Numerical methods for reconstruction of source term of heat equation from the final overdetermination
aa r X i v : . [ m a t h . A P ] F e b Noname manuscript No. (will be inserted by the editor)
Numerical methods for reconstruction of source term of heatequation from the final overdetermination
Xiaoping Fang · Youjun Deng · Jing Li
Received: date / Accepted: date
Abstract
This paper deals with the numerical methods for the reconstruction of source term in lin-ear parabolic equation from final overdetermination. We assume that the source term has the form f ( x ) h ( t ) and h ( t ) is given, which guarantees the uniqueness of the inverse problem of determining thesource term f ( x ) from final overdetermination. We present the regularization methods for reconstruc-tion of the source term in the whole real line and with Neumann boundary conditions. Moreover, weshow the connection of the solutions between the problem with Neumann boundary conditions and theproblem with no boundary condition (in the whole real line) by using extension method. Numericalexperiments are done for the inverse problem with the boundary conditions. Mathematics subject classification (MSC2000):
We consider the reconstruction of the source term in the following mathematical model u t = ku xx + f ( x ) h ( t ) 0 < x < , < t ≤ T,u ( x,
0) = µ ( x ) x ∈ (0 , l ) ,u x (0 , t ) = u x (1 , t ) = 0 t ∈ (0 , T ] , (1)where k is the heat conductivity, f ( x ) and h ( t ) relate to the source term. µ ( x ) is the initial status. l and T are finite numbers. If all those parameters are given then the direct problem (1) has a uniquesolution. The inverse problem here is the determination of the source term f ( x ) from the final stateobservation µ T ( x ) = u ( x, T ).The mathematical model (1) arises in various physical and engineering settings, in particular inhydrology [1], material sciences [17], heat transfer [19] and transport problems [20], etc. The inverseproblem in determination of source term has been studied intensively for decades (cf., e.g., [5,6,12,14, Fang is supported by NSF grants No. NSFC70921001 and No. 71210003. Deng and Li are supported by NSF grants No.NSFC11301040,X. FangPostdoctoral, Management Science and Engineering Postdoctoral Mobile Station, School of Business; School of Mathe-matics and Statistics, Central South University, Changsha, Hunan 410083, P. R. China.E-mail: [email protected]. DengCorresponding author. School of Mathematics and Statistics, Central South University, Changsha, Hunan 410083, P.R. China. E-mail: [email protected], dengyijun [email protected]. LiDepartment of Mathematics, Changsha University of Science and Technology, Changsha, Hunan 410004, P.R.China.E-mail: [email protected] Xiaoping Fang et al. f ( x ) isassumed to be the sum of a known function f ( x ) = P Ii =1 ρ ( x − a i ) with I different locations andthe locations a i are determined by three non-collinear measurement points. On the other hand, theinverse problems for parabolic equations with final overdetermination also have been considered bylots of authors (see [2–4, 11] and the references there in). However, numerical methods for uniquelysolving the inverse source term f ( x ) in (1) without using further data concerning the solution u ( x, t )are seldom. We shall provide the numerical solution for solving the inverse source problem (1) andmore importantly, we show the relationship between solution of the boundary problem (1) and itscorresponding no boundary problem (in the whole real line). We only consider the one dimensionalproblem here to simplify our calculation and point out the main idea. The method can in fact beimplemented in two dimensional or higher dimensional problem.In this paper, we shall first consider the heat conduction problem in the whole real line, which is (cid:26) u t = ku xx + f ( x ) h ( t ) x ∈ R , < t ≤ T,u ( x,
