Inverse problem of reconstruction of degenerate diffusion coefficient in a parabolic equation

Inverse Problems PAPER Inverse problem of reconstruction of degenerate diffusion coefficient in a parabolic equation To cite this article: Piermarco Cannarsa et al 2021 Inverse Problems 37 125002 View the article online for updates and enhancements. You may also like ON ASYMPTOTIC “EIGENFUNCTIONS” OF THE CAUCHY PROBLEM FOR A NONLINEAR PARABOLIC EQUATION V A Galaktionov, S P Kurdyumov and A A Samarski - Stabilization of solutions of quasilinear second order parabolic equations in domains with non-compact boundaries Ruslan Kh Karimov and Larisa M Kozhevnikova - Reconstruction of degenerate conductivity region for parabolic equations Piermarco Cannarsa, Anna Doubova and Masahiro Yamamoto - This content was downloaded from IP address 150.214.182.233 on 05/12/2024 at 11:47 Inverse Problems Inverse Problems 37 (2021) 125002 (56pp) https://doi.org/10.1088/1361-6420/ac274b Inverse problem of reconstruction of degenerate diffusion coefficient in a parabolic equation Piermarco Cannarsa1, Anna Doubova2,∗and Masahiro Yamamoto3,4 1University of Rome ‘Tor Vergata’, Italy 2Universidad de Sevilla, Dpto. EDAN e IMUS, Spain 3The University of Tokyo, Japan 4Honorary Member of Academy of Romanian Scientists, Correspondence member of Accademia Peloritana dei Pericolanti, Messina, Italy E-mail: [email protected],[email protected] and [email protected]p Received 12 June 2021, revised 25 August 2021 Accepted for publication 16 September 2021 Published 2 November 2021 Abstract We consider the inverse problem of identification of degenerate diffusion coefficient of the form xαa(x) in a one dimensional parabolic equation by some extra data. We first prove by energy methods the uniqueness and Lipschitz stability results for the identification of a constant coefficient aand the power αby knowing interior data at some time. On the other hand, we obtain the uniqueness result for the identification of a general diffusion coefficients a(x) andalsothepowerαform boundary data on one side of the space interval. The proof is based on global Carleman estimates for a hyperbolic problem and an inversion of the integral transform similar to the Laplace transform. Finally, the theoretical results are satisfactory verified by numerically experiments. Keywords: inverse problems, degenerate parabolic equations, numerical reconstruction (Some figures may appear in colour only in the online journal) 1. Introduction In this paper, we will address several inverse problems for parabolic equations with degeneracy in the diffusion coefficient. More precisely, we will consider the weakly degenerate ∗Author to whom any correspondence should be addressed. 1361-6420/21/125002+56$33.00 © 2021 IOP Publishing Ltd Printed in the UK 1 Inverse Problems 37 (2021) 125002 P Cannarsa et al Cauchy–Dirichlet problem ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ ∂tu−∂x(xαa(x)∂xu)=f(x,t), (x,t)∈(0, )×(0, T), u(0, t)=0, u(,t)=0, t∈(0, T), u(x,0)=u0(x), x∈(0, ), (1) where 0 <α<1, as well as strongly degenerate problems of the form ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ ∂tu−∂x(xαa(x)∂xu)=f(x,t), (x,t)∈(0, )×(0, T), (xα∂xu(x,t)|x=0=0, u(,t)=0, t∈(0, T), u(x,0)=u0(x), x∈(0, ), (2) where 1 ⩽α<2. Here, T>0, >0, u0∈L2(0, ), u0=0, f∈L2((0, )×(0, T)) are given, and a∈C1([0, ]) satisfies 0 <a⩽a(x)⩽a<+∞for all x∈(0, ), with aand apositive constants. Notice that the boundary conditions in (1)and(2) are different. In fact, weakly degenerate parabolic problems can be studied either with Dirichlet or Neumann data whereas, when solving strongly degenerate equations in L2(0, ), the latter turns out to be the natural condition to impose on the part of the boundary where degeneracy occurs, see remark 2.1 below. Our goal is to determine or estimate the diffusion coefficient for the considered degenerate problems. Although we can discuss more general cases, we concentrate here on the onedimensional linear equation. The following three types of inverse problems of the identification of the unknown diffusion coefficient will be considered and analyzed. Linear case: given α=1, we will deal with the inverse problem of identification of a constant coefficient, a(x)≡a,whereais an unknown positive constant, from given suitable internal data at some time (see section 3.1). Power-like case: given a(x)=1, the inverse problem will be the identification of the power α∈(0, 2) from suitable internal observations at some time (see section 3.2). General case: the inverse problem of the identification of a general diffusion coefficient xαa(x) from one boundary observation (see section 4). In recent years, degenerate parabolic equations have received increasing attention, in view of the important related theoretical analysis and practical applications such as climate science (see Sellers [35], Díaz [12], Ji and Huang [15]), populations genetics (Ethier [14]), vision (Citti and Manfredini [9]), financial mathematics (Black and Scholes [5]), and fluid dynamics (Oleinik and Samokhin [31]). See also Cannarsa et al [7] and references therein. Inverse problems are the kind of problems known as not well-posed in the sense of Hadamard (see [16]). This means that, either a solution does not exist, or it is not unique and/or small errors in known data lead to large errors in the computation of the solution(s). In particular, the first theoretical issues that we analyze in our paper are uniqueness and stability. The main techniques used here are usual in control and parameter identification theory: tools from real and complex analysis, energy methods and Hardy’s inequality, Laplace and other similar integral transforms, and Carleman inequalities. The literature related to inverse problems for degenerate parabolic PDEs, despite their fundamental importance and practical applications, is rather recent and scarce. For example, the inverse source problem was considered in Tort [38], Cannarsa et al [8], Cannarsa et al [7], Deng et al [10] and Hussein et al [18], where numerical reconstruction was also considered. 