DSpace at VNU: On an inverse boundary value problem of a nonlinear elliptic equation in three dimensions

Accepted Manuscript On an inverse boundary value problem of a nonlinear elliptic equation in three dimensions Nguyen Huy Tuan, Le Duc Thang, Vo Anh Khoa, Thanh Tran PII: DOI: Reference: S0022-247X(14)01178-0 10.1016/j.jmaa.2014.12.047 YJMAA 19103 To appear in: Journal of Mathematical Analysis and Applications Received date: 22 September 2014 Please cite this article in press as: N.H Tuan et al., On an inverse boundary value problem of a nonlinear elliptic equation in three dimensions, J Math Anal Appl (2015), http://dx.doi.org/10.1016/j.jmaa.2014.12.047 This is a PDF file of an unedited manuscript that has been accepted for publication As a service to our customers we are providing this early version of the manuscript The manuscript will undergo copyediting, typesetting, and review of the resulting proof before it is published in its final form Please note that during the production process errors may be discovered which could affect the content, and all legal disclaimers that apply to the journal pertain On an inverse boundary value problem of a nonlinear elliptic equation in three dimensions Nguyen Huy Tuana,∗, Le Duc Thangb , Vo Anh Khoaa,d , Thanh Tranc a Department of Mathematics, University of Science, Vietnam National University, 227 Nguyen Van Cu Street, District 5, Ho Chi Minh City, Viet Nam b Faculty of Basic Science, Ho Chi Minh City Industry and Trade College, Distric 9, Ho Chi Minh city, VietNam c School of Mathematics and Statistics, The University of New South Wales, Sydney 2052, Australia d Mathematics and Computer Science Division, Gran Sasso Science Institute, Viale Francesco Crispi 7, 67100, L’Aquila, Italy Abstract This work considers an inverse boundary value problem for a 3D nonlinear elliptic partial differential equation in a bounded domain In general, the problem is severely ill-posed The formal solution can be written as a hyperbolic cosine function in terms of the 2D elliptic operator via its eigenfunction expansion, and it is shown that the solution is stabilized or regularized if the large eigenvalues are cut off In a theoretical framework, a truncation approach is developed to approximate the solution of the ill-posed problem in a regularization manner Under some assumptions on regularity of the exact solution, we obtain several explicit error estimates including an error estimate of Hăolder type A local Lipschitz case of source term for this nonlinear problem is obtained For numerical illustration, two examples on the elliptic sine-Gordon and elliptic Allen-Cahn equations are constructed to demonstrate the feasibility and efficiency of the proposed methods Keywords and phrases: Nonlinear elliptic equation, Ill-posed problem, Regularization, Truncation method Mathematics subject Classification 2000: 35K05, 35K99, 47J06, 47H10 Introduction In this paper, we consider the problem of reconstructing the temperature of a body from interior measurements In fact, in many engineering contexts (see, e.g., [4]), we cannot attach a temperature sensor at the surface of a body (e.g., the skin of a missile) Hence, to get the temperature distribution on the surface, we have to use the temperature measured inside the body Let L be a positive real number and Ω = (0, π) ×(0, π) We are interested in the following inverse boundary value problem: Find u(x, y, 0) for (x, y) ∈ Ω where u(x, y, z) satisfies the following nonlinear elliptic equation: Δu = F(x, y, z, u(x, y, z)), subject to the conditions (x, y, z) ∈ Ω × (0, +∞), ⎧ ⎪ u(x, y, z) = 0, (x, y, z) ∈ ∂Ω × (0, +∞), ⎪ ⎪ ⎪ ⎪ ⎨ u(x, y, L) = ϕ(x, y), (x, y) ∈ Ω, ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ uz (x, y, L) = 0, (x, y) ∈ Ω (1.1) (1.2) Here Δu = ∂2 u/∂x2 + ∂2 u/∂y2 + ∂2 u/∂z2 , the