0) = µ ( x ) x ∈ R . (2)where we suppose f ( x ) , µ ( x ) ∈ L ( R ) and h ( t ) ∈ L (0 , T ). It is easy to see that the solution of (2)(cf. [10]) is u ( x, t ) = Z ∞−∞ √ kπt e − ( x − y )24 kt µ ( y ) dy + Z t Z ∞−∞ p kπ ( t − s ) e − ( x − y )24 k ( t − s ) f ( y ) h ( s ) dyds. (3)By taking the Fourier transform with respect to x we can immediately getˆ u t ( ξ, t ) = − kξ ˆ u ( ξ, t ) + ˆ f ( ξ ) h ( t )and by initial condition in (2) there holdsˆ u ( ξ, t ) = ˆ µ ( ξ ) e − kξ t + ˆ f ( ξ ) Z t h ( s ) e − kξ ( t − s ) ds. (4)For the inverse problem with µ T ( x ) being measured, from (4) we haveˆ µ T ( ξ ) = ˆ µ ( ξ ) e − kξ T + ˆ f ( ξ ) Z T h ( s ) e − kξ ( T − s ) ds, and so ˆ f ( ξ ) = ˆ µ T ( ξ ) − ˆ µ ( ξ ) e − kξ T R T h ( s ) e − kξ ( T − s ) ds . (5)Furthermore, the solution u ( x, t ) can also be written asˆ u ( ξ, t ) = ˆ µ ( ξ ) e − kξ t + (ˆ µ T ( ξ ) − ˆ µ ( ξ ) e − kξ T ) R t h ( s ) e − kξ ( t − s ) ds R T h ( s ) e − kξ ( T − s ) ds = ˆ µ ( ξ ) e − kξ t R Tt h ( s ) e kξ s ds R T h ( s ) e kξ s ds + ˆ µ T ( ξ ) R t h ( s ) e − kξ ( t − s ) ds R T h ( s ) e − kξ ( T − s ) ds . (6)The relation (5) tells that if h ( t ) is appropriately given in [0 , T ] such that the denominator in (5) isnonzero for every ξ (or be nonzero in the distribution meaning), and µ ( x ) ∈ L ( R ) and µ T ( x ) ∈ L ( R )are given for x ∈ R , then ˆ f ( ξ ) and so f ( x ) can be reconstructed uniquely. The argument is also suitablefor (1) as we shall see.In this paper, we consider the determination of source term f ( x ) in both (1) and (2). For thesake of simplicity, we assume that h ( t ) is identically non-positive or non-negative function in [0 , T ]and C h := R T | h ( s ) | ds >
0. This assumption is quite nature. For example, in the case where the heat umerical methods for reconstruction of source term of heat equation from the final overdetermination is provided by a single kind of radioactive isotope, we can set h ( t ) = e − λt with a constant λ > f ( x ) ∈ H p ( R ), p ≥
0, where k f k p is defined as the norm of f ( x )in H p ( R ) k f k p := ( Z ∞−∞ (1 + ξ ) p | ˆ f ( ξ ) | dξ ) / . (7)This paper is organized as follows. In section 2, we analyze the severe ill-posedness of the reconstructionof the source term in (2), then we introduce the iterative method to solve the inverse problem.Convergence rates are given under both a priori and a posteriori stopping rules. In section 3, we usethe same fundamental solution method to get the solution of (1), by extending the source term andthe initial state to the whole real axis and then show the solution of the Neumann boundary problemis actually another form of the solution to (1) by separating variables method. In fact, the solutionto any kind of boundary problem can be got by the method extending the source and initial termsto the whole region other than by separating variables method. We only use the Neumann boundaryproblem as an example. We then give the frequency cut-off technique to solve the inverse problem.Numerical experiments for boundary problem are done in section 4 and show attractive results. In this section, we discuss the reconstruction of f ( x ) in L ( R ). We shall keep the denotation of normin L ( R ) as k · k . The direct problem is (cid:26) u t = ku xx + f ( x ) h ( t ) ( x, t ) ∈ R × (0 , T ] ,u ( x,
0) = µ ( x ) x ∈ R , (8)and the inverse problem is reconstruction of f ( x ) with final overdetermination u ( x, T ) = µ T ( x ). Wehave already got ˆ f ( ξ ) in (5). However, it is a severely ill-posed problem. In fact, the denominator in(5) decrease to zero exponentially as ξ → ∞ . Thus small perturbations in the measured data µ T ( x )may produce high frequency parts in ˆ f ( ξ ) and make the reconstruction quite unstable. Suppose thatthe measured final overspecified data µ δT ( x ) satisfies k µ δT − µ T k ≤ δ. (9)We do not assume measurement error in µ ( x ) because the perturbation of µ ( x ) will affect very littlein the reconstruction of f ( x ) and u ( x, t ). Denote v ( ξ, t ) = e − kξ t and rewrite (5) asˆ g ( ξ ) := ˆ µ ( ξ ) + ˆ f ( ξ ) Z T h ( s ) e kξ s ds = ˆ µ T ( ξ ) e kξ T = ˆ µ T ( ξ ) v ( ξ, T ) . (10)We can now introduce similar iterative method to [7, 8] for solving ˆ g ( ξ )ˆ g δn ( ξ ) = (1 − λ )ˆ g δn − ( ξ ) + λv ( ξ, T ) χ ϑ ˆ µ δT ( ξ ) + λ (1 − χ ϑ )ˆ µ ( ξ ) , (11)where λ = N p v ( ξ, T ) < N is a nature number, χ ϑ denotes the characteristic function of interval[ − ϑ , ϑ ] and ϑ is large enough.2.1 A priori stopping ruleBy using the a priori stopping rule we have the following convergence theorem Xiaoping Fang et al.
