u(T,x;hδ)-u~, respectively. We have(133)J′hδ-J′hL2Q2=∫0T∫01δv+εδh2dxdt≤2∫0T∫01δv2dxdt+2ε2δhL2Q2,where δv=δv(t,x;h)≔v(t,x;hδ)-v(t,x;h) is the solution of the backward parabolic problem(134)∂tδv+∂xa∂xδv+λbδv=0,t,x∈Q,δv0=δv1=0,t∈0,T,δvT,x=δuT,x;h,x∈0,1.Multiplying (134) by δv, integrating over (0,1), and using the boundary conditions, we get (135)12ddt∫01δvt,x;h2dx=∫01a∂xδvt,x;h2-λbδvt,x;h2dx.Using again (15), we have (136)12ddt∫01δvt,x;h2dx≥Λ∫01a∂xδvt,x;h2dx.Let (137)Ft≔∫01δvt,x;h2dx.Since F′(t)≥0, F is increasing on (0,T], and thus(138)Ft≤FT,∀t∈0,T.By (138), we obtain that (139)∫01δvt,x;h2dx≤∫01δvT,x;h2dx=∫01δuT,x;h2dx. Finally, by (111) it follows that(140)∫01δvt,x;h2dx≤eTδhL2Q2.Combining (133) and (140), we get the inequality (131).An important application of this theorem is the following lemma (see [23, Lemmas 3.4.4 and 3.4.5]).Lemma 24.
Let the Fréchet gradient of the Tikhonov functional (104) defined on the set of admissible source terms U be Lipschitz continuous with the Lipschitz constant L>0. Denote by {hn}⊂U the sequence of iterations obtained by the Landweber iteration algorithm (129). If the relaxation parameter tn∈(0,2/L), then the following statements hold: (i)
{J(hn)} is a monotone decreasing sequence;(ii)
limn→∞J′(hn)L2(Q)=0;(iii)
for the rate of convergence of the sequence{J(hn)}, the following estimate holds: (141)0≤Jhn-J⋆≤2Ldn-1,d>0,n=1,2,3,…,whereJ⋆≔infh∈U⁡J(h).


## 5. Conclusion

In this paper, we have considered an inverse source problem for a class of degenerate and singular parabolic equations. Based on Carleman estimates, global Lipschitz stability result is proved. Then, the identification of the source term is formulated as a minimization problem combining the output least squares and the Tikhonov regularization. It is proved that the Fréchet derivative of the cost functional can be formulated via the solution of the adjoint parabolic problem. Lipschitz continuity of the gradient functional was also proved, which implies the monotonicity of the numerical sequence of iterations obtained by the Landweber iteration algorithm. Some applications with numerical implementations are in progress.



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*Source: 2067304-2018-12-09.xml*