In this paper, we prove a stability estimate of H¨older type and propose a regularization method with error estimates of H¨older type for an inverse heat source problem in the Banach space L1(R n ).
Trường Đại học Vinh Tạp chí khoa học, Tập 49 - Số 1A/2020, tr 5-13 STABILITY RESULTS FOR IDENTIFYING AN UNKNOWN SOURCE TERM OF A HEAT EQUATION IN THE BANACH SPACE L1 (Rn ) Nguyen Van Duc (1) , Luong Duy Nhat Minh (2) School of Natural Sciences Education, Vinh University, Vinh City, Nghe An Province Nam Ly Secondary School, Dong Hoi City, Quang Binh Province Received on 03/01/2020, accepted for publication on 7/4/2020 Abstract: In this paper, we prove a stability estimate of Hăolder type and propose a regularization method with error estimates of Hăolder type for an inverse heat source problem in the Banach space L1 (Rn ) Keyword: Stability estimate; regularization method; inverse source problem Introduction Inverse problems typically lead to mathematical models that are not well-posed in the sense of Hadamard [5] This means especially that their solution is unstable under data perturbations The two topics of concern in the field of inverse problems are the establishment of stability estimates ([1], [3], [8], [9], [10]) and the proposal of regularization methods ([2], [3], [6], [7], [9], [10]) In this paper, we establish stability estimates of Hăolder type and propose a regularization method with error estimates of Hăolder type for the ndimensional inverse source problem of finding a pair of functions {u(x, t), f (x)} with u(·, t) ∈ L1 (Rn ), ∀t ∈ [0, T ], f (·) ∈ L1 (Rn ) satisfying: n ut = ∆u + f (x)h(t), x ∈ R , t ∈ (0, T ) u(x, 0) = 0, x ∈ Rn , u(x, T ) = g(x), x ∈ Rn , (1) where ∆ denotes the Laplace operator, g and h are given functions such that g(·) ∈ L1 (Rn ), T h : [0, T ] → R is continuous on [0, T ], and h(s)ds = We would like to emphasize that although there are many results for inverse source problems of the heat equation in Hilbert spaces, there are very few results for this problem in Banach spaces, see, e.g., [1], [4], [11] and the references therein 1) Email: nguyenvanducdhv@gmail.com (N V Duc) N V Duc, L D N Minh / Stability results for identifying an unknown source term With q(·) ∈ L1 (Rn ), we denote: q + := max{q, 0} q − := max{−q, 0} m(q) := max{ q + , q − } n(q) := min{ q + , q − } Remark 1.1 We have q + = (−q)− q − = (−q)+ Definition 1.2 With any real number k 0, we denote by Sk the following set of functions: Sk = {q : Rn → R, q ∈ L1 (Rn ) such that m(q) (1 + k)n(q)} Lemma 1.3 The set Sk has the following properties: a) In case k = then S0 ≡ L1 (Rn ), b) ∈ Sk , ∀k 0, c) If q ∈ L1 (Rn ) and q does not change sign over Rn then q ∈ Sk for all k d) If q ∈ Sk then −q ∈ Sk , e) If q ∈ Sk then λq ∈ Sk , ∀λ ∈ R (2) 0, Proof a) Clearly, for all q ∈ L1 (Rn ) we have max{ q + , q − } min{ q + , q − } or m(q) n(h) = (1 + 0)n(q) That means q ∈ S0 Therefore S0 ≡ L1 (Rn ) b) If f = then f + = f − = Therefore f + = f − = That implies