Theorem 1
Let u ( x, t ) be the exact temperature history of (8), h ( t ) is identically nonpositiveor nonnegative in [0 , T ] and µ δT ( x ) be the measured final temperature satisfying (9). f ( x ) satisfies k f k p ≤ M . Let ˆ g δk ( ξ ) be the k -th iteration solution defined by (11) with ˆ g δ ( ξ ) = ˆ µ ( ξ ) , where ϑ ∼ r σ ) kT h ln Mδ ( ln Mδ ) − σ p i , σ ≥ . Suppose u δn ( x, t ) , f δn ( x ) are solved via (4), (5) and the inverseFourier transform for every ˆ g δk , respectively. If we choose n ∼ ⌊ N q Mδ ⌋ , then there holds k f δn − f k ≤ C ( ln Mδ ) − p M σ δ σ σ + M ( ln Mδ ln Mδ ( ln Mδ ) − σ p ) p ! , (12) and k u δn ( · , t ) − u ( · , t ) k ≤ C h ( t )( ln Mδ ) − p M σ δ σ σ + M ( ln Mδ ln Mδ ( ln Mδ ) − σ p ) p ! , (13) for δ → , where C is a constant independent of δ and M and C h ( t ) = C ( R t | h ( s ) | ds ) . Proof.
By the iteration (11) we haveˆ g δn ( ξ ) = (1 − λ )ˆ g δn − ( ξ ) + λv ( ξ, T ) χ ϑ ˆ µ δT ( ξ ) + λ (1 − χ ϑ )ˆ µ ( ξ )= (1 − λ ) n ˆ µ ( ξ ) + n − X i =0 (1 − λ ) i (cid:20) λv ( ξ, T ) χ ϑ ˆ µ δT ( ξ ) + λ (1 − χ ϑ )ˆ µ ( ξ ) (cid:21) = (1 − λ ) n ˆ µ ( ξ ) + n − X i =0 (1 − λ ) i λ (1 − χ ϑ )ˆ µ ( ξ ) + n − X i =0 (1 − λ ) i λv ( ξ, T ) χ ϑ ˆ µ δT ( ξ ) . Set p n ( λ ) = P n − i =0 (1 − λ ) i , r n ( λ ) = 1 − λp n ( λ ) = (1 − λ ) n , we have the elementary results (cf. [18]) p n ( λ ) λ µ ≤ n − µ , for all 0 ≤ µ ≤ ,r n ( λ ) λ v ≤ θ v ( n + 1) − v , where θ v = (cid:26) , ≤ v ≤ v v , v > . ˆ g δn ( ξ ) − ˆ g ( ξ ) = r n ( λ )ˆ µ ( ξ ) + p n ( λ ) λ (1 − χ ϑ )ˆ µ ( ξ ) + p n ( λ ) λv ( ξ, T ) χ ϑ ˆ µ δT ( ξ ) − ˆ g ( ξ )= p n ( λ ) λv ( ξ, T ) [ χ ϑ ˆ µ δT ( ξ ) − ˆ µ T ( ξ ) + v ( ξ, T )(1 − χ ϑ )ˆ µ ( ξ )] − r n ( λ )[ˆ g ( ξ ) − ˆ µ ( ξ )]= p n ( λ ) λv ( ξ, T ) [ χ ϑ (ˆ µ δT ( ξ ) − ˆ µ T ( ξ )) + v ( ξ, T )(1 − χ ϑ ) ˆ f ( ξ ) Z T h ( s ) e kξ s ds ] − r n ( λ ) ˆ f ( ξ ) Z T h ( s ) e kξ s ds. Thus k ˆ f δn − ˆ f k = Z ∞−∞ ˆ g δn ( ξ ) − ˆ g ( ξ ) R T h ( s ) e kξ s ds ! dξ ≤ Z ∞−∞ p n ( λ ) λv ( ξ,T ) [ χ ϑ (ˆ µ δT ( ξ ) − ˆ µ T ( ξ )) + v ( ξ, T )(1 − χ ϑ ) ˆ f ( ξ ) R T h ( s ) e kξ s ds ] R T h ( s ) e kξ s ds dξ +2 Z ∞−∞ (cid:16) r n ( λ ) ˆ f ( ξ ) (cid:17) dξ := 2 I + 2 I . umerical methods for reconstruction of source term of heat equation from the final overdetermination Next, we give separated evaluation for I and I . We have I = Z ∞−∞ p n ( λ ) λv ( ξ,T ) [ χ ϑ (ˆ µ δT ( ξ ) − ˆ µ T ( ξ )) + v ( ξ, T )(1 − χ ϑ ) ˆ f ( ξ ) R T h ( s ) e kξ s ds ] R T h ( s ) e kξ s ds dξ = Z | ξ |≤ ϑ (cid:16) p n ( λ ) λv ( ξ,T ) (cid:2) χ ϑ (ˆ µ δT ( ξ ) − ˆ µ T ( ξ )) (cid:3)(cid:17) ( R T h ( s ) e kξ s ds ) dξ + Z | ξ | >ϑ ( p n ( λ ) λ ˆ f ( ξ )) dξ ≤ C h e kϑ T δ + Z | ξ | >ϑ ˆ f ( ξ ) dξ ≤ C h M σ δ σ σ ( ln Mδ ) − p + ϑ − p M ≤ C h M σ δ σ σ ( ln Mδ ) − p + CM (cid:20) ln Mδ ( ln Mδ ) − σ p (cid:21) − p ≤ ( ln Mδ ) − p C h M σ δ σ σ + CM ( ln Mδ ln Mδ ( ln Mδ ) − σ p ) p ! where C is the general constant depending on k and T . I = Z ∞−∞ ( r n ( λ ) ˆ f ( ξ )) dξ = Z | ξ |≤ ϑ ( r n ( λ ) ˆ f ( ξ )) dξ + Z | ξ | >ϑ ( r n ( λ ) ˆ f ( ξ )) dξ ≤ Z | ξ |≤ ϑ ( r n ( λ ) λ N ˆ f ( ξ ) v ( ξ, T ) ) dξ + ϑ − p M ≤ N N ( n + 1) − N e kϑ T M + ϑ − p M ≤ CN N M σ δ σ σ ( ln Mδ ) − p + CM (cid:20) ln Mδ ( ln Mδ ) − σ p (cid:21) − p ≤ ( ln Mδ ) − p CN N M σ δ σ σ + CM ( ln Mδ ln Mδ ( ln Mδ ) − σ p ) p ! . Finally we have k ˆ f δn − ˆ f k ≤ C ( ln Mδ ) − p M σ δ σ σ + M ( ln Mδ ln Mδ ( ln Mδ ) − σ p ) p ! and k ˆ u δn ( · , t ) − ˆ u ( · , t ) k = Z ∞−∞ (cid:20) ( ˆ f δn ( ξ ) − ˆ f ( ξ )) Z t h ( s ) e − kξ ( t − s ) ds (cid:21) dξ ≤ ( Z t | h ( s ) | ds ) ( ln Mδ ) − p M σ δ σ σ + M ( ln Mδ ln Mδ ( ln Mδ ) − σ p ) p ! . The proof is completed by using the Parseval equality. ✷ Remark 1
We see in Theorem 1 that if p = 0 then the second term in (22) and (23) is just a boundedterm and does not converge when δ →
0. However, the fact that the second term turns to zero when δ → R | ξ | >ϑ ( r n ( λ ) ˆ f ( ξ )) dξ definitely turns to zero (since ϑ → ∞ and f ∈ L ( R )).Thus if p = 0, which means f ( x ) ∈ L ( R ) and f ( x ) H p ( R ) , p >
0, and by choosing σ >
Xiaoping Fang et al. k ˆ µ δT − v ( · , T )ˆ g δn ∗ ϕ k ≤ τ δ σ < k ˆ µ δT − v ( · , T )ˆ g δn ∗ ϕ k , for 0 ≤ n < n ∗ , (14)where ϕ ( ξ ) = R T h ( s ) e kξs ds and n ∗ is the first iteration step which satisfies the left inequality of (14).With the discrepancy principle, we have similar convergence results Theorem 2
Let u ( x, t ) be the exact temperature history of (8), h ( t ) is identically nonposi-tive or nonnegative in [0 , T ] and µ δT ( x ) be the measured final temperature satisfying (9). f ( x ) sat-isfies k f k p ≤ M . Let ˆ g δk ( ξ ) denote the iterates defined by (11) with ˆ g δ ( ξ ) = ˆ µ ( ξ ) , where ϑ ∼ r σ ) kT h ln Mδ ( ln Mδ ) − σ p i . Suppose u δn ( x, t ) , f δn ( x ) are solved via (4), (5) and the inverse Fouriertransform for every ˆ g δk , respectively. If we select (14) as the a posteriori stopping rule, then there holds k f δn − f k ≤ C ( ln Mδ ) − p M σ δ σ σ + M ( ln Mδ ln Mδ ( ln Mδ ) − σ p ) p ! , (15) and k ˆ u δn ( · , t ) − ˆ u ( · , t ) k ≤ C h ( t )( ln Mδ ) − p M σ δ σ σ + M ( ln Mδ ln Mδ ( ln Mδ ) − σ p ) p ! , (16) for δ → , where C is a constant independent of δ and M and C h ( t ) = C ( R t | h ( s ) | ds ) . Proof.
