A new regularization method for a parameter identification problem in a non-linear partial differential equation
aa r X i v : . [ m a t h . NA ] A p r A NEW REGULARIZATION METHOD FOR A PARAMETER IDENTIFICATIONPROBLEM IN A NON-LINEAR PARTIAL DIFFERENTIAL EQUATION
M. THAMBAN NAIR AND SAMPRITA DAS ROY
Abstract.
We consider a parameter identification problem associated with a quasi-linear ellipticNeumann boundary value problem involving a parameter function a ( · ) and the solution u ( · ), wherethe problem is to identify a ( · ) on an interval I := g (Γ) from the knowledge of the solution u ( · ) as g on Γ, where Γ is a given curve on the boundary of the domain Ω ⊆ R of the problem and g is a continuous function. The inverse problem is formulated as a problem of solving an operatorequation involving a compact operator depending on the data, and for obtaining stable approximatesolutions under noisy data, a new regularization method is considered. The derived error estimatesare similar to, and in certain cases better than, the classical Tikhonov regularization considered inthe literature in recent past. Keywords:
Ill-posed, regularization, parameter identification.
AMS Subject Classification:
MSC 2010: 35R30, 65N30, 65J15, 65J20, 76S05
Contents
1. Introduction 12. Operator Theoretic Formulation 33. The New Regularization 53.1. Error estimates under exact data 103.2. Error estimates under noisy data 114. Relaxation of assumption on perturbed data 205. With exact solution having non-zero value at g Introduction
Let Ω be a bounded domain in R with C , boundary. Consider the problem of finding a weaksolution u ∈ H (Ω) of the partial differential equation(1.1) − ∇ . ( a ( u ) ∇ u ) = 0 in Ωwith boundary condition(1.2) a ( u ) ∂u∂ν = j on ∂ Ω , where a ∈ H ( R ) and j ∈ L ( ∂ Ω). It is known that such a solution u exists if a ≥ κ > κ and R ∂ Ω j = 0 (see [10], [7]). It is also known that, under an additional assumptionthat j ∈ W (1 − /p ) ,p ( ∂ Ω) with p > u ∈ C (Ω) (cf. [4]). One can come across this type of problemsin the steady state heat transfer problem with u being the temperature, a the thermal conductivitywhich is a function of the temperature, and j the heat flux applied to the surface. In this paper we consider one of the inverse problems associated with the above direct problem,namely the following:
Problem (P):
Let γ : [0 , → ∂ Ω be a C - curve on ∂ Ω and Γ be its range, that is, Γ := γ ([0 , .Given g : Γ → R such that g ◦ γ ∈ C ([0 , and j ∈ W (1 − /p ) ,p ( ∂ Ω) with p > and R ∂ Ω j = 0 , theproblem is to identify an a ∈ H ( R ) on I := g (Γ) such that the corresponding u satisfies (1.1)-(1.2)along with the requirement (1.3) u = g on Γ . It is known that, with only the knowledge of u = g on Γ, the parameter a can be identified uniquelyonly on I (cf.[2]). In the following we shall use the same notation for a for a ∈ H ( R ) and its restriction a I ∈ H ( I ).We shall see that Problem (P) is ill-posed, in the sense that the solution a | I does not dependcontinuously on the data g and j (see Sections 2). To obtain a stable approximate solution forProblem (P), we use a new regularization method which is different from some of the standard onesin the literature. We discuss this method in Section 3.The existence and uniqueness of the solution for Problem (P) is known under some additionalconditions on γ and g , as specified in Section 2 (c.f. [2, 4]). In [7] and [4] the problem of findinga stable approximate solution of the problem is studied by employing Tikhonov regularization withnoisy data. In [7], with the noisy data g δ , in place of g , satisfying k g − g δ k L (Γ) ≤ δ, convergence rate k a − a δ k H ( I ) = O ( √ δ ) is obtained whenever a ∈ H ( I ) and its trace is Lipschitz on ∂ Ω, where a δ isthe approximate solution obtained via Tikhonov regularization. In [4], the rate k a − a δ k L ( I ) = O ( √ δ )is obtained without the additional assumption on a . Moreover, here the noisy data j δ belonging to W − /p,p ( ∂ Ω) with p >
3, and satisfying k j − j δ k L ( ∂ Ω) ≤ δ , is also considered along with the noisydata g δ as considered in [7]. It is stated in [4] that “the rate O ( √ δ ) is possible with respect to H -norm, provided some additional smoothness conditions are satisfied”; however, the details of theanalysis is missing.Under our newly introduced method, we obtain the above type of error estimates using appropriatesmoothness assumptions. In particular we prove that, if g ∈ R is such that I = [ g , g ] and if a ( g )is known or is approximately known, and the perturbed data j δ and g δ belong to W − /p,p ( ∂ Ω) for p > C (Γ), respectively, satisfying k j − j δ k L ( ∂ Ω) ≤ δ, k g − g δ k W , ∞ (Γ) ≤ δ, then the convergence rate is O ( √ δ ) with respect to L -norm. With additional assumption that theexact solution is in H ( I ) we obtain a convergence rate O ( δ / ) with respect to L -norm. Again,in particular, if g ◦ γ is in H ([0 , O ( δ / ) with respect to L -norm is obtained under aweaker condition on perturbed data g δ , namely, g δ ∈ L (Γ) with k g − g δ k L (Γ) ≤ δ . Also, in the newmethod we do not need the assumption on g δ made in [4] which is g δ (Γ) ⊂ g (Γ). Thus some of theestimates obtained in this paper are improvements over the known estimates, and are also better thanthe expected best possible estimate, namely O ( δ / ), in the context of Tikhonov regularization, asmentioned in [4].The paper is organized as follows: In Section 2 we present a theorem which characterize the solutionof the inverse Problem (P) in terms of the solution of the Laplace equation with an appropriate Neu-mann condition. Also, the inverse problem is represented as the problem of solving a linear operatorequation, where the operator is written as a composition of three injective bounded operators one ofwhich is a compact operator, and prove some properties of these operators. The new regularizationmethod is defined in Section 3, and error estimates with noisy as well as exact data are derived. InSection 4 we present error analysis with some relaxed conditions on the perturbed data. In Section5 a procedure is described to relax a condition on the exact data and corresponding error estimate ONLINEAR INVERSE PROBLEM 3 is derived. In Section 6 we illustrate the procedure of obtaining a stable approximate solution toProblem (P). 2.
Operator Theoretic Formulation
Throughout the paper we denote by I the range of the function g : Γ → R , and write it as I = [ g , g ], that is g and g are the left and right end-points of the closed interval g ( γ ([0 , Theorem 2.1.