function ϕ ∈ L2 (Ω) is known, and F is called the source function to be defined later Having found u(x, y, 0) a forward problem can be solved to find u(x, y, z) for all (x, y, z) ∈ Ω × (0, L) ∗ Corresponding author: Nguyen Huy Tuan; Email: thnguyen2683@gmail.com; Preprint submitted to Elsevier December 29, 2014 It is widely recognized nowadays that Cauchy problems for the Poisson equation, and more generally for elliptic equations, has a central position in all inverse boundary value problems which are encountered in many practical applications such as electrocardiography [25], astrophysics [9] and plasma physics [3, 21] These problems are also closely related to inverse source problems arising from, e.g., electroencephalography and magnetoencephalography [26] The continued interest in this kind of problems is evidenced by the number of publications on this topic We refer to the monograph [25] for further reading on Cauchy problems for elliptic equations It is well-known that inverse boundary value problems are exponentially ill-posed in the sense of Hadamard Existence of solutions and their stability with respect to given data not hold even if the data are very smooth In fact, the problems are extremely sensitive to measurement errors; hence, even in the case of existence, a solution does not depend continuously on the given data This, of course, implies that a properly designed numerical treatment is required Inverse boundary problems for linear elliptic equations have been studied extensively, see, e.g., [1, 3] Indeed, in the case F = in (1.1) with the following conditions ⎧ ⎪ u(x, y, z) = 0, (x, y, z) ∈ ∂Ω × (0, +∞), ⎪ ⎪ ⎪ ⎪ ⎨ u(x, y, L) = ϕ(x, y), (x, y) ∈ Ω, (1.3) ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ lim uz (x, y, L) = 0, (x, y) ∈ Ω, z→∞ the problem is studied in [10, 26, 33] In these studies, the algebraic invertibility of the inverse problem is established However, regularization is not investigated In [24], the authors apply the nonlocal boundary value method to solve an abstract Cauchy problem for the homogeneous elliptic equation Eld´en et al develope useful numerical methods to solve the homogeneous problem; see for example [14, 15, 16, 17] Level set type methods are also proposed [32] for Cauchy problems for linear elliptic equations Although there are many works on Cauchy problems for linear elliptic equations, to the best of our knowledge the literature on the nonlinear case is very few In the abstract framework of operators on Hilbert spaces, regularization techniques are developed by B Kaltenbacher and her coauthors in [2, 27, 28, 29] The present paper serves to develop necessary theoretical bases for a regularization of problem (1.1)–(1.2) Our approach can be summarized as follows Let ϕ and ϕ be the exact and measured data at z = L, respectively, which satisfy ϕ − ϕ L2 (Ω) ≤ Assume that problem (1.1)–(1.2) has a unique solution u(x, y, z) By using the method of separation of variables, one can show that √ ⎤ ⎡ ∞ ∞ ⎢⎢ ⎥⎥⎥ L sinh (τ − z) m2 + n2 ⎢⎢⎢ 2 u(x, y, z) = Fmn (u)(τ)dτ⎥⎥⎥⎥⎦ φmn (x, y).(1.4) √ ⎢⎢⎣cosh (L − z) m + n ϕmn + z m2 + n2 m=1 n=1 Indeed, let u(x, y, z) = ∞ ∞ m=1 n=1 umn (z)φmn (x, y) be the Fourier series in L2 (Ω) with φmn (x, y) = π sin(mx) sin(ny) From (1.1)–(1.2) , we can obtain the following ordinary differential equation with given data at z = L ⎧ ⎪ d ⎪ ⎪ ⎪ u (z) − m2 + n2 umn (z) = Fmn (u)(z), z ∈ (0, +∞), ⎪ mn ⎪ ⎪ dz ⎪ ⎪ ⎨ (1.5) umn (L) = ϕmn , ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ d ⎪ ⎪ ⎩ umn (L) = 0, dz where Fmn (u)(z) = Ω F(x, y, z, u(x, y, z))φmn (x, y)dxdy It is easy to see that the solution of problem (1.5) is given by L umn (z) = cosh (L − z) m2 + n2 ϕmn + z √ sinh (τ − z) m2 + n2 Fmn (u)(τ)dτ √ m2 + n2 (1.6) Then the solution of (1.1)–(1.2) satisfies the integral equation (1.4) Since z < L, we