It suffices to prove that n ∗ ∼ ⌊ N q Mδ ⌋ . In fact, by (11) we have k v ( · , T )ˆ g δn − ˆ µ δT ϕ k = k (1 − λ )[ v ( · , T )ˆ g δn − − χ ϑ ˆ µ δT ] + λ (1 − χ ϑ ) v ( · , T )ˆ µ + (1 − χ ϑ )ˆ µ δT ϕ k = k r n ( λ )[ v ( · , T )ˆ µ − χ ϑ ˆ µ δT ] + p n ( λ ) λ (1 − χ ϑ ) v ( · , T )ˆ µ + (1 − χ ϑ )ˆ µ δT ϕ k≤ k r n ( λ ) χ ϑ (ˆ µ T − ˆ µ δT ) + (1 − χ ϑ )(ˆ µ T − ˆ µ δT ) ϕ k + k r n ( λ ) χ ϑ (ˆ µ T − v ( · , T )ˆ µ ) + (1 − χ ϑ )(ˆ µ T − v ( · , T )ˆ µ ) ϕ k≤ δC h + k r n ( λ ) χ ϑ v ( · , T ) ˆ f + (1 − χ ϑ ) v ( · , T ) ˆ f k≤ δC h + N N ( n + 1) − N M + e − kϑ T ϑ − p M ≤ δC h + N N ( n + 1) − N M + CM σ σ δ σ ( ln Mδ ln Mδ ( ln Mδ ) − σ p ) p = δ σ ( C M σ σ + 1 C h δ σ σ ) + N N ( n + 1) − N M, where C = ( ln Mδ ln Mδ ( ln Mδ ) − σ p ) p is a bounded term. We come to the conclusion if we choose τ = C M σ σ + C h +1 C h δ σ σ . umerical methods for reconstruction of source term of heat equation from the final overdetermination In this section, we consider the reconstruction of source term f ( x ) in (1) with final overdetermination µ T ( x ). For the sake of simplicity we still use the sign k·k for norm in L [0 , u ( x, t ) = Z ∞−∞ √ kπt e − ( x − y )24 kt µ ( y ) dy + Z t Z ∞−∞ p kπ ( t − s ) e − ( x − y )24 k ( t − s ) f ( y ) h ( s ) dyds, (17)where µ ( x ) and f ( x ) are extended to the following form µ ( x ) = Z µ ( y ) dy + 2 ∞ X m =1 cos( mπx ) Z µ ( y ) cos( mπy ) dy,f ( x ) = Z f ( y ) dy + 2 ∞ X m =1 cos( mπx ) Z f ( y ) cos( mπy ) dy. It is easy to verify that under the above extensions of µ ( x ) and f ( x ), the boundary conditions aresatisfied. Furthermore the solution (17) is similar to (3). But we can not directly use the iterativemethod mentioned in section 2 to solve the inverse problem to get f ( x ) and u ( x, t ). In order to usethe iterative method (11) to solve the inverse problem we need further extend µ T ( x ) as µ T ( x ) = Z µ T ( y ) dy + 2 ∞ X m =1 cos( mπx ) Z µ T ( y ) cos( mπy ) dy. However, the extended functions µ ( x ) and µ T ( x ) are no longer L integrable functions in R . Thussome further analysis of (17) is needed.In fact (17) can be changed into another solution form. To explain this, firstly we give the followinglemma. Lemma 1
There holds the following identity Z ∞−∞ cos( mπy ) 1 √ kπt e − ( x − y )24 kt dy = cos( mπx ) e − m π kt . Proof.
First, by Taylor expansion of cos( mπy ) we have Z ∞−∞ cos( mπy ) 1 √ kπt e − y kt dy = Z ∞−∞ ∞ X i =0 ( − i ( mπy ) i (2 i )! 1 √ kπt e − y kt dy = ∞ X i =0 ( − i ( mπ ) i (2 i )! Z ∞−∞ y i √ kπt e − y kt dy = ∞ X i =0 ( − i ( mπ ) i (2 i )! (4 kt ) i i Y j =1 j − ∞ X i =0 ( − i ( mπ ) i i ! ( kt ) i = e − m π kt . Xiaoping Fang et al.