m(f ) = n(f ) = Hence the inequality m(f ) (1 + k)n(f ) holds true for all k So ∈ Sk , ∀k c) If q ∈ L1 (Rn ) and q does not change sign over Rn then q + = max{q, 0} = or q − = max{−q, 0} = That deduces min{ q + , q − } = Therefore for all k m(q) = max{ q + , q − } So q ∈ Sk for all k we have = (1 + k) min{ q + , q − } = (1 + k)n(q) d) Since q + = (−q)− , q − = (−q)+ then we have m(q) = max{ q + , q − } = max{ (−q)− , (−q)+ } = max{ (−q)+ , (−q)− } = m(−q), n(q) = min{ q + , q − } = min{ (−q)− , (−q)+ } = min{ (−q)+ , (−q)− } = n(−q) Trường Đại học Vinh Tạp chí khoa học, Tập 49 - Số 1A/2020, tr 5-13 If q ∈ Sk then q ∈ L1 (Rn ) and m(q) (1 + k)n(q) So that we have −q ∈ L1 (Rn ) and m(−q) (1 + k)n(−q) That implies −q ∈ Sk e) We consider the following cases: Case 1: If λ = then λq = ∈ Sk , ∀k Case 2: If λ > then we have follow b) (λq)+ (x) = max{(λq)(x), 0} = max{λq(x), 0} = λ max{q(x), 0} = λ.q + (x), (λq)− (x) = max{−(λq)(x), 0} = max{−λq(x), 0} = λ max{−q(x), 0} = λ.q − (x) We infer (λq)+ = λ.q + = |λ| q + = λ q+ 1, (λq)− = λ.q − = |λ| q − = λ q− Then we have m(λq) = max{ (λq)+ , (λq)− } = max{λ q + , λ q − } = λ max{ q + , q − } = λm(q), n(λq) = min{ (λq)+ , (λq)− } = min{λ q + , λ q − } = λ min{ q + , q − } = λn(q) Because q ∈ Sk then we have m(q) (1+k)n(q) Multiplying the two sides of this inequality by λ > we have λm(q) (1 + k)λn(q) or m(λq) (1 + k)n(λq) So λq ∈ Sk for all λ > Case 3: If λ < then −λ > Because q ∈ Sk then from d) we have −q ∈ Sk The result of Case showed that (−λ).(−q) ∈ Sk or λq ∈ Sk Stability Estimates We now provide a stability estimate for the solution of problem (1) on the class of function Sk with k > Theorem 2.1 Suppose that {ui , fi }, i = 1, where f1 − f2 ∈ Sk are solutions of problem (1) corresponding to the data gi ∈ L1 (Rn ), i = 1, 2, satisfying g1 − g2 δ, (3) Cδ, (4) then we have the following estimate f1 − f2 with C = T h(s)ds 1+ k N V Duc, L D N Minh / Stability results for identifying an unknown source term Proof We set |x|2 − 4t √ , x ∈ Rn , t > e (2 πt)n f = f1 − f2 K(x, t) = g = g1 − g2 u = u1 − u2 Then f ∈ Sk , g ∈ L1 (Rn ) and u, f, g satisfy problem (1), then we have (see [10]) t K(x − y, t − s)f (y)h(s)dyds u(x, t) = (5) Rn Replacing t by T in (5) we have T K(x − y, T − s)f (y)h(s)dyds g(x) = u(x, T ) = (6) Rn Integrating both sides of (6) and then changing the order of integration we have T K(x − y, T − s)f (y)h(s)dyds dx g(x)dx = Rn Rn Rn T K(x − y, T − s)dx h(s)ds f (y) dy = Rn Notice that Rn (7) Rn K(x − y, T − s)dx = 1, from (7) we have T g(x)dx = Rn h(s)ds f (y) dy Rn T = T h(s)ds f (y)dy = Rn h(s)ds f (x)dx Rn Then we have f (x)dx = Rn Because f ∈ Sk then we have m(f ) max{ f + , f − } T h(s)ds g(x)dx (1 + k)n(f ) or (1 + k) min{ f + , f − } ⇔ max{ f + , f − } − min{ f + , f − } k min{ f + , f − } ⇔ min{ f + , f − } max{ f + , f − } − min{ f + , f − } k ⇔ min{ f + , f − } max{ f + , f − } − min{ f + , f − } k (8) Rn (9) Trường Đại học Vinh Tạp chí khoa học, Tập 49 - Số 1A/2020, tr 5-13 