Let j , g and γ be as specified in Problem(P). Then, Problem(P) has a unique solution a ∈ H ( I ) , and it is the unique a ∈ H ( I ) such that (2.1) v ( γ ( s )) = Z g ( γ ( s )) g a ( t ) dt ∀ s ∈ [0 , , where v ∈ C (Ω) is the unique function which satisfies (2.2) − △ v = 0 in Ω , (2.3) ∂v∂ν = j on ∂ Ω and (2.4) Z Ω v = 0 . It is known that if j ∈ W − /p,p ( ∂ Ω) for p >
3, then v satisfying (2.2)-(2.3) belongs to W ,p (Ω),and(2.5) k v j k W ,p (Ω) ≤ C k j k L ( ∂ Ω) for some constant C > j and g as inProblem (P), let v ∈ C (Ω) be the function satisfying (2.2), (2.3) and (2.4). Then, a ∈ H ( I ) is thesolution of Problem(P) if and only if Z g ( γ ( s )) g a ( t ) dt = v ( γ ( s )) , s ∈ [0 , . The above equation can be represented as an operator equation(2.6)
T a = v j ◦ γ, where v j is the solution of (2.2)-(2.4) and the operator T : L ( I ) → L [0 ,
1] is defined by(2.7) (
T w )( s ) = Z g ( γ ( s )) g w ( t ) dt, w ∈ L ( I ) , s ∈ [0 , . Theorem 2.2.
The operator T defined in (2.7) is an injective compact operator of infinite rank. Inparticular, T : H ( I ) → L [0 , is a compact operator of infinite rank.Proof. Note that for every w ∈ L ( I ) and for every s, τ ∈ [0 , | ( T w )( s ) − ( T w )( τ ) | ≤ k w k L ( I ) | ( g ◦ γ )( s ) − ( g ◦ γ )( τ ) | / . Since g ◦ γ is continuous, the set { T w : k w k L ( I ) ≤ } is equicontinuous and uniformly bounded in C [0 , T is a compact operator from L ( I ) to C [0 , C [0 , ⊆ L [0 , T : L ( I ) → L [0 ,
1] is also a compact operator. We note that T isinjective. Hence, T is of infinite rank. (cid:3) M. THAMBAN NAIR AND SAMPRITA DAS ROY
It is to be observed that the compact operator T defined in (2.7) depends on g . Thus, the problemof solving operator equation (2.6) based on the data ( g, j ) is non-linear as well as ill-posed.In order to consider our new regularization method for obtaining stable approximate solutions, werepresent the operator T in (2.6) as a composition of three operators, that is, T = T T T , where, for r ∈ { , } , T : H r ( I ) → H r +1 ( I ) , T : H r +1 ( I ) → L ( I ) , T : L ( I ) → L ([0 , T ( w )( τ ) := Z τg w ( t ) dt, w ∈ H r ( I ) , τ ∈ I, (2.8) T ( w ) := w, w ∈ H r +1 ( I ) , (2.9) T ( w ) := w ◦ g ◦ γ, w ∈ L ( I ) . (2.10)Clearly, T , T , T are linear operators and( T T T w )( s ) = Z g ( γ ( s )) g w ( t ) dt = ( T w )( s ) , s ∈ [0 , . Here, we used the convention that H ( I ) := L ( I ).By the above representation of T , the operator equation (2.6) can be split into three equations: T ( ζ ) = v j ◦ γ, (2.11) T ( b ) = ζ, (2.12) T ( a ) = b. (2.13)To prove some properties of the operators T , T , T , we specify the requirements on j, g and γ , namelythe following. Assumption 2.3.
Let j ∈ W (1 − /p ) ,p ( ∂ Ω) with p > R ∂ Ω j = 0. Let γ : [0 , → ∂ Ω be a C -curve on ∂ Ω and g : Γ → R be such that g ∈ C (Γ),(2.14) C γ ≤ | γ ′ ( s ) | ≤ C ′ γ ∀ s ∈ [0 , , (2.15) C g ≤ | g ′ ( γ ( s )) | ≤ C ′ g ∀ s ∈ [0 , , for some positive constants C γ , C ′ γ , C g and C ′ g .Next we state a result from analysis which will be used in the next result and also in many otherresults that follow. Lemma 2.4.
Let h and h be two continuous functions on intervals J and J respectively, such that h ( J ) = J . Also, let h ′ be continuous with h ′ = 0 . Then, we have the following. Z J h ( h ( x )) dx = Z J h ( y ) | h ′ ( h − ( y )) | dy. We shall also make use of the following proposition.
Proposition 2.5.
Let C g , C γ , C ′ g C ′ γ be as in Assumption 2.3. Then for any w ∈ L ( I ) , (2.16) C g C γ Z | w ( g ( γ ( s ))) | ds ≤ Z I | w ( y ) | dy ≤ C ′ g C ′ γ Z | w ( g ( γ ( s ))) | ds. ONLINEAR INVERSE PROBLEM 5
Proof.
By Lemma 2.4 and the inequalities (2.14) and (2.15) in Assumption 2.3, we have Z | w ( g ( γ ( s ))) | ds = Z g g | w ( y ) | | g ′ ( g − ( y )) γ ′ ( γ − ( g − ( y ))) | dy ≤ C g C γ Z I | w ( y ) | dy, Z g g | w ( y ) | dy = Z | w ( g ( γ ( s ))) | | g ′ ( γ ( s )) γ ′ ( s ) | ds ≤ C ′ g C ′ γ Z | w ( g ( γ ( s ))) | ds. From the above, we obtain the required inequalities in (2.16). (cid:3)
Theorem 2.6.