Thus we have Z ∞−∞ cos( mπy ) 1 √ kπt e − ( x − y )24 kt dy = Z ∞−∞ cos( mπ ( x − y )) 1 √ kπt e − y kt dy = Z ∞−∞ [cos( mπx ) cos( mπy ) + sin( mπx ) sin( mπy )] 1 √ kπt e − y kt dy = cos( mπx ) Z ∞−∞ cos( mπy ) 1 √ kπt e − y kt dy = cos( mπx ) e − m π kt . ✷ By Lemma 1 and (17) one can easily get another solution form of (1) u ( x, t ) = ∞ X m =0 (cid:20) e − m π kt a m cos( mπx ) + b m cos( mπx ) Z t h ( s ) e − m π k ( t − s ) ds (cid:21) , (18)where for m = 0 there hold a = R µ ( x ) dx , b = R f ( x ) dx , and for m > a m =2 R µ ( x ) cos( mπx ) dx , b m = 2 R f ( x ) cos( mπx ) dx . By using the final data µ T ( x ) = u ( x, T ) we have µ T ( x ) = ∞ X m =0 " e − m π kT a n cos( mπx ) + b n cos( mπx ) Z T h ( s ) e − m π k ( T − s ) ds . Denoting c = R µ T ( x ) dx and c n = 2 R µ T ( x ) cos( mπx ) dx, m > mπx ) we obtain b m = e m π kT c m − a m R T h ( s ) e m π ks ds . (19)We can use the singular decomposition to solve b m . In fact, if we define linear operator K as Kf ( x ) := ∞ X n =0 R T h ( s ) e − m π k ( T − s ) ds R T h ( x ) ds b m cos( mπx )then we have the singular values (or eigenvalues) { σ m } with σ m = R T h ( s ) e − m π k ( T − s ) ds R T h ( x ) ds and corre-sponding eigenvector { cos( mπx ) } . Since | σ m | ≤ m →∞ σ m = 0. The problem is a general linearoperator equation in inverse problem. Lots of regularization methods, such as Tikhonov regularization,Landerweber iteration, etc., can be used to solve this problem. In this paper we shall discuss about thefrequency cut-off method to solve the problem, references for other methods can be found in [9,13]. Infact, we only need to solve b m in (19). We denote c δ = R µ δT ( x ) dx , c δm = R µ δT ( x ) cos( mπx ) dx, m > k µ δT ( · ) − µ T ( · ) k = Z " ∞ X m =0 ( c δm − c m ) cos( nπx ) dx ≤ δ. (20)We introduce the frequency cut-off method to solve b δm by b δm = χ ϑ ( e m π kT c δm − a m ) R T h ( s ) e m π ks ds , (21) umerical methods for reconstruction of source term of heat equation from the final overdetermination where χ ϑ is the discrete version of characteristic function defined in section 2, that is χ ϑ = 1 for m ≤ ϑ ∈ N and χ ϑ = 0 for m > ϑ . Thus we can reconstruct f ( x ) with f δ ( x ) f δ ( x ) = ϑ X m =0 b δm cos ( mπx )and u δ ( x, t ) = ∞ X m =0 e − m π kt a m cos ( mπx ) + ϑ X m =0 b δm cos ( mπx ) Z t h ( s ) e − m π k ( t − s ) ds. Based on (21) we have the convergence theorem as follows
Theorem 3
Let u ( x, t ) be the exact temperature history of (1), h ( t ) is identically nonpositiveor nonnegative in [0 , T ] and µ δT ( x ) be the measured final temperature satisfying (20). f ( x ) satisfies k f k H p (0 , ≤ M , where k f k H p (0 , is defined as k f k H p (0 , = ∞ X m =0 (1 + m ) p b m ! . Let b δm defined by (21) and ϑ ∼ ⌊ r σ ) kT h ln Mδ ( ln Mδ ) − σ p i ⌋ , σ ≥ then there holds k f δ − f k ≤ C ( ln Mδ ) − p M σ δ σ σ + M ( ln Mδ ln Mδ ( ln Mδ ) − σ p ) p ! , (22) and k u δ ( · , t ) − u ( · , t ) k ≤ C h ( t )( ln Mδ ) − p M σ δ σ σ + M ( ln Mδ ln Mδ ( ln Mδ ) − σ p ) p ! , (23) for δ → , where C is a constant independent of δ and M and C h ( t ) = C ( R t | h ( s ) | ds ) . Proof. k f δ − f k = k ϑ X m =0 ( b δm − b m ) cos ( mπ · ) − ∞ X m = ϑ +1 b m cos ( mπ · ) k ≤ k ϑ X m =0 ( b δm − b m ) cos ( mπ · ) k + 2 k ∞ X m = ϑ +1 b m cos ( mπ · ) k = 2 k ϑ X m =0 e m π kT ( c δm − c m ) R T h ( s ) e m π ks ds cos ( mπ · ) k + 2 k ∞ X m = ϑ +1 b m cos ( mπ · ) k ≤ C h e ϑ π kT δ + 2 ϑ − p M ≤ ln Mδ ) − p C h M σ δ σ σ + CM ( ln Mδ ln Mδ ( ln Mδ ) − σ p ) p ! . k u δ ( · , t ) − u ( · , t ) k = k ( f δ − f ) Z t h ( s ) e − m π k ( t − s ) ds k ≤ Z t | h ( s ) | ds ) ( ln Mδ ) − p M σ δ σ σ + M ( ln Mδ ln Mδ ( ln Mδ ) − σ p ) p ! , which proves the theorem. ✷ Remark 2