2 Inverse Problems 37 (2021) 125002 P Cannarsa et al The inverse problem of recovering the first-order coefficient of degenerate parabolic equations is analyzed in Deng and Yang [11] and Kamynin [24]. An inverse diffusion problem was consideredin[39], where a constant diffusion coefficient was recovered from measurement of second order derivatives by means of Carleman estimates provided in Cannarsa et al [8]. As for inverse problems of determining spatially varying coefficients such as conductivities and source terms for non-degenerate parabolic equations, we note that there has been a substantial amount of works. Here we refer to chapters of the book by Isakov [23], and Yamamoto [40] (see also the references therein). In particular, for the same type of inverse problems for one-dimensional uniformly parabolic equations we refer to Murayama [30], see also Pierce [32], Suzuki and Murayama [36]. On the other hand, most of the techniques that have been developed to treat non-degenerate parabolic equations cease to be applicable in the degenerate case. For instance, for strongly degenerate operators the trace of the co-normal derivate must vanish on the part of the boundary where ellipticity fails, yielding no useful measurement5. The second main issue of this paper is the reconstruction method for our inverse problems. The goal is to compute approximations of the diffusion coefficient as solutions to the inverse problems using the observation data. To our best knowledge, there are no works combining the theoretical studies on the uniqueness, stability and the numerical reconstruction of the diffusion coefficient for a degenerate parabolic equation. An efficient technique to achieve this, as shown below, relies on a reformulation of the search of the diffusion coefficient as an extremal problem. This is classical nowadays and has been applied in a lot of situations; see for instance Lavrentiev et al [28], Samarskii and Vabishchevich [34], and Vogel [37]. The paper is organized as follows. In section 2, we consider the well-posedness of the direct problem (1)and(2). In section 3.1 we prove, using the energy methods, Lipschitz stability and uniqueness results for the linear case corresponding to the constant coefficient a. Section 3.2 will be devoted to the power-like case of the identification of the power αin (1)and(2). Again, Lipschitz stability and uniqueness will be studied by the energy methods. In section 4a case of the reconstruction of the unknown general diffusion coefficient xαa(x) will be treated by means of the Carleman estimates, obtaining the uniqueness result for both αand a(x). Finally, in section 5, in order to illustrate theoretical results obtained in the previous sections, we perform satisfactory numerical experiments corresponding to the considered inverse problems. 2. Preliminaries In this section we will consider the well-posedness of the direct problem related to (1)and(2). We begin with recalling the natural functions spaces where the problem can be set. For any >0, let us call H=L2(0, ). For α∈(0, 2) we will consider the following function spaces: H1 α(0, )=⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ u∈H| 0 xα|∂xu|2dx<∞,u(0) =u()=0,0<α<1, u∈H| 0 xα|∂xu|2dx<∞,u()=0,1⩽α<2 (3) 5Although in this paper we consider equations that degenerate on a strict subset of the boundary, our approach applies unaltered when degeneracy occurs on the whole boundary, like in the Budyko–Sellers model [35]. 3 Inverse Problems 37 (2021) 125002 P Cannarsa et al and H2 α(0, )={u∈H1 α(0, )|x→ xα∂xu∈H1(0, )},0<α<2. Remark 2.1 (Neumann boundary condition). For u∈H2 α(0, ) with 1 ⩽α<2, we have xα∂xu|x=0=0. Indeed, if xα∂xu(x)→Lwhen x→0, then xα|∂xu(x)|2∼L2/xαand, therefore L=0otherwiseu/∈H1 α(0, ). We now adapt the classical Poincar´ e inequality to the above weighted spaces. Lemma 2.2 (Poincar´ e’s inequality). For all ε>0, there exists a constant Cp= Cp(ε,)>0such that for all α∈[0, 2 −ε], the following holds:  0|u|2dx⩽Cp 0 xα|∂xu|2dx∀u∈H1 α(0, ).(4) Proof of lemma 2.2. Let >0 be given. We will distinguish two cases. (a) Assume α=1, for all x∈(0, ), we have |u(x)|⩽ x|∂xu|dy⩽ x y|∂xu(y)|2dy1/2 x 1 ydy1/2 ⩽log 1 −log 1 x1/2 x y|∂xu(y)|2dy1/2 . Therefore, we obtain  0|u(x)|2dx⩽ 0 log 1 xdx 0 x|∂xu(x)|2dx =(1 −log ) 0 x|∂xu(x)|2dx. (b) Assume now that α=1. We can write |u(x)|⩽ x yα/2u(y)dy yα/2 ⩽ x dy yα/21/2 0 yα|u(y)|2dy1/2 ⩽1−α−x1−α 1−α1/2 0 yα|u(y)|2dy1/2 . Therefore, we obtain  0|u(x)|2dx= 0 1−α−x1−α 1−αdx 0 xα|∂xu(x)|2dx =2−α 2−α 0 xα|∂xu(x)|2dx=Cp 0 xα|∂xu(x)|2dx, here we have used that α<2. This ends the proof.  For given constants 0 <a<a, we assume that a∈C1([0, ]), 0 <a⩽a(x)⩽a<+∞∀x∈(0, ).(5) 4 Inverse Problems 37 (2021) 125002 P Cannarsa et al Problems (1)and(2) can be recast in the abstract form u(t)=Au(t)t⩾0 u(0) =u0 (6) by introducing the linear operator A:D(A)⊂H→Hdefined by D(A)=H2 α(0, ), Au =∂x(xαa(x)∂xu)∀u∈D(A).(7) Observe that the degenerate coefficient ˜a(x)=xαa(x)isofclassC1([0, ]) and satisfies lim x→0x˜a(x)˜a(x)=lim x→0 xα+1a(x)+αxαa(x) xαa(x)=α∈(0, 2).(8) Therefore, we can appeal to [29] to obtain a well-posedness result for (6)aswellasforthe nonhomogeneous problem u(t)=Au(t)+f(t)t⩾0 u(0) =u0 .(9) So, one can prove the following, where we recall that D((−A)1/2)=H1 α(0, ) for the operator Adefined in (7). Theorem 2.3. Assume (5). Then, A is the infinitesimal generator of a strongly continuous semigroup of contractions, etA. Moreover, etA is analytic. Therefore, for any u0∈H,the solution u(x,t)=(etAu0)(x)of problem (6)satisfies: (a)u∈C([0, ∞); H), (b)t→ u(·,t)is analytic as a map (0, +∞)→H, (c)u(t)∈n⩾1D(An)∀t>0. Furthermore, if u0∈D((−A)1/2), then for any f ∈L2(0, T;H)the mild solution u(t):=etAu0+t 0 e(t−s)Af(s)ds(t∈[0, T]) of problem (9)belongs to H1(0, T;H)∩C([0, T]; D((−A)1/2)) ∩L2(0, T;D(A)) and satisfies the equation in (9)for a.e. t ∈[0, T]. Consequently,the equationsin (1)and(2) are satisfied in classical sense on (0, )×(0, +∞), as well as boundary conditions, taking into account remark 2.1. As for the initial condition, we recall that u0∈L2(0, ) implies that u∈C0([0, +∞); L2(0, )) and u0∈H1 α(0, ) implies that u∈C0([0, +∞); H1 α(0, )). Hereafter, we will assume solutions as smooth as required. 