Adding the two sides of the inequality (9) with the quantity max{ f + , f − }−min{ f + , f − } we have max{ f + , f − } + min{ f + , f − } k 1+ max{ f + , f − } − min{ f + , f − } or f+ + f− 1+ k max{ f + , f − } − min{ f + , f − } (10) On the other hand, we have f (f + (x) + f − (x))dx |f (x)|dx = = Rn Rn f − (x)dx f + (x)dx + = Rn Rn |f − (x)|dx |f + (x)|dx + = Rn Rn = f+ + f− (11) and (f + (x) − f − (x))dx f (x)dx = Rn Rn Rn Rn |f − (x)|dx |f + (x)|dx − = = f − (x))dx f + (x)dx − = Rn + Rn f − f− = max{ f + 1, f − 1} − min{ f + , f − } (12) From (8), (10) and (12) we have the estimate: f 1+ = 1+ k k T = h(s)ds T h(s)ds f (x)dx Rn T g(x)dx h(s)ds 1+ k 1+ k Rn |g(x)|dx Rn g Cδ The theorem is proved N V Duc, L D N Minh / Stability results for identifying an unknown source term Regularization We set K1 (x, T ) = T K(x, T − s)h(s)ds, x ∈ Rn From (6) we have g = K1 ∗ f (13) According to the proof of Theorem 2.1, if f ∈ Sk then we have the following estimate: f 1 T 1+ h(s)ds k g (14) From (13) and (14) we have estimation f 1 T k 1+ h(s)ds K1 ∗ f (15) Then we have the lemma Lemma 3.1 If q ∈ Sk with k > then the following inequality holds: q with C = T h(s)ds 1+ C K1 ∗ q (16) k Now, with each k > 0, let Mk denote a subset of L1 (Rn ) such that: if q ∈ Mk , q ∈ Mk then q − q ∈ Sk Let us first take an example of a set satisfying the above property For any functions p, q such that p ∈ L1 (Rn ) and q ∈ Sk , we define the set Mk as follows Mk = {p + αq : α ∈ R} In this case, if w ∈ Mk , w ∈ Mk there exist real numbers α1 , α2 so that w = p + α1 q w = p + α2 q Hence we have w − w = (p + α1 q) − (p + α2 q) = (α1 − α2 )q Since q ∈ Sk then from Lemma 1.3, e) we have (α1 − α2 )q ∈ Sk or w − w ∈ Sk Suppose that there exists a pair of functions {u, f } with u(·, t) ∈ L1 (Rn ), ∀t ∈ [0, T ], f ∈ Mk satisfying problem (1) in the case that function g(·) ∈ L1 (Rn ) is not known but we know an approximation g δ (·) ∈ L1 (Rn ) such that g − gδ δ The goal of the problem is to determine the function f from the noisy version g δ of g We set J(q) := K1 ∗ q − g δ 10 2, q ∈ Mk We obtain the following theorem: (17) Trường Đại học Vinh Tạp chí khoa học, Tập 49 - Số 1A/2020, tr 5-13 Theorem 3.2 Let τ > be a fixed real number Choose q¯ ∈ Mk such that inf J(h) + τ δ J(¯ q) (18) q∈Mk Then we have the following estimate q¯ − f with C1 = T 1+ h(s)ds k √ (1 + C1 δ (19) + τ ) Proof From (18) we get K1 ∗ q¯ − g δ inf J(q) + τ δ = J(¯ q) q∈Mk J(f ) + τ δ = K1 ∗ f − g δ = g − gδ 2 + τ δ2 + τ δ2 δ + τ δ = (1 + τ )δ , (20) From estimate (20) it implies that K1 ∗ q¯ − g δ √ 1 + τ δ (21) From estimate (21) we have K1 ∗ (¯ q − f) = K1 ∗ q¯ − K1 ∗ f δ = K1 ∗ q¯ − g δ = K1 ∗ q¯ − g + g − g δ 1 δ K1 ∗ q¯ − g + g − g √ √ + τ δ + δ = (1 + + τ )δ (22) Because q¯ ∈ Mk and f ∈ Mk , we have q¯ − f ∈ Sk Applying Lemma 3.1 we obtain q¯ − f C K1 ∗ (¯ q − f ) (23) From (22) and (23) we conclude that q¯ − f C(1 + √ + τ )δ = C1 δ The theorem is proved 11 N V Duc, L D N Minh / Stability results for