know from (1.4) that, when m, n become large, the terms √ sinh (τ − z) m2 + n2 cosh (L − z) m2 + n2 and √ m2 + n2 increase rather quickly Thus, these terms are the cause for instability In order to regularize problem (1.1)–(1.2) , we stabilize problem (1.4) by filtering the high frequencies with a suitable method The essence of our regularization method is to eliminate all high frequencies from the solution, and consider √ (1.4) only for m, n satisfying m2 + n2 ≤ C Here C is a constant which will be selected appropriately as a regularization parameter which satisfies lim →0 C = ∞ Such a method is the called Fourier truncated method We shall use the following well-posed problem ⎧ ⎪ Δu = PC F(x, y, z, u (x, y, z)), (x, y, z) ∈ Ω × (0, +∞), ⎪ ⎪ ⎪ ⎪ ⎨ (x, y, z) ∈ ∂Ω × (0, +∞), u (x, y, z) = 0, (1.7) ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ u (x, y, L) = ϕ (x, y), uz (x, y, L) = 0, (x, y) ∈ Ω, √ where PC is the orthogonal projection from L2 (Ω) onto the eigenspace span{φmn (x, y)| m2 + n2 ≤ C } , i.e w, φmn φmn for all w ∈ L2 (Ω) PC w = √ m,n≥1 m2 +n2 ≤C The Fourier truncation method is useful and convenient for dealing with ill-posed problems The method is effective for linear backward problems; see e.g [19, 36] It has also been successfully applied to some other ill-posed problems [15, 39] In the present paper, by using the truncation regularization method, we show that the approximate solution u of problem (1.7) satisfies the following integral equation u (x, y, z) = ⎡ ⎢⎢⎢ ⎢⎢⎢cosh (L − z) m2 + n2 ϕ + mn ⎢⎣ √ m,n≥1 m2 +n2 ≤C L z √ ⎤ ⎥⎥⎥ sinh (τ − z) m2 + n2 Fmn (u )(τ)dτ⎥⎥⎥⎥⎦ φmn (x, y) √ m2 + n2 (1.8) where ϕmn = Ω ϕ (x, y)φmn (x, y)dxdy and Fmn (u )(τ) = Ω F(x, y, τ, u (x, y, τ))φmn (x, y)dxdy We then prove that under some suitable conditions of the exact solution u, the approximate solution u converges to u as → To the best of our knowledge, this is the first result on convergence rate when a regularization method is used to solve inverse boundary value problems for elliptic equations with locally Lipschitz source terms Moreover, error estimates in higher Sobolev space is presented for the first time in this paper In particular, in this paper we will present regularized solutions for two cases of the source function F: Case F is a global Lipschitz function Case F is a locally Lipschitz function Our method is in principle not restricted to the Poisson equation on a rectangular domain In fact, the method works for more general operators defined on any bounded Euclidean domain Indeed, the analysis presented in this paper will be particularly derived from a general abstract problem in a Hilbert space H ⎧ ⎪ ⎪ ⎨uzz = Au + F(z, u(z)), z ∈ (0, +∞), (1.9) ⎪ ⎪ ⎩u(L) = ϕ, uz (L) = 0, where u is a mapping from [0, +∞) to H and A : D(A) → H is a positive self-adjoint unbounded operator The paper is organized as follows In Section 2, we present the main results on regularization theory for both cases: global and local Lipschitz source functions Section is devoted to numerical experiments which show the efficacy of the proposed methods We finish the paper with some concluding remarks in Section The main results Definition 2.1 (Gevrey-type space (see [7, 35])) The Gevrey class of functions of order s > and index σ > is denoted by Gσs/2 and is defined as ∞ ∞ Gσs/2 := f ∈ L2 (Ω) : s (m2 + n2 ) exp 2σ m2 + n2 | f, φmn |2 < ∞ m=1 n=1 It is a Hilbert space equipped with the norm ∞ ∞ f Gσs/2 s (m2 + n2 ) exp 2σ m2 + n2 | f, φmn |2 < ∞ = m=1 n=1 For a Hilbert space B, we denote L∞ (0, L; B) = f : [0, L] → B ess sup f (z) 0≤z≤L B independent of x, y, z, w, v In this paper, we shall write u(z) = u(·, ·, z) for short The following theorem provides an error estimate in the L2 -norm when the exact solution belongs to the Gevrey space Theorem 2.1 Let C([0, L]; L2 (Ω)) > and let F