3. Reconstruction by energy methods 3.1. The linear case: uniqueness and Lipschitz stability Let us consider the linear case in which a(x)≡a,whereais a constant such that 0 <a⩽a⩽a and α=1. This corresponds to the problem (2) with strong degeneracy, that is written as 5 Inverse Problems 37 (2021) 125002 P Cannarsa et al follows: ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ ∂tu−a∂x(x∂xu)=f(x,t), x∈(0, ), t∈(0, T), x∂xu(x,t)|x=0=0u(,t)=0, t∈(0, T), u(x,0)=u0(x), x∈(0, ). (10) In this section we will analyze uniqueness and Lipschitz stability for the inverse problem of the identification of the constant coefficient ain (10) from measurements ∂tu(x,t0)andx∂xu(x,t0), for some t0∈(0, T]andforallx∈(0, ), as is indicated in theorem 3.2. In section 5.1 we will perform numerical reconstructions of arelated to this inverse problem. Remark 3.1. On the other hand, let us note that it is also possible to consider other observations, for example, using the boundary data ∂xu(,t)forallt∈(0, T). The related inverse problem will be the following: given η=∂xu(,·), find asuch that the corresponding solution uaof (10) satisfies ∂xu(,t)=η(t), for all t∈(0, T). However, uniqueness and stability for this second inverse problem are, to our knowledge, open questions that we will consider in a forthcoming paper. Nonetheless, we have satisfactory numerical approaches for this second inverse problem, that we will present in the section 5.2. Let ui,i=1, 2 be the solution of (10) with a=ai,thatistosay ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ ∂tui−ai∂x(x∂xui)=f(x,t), x∈(0, ), t∈(0, T), x∂xui(x,t)|x=0=0ui(,t)=0, t∈(0, T), ui(x,0)=u0(x), x∈(0, ). (11) The main result of this section is the following: Theorem 3.2. Assume that a1,a2∈Rsatisfy 0<a⩽a1,a2⩽a<+∞(with a and a given positive constants),u 0∈L2(0,),u0=0, and f ∈L2((0, )×(0, T)).Letu 1,u2be solutions of (11)satisfying the following hypothesis: for some t0∈(0, T]and μ>0such that  0 x|∂xui(x,t0)|2dx⩾μ,i=1, 2.(12) Then there exists a positive constant C =C(a ¯,μ)such that |a2−a1|⩽C 0|∂tu1(x,t0)−∂tu2(x,t0)|2+x|∂xu2(x,t0)−∂xu1(x,t0)|2dx1/2 .(13) Remark 3.3. Notice that, as we have mentioned before, we can obtain a similar uniqueness result for more general equations containing zero order terms. Remark 3.4. Recall that, since etA is analytic, ui(t,·)∈H2 1(0, ) for a.e. t>0, i=1, 2. Moreover, assuming in addition that f≡0, we have the following. (a) By backward uniqueness we know that, if u0=0, then the solution uof (10) satisfies  0 x|∂xu(x,t)|2dx>0∀t∈[0, T].(14) This explains the role played by assumption (12) above. 6 Inverse Problems 37 (2021) 125002 P Cannarsa et al (b) Since Ais dissipative, we have that  0 x|∂xui(x,t)|2dx⩾ 0 x|∂xui(x,T)|2dx∀t∈[0, T]. Therefore, μ=min i=1,2 0 x|∂xui(x,T)|2dx is a possible choice of μfor every t0∈(0, T]. (b) Another way to construct μ>0 satisfying (12)istotakeu0∈D(A), observing that  0 x|∂xu(t,x)|2dx⩾ 0 x|∂xu0(x)|2dx−2t 0|(x∂xu0)x|2dx. Therefore, (12) holds with μ=1 2 0x|∂xu0(x)|2dxfor all 0<t0<1 4 0x|∂xu0(x)|2dx  0|(x∂xu0)x|2dx. Proof of theorem 3.2.We assume that a2⩾a1and set w=u1−u2. Then, we can write ∂tw−a1∂x(x∂xw)=(a1−a2)∂x(x∂xu2). Multiplying by u2and integrating by parts this equality, we get (a2−a1) 0 x|∂xu2|2dx= 0 u2∂twdx+a1 0 x∂xw∂xu2dx ⩽ 0|u2|2dx1/21 0|∂tw|2dx1/2 +a1 0 x|∂xw|2dx1/2 0 x|∂xu2|2dx1/2 . So, by (4), we have |a2−a1| 0 x|∂xu2|2dx ⩽Cp 0|∂tw|2dx1/2 +a1 0 x|∂xw|2dx1/2 × 0 x|∂xu2|2dx1/2 , (15) which in turn yields the the conclusion in view of (12) and ends the proof of this theorem.  It is clear that from the proof of this theorem we can deduce the following uniqueness result. Corollary 3.5. Let us assume that a1,a2⩾a>0,f≡0, and u0∈L2(0, ), u0=0.If u1,u2∈C0([0, T]; L2(0, )) are solutions of (11)such that for some t0∈(0, T]we have ∂tu1(x,t0)=∂tu2(x,t0)and ∂xu2(x,t0)=∂xu1(x,t0)for all x ∈(0, ), then a1=a2. 7 Inverse Problems 37 (2021) 125002 P Cannarsa et al Indeed, it is enough to realize that we do not need hypothesis (12) for uniqueness, because we do not need the constant in (15) to be uniform. Therefore, form (15) and thanks to the inequality (14), we obtain uniqueness. 3.2. The power-like case: uniqueness and Lipschitz stability In this section we will consider the inverse problem of identification of the power αin the degenerate parabolic problems (1)and(2), assuming for simplicity that a(x)≡1andf≡0. That is to say, that we will consider the following problems: with weak degeneracy(0 ⩽α<1) ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ ∂tu−∂x(xα∂xu)=0, (x,t)∈(0, )×(0, T), u(0, t)=0, u(,t)=0, t∈(0, T), u(x,0)=u0(x), x∈(0, ), (16) and with strong degeneracy (1 ⩽α<2) ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ ∂tu−∂x(xα∂xu)=0, (x,t)∈(0, )×(0, T), xα∂xu(x,t)|x=0=0, u(,t)=0, t∈(0, T), u(x,0)=u0(x), x∈(0, ). (17) Similarly to what we have done in the section 3.1, we will analyze uniqueness and stability for the inverse problem of identification of the power αin the above problems from measurements ∂tu(x,t0)andx∂xu(x,t0), for some t0∈(0, T]andforallx∈(0, ). In section 5.3 we will perform the numerical reconstruction of αfor the problems (16)and(17). As we have noted in the previous section, it is also possible to consider the inverse problem of identification of the power αwith only one boundary observation, for example ∂xu(,t)for all t∈(0, T). However, the uniqueness and stability related to this second inverse problem are, to our knowledge, open questions. Nevertheless, we have satisfactory numerical approaches for this second inverse problem, that we will expose in section 5.4. In this section, we analyze uniqueness and stability for the power αin (17)and(16). Let us start with a uniqueness result. Theorem 3.6. Let assume that 0<⩽1, u0∈L2(0, ), u0=0and ui,i=1, 2, be the solution of (17)(or (16)) corresponding to αisuch that for some t0∈(0, T]we have ∂tu1(x,t0)= ∂tu2(x,t0)and ∂xu1(x,t0)=∂xu2(x,t0)for all x ∈(0, ), then α1=α2. Proof of theorem—we can assume, without loss of generality, that α1<α 2. Setting w(x,t)=u2(x,t)−u1(x,t), we have ∂tw−∂x(xα2∂xw)=∂x((xα2−xα1)∂xu1).