Let r ∈ { , } , and let T : H r ( I ) → H r +1 ( I ) , T : H r +1 ( I ) → L ( I ) , T : L ( I ) → L ([0 , be defined as in (2.8), (2.9) and (2.10), respectively. Then, T is a compact operator, and for every w ∈ L ( I ) , (2.17) k w k H r ( I ) ≤ k T ( w ) k H r +1 ( I ) ≤ (1 + √ g − g ) k w k H r ( I ) , (2.18) C g C γ k T ( w ) k L ( I ) ≤ k w k L ( I ) ≤ C ′ g C ′ γ k T ( w ) k L ([0 , , In particular, T and T are bounded operators with bounded inverse from their ranges.Proof. Since H ( I ) and H ( I ) are compactly embedded in L ( I ) (cf. [6]), T is a compact operatorof infinite rank. Now, let w ∈ H ( I ) and τ ∈ I . Then | T ( w )( τ ) | ≤ Z τg | w ( t ) | dt ≤ k w k L ( I ) √ g − g , so that k T ( w ) k L ( I ) ≤ k w k L ( I ) √ g − g . Hence, using the fact that ( T ( w )) ′ = w and ( T ( w )) ′′ = w ′ , we have k w k L ( I ) ≤ k T ( w ) k L ( I ) + k w k L ( I ) ≤ (1 + √ g − g ) k w k L ( I ) so that k w k L ( I ) ≤ k T ( w ) k H ( I ) ≤ (1 + √ g − g ) k w k L ( I ) , k w k H ( I ) ≤ k T ( w ) k H ( I ) ≤ (1 + √ g − g ) k w k H ( I ) , Thus, (2.17) is proved.By the inequalities in (2.16) we obtain(2.19) C g C γ k T ( w ) k L ([0 , ≤ k w k L ( I ) ≤ C ′ g C ′ γ k T ( w ) k L ([0 , for every w ∈ L ( I ). The inequalities in (2.17) and (2.19) also show that T and T are boundedoperator with bounded inverse from their ranges. (cid:3) The New Regularization
We know that Problem (P) is ill-posed. We may also recall that the operator equation (2.6) isequivalent to the system of of operator equations (2.11)-(2.13), wherein equation (2.12) is ill-posed,since T is a compact operator of infinite rank. Thus, in order to regularize (2.6), we shall replacethe equation (2.12) by a regularized form of it using a family of bounded operators T α , α >
0, whichapproximates the compact operator T in norm.Note that T : H ( I ) → L ( I ) is defined by T ( w ) = w, w ∈ H ( I ) . We consider T α as a perturbed form of T , namely, T α : H ( I ) → L ( I ), defined by(3.1) T α ( w ) = w − αw ′′ , w ∈ H ( I )for each α > M. THAMBAN NAIR AND SAMPRITA DAS ROY
Theorem 3.1.
For α > , let T α : H ( I ) → L ( I ) be defined as in (3.1) Then, k T α ( w ) k L ( I ) ≤ max { , α }k w k H ( I ) , w ∈ H ( I ) . In particular, T α is a bounded operator with k T α k ≤ max { , α } . Further, k T α − T k → as α → . Proof.
We observe that, for any w ∈ H ( I ), k T α ( w ) k L ( I ) = k w − αw ′′ k L ( I ) ≤ k w k L ( I ) + α k w ′′ k L ( I ) ≤ max { , α }k w k H ( I ) . Thus, T α is a bounded operator with k T α k ≤ max { , α } for all α >
0. Further, k ( T α − T )( w ) k L ( I ) = k αw ′′ k L ( I ) ≤ α k w k H ( I ) . Hence, we also have k T α − T k → α → . (cid:3) In order to define a regularization family for T , we introduce the space(3.2) W := { w ∈ H ( I ) : w ( g ) = 0 , w ′ ( g ) = 0 } . Note that, for w ∈ H ( I ), w ∈ W if and only if w ( t ) = Z tg ξ ( s ) ds for some ξ ∈ H ( I ) satisfying ξ ( g ) = 0 . We prove that W is a closed subspace of H ( I ) and T α asan operator from W to L ( I ) is bounded below with respect to H ( I ) norm. Proposition 3.2.
The space W defined in (3.2) is a closed subspace of H ( I ) and ( T α | W ) ∗ = Q ( T α ) ∗ , where Q : H ( I ) → H ( I ) is the orthogonal projection onto W .Proof. Let ( w n ) in W be such that w n → w in H ( I ) for some w ∈ H ( I ). By a Sobolev imbeddingTheorem (cf. [6]), H ( I ) is continuously imbedded in the space C ( I ) with C -norm. Therefore, w ∈ C ( I ), and sup t ∈ I {| w n ( t ) − w ( t ) | + | w ′ n ( t ) − w ′ ( t ) |} → n → ∞ . Also, | w n ( g ) − w ( g ) | ≤ sup t ∈ I {| w n ( t ) − w ( t ) | + | w ′ n ( t ) − w ′ ( t ) |} ∀ n ∈ N and | w ′ n ( g ) − w ′ ( g ) | ≤ sup t ∈ I {| w n ( t ) − w ( t ) | + | w ′ n ( t ) − w ′ ( t ) |} ∀ n ∈ N . Thus, since w n ∈ W , in particular | w ( g ) | = lim n →∞ w n ( g ) = 0 and | w ′ ( g ) | = lim n →∞ w ′ n ( g ) = 0 . Hence w ∈ W . Thus W is closed. Now, let Q : H ( I ) → H ( I ) be the orthogonal projection onto W . Then, for y ∈ L ( I ) and w ∈ W we have, h Q ( T α ) ∗ ( y ) , w i H ( I ) = h y, ( T α ) Qw i L ( I ) = h y, ( T α | W ) w i L ( I ) = h ( T α | W ) ∗ y, w i H ( I ) Hence we have ( T α | W ) ∗ = Q ( T α ) ∗ . (cid:3) Let us see some other properties of the space W which shall be used in order to construct theregularization method. ONLINEAR INVERSE PROBLEM 7
Proposition 3.3.
Let α > . Let L : H ( I ) → H ( I ) be defined by Lx ( t ) = x ′ ( g ) √ α " e ( t − g √ α ) − e − ( t − g √ α ) e ( g − g √ α ) + e − ( g − g √ α ) + x ( g ) " e ( t − g √ α ) + e − ( t − g √ α ) e ( g − g √ α ) + e − ( g − g √ α ) for every x ∈ H ( I ) , t ∈ I . Then we have the following. (i) For any x ∈ H ( I ) , Lx ∈ C ∞ ( I ) ⊂ H ( I ) , α ( Lx ) ′′ = Lx and Lx ∈ N ( T α ) . (ii) L is a bounded linear operator. (iii) The map id − L is a projection onto W , where id is the identity map on H ( I ) .Proof. Clearly, L is a linear operator, and for any x ∈ H ( I ), we have Lx ∈ C ∞ ( I ) ⊂ H ( I ) and α ( Lx ) ′′ = Lx . To show that L is continuous, let ( x n ) be a sequence in H ( I ) such that k x n − x k H ( I ) → x ∈ H ( I ). By a Sobolev imbedding Theorem (cf. [6]), H ( I ) is continuously imbeddedin the space C ( I ) with C -norm, and so we have | x n ( g ) − x ( g ) | → | x ′ n ( g ) − x ′ ( g ) | → n → ∞ . Using this, it can be shown that L is continuous. Now again by definition of L , for any x ∈ H ( I ) we have ( x − Lx )( g ) = x ( g ) − Lx ( g ) = x ( g ) − x ( g ) = 0 , ( x − Lx ) ′ ( g ) = x ′ ( g ) − ( Lx ) ′ ( g ) = x ′ ( g ) − x ′ ( g ) = 0 , so that ( id − L )( x − Lx ) = x − Lx − L ( x − Lx ) = x − Lx.