satisfy (2.14) Then the problem (1.7) has a unique solution u ∈ Assume that u satisfies (2.10) If C > is chosen such that eLC is bounded, then we obtain u (z) − u(z) L2 (Ω) ≤ 2I12 + e2LC e2KF L(L−z) e−zC Moreover, for sufficiently small, there exists z ∈ [0, L] such that lim u (z ) − u(0) L2 (Ω) ≤ →0 z = and 2I12 + e2LC e2KF L(L−z) + uz As a consequence, if we choose C = L L∞ (0,L;L2 (Ω)) ln(C ) C ln( ) then ⎧ 2 z ⎪ ⎪ ⎪ u (z) − u(z) L2 (Ω) ≤ 2I12 + 4e2KF L L , for z ∈ (0, L] ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ L ln L1 ln( ⎪ ⎪ ⎪ + 4e2KF2 L2 + I ⎪ ⎪ u (z ) − u(0) ≤ 2I ⎪ L (Ω) ⎩ ln( ) Assume that u satisfies (2.11) If C is chosen such that lim u (z) − u(z) L2 (Ω) ≤ 2C −2β I2 L2 (Ω) ≤ eLC = 0, then we have →+∞ + e2LC e2KF L(L−z) e−zC Assume that u satisfies (2.12) If C is chosen such that lim u (z) − u(z) (2.15) eLC = 0, then we have →+∞ 2e−2αC I32 + e2LC e2KF L(L−z) e−zC z Remark 2.1 In part 1, Theorem 2.1, if we choose C = L1 ln( ) then the error is of order L This error gives no information on the continuous dependence of the solution on the data at z = To improve this, we need a stronger condition of u as in Part 2, Part In part 2, Theorem 2.1, if we choose C = order ln( ) −β γ L , ln < γ < 1, then the error is of logarithmic In part 3, Theorem 2.1, if we choose C = L+α ln , then the error is of Hăolder order L+ Before proving Theorem 2.1, we prove the following lemmas Lemma 2.2 The problem (1.8) has unique a weak solution u (x, y, z) which is in C([0, L]; L2 (Ω) ∩ L2 (0, L; H01 (Ω)) ∩ C (0, L; H01 (Ω)) Proof Put ⎛ ⎜⎜⎜ ⎜⎜⎜ ⎜⎝ H(u )(x, y, z) = Ψ(x, y, z) + √ m,n≥1 L z √ ⎞ ⎟⎟⎟ sinh (τ − z) m2 + n2 Fmn (u )(τ)dτ⎟⎟⎟⎟⎠ φmn (x, y), √ m2 + n m2 +n2 ≤C where Ψ(x, y, z) = cosh (L − z) m2 + n2 ϕmn φmn (x, y) √ m,n≥1 m2 +n2 ≤C We claim that p H p (v )(z) − H p (w )(z) L2 (Ω) KF2 L exp(2LC ) max{L, 1} ≤ p! |||v − w |||, (2.16) for p ≥ 1, where |||.||| is the sup norm in C([0, L]; L2 (Ω)) We shall prove the above inequality by induction For p = 1, using the inequality √ ⎤2 L⎡ ⎢⎢⎢ sinh (τ − z) m2 + n2 ⎥⎥⎥ L ⎢⎢⎢ ⎥⎥⎥ dτ ≤ exp m2 + n2 (τ − z) dτ ≤ exp(2 m2 + n2 L)L, √ ⎢⎣ ⎥⎦ 2 z m +n z and the Lipschitz property of F, we have √ ⎤2 ⎡ ⎥⎥⎥ ⎢⎢⎢ L sinh (τ − z) m2 + n2 ⎥⎥⎥ ⎢⎢⎢⎢ H(v )(z) − H(w )(z) = (F (v )(τ) − F (w )(τ))dτ √ mn mn ⎥⎦ ⎣ z L (Ω) + n2 m m,n≥1 √ m2 +n2 ≤C L ≤ √ m,n≥1 m2 +n2 ≤C z √ ⎡ ⎢⎢⎢ sinh (τ − z) m2 + n2 ⎢⎢⎢⎢ √ ⎣ m2 + n ⎡ L ⎢⎢⎢ ⎢⎢⎢ ≤ exp(2LC )L ⎢⎢⎢⎢ ⎢⎢⎣ √ z L ≤ exp(2LC )L ⎡ ⎢⎢⎢ ⎢⎢⎣ m,n≥1 m2 +n2 ≤C ∞ ∞ ⎤2 ⎥⎥⎥ ⎥⎥⎥⎥ dτ ⎦ L Fmn (v )(τ) − Fmn (w )(τ) dτ z ⎤ ⎥⎥⎥ ⎥⎥⎥ Fmn (v )(τ) − Fmn (w )(τ) ⎥⎥⎥⎥ dτ ⎥⎥⎦ ⎤ Fmn (v )(τ) − Fmn (w )(τ) ⎥⎥⎥⎦ dτ ⎥⎥ m=1 m=1 z L F(x, y, τ, v (x, y, τ)) − F(x, y, τ, u (x, y, τ)) dτ = exp(2LC )L z ≤ max{L, 1}KF2 exp(2LC )L2 |||v − w |||2 Thus (2.16) holds for p = Suppose that (2.16) holds for p = j We prove that (2.16) holds for p = j+1 We have H j+1 (v )(z) − H j+1 (w )(z) L (Ω) √ ⎤2 ⎡ ⎥⎥⎥ ⎢⎢⎢ L sinh (τ − z) m2 + n2 ⎢⎢⎢ Fmn (H j (v ))(τ) − Fmn (H j (w ))(τ) dτ⎥⎥⎥⎥⎦ = √ ⎢⎣ z m2 + n2 √ m,n≥1 m2 +n2 ≤C L ≤ exp(2LC )L z ⎛ ⎜⎜⎜ ⎜⎜⎜ ⎜⎜⎜ ⎜⎜⎜ ⎝√ ⎞ ⎟⎟⎟ ⎟⎟ 2⎟ j j Fmn (H (v ))(τ) − Fmn (H (w ))(τ) ⎟⎟⎟⎟ dτ ⎟⎟⎠ m,n≥1 m2 +n2 ≤C L ≤ exp(2LC )L F(τ, H j (v )(τ)) − F(τ, H j (w )(τ)) dτ z L ≤ exp(2LC )LKF2 H j (v )(τ) − H j (w )(τ) dτ z L ≤ exp(2LC 2( j+1) ≤ KF 2j )LKF2 KF exp(2L jC exp(2L( j + 1)C ) ) z j+1 z) (L − ( j + 1)! j (L − τ) j dτ max{A, 1} |||v − w |||2 j! max{L, 1} j+1 |||v − w |||2 Therefore, we get p H p (v )(z) − H p (w )(z) L2 (Ω) ≤ KF2 L exp(2LC ) max{L, 1} p! |||v − w ||| (2.17) for all v , w ∈ C([0, L]; L2 (Ω)) Now we consider H : C([0, L]; L2 (Ω)) → C([0, L]; L2 (Ω)) It can be shown that p KF2 Le2LC max{L, 1} lim p! p→∞ = As a consequence, there exists a positive integer number p0 such that H p0 is a contraction It follows that the equation H p0 (u) = u has a unique solution u ∈ C([0, L]; L2 (Ω)) We claim that H(u ) = u In fact, one has