We see from Theorem 3, that the convergence results are similar to the results in Section2. Actually, we can also design the similar iterative method like we design in Section 2. And we canintroduce the discrepancy principle as the a posteriori stopping rule. If (17) is used for reconstructionthen the corresponding functions are extended to periodical functions. Thus in the numerical calcu-lation, by using iterative method in section 2 the fast Fourier transform (FFT) can be considered forthe periodical function in R . In this section, we present some numerical experiments on reconstruction of the source term with thefinal measurement µ δT ( x ) for T = 1. We separate the span [0 ,
1] for x variable into an equidistancegrid 0 = x < · · · < x i < · · · < x N = 1 ( x i = ih, h = 0 . , N = 50), and the span [0 ,
1] for t variableinto an equidistance grid 0 = t < · · · < t j < · · · < t N = 1 ( t j = jl, l = 0 . , N = 20). We producethe random noise as follows µ δT ( x i ) = µ δT ( x i ) + 2(rand(0 , − . ∗ noiselv ∗ µ δT ( x i ) , where rand(0,1) denotes the uniformly distributed pseudo-random numbers in [0,1] generated byMatlab software and noiselv is a positive number between 0 and 1 for noise level. The noise δ iscalculated by numerical calculation of L (0 ,
1) norm (by first approximating the function with splineinterpolation and then using the integral algorithm).We only consider the numerical implementation of reconstruction of the source term in problem (1)although numerical method for (2) can be similarly implemented.4.1 Example 1Set f ( x ) = 1 + cos(3 πx ) + 2 cos(5 πx ), µ ( x ) = cos(2 πx ), h ( t ) = t and k = 1 then the solution of (1) is u ( x, t ) = cos(2 πx ) e − π t + t π t − e − π t π cos(3 πx ) + 2 25 π t − e − π t π cos(5 πx ) , and the final measurement at T = 1 is µ T ( x, t ) = 12 + cos(2 πx ) e − π + 9 π − e − π π cos(3 πx ) + 2 25 π − e − π π cos(5 πx ) . We choose p = 1 / σ = 0 . P ∞ m =0 e − m π t cos( mπx ) R µ ( y ) cos( mπy ) dy , we choose a sufficiently largenumber m < ϑ to numerically approximate it. Thus the numerical implementation of u δ ( x, t ) is u δ ( x, t ) = ϑ X m =0 e − m π t a m cos ( mπx ) + ϑ X m =0 b δm cos ( mπx ) Z t h ( s ) e − m π ( t − s ) ds. Table 1 shows the numerical results for different choice of ϑ and error level. We see from the tablethat the reconstruction of the solution u δ has more accuracy than the source f ( x ). Fig. 1 shows theperformance of reconstruction of source term f ( x ) under different final measurements µ δT ( x ) while Fig.2 gives the comparison between the true solution and the numerical solution with noise level 1%. It isclearly in this figure that the solution is not affected that much compared with the source term underthe measurement noise of µ T ( x ). Fig. 3 shows the numerical results under the noise level 20%. Sincethe noise level is quite high, the numerical method can not produce good approximation solution. umerical methods for reconstruction of source term of heat equation from the final overdetermination Table 1
Convergence results. The parameter p = 1 / σ = 0 . M = 1 . noise level (1%) noise level (5%) δ ϑ = 12 — 24 — 36 6 — 12 — 18 k f δ − f k k u δ ( · , t ) − u ( · , t ) k noise level (10%) noise level (20%) δ ϑ = 6 — 12 — 18 6 — 12 — 18 k f δ − f k k u δ ( · , t ) − u ( · , t ) k µ T (x))measured final data ( µ T δ (x) 0 0.2 0.4 0.6 0.8 1−2−1012345 x exact source (f(x))reconstruct source (f δ (x))0 0.2 0.4 0.6 0.8 10.440.460.480.50.520.540.560.580.6 x true final data ( µ T (x))measured final data ( µ T δ (x) 0 0.2 0.4 0.6 0.8 1−3−2−101234 x exact source (f(x))reconstruct source (f δ (x)) Fig. 1