Hence, using the definition of the space W ,we have id − L is a projection onto W . (cid:3) We shall use the notation(3.3) C L := k id − L k , where L is the bounded operator as in Proposition 3.3. Theorem 3.4.
Let < α < . Then, for every w ∈ W , (3.4) k T α ( w ) k L ( I ) ≥ α k w k H ( I ) , (3.5) k T α ( w ) k L ( I ) ≥ √ α k w k H ( I ) . Proof.
First we observe, by integration by parts, that for w , w ∈ W , R I w w ′′ = − R I w ′ w ′ . Hence,for every w ∈ W , k T α ( w ) k L ( I ) = Z g g | w − αw ′′ | = Z g g | w | + α Z g g | w ′′ | − α Z g g ww ′′ = Z g g | w | + α Z g g | w ′′ | + 2 α Z g g | w ′ | . Since 0 < α <
1, for every w ∈ W , Z g g | w | + α Z g g | w ′′ | + 2 α Z g g | w ′ | ≥ α k w k H ( I ) , Z g g | w | + α Z g g | w ′′ | + 2 α Z g g | w ′ | ≥ α k w k H ( I ) . This completes the proof. (cid:3)
At this point let us note that, by (3.4), T α is is bounded below on W . Henceforth, we shall use thesame notation for T α and its restriction to W , that is,(3.6) T α ( w ) = w − αw ′′ , w ∈ W and the adjoint of this operator will be denoted ( T α ) ∗ . M. THAMBAN NAIR AND SAMPRITA DAS ROY
In the following, we use the notations R ( S ) and N ( S ) for the range and null space, respectively, ofthe operator S . Lemma 3.5.
Let H and H be Hilbert spaces and let S : H → H be a bounded linear operator withclosed range. Then, (3.7) R ( S ∗ S ) = R ( S ∗ ) Suppose, in addition, that there exist c > such that k Sx k ≥ c k x k for all x ∈ H . Then (3.8) k S ∗ Sx k ≥ c k x k ∀ x ∈ H , Further, if k · k is any norm on H and if c > is such that k Sx k ≥ c k x k for all x ∈ H , then (3.9) k S † y k ≤ c k y k ∀ y ∈ H , where S † := ( S ∗ S ) − S ∗ , the generalized inverse of S .Proof. Clearly, R ( S ∗ S ) ⊆ R ( S ∗ ). Now, let x ∈ R ( S ∗ ), and let y ∈ H be such that x = S ∗ y . Let y ∈ N ( S ∗ ) and y ∈ N ( S ∗ ) ⊥ be such that y = y + y . Hence, x = S ∗ y . Since R ( S ) is closed, N ( S ∗ ) ⊥ = R ( S ). Hence, there exists x ∈ H such that y = Sx . So, x = S ∗ Sx ∈ R ( S ∗ S ). Thus, R ( S ∗ ) ⊆ R ( S ∗ S ). Thus, we have proved (3.7).Next, suppose that there exist c > k Sx k ≥ c k x k for all x ∈ H . Then for every x ∈ H , k S ∗ Sx k k x k ≥ h S ∗ Sx, x i H = k Sx k ≥ c k x k . Thus, we obtain (3.8).By (3.8), R ( S ∗ S ) is closed and S ∗ S has a bounded inverse from its range and hence, by (3.7),( S ∗ S ) − S ∗ is well defined as a bounded operator from H to H . Since R ( S ) is closed, it is knownthat for every y ∈ H , there exists x ∈ H such that(3.10) ( S ∗ S ) x = S ∗ y and Sx = P y, where P : H → H is the orthogonal projection onto R ( S ) = R ( S ), and this x is unique since S and S ∗ S are bounded below (see, e.g. [8]). Now, assume that k · k is any norm on H such that k Sx k ≥ c k x k for all x ∈ H for some c >
0. For y ∈ H , if x is as in (3.10), then k ( S ∗ S ) − S ∗ y k = k x k ≤ c k Sx k = 1 c k P y k ≤ c k y k . Thus, we obtain (3.9). (cid:3)
Corollary 3.6.
Let < α < and T α be as in (3.6). Then for every y ∈ L ( I ) , k (( T α ) ∗ T α ) − ( T α ) ∗ y k H ( I ) ≤ α k y k L ( I ) , (3.11) k (( T α ) ∗ T α ) − ( T α ) ∗ y k H ( I ) ≤ √ α k y k L ( I ) , (3.12) Proof.
Taking H = W and H = L ( I ) in Lemma 3.5, the inequalities in (3.11) and (3.12) followfrom (3.9) by taking the norm k · k as k · k H ( I ) and k · k H ( I ) respectively, on W and by using (3.4)and (3.5), respectively. (cid:3) Let R α : L ( I ) → W for α > R α := (( T α ) ∗ ( T α )) − ( T α ) ∗ , α > . We note that, by Corollary 3.6, R α is a bounded operator from L ( I ) to W (with respect to the norm k · k H ( I ) ), for each α >
0. Since ( T − T α )( w ) = αw ′′ , we have(3.14) R α T w − w = αR α ( w ′′ ) . Next, we prove that { R α } α> , defined as in (3.13), is a regularization family for T : W → L ( I ).Towards this aim, we first prove the following theorem. ONLINEAR INVERSE PROBLEM 9
Theorem 3.7.