(18) Then, multiplying this equality by u1and integrating by parts, we obtain  0 (xα1−xα2)|∂xu1|2dx= 0 (u1∂tw+xα2∂xw∂ xu1)dx.(19) 8 Inverse Problems 37 (2021) 125002 P Cannarsa et al for arbitrary μ∈(0, 1). We can verify that Aa,αusatisfies (25) in the sense of distributions. The uniqueness of solution to (25) yields v=Aa,αu∈C1([0, ∞); H1(μ, 1)), and so u∈C1([0, ∞); H3(μ,1)). Thus the proof of lemma 4.5 is complete.  Next, we introduce an integral transform called the Reznitskaya transform (see [33], pp 213–215) as follows: (Kη)(t):=∞ 0 η(τ)G(t,τ)dτ,η∈L∞(0, ∞), (36) where G(t,τ)=1 √πte−τ2 4t,t,τ>0. We note that ∂k τG(t,·)∈L∞(0, ∞)∩L1(0, ∞), ∂k tG(·,τ)∈L∞(0, ∞)forallt,τ>0andk∈N. (37) Then, the following result can be proved. Lemma 4.6. Let η∈W2,∞(0, ∞)satisfy η(0) =0.Then d dt(Kη)(t)=Kd2η dτ2(t), t>0. Proof of lemma 4.6.The proof is given in [33], pp 213–214, and for completeness we will here present the proof. Since ∂tG(t,τ)=∂2 τG(t,τ)fort>0andτ>0, and η∈W2,∞(0, ∞), we have d dt(Kη)(t)=∞ 0 η(τ)∂tG(t,τ)dτ=∞ 0 η(τ)∂2 τG(t,τ)dτ =η(τ)∂τG(t,τ)−η(τ)G(t,τ)τ=∞ τ=0+∞ 0 η(τ)G(t,τ)dτ,t>0. Using that η(0) =0and∂τG(t,0)=0fort>0, we obtain d dt(Kη)(t)=∞ 0 η(τ)G(t,τ)dτ=Kd2η dt2(t), t>0. The proof of lemma 4.6 is complete.  15 Inverse Problems 37 (2021) 125002 P Cannarsa et al We will also need the following result. Lemma 4.7. (i)Let ˜ua,αsatisfy (32)with (a,α,u0)∈A(ε0,ε1,δ0,δ1), and let us set Va,α(x,t):=(K˜ua,α(x,·))(t)=∞ 0 ˜ua,α(x,τ)G(t,τ)dτ,0<x<1, t>0.(38) Then ∂tVa,α(x,t)=∂x(xαa(x)∂xVa,α(x,t)) for almost all x ∈(0, 1) and t >0 (39) and Va,α(0, t)=Va,α(1, t)=0, t>0, Va,α(x,0)=u0(x), for almost all x ∈(0, 1). (40) (ii)Let ua,αbe the solution to (25)where (a,α,u0)∈A(ε0,ε1,δ0,δ1).Then ua,α(x,t)=∞ 0 ˜ua,α(x,τ)G(t,τ)dτ,0<x<1, t>0.(41) Proof of lemma 4.7. Part (ii) follows readily from part (i) by the uniqueness of the solution ua,α(see theorem 2.3 in section 2). Therefore, it suffices to prove part (i). Henceforth, we set (g,h):=1 ε2 g(x)h(x)dx, with arbitrary ε2>0. For θ∈C∞ 0(ε2,1),weset ˜w(t)=(˜ua,α(·,t), θ)=1 ε2 ˜ua,α(x,t)θ(x)dxand w(t)=(Va,α(·,t), θ)fort>0. Then, the initial condition in (32) and the regularity (35) yield ˜w∈W2,∞(0, ∞), ∂t˜w(0) =0 and w(t)=(Va,α(·,t), θ)=1 ε2 Va,α(x,t)θ(x)dx =∞ 0 ˜w(τ)G(t,τ)dτ=(K˜w)(t), t>0. Therefore, lemma 4.6 implies dw dt(t)=d dt(K˜w)(t)=Kd2˜w dτ2(t)=∞ 0 d2 dτ2˜w(τ)G(t,τ)dτ, 16 Inverse Problems 37 (2021) 125002 P Cannarsa et al that is, d dt(Va,α(·,t), θ)=∞ 0 d2 dτ2(˜ua,α(·,τ), θ)G(t,τ)dτ =∞ 01 ε2 ∂2˜ua,α(x,τ) ∂τ2θ(x)dxG(t,τ)dτ =1 ε2∞ 0 ∂2˜ua,α ∂τ2(x,τ)G(t,τ)dτθ(x)dx =K∂2˜ua,α ∂τ2(·,t), θ. Therefore, ∂Va,α ∂t(·,t), θ=K∂2˜ua,α ∂τ2,θ(42) for all θ∈C∞ 0(ε2,1). Recall that Aa,αw(x,t)=∂x(xαa(x)∂xw(x,t)), 0 <x<1, t>0. By (33), (35)and(37), we can justify the exchange of the operation by Aa,αand ∞ 0...dτin (38), and obtain Aa,αVa,α(x,t)=∞ 0 (Aa,α˜ua,α)(x,τ)G(t,τ)dτ =K(Aa,α˜ua,α)(x,t), ε2<x<1, t>0. Hence, using (42) and since ˜ua,αis the solution of (32), we have ∂Va,α ∂t(·,t)−Aa,αVa,α(·,t), θ=K∂2˜ua,α ∂τ2−Aa,α˜ua,α(·,t), θ=0 for all θ∈C∞ 0(ε2,1)andt>0. Consequently, since ε2>0 is arbitrary, we reach ∂Va,α ∂t(x,t)=∂ ∂xxαa(x)∂Va,α ∂x(x,t) for almost all x∈(0, 1) and t>0. Therefore, (39) holds. By ˜ua,α(0, t)=˜ua,α(1, t)=0fort>0, the first condition in (40) is directly seen. By the change of variables ξ=τ 2√t,wehave Va,α(x,t)=∞ 0 ˜ua,α(x,τ)G(t,τ)dτ=2 √π∞ 0 ˜ua,α(x,2ξ√t)e−ξ2dξ,t>0. (43) Since ˜ua,α∈C([0, ∞); C([ε2, 1])), and using that ∂t˜ua,α∈L∞(0, ∞;H1(ε2,1))in(35), we have lim t→0 ˜ua,α(x,2ξ√t)=˜ua,α(x,0) 17 Inverse Problems 37 (2021) 125002 P Cannarsa et al for almost all x∈(ε2,1)andξ>0, and by ˜ua,α∈L∞(0, ∞;H1(ε2,1))in(35), we apply the Lebesgue convergence theorem to (43) and obtain Va,α(x,0)=2 √π∞ 0 ˜ua,α(x,0)e −ξ2dξ=˜ua,α(x,0) 2 √π∞ 0 e−ξ2dξ=u0(x), for almost all x∈(ε2, 1), which is the second condition in (40). Since ε2>0 is arbitrary, the proof of lemma 4.7 is complete.  Moreover, we prove Lemma 4.8. Let us assume that ∂xua,α(1, t)=∂xub,β(1, t)for 0<t<t0. Then, ∂x˜ua,α(1, t)=∂x˜ub,β(1, t), 0 <t<∞. Proof of lemma 4.8. Since ua,α(x,·) is analytic in t>0 with fixed x∈[ε2, 1], by theorem 2.3(b) in section 2, we see that ∂xua,α(1, t)=∂xub,β(1, t)forallt>0. Therefore, by (41)we have ∞ 0 (∂x˜ua,α−∂x˜ub,β)(1, τ)G(t,τ)dτ=0, t>0.(44) On the other hand, in the definition (36)ofK, we change the variables in the expression of K by r:=1 4tand τ:=√z,andwesee (Kη)1 4r=r π∞ 0 e−rz η(√z) √zdz,r>0, for η∈L∞(0, ∞). Thus, the uniqueness of the Laplace transform implies that η≡0if(Kη)(t)=0fort>0 and η∈L∞(0, ∞). Since (∂x˜ua,α−∂x˜ub,β)(1, ·)∈L∞(0, ∞)by(35), it follows from (44)that ∂x˜ua,α(1, t)=∂x˜ub,β(1, t)fort>0. Therefore, the proof of lemma 4.8 is complete.  4.2.2. Second step: uniquenessfor the linearized inverse hyperbolic problem. Here we establish the uniqueness result for the following hyperbolic inverse source problem. We recall that a sufficiently small constant ε1>0 is given. Moreover let p0>0 be arbitrarily given. Given p∈C2([ε1, 1]) with p⩾p0>0on[ε1,1],lety=y(x,t) satisfy ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ ∂2 ty=∂x(p(x)∂xy)+∂x(f(x)∂xR(x,t)), ε1<x<1, t>0, y(1, t)=0, t>0, y(x,0)=∂ty(x,0)=0, ε1<x<1, (45) where R(x,t) is a given function. In what follows, we show the uniqueness for an inverse problem of determining f(x)for ε1<x<1 by boundary data ∂xy(1, t)forallt>0. 