identifying an unknown source term REFERENCES [1] A Ashyralyev, A S Erdogan and O Demirdag, “On the determination of the right-hand side in a parabolic equation”, Applied Numerical Mathematics, 62, pp 1672–1683, 2012 [2] N V Duc, “An a posteriori mollification method for the heat equation backward in time”, Journal of Inverse and Ill-Posed Problems, 25(4), pp 403-422, 2017 [3] N V Duc, N V Thang, “Stability results for semi-linear parabolic equations backward in time”, Acta Mathematica Vietnamica, 42(1), pp 99–111, 2017 [4] D Guidetti, “Determining the Source Term in an Abstract Parabolic Problem From a Time Integral of the Solution”, Mediterranean Journal of Mathematics, 9, pp 611–633, 2012 [5] J Hadamard, Lectures on the Cauchy Problem in Linear Partial Differential Equations, Yale University Press, New Haven, 1923 [6] D N Hao and N V Duc, “Regularization of backward parabolic equations in Banach spaces”, Journal of Inverse and Ill-Posed Problems, 20, pp 745-763, 2012 [7] D N Hao and N V Duc, “A non-local boundary value problem method for semi-linear parabolic equations backward in time”, Applicable Analysis, 94(3), pp 446–463, 2015 [8] D N Hao, N V Duc and N V Thang, “Stability estimates for Burgers-type equations backward in time”, Journal of Inverse and Ill-Posed Problems, 23 (1), pp 41–49, 2015 [9] D N Hao, N V Duc and N V Thang, “Backward semi-linear parabolic equations with time-dependent coefficients and local Lipschitz source”, Inverse Problems, 34(5), 0550109 (33pp), 2018 [10] D N Hào, J Liu, N V Duc and N V Thang, “Stability estimates backward timefractional parabolic equations”, Inverse Problems, 35, 125006 (25pp), 2019 [11] A Prilepko, S Piskarev and S Y Shaw, “On approximation of inverse problem for abstract parabolic differential equations in Banach spaces”, Journal of Inverse and Ill-posed Problems, 15(8), pp 831–851, 2008 [12] V P Mikhailov, Partial Differential Equations, Mir Publishers, 1978 12 Trường Đại học Vinh Tạp chí khoa học, Tập 49 - Số 1A/2020, tr 5-13 TÓM TẮT CÁC KẾT QUẢ ỔN ĐỊNH CHO BÀI TỐN XÁC ĐỊNH NGUỒN CỦA MỘT PHƯƠNG TRÌNH NHIỆT TRONG KHÔNG GIAN BANACH L1 (Rn ) Trong báo này, chỳng tụi chng minh ỏnh giỏ n nh kiu Hăolder đề xuất phương pháp chỉnh hóa với đánh giỏ sai s kiu Hăolder cho mt bi toỏn ngun nhiệt ngược không gian Banach L1 (Rn ) Từ khóa: Đánh giá ổn định; phương pháp chỉnh hóa; toán xác định nguồn 13 ... identifying an unknown source term REFERENCES [1] A Ashyralyev, A S Erdogan and O Demirdag, “On the determination of the right-hand side in a parabolic equation? ??, Applied Numerical Mathematics,... semi-linear parabolic equations backward in time”, Acta Mathematica Vietnamica, 42(1), pp 99–111, 2017 [4] D Guidetti, “Determining the Source Term in an Abstract Parabolic Problem From a Time Integral... (25pp), 2019 [11] A Prilepko, S Piskarev and S Y Shaw, “On approximation of inverse problem for abstract parabolic differential equations in Banach spaces”, Journal of Inverse and Ill-posed Problems,