For α > , let R α be as in (3.13), and let C L be as in (3.3). Then the followingresults hold. (i) k R α T w k H ( I ) ≤ k w k H ( I ) for all w ∈ W . (ii) k R α T w − w k H ( I ) ≤ (1 + C L ) α k w ′′ k H ( I ) for all w in W ∩ H ( I ) . (iii) k R α T w − w k H ( I ) ≤ √ α k w ′′ k L ( I ) for all w in W .Proof. (i) Let w ∈ W . By (3.14), we have k R α T w k H ( I ) = k w − [ w − R α T ( w )] k H ( I ) = k w + αR α ( w ′′ ) k H ( I ) . Hence, using (3.11), k R α T w k H ( I ) ≤ k w k H ( I ) + α k R α ( w ′′ ) k H ( I ) ≤ k w k H ( I ) + k w ′′ k L ( I ) . Thus, k R α T w k H ( I ) ≤ k w k H ( I ) for every w ∈ W .(ii) Let w ∈ W ∩ H ( I ). Let us note that w ′′ is in the domain of T and hence is in H ( I ) (maynot be in W ). By Proposition 3.3, w ′′ − Lw ′′ ∈ W and Lw ′′ ∈ N ( T α ). Thus, using the above fact,along with the fact that w ′′ is in the domain of T , by (3.14) and (i) above, we have k R α T w − w k H ( I ) = α k R α ( w ′′ ) k H ( I ) = α k R α T ( w ′′ ) k H ( I ) = α k R α [ T α ( w ′′ ) + αw ′′′′ ] k H ( I ) ≤ α k R α ( w ′′′′ ) k H ( I ) + α k R α T α ( w ′′ ) k H ( I ) = α k R α ( w ′′′′ ) k H ( I ) + α k R α T α ( Lw ′′ ) + R α T α [( id − L )( w ′′ )] k H ( I ) = α k R α ( w ′′′′ ) k H ( I ) + α k R α T α [( id − L )( w ′′ )] k H ( I ) = α k R α ( w ′′′′ ) k H ( I ) + α k ( id − L )( w ′′ ) k H ( I ) ≤ α [ k w ′′′′ k L ( I ) + k ( id − L )( w ′′ ) k H ( I ) ] . Now, since k w ′′′′ k L ( I ) ≤ k w ′′ k H ( I ) and k ( id − L )( w ′′ ) k H ( I ) ≤ C L k w ′′ k H ( I ) , we obtain the requiredinequality.(iii) For w ∈ W , using (3.12), we have k R α T w − w k H ( I ) = α k R α ( w ′′ ) k H ( I ) ≤ √ α k w ′′ k L ( I ) . Thus, the proof is complete. (cid:3)
Lemma 3.8.
The space
W ∩ H ( I ) is dense in W .Proof. Let w ∈ W . Since H ( I ) is dense in H ( I ) as a subspace of H ( I ) (cf. [6]), there exists asequence ( w n ) in H ( I ) such that(3.15) k w n − w k H ( I ) → n → . Now, define P : H ( I ) → W by P ( w )( t ) = w ( t ) − w ( g ) − w ′ ( g )( t − g ) , w ∈ H ( I ) and t ∈ I. Since H ( I ) is continuously imbedded in C ( I ) (cf. [6]), (3.15) implies that | w n ( g ) − w ( g ) | → | w ′ n ( g ) − w ′ ( g ) | → n → . Thus, as I is bounded we have(3.16) k P ( w n ) − P ( w ) k H ( I ) → n → . Again by definition of P and W we have P w n ∈ W ∩ H ( I ) and P w = w . Hence from (3.15) and(3.16) we have the proof. (cid:3) Theorem 3.9.
Let w ∈ W , and let { R α } α> be as in (3.13). Then k R α T w − w k H ( I ) → as α → . In particular, { R α } α> is a regularization family for T . Proof.
By Theorem 3.7, ( R α T ) is a uniformly bounded family of operators from W to W and k R α T w − w k H ( I ) → α → x ∈ W ∩ H ( I ). Since W ∩ H ( I ) is dense in W (see Lemma3.8), by a result in functional analysis (see Theorem 3.11 in [8]), we obtain k R α T w − w k H ( I ) → α → w ∈ W . Thus { R α } α> is a regularization family for T . (cid:3) Throughout, we assume that a ∈ H ( I ) is the unique solution of Problem (P). Thus, equations(2.11)-(2.13) have solutions namely, ζ , b and a , respectively. That is, T ( ζ ) = v j ◦ γ, (3.17) T ( b ) = ζ , (3.18) T ( a ) = b . (3.19)Having obtained the regularization family { R α } α> for T as in (3.13), we may replace the solution b of the equation (2 .
12) by b α := R α ζ . Thus, we may define the regularized solution a α for Problem (P) as the solution of (2.13) with b replaced by b α . Thus the regularized solution a α for Problem (P) is defined along the following lines: T ( ζ ) = v j ◦ γ, (3.20) ( T α ) ∗ T α ( b α ) = ( T α ) ∗ ζ , (3.21) T ( a α ) = b α . (3.22)Since b α ∈ W ⊂ R ( T ), each of the above equations has unique solution. In fact ζ = T b with b = T a , where a is the unique solution of (2.6). Note that, the operator equation (3.21) has aunique solution because T α is bounded below, and (3.22) has a unique solution as T is injective withrange W , and b α ∈ W . Hence we have, a α ( g ) = 0. Thus to obtain convergence of { a α } to a as α →
0, it is necessary that a ( g ) = 0. Therefore, in this section, we assume that,(3.23) a ( g ) = 0 . We shall relax this condition in Section 5, by appropriately redefining regularized solutions.3.1.
Error estimates under exact data.
For α >
0, let a α be defined via equations (3.20)-(3.22).Also, let a be the unique solution to Problem(P) satisfying (3.23). Then, we look at the estimatesfor the error term ( a − a α ) in both L ( I ) and H ( I ) norms in the following theorem. Theorem 3.10.
The following results hold. (1) k a − a α k H ( I ) → as α → . (2) k a − a α k L ( I ) ≤ √ α k a ′ k L ( I ) . (3) If a ∈ H ( I ) , then with C L is as in (3.3), k a − a α k H ( I ) ≤ (1 + C L ) α k a ′ k H ( I ) .Proof. By our assumption, a ( g ) = 0. Therefore, by definition of T and the space W , we have b = T ( a ) ∈ W . Now let us first observe that, by the definition of b α T ( a ) − T ( a α ) = b − b α = b − R α ζ = b − R α T b . Hence, by the inequality (2.17), for r ∈ { , } , we have,(3.24) k a − a α k H r ( I ) ≤ k T ( a ) − T ( a α ) k H r +1 ( I ) = k b − R α T b k H r +1 ( I ) , and hence, by Theorem 3.9, k a − a α k H ( I ) → α →
0. Thus we have proved (1).Also, since b ∈ W , from (3.24) and Theorem 3.7(iii), we have k a − a α k L ( I ) ≤ k T ( a ) − T ( a α ) k H ( I ) = k b − R α T b k H ( I ) ≤ √ α k b ′′ k L ( I ) = √ α k a ′ k L ( I ) . which proves (2). Now, let a ∈ H ( I ). Then b ∈ H ( I ). Since b ∈ W , we have b ∈ W ∩ H ( I ).Hence proof of (3) follows from (3.24) and Theorem 3.7 (ii). (cid:3) ONLINEAR INVERSE PROBLEM 11
Error estimates under noisy data.