18 Inverse Problems 37 (2021) 125002 P Cannarsa et al Proposition 4.9. Let p ∈C2([ε1,1]) with p >0on [ε1,1] and let f ∈H1(ε1,1) satisfy f(1) =0. We further assume that there exists a constant r0>0such that ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ ∂tR(x,0)=0, |∂xR(x,0)|⩾r0>0, ε1⩽x⩽1, ∂t∂xR,∂t∂2 xR∈L∞ loc((ε1,1)×(0, ∞)), ∂2 xR(·,0)∈L∞(ε1,1). (46) If y ∈H2 loc((ε1,1)×(0, ∞)) satisfies (45),∂ty∈H2 loc((ε1,1)×(0, ∞)) and ∂xy(1, t)=0, t>0, then f(x)=0for ε1<x<1. Notice that in this proposition, as is seen by the proof, we need not assume ∂xy(1, t)=0for all t>0, but ∂xy(1, t)=0for0<t<T∗with some T∗>0 which is determined by p(x), is sufficient. However for our purpose we can assume ∂xy(1, t)=0forallt>0. The rest of this subsection is devoted to the proof of proposition 4.9. Proof of proposition 4.9.The methodology for the proof is based on Bukhgeim and Klibanov method from [6], and also on the arguments by Imanuvilov and Yamamoto from [21,22]. Recently, Huang et al in [17] have simplified those arguments and omit the cut-off procedures. Here we apply such argument form [17]. The key idea is that we take a suitable weight function that takes smaller values on the boundary of a domain in (x,t) where data are not given, so that the weight function can well control such unknown boundary values and therefore the cut-off function is not necessary. The proof will be divided into three steps. First step of proposition 4.9: Carleman estimates. Henceforth, with arbitrarily chosen constants M>0andp0>0, we assume p∈C2([ε1, 1]), p⩾p0>0on[ε1,1], pC2([ε1,1]) ⩽M. Furthermore, we fix a sufficiently small constant γ>0 and a sufficiently large constant λ>0. Here and henceforth, for Mand r0, we choose θ∈(0, 1) sufficiently close to 1 and sufficiently small δ>0 such that ε1<θ+δ<1 is always satisfied. We set ψ(x,t):=(x−θ)2−γt2,ϕ(x,t)=eλψ(x,t) and Qδ:={(x,t); −∞ <t<∞,θ+δ<x<1, ψ(x,t)>δ 2}, T0:=(1 −θ)2−δ2 γ. 19 Inverse Problems 37 (2021) 125002 P Cannarsa et al Figure 1. Set Qδ. Figure 2. Set Q− δ. Then, we note that Qδis bounded by a hyperbolic curve (x−θ)2−γt2=δ2and x=1, and (x,t)∈Qδimplies θ+δ<x<1and|t|<T0,Qδ∩{t=0}=(θ+δ, 1), (47) see figure 1. Henceforth C,C0>0, etc will denote generic constants which are independent of the parameter s>0, but can be dependent on other parameters such as λ>0. Since other parameters such as λare fixed, we do not specify the dependence of C,C0, etc on them. We show the following result: 20 Inverse Problems 37 (2021) 125002 P Cannarsa et al Lemma 4.10 (Hyperbolic Carleman estimate). There exist constants C >0and let it be s0>0depending on M and r0,but independent of s and the choices of p,such that Qδ (s|∂xy|2+s|∂ty|2+s3|y|2)e2sϕdxdt ⩽CQδ|F(x,t)|2e2sϕdxdt+C∂Qδ (s|∂xy|2+s|∂ty|2+s3|y|2)e2sϕdS for all s ⩾s0and y ∈H2(Qδ)satisfying ∂2 ty=∂x(p(x)∂xy)+FinQ δ with some F ∈L2(Qδ). In particular, if y(1, t)=∂xy(1, t)=0for −T0<t<T0,then Qδ (s|∂xy|2+s|∂ty|2+s3|y|2)e2sϕdxdt ⩽CQδ|F(x,t)|2e2sϕdxdt+Cs3e2seλδ2 y2 H2(Qδ)(48) for all s ⩾s0and y ∈H2(Qδ). Proof of lemma 4.10.This is a classical Carleman estimate: the proof can be done similarly to theorem 4.2 in Bellassoued and Yamamoto [6] or Imanuvilov [19]. Here we omit the details of the proof. Estimate (48) can follow because ψ(x,t)=δ2for (x,t)∈∂Qδ\{x=1}and ∂xyL2(∂Qδ)+∂tyL2(∂Qδ)+yL2(∂Qδ) ⩽C(∂xyH1(Qδ)+∂tyH1(Qδ)+yH1(Qδ))⩽CyH2(Qδ) by the trace theorem.  Remark 4.11. For Carleman estimates for hyperbolic equations, we usually have to assume a certain extra condition for the principal part p(x) e.g., about the wave speed, and the extra condition is restrictive in higher spatial dimensions. However, the one spatial dimension case is exceptional and does not require any extra condition thanks to the smallness of the domain Qδ. More precisely, in the one dimensional case, such an extra condition is described as  p(x) 2p(x)(x−θ) <1, θ+δ⩽x⩽1 (see e.g., proposition 2.1 in Imanuvilov and Yamamoto [22]). Since maxθ+δ⩽x⩽1|x−θ|= 1−θis sufficiently small, we see that  p(x) 2p(x)(x−θ) ⩽M 2r0 (1 −θ)<1.(49) Hence, there exists a constant μ∗(r0,M)>0suchthatif0<1−θ< μ ∗(r0,M), then (49)and so the Carleman estimate (48) hold. 21 Inverse Problems 37 (2021) 125002 P Cannarsa et al In view of (49), the key Carleman (48) estimate holds near x=1 where Cauchy data are given, that is, for θ+δ<x<1. Thus, we first apply the Carleman estimate only for θ+ δ<x<1 to prove that f=0in(θ+δ, 1). This will be done in second step of the proof of proposition 4.9. Next, we repeat the same argument by replacing x=1byx=θ, and this will be done in third step of the proof of proposition 4.9. Before proceeding to second step, we prove a Carleman estimate for a simple operator d dx. We set ϕ0(x)=eλ(x−θ)2 with fixed large constant λ>0. Lemma 4.12. There exist constants s0>0and C >0such that s1 θ+δ|f(x)|2e2sϕ0(x)dx⩽C1 θ+δ|f|2e2sϕ0dx, for all s ⩾s0and all f ∈H1(θ+δ,1)satisfying f(1) =0. Proof of lemma 4.12.The proof relies on the estimation of h:=fesϕ0and the integration by parts, which is a conventional proof of the Carleman estimate. We set h(x):=f(x)esϕ0(x),Lh(x):=esϕ0(x)d dx(e−sϕ0(x)h(x)) =h(x)−sϕ 0(x)h(x). Then, integrating by parts and taking into account that f(1) =0, we have that there exists a positive constant Csuch that 1 θ+δ|f|2e2sϕ0dx=1 θ+δ|Lh|2dx=1 θ+δ|h(x)−sϕ 0(x)h(x)|2dx ⩾−2s1 θ+δ ϕ 0(x)h(x)h(x)dx=−s1 θ+δ ϕ 0(x)d dxh2(x)dx =sϕ 0(x)h2(x)x=θ+δ x=1+s1 θ+δ ϕ 0(x)h2(x)dx =2λsδeλδ2h2(θ+δ)+s1 θ+δ (2λϕ0(x) +4λ2(x−θ)2ϕ0(x)h2(x)dx ⩾s1 θ+δ 2λeλ(x−θ)2h2(x)dx⩾Cs1 θ+δ|f|2e2sϕ0dx. Thus, the proof of lemma 4.12 is complete.  Second step of the proof of proposition 4.9. 22 Inverse Problems 37 (2021) 125002 P Cannarsa et al We take the even extension in tof the solution yof (45)to(−∞, 0), and we denote it by the same letter: y(x,t)=y(x,t), t⩾0, y(x,−t), t<0. Also, for R,wemaketheevenextensionint<0. Then, using that y(·,0)=∂ty(·,0)=0and ∂tR(·,0)=0on[ε1,1]by(46), we can verify that ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ ∂ty,y∈H2 loc((ε1,1)×(−∞,∞)), ∂tR,R∈L2 loc(−∞,∞;H2(ε1, 1)), ∂t∂xR,∂t∂2 xR∈L∞ loc((ε1,1)×(−∞,∞)) (50) and ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ ∂2 ty=∂x(p(x)∂xy)+∂x(f(x)∂xR(x,t)), ε1<x<1, −∞ <t<∞, y(1, t)=∂xy(1, t)=0, −∞ <t<∞, y(x,0)=∂ty(x,0)=0, ε1<x<1. (51) In (51), we see that y,∂ty∈H2((ε1,1)×(−T,T)) with arbitrarily given T>0. Setting z:=∂ty∈H2((ε1,1)×(−T,T)), and using (51), we have ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ ∂2 tz=∂x(p(x)∂xz)+∂x(f(x)∂t∂xR(x,t)), ε1<x<1, −∞ <t<∞, z(1, t)=∂xz(1, t)=0, −∞ <t<∞, z(x,0)=0, ∂tz(x,0)=∂x(f(x)∂xR(x, 0)), ε1<x<1. (52) Notice that the second and the third conditions in (52) are obtained by (51) taking into account that ∂2 ty∈H1(−T,T;L2(ε1, 1)) and the Sobolev embedding such as H1(−T,T;L2(ε1,1))⊂ C0([−T,T]; L2(ε1, 1)). Here, we also note that ∂t∂2 xy,∂t∂2 xR∈L2(−T,T;L2(ε1, 1)), and so ∂2 xy,∂2 xR∈C0([−T,T]; L2(ε1, 1)) by the Sobolev embedding. Setting M0:=zH2((ε1,1)×(−T,T)) =∂tyH2((ε1,1)×(−T,T)), by the regularity assumption (50)ony, we