Comparison of convergence results. The top two with the noise level 1% and ϑ = 12, while the bottom twowith the noise level 5% and ϑ = 18. The left are the final data and corresponding measurement error data. The otherparameters are p = 1 / σ = 0 . u δ (x,t))x u Fig. 2
Comparison of true solution and reconstruct solution. The noise level is 1% and ϑ = 12, p = 1 / σ = 0 . µ T (x))measured final data ( µ T δ (x) 0 0.2 0.4 0.6 0.8 1−6−4−20246 x exact source (f(x))reconstruct source (f δ (x))0 0.2 0.4 0.6 0.8 100.20.40.60.81−1−0.500.51 treconstrut solution (u δ (x,t))x u Fig. 3
Convergence results with noise level 20% and ϑ = 18, p = 1 / σ = 0 . umerical methods for reconstruction of source term of heat equation from the final overdetermination f ( x ) = (1 − x ) x | x − | + , µ ( x ) = 0, h ( t ) = 5 sin(2 πt ) + 1 and k = 1. The true solution and the final datacan be calculated by (18) (Fig. 4). Since the final data is nearly a constant function, small noise levelcan still produce striking different measurement data comparing with the exact data. Thus here wechoose noise level with 1% and 0 . h ( t ) does not satisfy the identicallynon-positive or non-negative property. However, we can still get the convergence results since h ( t ) isnot too ’bad’. Fig. 5 shows the reconstructed solution and the exact solution. µ T (x))measured final data ( µ T δ (x) 0 0.2 0.4 0.6 0.8 1−0.100.10.20.30.40.50.6 x exact source (f(x))reconstruct source (f δ (x))0 0.2 0.4 0.6 0.8 10.2470.2480.2490.250.2510.2520.2530.254 x true final data ( µ T (x))measured final data ( µ T δ (x) δ (x)) Fig. 4
Comparison of convergence results. The top two with the noise level 1%, while the bottom two with the noiselevel 0 . ϑ = 12, p = 1 / σ = 0 . f ( x ) = x < . x . ≤ x ≤ . − x . < x ≤ . x > . µ ( x ) = 0, h ( t ) = e t + 6 sin(4 πt ) + t + 1 and k = 1. The source f ( x ) has two discontinuouspoints ( x = 0 . , .
8) and one non-differentiable point ( x = 0 . u δ (x,t))x u Fig. 5
Comparison of true solution and reconstruct solution. The noise level is 1% and ϑ = 12, p = 1 / σ = 0 . µ T (x))measured final data ( µ T δ (x) 0 0.2 0.4 0.6 0.8 1−0.100.10.20.30.40.50.6 x exact source (f(x))reconstruct source (f δ (x))0 0.2 0.4 0.6 0.8 10.630.640.650.660.670.680.69 x true final data ( µ T (x))measured final data ( µ T δ (x) 0 0.2 0.4 0.6 0.8 1−0.100.10.20.30.40.50.6 x exact source (f(x))reconstruct source (f δ (x)) Fig. 6
Comparison of convergence results. The top two with the noise level 1%, while the bottom two with the noiselevel 0 . ϑ = 12, p = 1 / σ = 0 . In this paper, the numerical methods for reconstruction of source term in both no boundary andNeumann boundary conditions are presented. The convergence rate has been proved for both a pri-ori and a posteriori stopping rules. More importantly, we show that the solution of the boundaryconditions problem has the form of solution for the no boundary problem, which can be applied forboth Neumann and Drichlet boundary conditions. The numerical experiments have shown that thefrequency cut-off technique method applies well for the boundary conditions problem, although for umerical methods for reconstruction of source term of heat equation from the final overdetermination u δ (x,t))x u Fig. 7
Comparison of true solution and reconstruct solution. The noise level is 0 .
1% and ϑ = 12, p = 1 / σ = 0 . more accurate results we may implement the iterative methods together with the a posteriori stoppingrule. The numerical methods can be moved parallel to the two dimensional inverse source problem. References
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