In practical situations the observations of the data j and g may not be known accurately and we may have some noisy data instead. In this section we assumethat the noisy data g ε and j δ are such that(3.25) g ε ∈ C (Γ) , j δ ∈ W − /p,p ( ∂ Ω) , p > k g − g ε k W , ∞ (Γ) ≤ ε, (3.27) k j − j δ k L ( ∂ Ω) ≤ δ for some known noise level ε and δ , respectively. At this point let us note that a weaker conditionon perturbed data j δ , for example j δ ∈ L ( ∂ Ω), is not very feasible to work with, in this problem.This is because, in that case the corresponding solution v j δ of (2.3)-(2.4) with j δ in place of j , isnot continuous and hence its restriction on Γ does not make sense. In practical situations if such aperturbed data arise we may work with its appropriate approximation which is in W − /p,p ( ∂ Ω) with p >
3. For the perturbed data g ε , in the next section we consider the case when it is in a more generalspace which is L (Γ).Corresponding to the data j, j δ as above, we denote(3.28) f j := v j ◦ γ, f j δ := v j δ ◦ γ. Lemma 3.11.
Let γ be a C curve on R and let Γ = { ( x, γ ( x )) ∈ R : d ≤ x ≤ d } for some d , d in R with d < d . Then (3.29) k w k L (Γ ) ≤ k w k H ( R ) , ∀ w ∈ H ( R ) . Proof.
Let w ∈ C ∞ c ( R ). Then, using H¨older’s inequality we have k w k L (Γ ) = Z Γ ( w ( z )) dz = Z d d ( w ( x, γ ( x ))) dx = Z d d " − Z ∞ γ ( x ) ∂∂t ( w ( x, t )) dt ! dx = Z d d Z ∞ γ ( x ) ( − w ( x, t ) ∂∂t w ( x, t )) dt ! dx ≤ Z d d Z ∞ γ ( x ) | w ( x, t ) | dt + Z ∞ γ ( x ) | ∂∂t ( w ( x, t ) | dt ! dx ≤ k w k L ( R ) + k∇ w k L ( R ) ≤ k w k H ( R ) . Hence, C ∞ c ( R ) being dense in H ( R ), we have the proof. (cid:3) Lemma 3.12.
Let w ∈ H ( ∂ Ω) and γ be a curve on ∂ Ω such that | γ ′ ( t ) | is bounded away from asin (2.14). Then there exists C > such that k w ◦ γ k L ([0 , ≤ C k w k H ( ∂ Ω) . Proof.
Let w ∈ H ( ∂ Ω). Since Ω is with C boundary,(3.30) k w k H ( ∂ Ω) := m X i =1 k ω i k H ( R )2 M. THAMBAN NAIR AND SAMPRITA DAS ROY for some elements ω , ω , · · · , ω m ∈ H ( R ) (cf. [3], [6]). Also, there exists a set { σ , · · · , σ m } ofdiffeomorphisms from some neighbourhoods in ∂ Ω to R , which satisfies(3.31) k w ◦ γ k L ([0 , = m X i =1 k ω i ◦ σ i ◦ γ k L ([0 , . For any i ∈ { , · · · , m } , since σ i is a diffeomorphism σ i ◦ γ is a curve in R . Since | γ ′ | is boundedaway from 0, there exists constant C γ > | γ ′ ( t ) | ≥ C γ for all t ∈ [0 , σ ′ i (Γ) iscompact and σ i is one-one there exists constant C σ > | σ ′ i ( x ) | ≥ C σ for all x ∈ γ ([0 , ≤ i ≤ m . Hence, by Lemma 2.4 and (3.31), we obtain k w ◦ γ k L ([0 , = m X i =1 k ω i ◦ σ i ◦ γ k L ([0 , ≤ p C γ C σ m X i =1 k ω i k L ( σ i (Γ)) . Hence, using (3.29) and (3.30), we get k w ◦ γ k L ([0 , ≤ p C σ C γ m X i =1 k ω i k H ( R ) = 1 p C σ C γ k w k H ( ∂ Ω) . This completes the proof. (cid:3)
Proposition 3.13.
Let ˜ j ∈ W − /p,p ( ∂ Ω) . Let v ˜ j ∈ W ,p (Ω) be the solution of (2.3)-(2.4) with ˜ j inplace of j , such that it satisfies (2.1). Then there exists ˜ C γ > such that k v ˜ j ◦ γ k L ([0 , ≤ ˜ C γ k ˜ j k L ( ∂ Ω) . Proof.
Since ˜ j is in W − /p,p ( ∂ Ω), we know that v ˜ j ∈ W ,p (Ω) (cf. [3]) and(3.32) k v ˜ j k W ,p (Ω) ≤ C k ˜ j k L ( ∂ Ω) for some constant C > W (2 − /p ) ,p ( ∂ Ω) into W ,p ( ∂ Ω), we have v ˜ j | ∂ Ω ∈ W − /p,p ( ∂ Ω) ⊆ W ,p ( ∂ Ω)and(3.33) k v ˜ j | ∂ Ω k W ,p ( ∂ Ω) ≤ C k v ˜ j | ∂ Ω k W − /p,p ( ∂ Ω) ≤ C k v ˜ j k W ,p (Ω) for some constants C , C > p >
3, we have v ˜ j | ∂ Ω ∈ H ( ∂ Ω) and, there exists constant C > k v ˜ j | ∂ Ω k H ( ∂ Ω) ≤ C k v ˜ j | ∂ Ω k W ,p ( ∂ Ω) . Thus, using (3.32), (3.33) and with v ˜ j | ∂ Ω in place of w in Lemma 3.12 we have, k v ˜ j ◦ γ k L ([0 , ≤ p C σ C γ k v ˜ j | ∂ Ω k H ( ∂ Ω) ≤ C p C σ C γ k v ˜ j | ∂ Ω k W ,p ( ∂ Ω) ≤ ˜ C γ k ˜ j k L ( ∂ Ω) , where ˜ C γ = C C C / p C σ C γ . (cid:3) Corollary 3.14.
Let j be as in Assumption 2.3 and j δ satisfy (3.25) and (3.27). Let f and f j δ beas in (3.28). Then (3.34) k f j − f j δ k L ([0 , ≤ ˜ C γ δ, where ˜ C γ > is as in Proposition 3.13.Proof. By Proposition 3.13 we have k f j − f j δ k L ([0 , ≤ ˜ C γ k j − j δ k L ( ∂ Ω) ≤ ˜ C γ δ. Hence, k f j − f j δ k L ([0 , ≤ ˜ C γ δ. (cid:3) ONLINEAR INVERSE PROBLEM 13
Lemma 3.15.