see that M0<∞. Now, we apply the Carleman estimate (48)to(52) and we obtain that there exists a constant C>0, which is independent of s>0, such that 23 Inverse Problems 37 (2021) 125002 P Cannarsa et al Qδ (s|∂xz|2+s|∂tz|2+s3|z|2)e2sϕdxdt ⩽CQδ|∂x(f(x)∂x∂tR(x,t)|2e2sϕdxdt+Cs3e2seλδ2 M2 0, (53) for all large s>0. We set w:=(∂ty)esϕ=zesϕ. Then, direct calculations yield ∂2 tw=(∂3 ty)esϕ+2s(∂2 ty)(∂tϕ)esϕ+(s∂2 tϕ+s2(∂tϕ)2)) esϕ∂ty. On the other hand, p∂2 xw=p(∂2 x∂ty)esϕ+2ps(∂t∂xy)(∂xϕ)esϕ+p(s(∂2 xϕ)+s2(∂xϕ)2)) esϕ∂ty, and also p∂xw=p(∂x∂ty)esϕ+sp(∂xϕ)(∂ty)esϕ. Therefore, taking the derivative with respect to tin (45) and multiplying by esϕ, we can write ∂2 tw−∂x(p(x)∂xw)=esϕ∂x(f(x)∂x∂tR(x,t)) +Ps(w), (54) where the term Ps(w) satisfies |Ps(w)|⩽Cs(|∂2 ty|+|∂ty|+|∂x∂ty|)esϕ+Cs2|∂ty|esϕ ⩽C(s2|w|+s|∂tw|+s|∂xw|), (x,t)∈Qδ, (55) for all large s>0. For the last inequality, we have used that w=(∂ty)esϕand (∂2 ty)esϕ=∂tw−s(∂tϕ)(∂ty)esϕ=∂tw−s(∂tϕ)w, and (∂x∂ty)esϕ=∂xw−s(∂xϕ)w, which imply |∂2 ty|esϕ⩽|∂tw|+Cs|∂ty|esϕ and |∂x∂ty|esϕ⩽|∂xw|+Cs|∂ty|esϕ, respectively. On the other hand, it is not difficult to see that we can write (53)intermsofw=zesϕ= (∂ty)esϕ Qδ (s|∂xw|2+s|∂tw|2+s3|w|2)dxdt ⩽CQδ (∂x(f(x)∂x∂tR(x,t))|2e2sϕdxdt+Cs3e2seλδ2 M2 0, (56) 24 Inverse Problems 37 (2021) 125002 P Cannarsa et al 5. Numerical results In this section, we will consider numerical results for the previous inverse problems. We will carry out the reconstruction of the unknown function aand the power αthrough the resolution of some appropriate extremal problems. This strategy has been applied in some previous papers for other similar problems, see [2,3,13,26,27]. The results of the numerical tests that follow pretend to illustrate the theoretical results in the previous sections. We start by showing numerical experiments related to the theoretical result of theorem 3.2 from section 3.1,then we consider other type of observations as indicated in remark 3.1 (see sections 5.1 and 5.2). Similarly, we continue with the theoretical result from section 3.2 (see sections 5.3 and 5.4). Finally, we conclude with numerical reconstruction related to section 4(see section 5.5). 5.1. Numerical tests for the linear case I In this section we present the numerical reconstruction of the constant coefficient ain (10)by distributed measurements, related to the theoretical result from theorem 3.2. More precisely, we will deal with the following inverse problem: given t0∈(0, T)asin(12), u0=u0(x), γ= ∂tu(·,t0)andβ=u(·,t0), find a∈Ua ad such that the corresponding solution uaof (10) satisfies the additional conditions: ∂tua(x,t0)=γ(x), x∂xua(x,t0)=β(x), ∀x∈(0, ), and the admissible set is given by Ua ad ={a:0<a⩽a⩽a}. We reformulate the inverse problem as the following extremal problem: find a∈Ua ad such that J(a)⩽J(a), ∀a∈Ua ad, (66) where the functional J:a∈Ua ad → Ris given by J(a)=1 2 0|γ(x)−∂tua(x,t0)|2dx+1 2 0|β(x)−x∂xua(x,t0)|2dx, (67) with uathe solution of (10) corresponding to the unknown a. We have performed several numerical tests with different values of ato recover: a>1, a=1anda<1. In them, we fix T=5, =1andu0(x)=0.5x2(1 −x). In the first three tests we will take, for simplicity, f(x,t)=0. We will denote by adthe desired value of athat we would like to obtain and by aithe initial guess for the optimization algorithm. Using an optimization algorithm from Optimization Toolbox of MatLab, the fmincon function and interior-point gradient based algorithm, we have computed ac. Our goal is to obtain ac as close as possible to ad. Test 1. Let us take ai=0.7as an initial guess for the minimization algorithm and t0=0.2, so that we have computed numerically such that (12)holds. The desired value to recover is ad=1.7. After numerical resolution of the optimization problem (66), we found ac=1.7. The numerical results can be seen in figure 3, where the desired adand computed ac have been represented. In figure 4we have plotted the evolution of the cost at each iterations of the optimization algorithm represented by points. In table 1we can see the evolution of the cost when we introduce random noises in the target. 31 Inverse Problems 37 (2021) 125002 P Cannarsa et al Remark 5.1 (Numerical illustration of stability). In order to illustrate the stability of the considered inverse problem in the test 1, we have represented in figure 5as points the quotients Kj=|a−a∗| γ(x)−γ∗ j(x)+β(x)−β∗ j(x),j=1, ..., 50, (68) where γ(x)=∂tua(x,t0), β(x)=x∂xua(x,t0)andforall j,γ∗ j(x)=γ(x)+εj,β∗ j(x)=β(x)+ εjwith the εjrandom perturbations in (0, 0.05). We can see that all the quotients Kjare bounded. Test 2. Let us now take ai=0.2, t0=0.2and the desired value to recover ad=1.The optimization algorithm gives ac=1, see figure 6. In table 2we can see the evolution of the cost when we introduce random noises in the target. Test 3. Let us take ai=0.7as an initial guess for the minimization algorithm, t0=0.2and ad=0.2. The numerical resolution of the optimization problem gives ac=0.2(see figure 7 and table 3). Test 4. We consider a non-homogeneous problem (10)with f(x,t)=2xt. Let us take ai= 0.7and t0=0.4. The desired value is ad=1.7. The numerical resolution of the optimization problem gives ac=1.7. Figure 8shows the evolution of iterations of the optimization algorithm and figure 9the evolution of the cost. In table 4we present the results with random noises in the target and figure 10 represents the solution corresponding to the computed ac. 5.2. Numerical tests for the linear case II In this section we will present some numerical results for second inverse problem of identification of afrom one boundary observation ux(1, t), mentioned at the beginning of the section 3.1. Arguing as in the previous section, we reformulate the inverse problem as an optimization problem (66), where the funcional J:a∈Ua ad → Ris now given by J(a)=1 2T 0|η(t)−∂xua(1, t)|2dt, (69) where η=∂xu(1, ·) is given and uais the solution of (10) corresponding to the unknown a. In what follows we present some numerical tests for this inverse problem. We will consider the same data as in the previous section: T=5, =1, u0(x)=0.5x2(1 −x)andf(x,t)=0. In order to simplify the presentation, we will show just the evolution of the cost and the results with random noises. Test 5. We take ai=0.7as an initial guess for the minimization algorithm. The desired value is ad=1.7. The results of the numerical computations are presented in table 5and figure 11. Test 6. Taking ai=0.2and ad=1, we can see the numerical results in figure 12 and table 6. Test 7. Taking ai=0.7and ad=0.2, figure 13 and table 7show the numerical results. 