H(G p0 (u )) = H(u ) Hence H p0 (H(u )) = H(u ) By the uniqueness of the fixed point of H p0 , one has H(u ) = u , i.e., the equation H(u ) = u has a unique solution u ∈ C([0, L]; L2 (Ω)) Lemma 2.3 Let u be an exact solution to problem (1.1) and let u be as (1.8) Then we have the following estimate u (z) − PC u(z) L2 (Ω) ≤ exp 2(L − z)C ϕ −ϕ L2 (Ω) L + 2KF2 (L − z) exp 2(τ − z)C z u (τ) − u(τ) dτ L2 (Ω) Proof By using the Lipschitz property of F, we obtain u − PC u L2 (Ω) cosh (L − z) ≤2 √ m,n≥1 m2 +n2 ≤C L +2 √ z m,n≥1 + m2 n2 ϕmn − ϕmn √ sinh (τ − z) m2 + n2 Fmn (u )(τ) − Fmn (u)(τ) dτ √ m2 + n2 m2 +n2 ≤C exp (L − z) m2 + n2 ϕmn − ϕmn ≤2 √ m,n≥1 m2 +n2 ≤C L +2 √ exp (τ − z) m2 + n2 Fmn (u )(τ) − Fmn (u)(τ) dτ z m,n≥1 m2 +n2 ≤C ϕ −ϕ ≤ exp 2(L − z)C L2 (Ω) L + 2KF2 (L − z) exp 2(τ − z)C u (τ) − u(τ) dτ L2 (Ω) z This completes the proof of lemma We now prove Theorem 2.1 Proof Proof of Part 1: Since u ∈ G0z , Lemma 2.1 gives ≤ e−2zC u(z) L2 (Ω) u(z) − PC u(z) G0z Lemma 2.2 and the triangle inequality yield u (z) − u(z) L2 (Ω) ≤ u (z) − PC u(z) ≤ 2e−2zC u(z) G0z L2 (Ω) + u(z) − PC u(z) L2 (Ω) ϕ −ϕ + exp 2(L − z)C L2 (Ω) L + 4KF2 (L − z) exp 2(τ − z)C u (τ) − u(τ) dτ L2 (Ω) z This implies that L 2zC e u (z) − u(z) L2 (Ω) ≤ sup u(z) 0≤z≤L G0z + 4e 2LC + 4KF2 L e2τC u (τ) − u(τ) dτ L2 (Ω) z Applying Gronwall’s inequality, we obtain e2zC u (z) − u(z) L2 (Ω) ≤ sup u(z) 0≤z≤L G0z + 4e2LC 2 e4KF L(L−z) Therefore u (z) − u(z) L2 (Ω) ≤ 2I12 + e2LC e2KF L(L−z) e−zC (2.18) If is sufficiently small then C > L1 e L Consider the following equation e−zC = z The solution to this equation satisfies another equation h(z) = where h(z) = ln(z) + zC The function h is strictly increasing Moreover, limz→0+ h(z) = −∞ and ln(C ) = ln ln(C ) − ln(C ) + ln(C ) = ln ln(C ) > C h for small enough Thus the equation h(z) = has a unique solution z > which satisfies ln(C ) C z ≤ The continuity of uz gives, for sufficiently small , u(z ) − u(0) L2 (Ω) z = uz (ξ)dξ L2 (Ω) Combining (2.18) and (2.19) and noting that e−z C = z ≤ u (z ) − u(0) L2 (Ω) ≤ z sup uz (z) 0≤z≤L ln(C ) C , ≤ u (z ) − u(z ) ≤ sup u(z) = 2I12 + e2LC e2KF L(L−z) + uz 0≤z≤L (2.19) we obtain + u(z ) − u(0) L2 (Ω) L2 (Ω) L2 (Ω) + e2LC e2KF L(L−z) e−z C + z sup uz (z) 2 G0z 0≤z≤L L2 (Ω) ln(C ) C L∞ (0,L;L2 (Ω)) β Proof of Part 2: Since u ∈ Gz Lemma 2.1 gives u(z) − PC u(z) L2 (Ω) ≤C −2β −2zC e u(z) β Gz Lemma 2.2 and triangle inequality lead to u (z) − u(z) L2 (Ω) ≤ u − PC u L2 (Ω) ≤ 2C e−2zC u(z) 2β + u − PC u β Gz L2 (Ω) ϕ −ϕ + exp 2(L − z)C L2 (Ω) L + 4KF2 (L − z) exp 2(τ − z)C u (τ) − u(τ) dτ L2 (Ω) z This implies that L 2zC e u (z) − u(z) L2 (Ω) ≤ 2C −2β sup u(z) 0≤z≤L β Gz + 4e 2LC + 4KF2 L e2τC u (τ) − u(τ) z Applying Gronwall’s inequality, we obtain e2zC u (z) − u(z) L2 (Ω) ≤ 2C −2β sup u(z) 0≤z≤L β Gz + 4e2LC 2 e4KF L(L−z) Therefore u (z) − u(z) L2 (Ω) ≤ 2C −2β I2 + e2LC e2KF L(L−z) e−zC Part can be proved by using the same technique The proof of which is omitted dτ L2 (Ω) We deduce by using the Lipschitz property of F √ ⎤2 ⎡ ⎥⎥⎥ ⎢⎢⎢ hL sinh (τ − z) m2 + n2 ⎥⎥⎥ ⎢ ⎢⎢⎢ F (u)(τ) − F (W )(τ) dτ B2 L2 (Ω) = √ mn mn ⎥⎦ ⎣ z 2 m +n √ m,n≥1 m2 +n2 ≤C /2 ≤ √ √ ⎢⎢⎢ sinh (τ − z) m2 + n2 ⎢⎢⎢ √ ⎢⎣ m2 + n2 hL ⎡ m,n≥1 z m2 +n2 ≤C /2 ⎤2 ⎥⎥⎥ ⎥⎥⎥ ds ⎥⎦ L Fmn (W )(τ) − Fmn (u)(τ) dτ z hL ≤ exp((τ − z)C )(hL − z) √ m,n≥1 Fmn (W )(τ) − Fmn (u)(τ) dτ z m2 +n2 ≤C /2 hL ≤ KF2 (hL − z) exp (τ − z)C W (τ) − u(τ) L2 (Ω) dτ z It follows from (2.24) that u(z) − W (z) L2 (Ω) ≤ A2 L2 (Ω) L2 (Ω) + B2 hL ≤ exp(−LC ) sup u(z) 0≤z≤L G0L + 2KF2 hL exp((τ − z)C ) u(τ) − W (τ) dτ L2 (Ω) z Therefore, we get exp(zC ) u(z) − W (z) L2 (Ω) ≤ exp((z − L)C ) sup u(z) 0≤z≤L G0L hL + 2KF2 hL exp(τC ) u(τ) − W (τ) dτ L2 (Ω) z ≤ exp((hL − L)C ) sup u(z) 0≤z≤L G0L hL + 2KF2 hL exp(τC ) u(τ) − W (τ) dτ L2 (Ω) z Applying Gronwall’s inequality, we have u(z) − W (z) L2 (Ω) ≤ sup u(z) 