For ε > , (3.35) C g − ε ≤ | g ε ′ ( γ ( s )) | ≤ C ′ g + ε, where C g and C ′ g are as in (2.15). In particular, if < ε ≤ C g / then (3.36) C g ≤ | g ε ′ ( γ ( s )) | ≤ C ′ g ∀ s ∈ [0 , . Proof.
For any s in [0 , | g ′ ( γ ( s )) | − | g ′ ( γ ( s )) − g ε ′ ( γ ( s )) | ≤ | g ε ′ ( γ ( s )) | ≤ | g ε ′ ( γ ( s )) − g ′ ( γ ( s )) | + | g ′ ( γ ( s )) | . Since | g ′ ( γ ( s )) − g ε ′ ( γ ( s )) | ≤ k g − g ε k W , ∞ (Γ) < ε, by (2.15), we obtain (3.35). The relations in (3.36)are obvious by the assumption on ε . (cid:3) Remark 3.16.
Since, γ ′ satisfies (2.14), and, ( g ε ) ′ satisfies (3.36) for ε < C g / g ε (Γ) is a non-degenerate closed interval, that is, I ε := g ε (Γ) = [ g ε , g ε ] for some g ε , g ε with g ε < g ε . ♦ The following lemma will help us in showing that I ∩ I ε is a closed and bounded (non-degenerate)interval. Lemma 3.17.
Let φ , φ be in C ([ ξ , ξ ]) for some ξ and ξ in R , and let η > be such that (3.37) k φ − φ k L ∞ ([ ξ ,ξ ]) ≤ η. Let I := φ ([ ξ , ξ ]) = [ a , b ] and I := φ ([ ξ , ξ ]) = [ a , b ] for some a , b , a and b in R . Weassume that I and I are non-degenerate intervals, that is, a < b and a < b , and (3.38) 2 η < min { ( b − a ) , ( b − a ) } . Then (3.39) max {| a − a | , | b − b |} ≤ η and I ∩ I = [ a, b ] is a non-degenerate interval, that is, a < b .Proof. Suppose a < b and a < b . Since a = φ ( s ) , a = φ ( s ) , b = φ ( s ′ ) , b = φ ( s ′ ), forsome s , s , s ′ , s ′ ∈ [ ξ , ξ ], and since a ≤ φ ( s ) , a ≤ φ ( s ) , b ≥ φ ( s ′ ) and b ≥ φ ( s ′ ), we obtain(3.40) | a − a | ≤ k φ − φ k L ∞ ([ ξ ,ξ ]) ≤ η, (3.41) | b − b | ≤ k φ − φ k L ∞ ([ ξ ,ξ ]) ≤ η. Thus, (3.39) is proved.To prove the remaining, let us first consider the case a ≤ a . Then, I ∩ I = [ a , ˜ b ], where˜ b := min { b , b } . Note that, by (3.38) and (3.40), we have b − a = ( b − a ) − ( a − a ) ≥ η − η = η. Thus, b > a , and also, as b > a we have, I ∩ I = [ a , ˜ b ] with ˜ b > a . Next, let a > a . In this case, I ∩ I = [ a , ˜ b ], where ˜ b := min { b , b } . Note, again by (3.38) and(3.40), that b − a = ( b − a ) − ( a − a ) ≥ η − η = η. Thus, b > a , and also, as b > a we have, I ∩ I = [ a , ˜ b ] with ˜ b > a . Hence, combining both the cases, we have the proof. (cid:3)
Remark 3.18.
Let s and s in [0 ,
1] be such that g = g ( γ ( s )) and g = g ( γ ( s )). Let us recall that I := [ g , g ] and I ε := [ g ε , g ε ]. Since g and g ε are in C (Γ), we have g ◦ γ and g ε ◦ γ are in C ([0 , k g ◦ γ − g ε ◦ γ k L ∞ ([0 , ≤ k g − g ε k W , ∞ (Γ) ≤ ε. Thus, by Lemma 3.17, we have | g − g ε | < ε and | g − g ε | < ε. Hence, taking ε < ( g − g ) /
4, we have( g ε − g ε ) ≥ | g ε ( γ ( s )) − g ε ( γ ( s )) |≥ | g − g | − | g ( γ ( s )) − g ε ( γ ( s )) | − | g ( γ ( s )) − g ε ( γ ( s )) | > ε − k g − g ε k W , ∞ (Γ) > ε − ε = 2 ε, and thus, 2 ε < min { ( g − g ) , ( g ε − g ε ) } . Hence by Lemma 3.17, I ∩ I ε is a closed and boundednon-degenerate interval. Let us denote this interval by ˜ I ε . Thus,(3.42) ˜ I ε = I ∩ I ε = [˜ g ε , ˜ g ε ]for some ˜ g ε , ˜ g ε ∈ R with ˜ g ε < ˜ g ε . Also, by Lemma 3.17 we have, | g − ˜ g ε | ≤ | g − g ε | < ε and | g − ˜ g ε | ≤ | g − g ε | < ε. ♦ Next, we shall make use of the following lemma whose proof is given in the appendix.
Lemma 3.19.
There exists a constant
C > such that for any closed interval J , k y k L ∞ ( J ) ≤ C J k y k H ( J ) , where C J := C max { , (2 | J | + 1) } . In particular, for any interval J such that J ⊆ J , (3.43) k y k L ∞ ( J ) ≤ C J k y k H ( J ) . If y ∈ W , ∞ ( J ) then using (3.43) we obtain k y k L ∞ ( J ) ≤ ( C J ) (cid:20)Z J y + Z J ( y ′ ) (cid:21) ≤ ( C J ) | J | h k y k L ∞ ( J ) + k y ′ k L ∞ ( J ) i . Thus(3.44) k y k L ( J ) ≤ p | J |k y k L ∞ ([ a,c ]) ≤ | J |√ C J k y k W , ∞ ( J ) , and additionally if y ′′ ∈ L ∞ ( J ), then k y k L ( J ) ≤ | J | ( C J ) ( k y k L ∞ ( J ) + k y ′ k L ∞ ( J ) ) ≤ | J | ( C J ) h k y k L ∞ ( J ) + k y ′ k L ∞ ( J ) + k y ′ k L ∞ ( J ) + k y ′′ k L ∞ ( J ) i ≤ | J | ( C J ) k y k W , ∞ ( J ) which implies(3.45) k y k L ( J ) ≤ | J | / ( C J ) k y k W , ∞ ( J ) . Lemma 3.20.