32 Inverse Problems 37 (2021) 125002 P Cannarsa et al 5.3. Numerical tests for the power-like case I We will present in this section some numerical results of the reconstruction of αfor weak and strong degenerate cases (16)and(17) from data that appear in theorem 3.8.Moreprecisely, given t0∈(0, T)asin(20), u0=u0(x)and<1, in order to reconstruct αwe will solve numerically the following optimization problem: find α∈(0, 2) such that J(α)⩽J(α), ∀α∈(0, 2), (70) where the functional J: (0, 2) → Ris given by J(α)=1 21 0|γ(x)−∂tuα(x,t0)|2dx+1 21 0|β(x)−x2∂xuα(x,t0)|2dx, with uαthe corresponding solution of (16)or(17) associated to to the unknown power α. In the numerical tests we will take T=10, u0=0.3x2(1 −x)2and f(x,t)=0 and we will use the interior-point MatLab gradient free algorithm from Optimization Toolbox. 5.3.1. Weak degeneracy I. Test 8. Let us first assume that 0<α<1, corresponding to the weak degeneracy case (16). We take αi=0.8as an initial guess for the minimization algorithm and the desired value to recover is αd=0.4. In figure 14 we can see the computed values of αat each iteration of the optimization algorithm and figure 15 shows the evolution of the cost. In the table 8,wepresent the computed values of αcdepending on the values of close to 1. 5.3.2. Strong degeneracy I. Test 9. Let us consider the strong degeneracy case with 1⩽α<2, i.e. the problem (17). We take αi=1.6and αd=1.3. The numerical results can be seen in the figures 16 and 17.In the table 9, we present the computed values of αcdepending on the values of close to 1. Remark 5.2. From the results of the tests 8and 9we can see numerically that the hypothesis <1 from theorem 3.8 probably is not really needed for the reconstruction of α.Butthe question whether we can eliminate this condition form this theorem remains open. 5.4. Numerical tests for the power-like case II We will present in this section numerical results for the reconstruction of αfrom one observation of the form ∂xu(1, t). As before, we will analyze first weak and then strong degenerate cases. More precisely, given ,u0=u0(x)andη=∂xu(1, ·), we will solve numerically the following optimization problem (70), with the functional J: (0, 2) → Rgiven by J(α)=1 2T 0|η(t)−∂xuα(1, t)|2dt, where uαis the solution to (16)or(17) corresponding to α. In that follows, we will take T=10, =1, u0=0.3x2(1 −x)2and f(x,t)=0. The computations are performed using fmincon MatLab function formOptimization Toolbox,combined with interior-point gradient algorithm. 33 Inverse Problems 37 (2021) 125002 P Cannarsa et al 5.4.1. Weak degeneracy II. Test 10. Let us first consider the weak degeneracy case corresponding to the problem (16) with 0<α<1.Wetakeαi=0.2as an initial guess for the minimization algorithm and the desired value to recover αd=0.6. The numerical results can be seen in figures 18 and 19. Table 10 shows the computations of αcand the cost with random noises in the target. Remark 5.3 (On the order of convergence). We can analyze the order of convergence of αnwhich we obtain at each iteration nof the optimization algorithm to αd. More precisely, we look for positive real numbers Cand κsuch that |en+1|≈C|en|κ, where en=αn−αd. From figure 20, we see that this approximate formula is satisfied for κ=1.8064 for some finite value of C. 5.4.2. Strong degeneracy II. We will consider here the identification of αin the case 1 ⩽ α<2 for strong degeneracy case, that is to say for the problem (17). For simplicity of the presentation, the results with random noises in the targets are not presented, because they are similar to those of the previous sections. Test 11. Let us take =0.9and αi=1.6as an initial guess for the minimization algorithm. The desired value of αto recover is αd=1.3. The numerical results can be seen in the figures 21 and 22 and the computed value of αis αc=1.3. Test 12. Let us take =1and αi=1.6and αd=1.3. The numerical results can be seen in figures 23 and 24 and the computed value is αc=1.3. 5.5. Numerical tests for a general case In this section we will present numerical results of identification of a function a=a(x)in(25) from ∂xu(1, t) in order to illustrate the theoretical part form the section 4. Let us take =1, T=5, α=0.6andu0(x)=0.3x2(1 −x)2. In order to reconstruct a= a(x), again we reformulate the corresponding inverse problem as an optimization problem: find a∈Asuch that J(a)⩽J(a), ∀a∈A, (71) where the functional J:A → Ris given by J(a)=1 2T 0|η(t)−∂xua(1, t)|2dt, where uais the solution of (25) corresponding to aand Ais given by (27). In fact, for the numerical reconstruction we do not need to assume that ais known near 1 as in (27). In order to present some examples, in what follows we will analyze two particular cases of the coefficient a(x): linear and quadratic functions. The numerical experiments will be performed using a gradient based interior-point algorithm of fmincon MatLab Optimization Toolbox. 