0≤z≤L G0L exp KF2 hL(L − z) exp hLC − LC − zC (2.27) Combining (2.23) and (2.27), we obtain for ≤ z ≤ hL u(z) − w (z) L2 (Ω) ≤ u(z) − W (z) ≤2 L2 (Ω) sup u(z) 0≤z≤L + sup u(z) 0≤z≤L G0L G0L + w (z) − W (z) L2 (Ω) + e2LC e2KF L(L−hL) exp KF2 hL(hL − z) exp exp KF2 hL(L − z) exp hLC − LC − zC For fixed, it is easy to see that max e h∈[0,1] −hLC , e− (1−h)LC 14 =e −LC (−hL − z)C This leads to u(z) − w (z) ≤ R2 (z, u)e L2 (Ω) −zC e −LC Remark 2.2 If the same technique as in Theorem 2.1 is used, the order of convergence will be e−zC In particular, when z = there is no convergence If C = L1 , then (2.20) becomes ⎧ z ⎪ ⎪ z ∈ [ L2 , L], ⎨R1 (z, u) L , U (z) − u(z) L2 (Ω) ≤ ⎪ (2.28) z ⎪ ⎩R2 (z, u) L , z ∈ [0, L2 ) It is obvious that (2.28) is sharp and this error may be better than the previous one in Theorem 2.1 The next theorem provides an error estimate in the Sobolev space H p (Ω) which is equipped with a norm defined by ∞ g H p (Ω) ∞ m2 + n = p g, φmn m=1 n=1 Theorem 2.3 If u satisfies (2.12), then with C = u (z) − u(z) L+α ln we have ≤ R3 (z)e−αC + 2eKF L(L−z) eLC C e−zC , z ∈ [0, L], H p (Ω) where R3 (z) = √ p 2eKF L(L−z) I32 If u satisfies (2.13), then we can construct a regularized solution U satisfying ⎧ p −zC p ⎪ ⎪ + u L∞ (0,L;G0 )C e−LC , z ∈ [ L2 , L], ⎨R1 (z, u)C e L U (z) − u(z) |H p (Ω) ≤ ⎪ −zC −LC p −LC ⎪ ⎩R2 (z, u)C p e e + u ∞ , z ∈ [0, L2 ) L (0,L;G0 )C e L Proof Proof of Part 1: First, we have u (z) − PC u(z) H p (Ω) m2 + n = √ ≤C u (z) − u(z), φmn (x, y) m,n≥1 m2 +n2 ≤C 2p u (z) − u(z), φmn (x, y) √ ≤C p 2p m,n≥1 m2 +n2 ≤C u(z) − u (z) L2 (Ω) It follows from Theorem 2.1 that u (z) − PC u(z) H p (Ω) ≤ e2KF L(L−z) 2e−2αC sup u(z) 0≤z≤L G0z+α + e2LC C e−zC p (2.29) On the other hand, consider the function G(ξ) = ξ p e−Dξ , D > Since G (ξ) = ξ p−1 e−Dξ (p − Dξ), it follows that G is strictly decreasing when ξ ≥ p Thus if ≤ e i.e, 2(z + α)C ≥ p, then for m2 + n2 ≥ C m2 + n p exp − 2(z + α) m2 + n2 ≤ C e−2(z+α)C , 2p 15 −p(L+α) 2α and u(z) − PC u(z) H p (Ω) m2 + n = √ ≤C p exp − 2(z + α) m2 + n2 exp 2(z + α) m2 + n2 u(z), φmn (x, y) m,n≥1 m2 +n2 >C 2p exp(−2(z + α)C ) √ ≤ sup u(z) 0≤z≤L u(z), φmn (x, y) exp 2(z + α) m2 + n2 m,n≥1 m2 +n2 >C 2p C e−2(z+α)C G0z+α Therefore u(z) − PC u(z) H p (Ω) ≤ sup u(z) 0≤z≤L G0z+α C p −(z+α)C e (2.30) Combining (2.29) and (2.30), we get u (z) − u(z) H p (Ω) ≤ u (z) − PC u(z) √ + u(z) − PC u(z) 2e−2αC sup u(z) ≤ e2KF L(L−z) The inequality H p (Ω) 0≤z≤L G0z+α H p (Ω) + e2LC + sup u(z) 0≤z≤L G0z+α e −αC C e−zC p a2 + b2 ≤ a + b for a, b ≥ leads to Proof of Part 2: If ≤ e −p ≤ R3 (z)e−αC + 2eKF L(L−z) eLC C e−zC u (z) − u(z) H p (Ω) p i.e, 2LC ≥ p, then for m2 + n2 ≥ C m2 + n p exp − 2L m2 + n2 ≤ C e−2LC p and u(z) − PC u(z) H p (Ω) p m2 + n = √ ≤C exp − 2L m2 + n2 exp 2L m2 + n2 u(., z), φmn (x, y) m,n≥1 m2 +n2 >C 2p exp 2L m2 + n2 exp(−2LC ) √ u(z), φmn (x, y) m,n≥1 m2 +n2 >C ≤ sup u(z) 0≤z≤L 2p C e−2LC G0L Therefore, we obtain u(z) − PC u(z) H p (Ω) ≤ sup u(z) 0≤z≤L G0L C p −LC e (2.31) Combining (2.29) and (2.31), we claim that U (z) − u(z) H p (Ω) ≤ U (z) − PC u(z) p ≤ C u(z) − U (z) It follows from U (z) − u(z) L2 (Ω) H p (Ω) L2 (Ω) + u(z) − PC u(z) + sup u(z) 0≤z≤L ⎧ ⎪ ⎪ ⎨R1 (z, u)e−zC , ≤⎪ ⎪ ⎩R2 (z, u)e −zC2 e −LC4 , 16 G0L C H p (Ω) p −LC z ∈ [ L2 , L], z ∈ [0, L2 ), e that U (z) − u(z) |H p (Ω) ⎧ p −zC p ⎪ ⎪ + u L∞ (0,L;G0 )C e−LC , ⎨R1 (z, u)C e L ≤⎪ −zC −LC p −LC ⎪ ⎩R2 (z, u)C p e e + u ∞ , L (0,L;G0 )C e L z ∈ [ L2 , L], z ∈ [0, L2 ) 2.2 Results for locally Lipschitz source functions In this subsection, we assume that the function F : [0, π] × [0, π] × [0, L] × R → R, F = F(x, y, z, u) satisfies: for each M > and for any u, v satisfying |u|, |v| ≤ M, there holds |F(x, y, z, u) − F(x, y, z, v)| ≤ KF (M) |u − v| , (2.32) where (x, y, z) ∈ [0, π] × [0, π] × [0, L] and KF (M) := sup F(x, y, z, u) − F(x, y, z, v) : |u| , |v| ≤ M, u u−v v, (x, y, z) ∈ [0, π] × [0, π] × [0, L] < +∞ We note that KF (M) is increasing and lim KF (M) = +∞ Now, we outline our ideas to construct a M→+∞ regularization for problem (1.1)–(1.2) For all M > 0, we approximate F by F M defined by ⎧ ⎪ ⎪ F(x, y, z, M), ⎪ ⎪ ⎪ ⎨ F M (x, y, z, u(x, y, z)) = ⎪ F(x, y, z, u(x, y, z)), ⎪ ⎪ ⎪ ⎪ ⎩F(x, y, z, −M), For each problem > 0, we consider a parameter M → +∞ as ⎧ ⎪ ⎪ Δv = PC F M (x, y, z, v(x, y, z)), ⎪ ⎪ ⎪ ⎨ v(x, y, z) = 0, ⎪ ⎪ ⎪ ⎪ ⎪ ⎩v(x, y, L) = ϕ (x, y), vz (x, y, L) = 0, u(x, y, z) > M, −M ≤ u(x, y, z) ≤ M, u(x, y, z) < −M → Let u ,ϕ be a solution of the following (x, y, z) ∈ Ω × (0, +∞), (x, y, z) ∈ ∂Ω × (0, +∞), (x, y) ∈ Ω (2.33) The following theorem provides some error estimates in the L2 -norm when the exact solution belongs to the Gevrey space Theorem 2.4 Let > and let F be the function defined in (2.32) Then the problem (2.33) has a unique solution u ,ϕ ∈ C([0, L]; L2 (Ω)) Assume that u satisfies (2.10) If C and M are chosen such that eLC is bounded and ⎧ 2 ln(C ) ⎪ ⎪ ⎨lim →0 exp 2KF (M )L C = 0, ⎪ ⎪ ⎩lim →0 exp 2K (M )L2 e−zC = 0, z > 0, F then we obtain u ,ϕ (z) − u(z) L2 (Ω) ≤ 2I12 + e2LC exp 2KF2 (M )L2 e−zC Moreover, for sufficiently small, there exists z ≤ L, ln(C C u ,ϕ (z ) − u(0) L2 (Ω) ≤ ) such that 2I12 + e2LC exp 2KF2 (M )L2 + u 17 (2.34) L∞ (0,L;L2 (Ω)) ln(C ) C (2.35) Assume that u satisfies (2.11) If C and M are chosen such that lim lim exp 2KF2 (M )L2 C −β →+∞ eLC = and = lim exp 2KF2 (M )L2 eLC = 0, →0 →0 then we have u (z) − u(z) ,ϕ ≤ L2 (Ω) 2C −2β I2 + e2LC exp 2KF2 (M )L2 e−zC Assume that u satisfies (2.12) If C and M are chosen such that lim →+∞ (2.36) eLC = and lim exp 2KF2 (M )L2 e−αC = lim exp 2KF2 (M )L2 eLC = 0, →0 →0 then we have u ,ϕ (z) − u(z) L2 (Ω) ≤ 2e−2αC I32 + e2LC exp 2KF2 (M )L2 e−zC Before proving the theorem, we show the following lemmas Lemma 2.4 For u1 (x, y, z), u2 (x, y, z), we have F M (x, y, z, u2 (x, y, z)) − F M (x, y, z, u1 (x, y, z)) ≤ KF (M) u2 (x, y, z) − u1 (x, y, z) Proof If u1 (x, y, z) < −M and u2 (x, y, z) < −M then F M (x, y, z, u2 (x, y, z)) − F M (x, y, z, u1 (x, y, z)) = If u1 (x, y, z) < −M ≤ u2 (x, y, z) ≤ M then F M (x, y, z, u2 (x, y, z)) − F M (x, y, z, u1 (x, y, z)) = F M (x, y, z, u2 (x, y, z)) − F M (x, y, z, −M) ≤ KF (M) M + u2 (x, y, z) ≤ KF (M) u2 (x, y, z) − u1 (x, y, z) If u1 (x, y, z) < −M < M < u2 (x, y, z) then F M (x, y, z, u2 (x, y, z)) − F M (x, y, z, u1 (x, y, z)) = F M (x, y, z, M) − F M (x, y, z, −M) ≤ 2MKF (M) ≤ KF (M) u2 (x, y, z) − u1 (x, y, z) If −M ≤ u1 (x, y, z), u2 (x, y, z) ≤ M then F M (x, y, z, u2 (x, y, z)) − F M (x, y, z, u1 (x, y, z)) = F(x, y, z, u2 (x, y, z)) − F(x, y, z, u1 (x, y, z)) ≤ KF (M) u2 (x, y, z) − u1 (x, y, z) This completes the proof Lemma 2.5 Let u be exact solution to problem (1.1)–(1.2) Then we have the following estimate u ,ϕ (z) − PC u(z) L2 (Ω) ≤ exp 2(L − z)C ϕ −ϕ L2 (Ω) L + 2KF2 (M )(L − z) exp 2(τ − z)C z 18 u ,ϕ (τ) − u(τ) dτ L2 (Ω) Proof From the definitions of u u and u, we obtain L2 (Ω) (z) − PC u(z) ,ϕ ,ϕ ≤2 cosh (L − z) m2 + n2 ϕmn − ϕmn √ m,n≥1 m2 +n2 ≤C L +2 √ z m,n≥1 √ sinh (τ − z) m2 + n2 (F M )mn (u √ m2 + n2 )(τ) − Fmn (u)(τ) dτ ,ϕ m2 +n2 ≤C exp (L − z) m2 + n2 ϕmn − ϕmn ≤2 √ m,n≥1 m2 +n2 ≤C L +2 exp (τ − z) m2 + (F M )mn (u ,ϕ )(τ) − Fmn (u)(τ) dτ n2 z √ m,n≥1 m2 +n2 ≤C ϕ −ϕ ≤ exp 2(L − z)C L2 (Ω) L + 2(L − z) F M x, y, τ, u exp 2(τ − z)C ,ϕ (τ) − F(x, y, τ, u(τ)) L2 (Ω) (2.37) dτ z Since lim+ M = +∞, for a sufficiently small > 0, there is an M such that M ≥ u →0 L∞ ([0,L],L2 (Ω)) For this value of M we have F M (x, y, z, u(x, y, z)) = F(x, y, z, u(x, y, z)) Using the Lipschitz property of F M as in Lemma 2.4, we get F M x, y, τ, u ,ϕ (τ) − F(x, y, τ, u(τ)) L2 (Ω) ≤ KF (M ) u ,ϕ (τ) − u(τ)) L2 (Ω) (2.38) Combining (2.37) and (2.38), we complete the proof of Lemma 2.5 We now prove