34 Inverse Problems 37 (2021) 125002 P Cannarsa et al Figure 3. Test 1,adsolid line, acdashed line. Let us notice that we can also reconstruct numerically both αand afrom one measurement, but in order to simplify the presentations of the results, we will omit the details. Linear function: let us consider aof the form a(x)=bx +c,whereband care unknown constants. The optimization problem (71) is written in this case as follows: find b,c∈A, with A={(b,c):0<b<b<b,0<c<c<c}such that J(b,c)⩽J(b,c), ∀(b,c)∈A. Test 13. Let us take bi=1and ci=1as initial guess for the optimization algorithm and desired values to recover bd=5and cd=1.5. After numerical approximation of the optimization problem above we get the computed values bc=5and cc=1.5. The numerical results we have obtained are in figures 25 and 26. Table 11 shows the evolution of the cost during the iterations of optimization algorithm and computed bcand ccwith random noises in the target. Finally, in figure 27 the solution of (25)is represented for the computed values of the coefficients. Quadratic function: let us now consider aof the form a(x)=bx2+cx +h. The optimization problem (71) is written in this case as follows: find (b,c,h)∈A, with A={(b,c,h):0<b< b<b,0<c<c<c,0<h<h<h}such that J(b,c,h)⩽J(b,c,h), ∀(b,c,h)∈A. 35 Inverse Problems 37 (2021) 125002 P Cannarsa et al Figure 4. Test 1, the evolution of the cost. Table 1. Cost at the best stopping and computed acwith random noises in the target in test 1. % noise Cost Iterations ac 5% 1 ×10−517 1.700 831 33 3% 1 ×10−515 1.69 463 291 1% 1 ×10−717 1.704 511 53 0.1% 1 ×10−816 1.700 178 91 0.01% 1 ×10−10 16 1.700 176 65 0.001% 1 ×10−13 15 1.700 003 62 0% 1 ×10−27 20 1.7 Test 14. Let us take bi=3.5, ci=2.5and hi=0.5as initial guess for the optimization algorithm and the desired values to recover bd=4, cd=3and hd=1. The numerical results are shown in the figures 28 and 29, and the computed values are bc=4.065 621 97, cc= 2.886 136 22 and hc=1.049 674 27. The numerical results can be seen in figures 28 and 29. Table 12 shows the evolution of the cost during the iterations of optimization algorithm and computed bcand ccwith random noises in the target. 36 Inverse Problems 37 (2021) 125002 P Cannarsa et al Figure 5. Test 1, numerical illustration of stability. 37 Inverse Problems 37 (2021) 125002 P Cannarsa et al Figure 6. Test 2, the evolution of the cost. Table 2. Evolution of the cost and computed acwith random noises in the target in test 2. % noise Cost Iterations ac 5% 1 ×10−512 0.975 793 57 3% 1 ×10−613 0.996 128 53 1% 1 ×10−612 0.996 702 60 0.1% 1 ×10−813 1.000 127 71 0.01% 1 ×10−14 10 0.999 951 71 0.001% 1 ×10−13 14 0.999 994 88 0% 1 ×10−24 14 1 38 Inverse Problems 37 (2021) 125002 P Cannarsa et al Figure 7. Test 3, the evolution of the cost. Table 3. Evolution of the cost and computed acwith random noises in the target in test 3. % noise Cost Iterations ac 5% 1 ×10−511 0.198 499 99 3% 1 ×10−611 0.197 690 06 1% 1 ×10−711 0.200 461 40 0.1% 1 ×10−13 9 0.199 976 76 0.01% 1 ×10−11 16 0.199 998 93 0.001% 1 ×10−13 14 0.200 000 28 0% 1 ×10−27 16 0.2 39 Inverse Problems 37 (2021) 125002 P Cannarsa et al Figure 8. Test 4, iterations of the algorithm. Figure 9. Test 4, the evolution of the cost. 40 Inverse Problems 37 (2021) 125002 P Cannarsa et al Table 9. Computed αcfor closeto1intest9. Cost Iterations αcCost Iterations αc 0.9 1 ×10−24 18 1.3 1.000 01 1 ×10−22 23 1.300 000 01 0.99 1 ×10−22 24 1.300 000 01 1.0001 1 ×10−23 21 1.3 0.999 1 ×10−21 19 1.300 000 06 1.001 1 ×10−23 25 1.3 0.9999 1 ×10−23 22 1.299 999 99 1.01 1 ×10−24 32 1.3 0.999 99 1 ×10−22 30 1.300 000 01 1.1 1 ×10−22 18 1.299 999 98 11×10−23 27 1.300 000 00 1.2 1 ×10−22 20 1.3 Figure 18. Test 10, identification of α. 47 Inverse Problems 37 (2021) 125002 P Cannarsa et al Figure 19. Test 10, current function values. Table 10. Evolution of the cost and computed αcwith random noises in the target in test 10. % noise Cost Iterations αc 5% 1 ×10−714 0.594 706 02 3% 1 ×10−715 0.616 112 17 1% 1 ×10−815 0.596 894 97 0.1% 1 ×10−10 16 0.600 371 82 0.01% 1 ×10−12 16 0.599 976 72 0.001% 1 ×10−14 18 0.599 998 07 0% 1 ×10−21 17 0.6 48 Inverse Problems 37 (2021) 125002 P Cannarsa et al Figure 20. Test 10, order of convergence. Figure 21. Test 11, identification of α. 49 Inverse Problems 37 (2021) 125002 P Cannarsa et al Figure 22. Test 11, function values. Figure 23. Test 12, identification of α. 50 Inverse Problems 37 (2021) 125002 P Cannarsa et al Figure 24. Test 12, function values. Figure 25. Test 13,adsolid line, acdashed line. 51 Inverse Problems 37 (2021) 125002 P Cannarsa et al Figure 26. Test 13, evolution of the cost. Table 11. Evolution of the cost and computed bcand ccwith random noises in the target in test 13. %CostIt.bccc 51×10−627 4.991 568 07 1.415 068 91 31×10−729 5.070 602 96 1.405 216 65 11×10−829 5.022 230 88 1.474 061 14 0.1 1 ×10−10 29 5.006 369 35 1.495 233 26 0.01 1 ×10−12 29 4.999 897 52 1.500 322 99 0.001 1 ×10−14 29 4.999 940 89 1.500 119 71 01×10−25 39 5 1.5 52 Inverse Problems 37 (2021) 125002 P Cannarsa et al Figure 27. Test 13, computed solution. Figure 28. Test 14,adsolid, acdashed lines. 53 Inverse Problems 37 (2021) 125002 P Cannarsa et al Figure 29. Test 14, evolution of the cost. Table 12. Evolution of the cost and computed bcand ccwith random noises in the target in test 14. % Cost Iterations bccchc 5% 1 ×10−632 4.100 796 95 2.901 687 36 0.991 736 73 3% 1 ×10−734 4.037 814 43 2.866 748 57 1.020 339 48 1% 1 ×10−841 4.059 712 41 2.886 552 92 1.091 051 66 0.1% 1 ×10−10 33 4.068 066 62 2.886 854 41 1.042 748 87 0.01% 1 ×10−12 34 4.065 653 61 2.886 231 99 1.050 383 51 0.001% 1 ×10−12 35 4.065 676 62 2.886 153 26 1.049 531 32 0% 1 ×10−12 33 4.065 621 97 2.886 136 22 1.049 674 27 Acknowledgments The first author was partly supported by Istituto Nazionale di Alta Matematica (GNAMPA 2020 Research Projects) and by the MIUR Excellence Department Project awarded to the Department of Mathematics, University of Rome Tor Vergata, CUP E83C18000100006. The second author was partially supported by MICINN, under Grant MTM2016-76690-P.The third author was supported by Grant-in-Aid for Scientific Research (S) 15H05740 and Grant-in Aid (A) 20H00117 of Japan Society for the Promotion of Science and by the National Natural Science 54 Inverse Problems 37 (2021) 125002 P Cannarsa et al Foundation of China (No. 11771270, 91730303). This paper has been supported by the RUDN University Strategic Academic Leadership Program. Data availability statement All data that support the findings of this study are included within the article (and any supplementary files). ORCID iDs Anna Doubova https://orcid.org/0000-0002-5708-2700 References [1] Alabau-Boussouira F, Cannarsa P and Leugering G 2017 Control and stabilization of degenerate wave equations SIAM J. 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