Theorem 2.4 Proof Proof of Part 1: Since u ∈ G0z then using Lemma 2.1, we get u(z) − PC u(z) L2 (Ω) ≤ e−2zC u(z) G0z Lemma 2.3 and the triangle inequality lead to u ,ϕ (z) − u(z) L2 (Ω) ≤2 u ,ϕ L2 (Ω) (z) − PC u(z) ≤ 2e−2zC u(z) G0z + u(z) − PC u(z) + exp 2(L − z)C ϕ −ϕ L2 (Ω) L2 (Ω) L + 4KF2 (M )(L − z) exp 2(τ − z)C u ,ϕ (τ) − u(τ)) L2 (Ω) dτ z This implies that L e 2zC u ,ϕ (z) − u(z) L2 (Ω) ≤ sup u(z) 0≤z≤L G0z + 4e 2LC + 2KF2 (M e2τC u )L z 19 ,ϕ (τ) − u(τ)) L2 (Ω) dτ Applying Gronwall’s inequality, we obtain e2zC u ,ϕ (z) − u(z) L2 (Ω) G0z ≤ sup u(z) 0≤z≤L + 4e2LC exp 4KF2 (M )L(L − z) , which leads to the desired result u ,ϕ (z) − u(z) L2 (Ω) ≤ G0z sup u(z) 0≤z≤L + e2LC exp 2KF2 (M )L2 e−zC Part and Part can be proved by using the same technique The proof of which is omitted Remark 2.3 In part 1, Theorem 2.4, let us choose C = KF (M ) = 2L L ln( ) We can find M such that 1 ln( ) L ln (2.39) Then (2.34) becomes u ,ϕ (z) − u(z) L2 (Ω) ≤ 2L + sup u(z) 0≤z≤L G0z 1 ln( ) L z L (2.40) and (2.35) becomes u ,ϕ (z ) − u(0) L2 (Ω) ≤ sup u(z) 0≤z≤L G0z +4 L ln L ln( ) ln( ) + sup uz (z) 0≤z≤L L ln L2 (Ω) L ln( ) ln( ) (2.41) If F(u) = u − u3 , equation (1.1) becomes the elliptic Allen-Cahn type equation which models phase transitions The elliptic Allen-Cahn equation appears in the study of differential geometry related to, for example, interface area, interface curvature and minimal hypersurface A simple computation reveals KF (M ) = sup = sup F(u) − F(v) : |u| , |v| ≤ M , u v, u−v u − v − u3 + v : |u| , |v| ≤ M , u v, u−v = + 3M and togeter with (2.39) KF (M ) = + 3M = 2L ln 1 ln( ) L This implies L M = L ln( ) − In part 2, Theorem 2.4, let us choose C = KF (M ) = ln 2L γ L ln( ), for γ ∈ (0, 1) We can take M such that γ ln ln( ) , β−r L 20 for r ∈ (0, β) Then (2.36) becomes u ,ϕ (z) − u(z) L2 (Ω) −2β ≤ 2C = sup u(z) sup u(z) 0≤z≤L 0≤z≤L In part 3, Theorem 2.4, let us choose C = KF (M ) = L+α β Gz β Gz +4 + e2LC exp 2KF2 (M )L2 e−zC 2−2γ β γ ln( ) L −r zγ L ln( ) We can take M such that r−α ln( ), L+α 2L γ ln( ) L 0M, ∈ [−M , M ] , ,ϕ < −M , ,ϕ 25 M = √ ln (− ln ( )) − (a) Exact solution (b) Regularized solution Figure 4: 3-D graphs of exact solution at y = π , z = 0.98 and its regularized solution for = 10−3 in Example 2 The term ϕmn can be directly calculated For each q ∈ N, we define the sequence u starting with u0,ϕ ≡ −exp(2) sin x sin y The sequence satisfies q+1 u ,ϕ q ,ϕ q≥0 by recurrence, α(m,n)≤[C ] (x, y, z) = cosh ((1 − z) α (m, n)) ϕmn + m,n=1 z sinh ((τ − z) α (m, n)) q Fmn u ,ϕ (τ) dτ sin (mx) sin (ny) α (m, n) Similarly to Example 1, we present the numerical results in Table and Figure for the case L1 = L2 = L3 = 50 The 3D graphs of the solutions are presented in Figure In comparison with the Example 1, it can be predicted that the smoother the function, the better the convergence rate However, sufficient recurrence steps (e.g q = 10) are required to obtain fast convergence In the case = 10−5 and q = the errors are E1 = 2.4461E − 02 and E2 = 7.0160E − 03 Of course, a larger value for q incurs high computational costs Moreover, the numerical results of points far from the point z = L may be obtained when the meshsizes and are sufficiently small A better approximation is required, and this is a topic for future research Conclusion In this paper, we propose a regularization method based on the cut-off method for solving 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Viale Francesco Crispi 7, 67100, L’Aquila, Italy Abstract This work considers an inverse boundary value problem for a 3D nonlinear elliptic partial differential equation in a bounded domain In. .. equation, and more generally for elliptic equations, has a central position in all inverse boundary value problems which are encountered in many practical applications such as electrocardiography