Let J and J be closed intervals such that J ⊆ J and let C J be as in Lemma 3.19.Let y ∈ H ( J ) , then we have the following. (i) k y k L ( J \ J ) ≤ √ C J k y k W , ∞ ( J ) | J \ J | . (ii) If y ′′ ∈ L ∞ ( J ) then k y k L ( J \ J ) ≤ C J ) k y k W , ∞ ( J ) | J \ J | / . ONLINEAR INVERSE PROBLEM 15
Proof.
Let J = [ a, b ] and J = [ c, d ] for some a ≤ b and c ≤ d . If J = J then J \ J = ∅ , and inthat case the result holds trivially. So let us consider the cases when either a < c or d < b , or bothholds. Without loss of generality let us assume that a < c and d < b . Let y ∈ H ( J ). Then by(3.43) y and y ′ are in L ∞ ( J ). Thus taking J = [ a, c ] in (3.44) we have k y k L ([ a,c ]) ≤ ( c − a ) √ C J k y k W , ∞ ([ a,c ]) ≤ ( c − a ) √ C J k y k W , ∞ ( J ) and taking J = [ d, b ] in (3.44) we have k y k L ([ d,b ]) ≤ ( b − d ) √ C J k y k W , ∞ ([ d,b ]) ≤ ( b − d ) √ C J k y k W , ∞ ( J ) . Hence we have (i). Next, additionally if, y ′′ ∈ L ∞ ( J ), having J = [ a, c ] in (3.45) we obtain k y k L ([ a,c ]) ≤ c − a ) / ( C J ) k y k W , ∞ ([ a,c ]) ≤ c − a ) / ( C J ) k y k W , ∞ ( J ) and having J = [ d, b ] in (3.45) we obtain k y k L ( d,b ) ≤ b − d ) / ( C J ) k y k W , ∞ ([ d,b ]) ≤ b − d ) / ( C J ) k y k W , ∞ ( J ) . Hence we have (ii). (cid:3)
Lemma 3.21.
Let φ , φ , I , I and η be as in Lemma 3.17 satisfying all the assumptions there.Then, for any interval I ⊂ I ∩ I and y ∈ C ( I )(3.46) Z I | y ( φ ( ξ )) − y ( φ ( ξ )) | dξ ≤ k y ′ k L ∞ ( I ) k φ − φ k L ([ ξ ,ξ ]) . Assume, further, that φ , φ ∈ C ([ ξ , ξ ]) satisfying | φ ′ ( ξ ) | ≥ C φ and | φ ′ ( ξ ) | ≥ C φ for someconstants C φ , C φ > . Then, for y ∈ H ( I )(3.47) k y ◦ φ − ^ y ◦ φ k L ([ ξ ,ξ ]) ≤ C I k y k H ( I ) (cid:16) k φ − φ k L ([ ξ ,ξ ]) + 2 √ p C φ η (cid:17) , where ^ y ◦ φ ( ξ ) := (cid:26) ( y ◦ φ )( ξ ) if ξ ∈ [ ˜ ξ , ˜ ξ ] , if ξ ∈ [ ξ , ξ ] \ [ ˜ ξ , ˜ ξ ] , with [ ˜ ξ , ˜ ξ ] = ( φ ) − ( I ∩ I ) and C I is as in Lemma 3.19.Proof. By Lemma 3.17 we have I ∩ I to be a closed non-degenerate interval. Let I be an interval in I ∩ I . Then for y ∈ C ( I ) using fundamental theorem of calculus and H¨older’s inequality we have Z I | y ( φ ( ξ )) − y ( φ ( ξ )) | dξ = Z I "Z φ ( ξ )) φ ( ξ )) y ′ ( θ ) dθ dξ ≤ Z I k y ′ k L ∞ ( I ) | φ ( ξ )) − φ ( ξ )) | dξ ≤ k y ′ k L ∞ ( I ) k φ − φ k L ([ ξ ,ξ ]) . Hence we have (3.46). Next, let I ∩ I be equal to [ ˜ a , ˜ b ] for some ˜ a and ˜ b in R , with ˜ a < ˜ b . Since φ ∈ C ([ ξ , ξ ]) and | φ ′ ( ξ ) | ≥ C φ > φ is invertible from its image and the inverse is continuous.Thus ( φ ) − ( I ∩ I ) = [ ˜ ξ , ˜ ξ ] for some [ ˜ ξ , ˜ ξ ] ⊆ [ ξ , ξ ]. Also, by the properties of φ , we have, φ ([ ˜ ξ , ˜ ξ ]) = [ ˜ a , ˜ b ] for some ˜ a and ˜ b in I , with ˜ a < ˜ b . Thus using Lemma 3.17 with [ ˜ a , ˜ b ] and[ ˜ a , ˜ b ] in place of I and I respectively, we have | ˜ a − ˜ a | ≤ η and | ˜ b − ˜ b | ≤ η . Hence, using Lemma3.17 and definition of ˜ a and ˜ b we have,(3.48) | a − ˜ a | ≤ | a − ˜ a | + | ˜ a − ˜ a | ≤ | a − a | + | ˜ a − ˜ a | ≤ η, (3.49) | b − ˜ b | ≤ | b − ˜ b | + | ˜ b − ˜ b | ≤ | b − b | + | ˜ b − ˜ b | ≤ η. Thus by definition of g y ◦ φ , we have(3.50) k y ◦ φ − g y ◦ φ k L ([ ξ ,ξ ]) = Z ˜ ξ ˜ ξ | y ( φ ( ξ )) − y ( φ ( ξ )) | dξ + Z [ ξ ,ξ ] \ [ ˜ ξ , ˜ ξ ] | y ( φ ( ξ )) | dξ. For any ξ ∈ [ ξ , ξ ], | φ ′ ( ξ ) | ≥ C φ hold. Thus, by Lemma 2.4, Z [ ξ ,ξ ] \ [ ˜ ξ , ˜ ξ ] | y ( φ ( ξ )) | dξ ≤ C φ Z I \ φ ([ ˜ ξ , ˜ ξ ]) | y ( z ) | dz. Hence, as (3.48) and (3.49) hold and (3.38) is assumed, taking J = I and J = φ ([ ξ , ξ ]) = [ ˜ a , ˜ b ]in Lemma 3.20-(i) we obtain Z [ ξ ,ξ ] \ [ ˜ ξ , ˜ ξ ] | y ( φ ( ξ )) | dξ ≤ C φ k y k L ( I \ φ ([ ˜ ξ , ˜ ξ ])) ≤ C I ) C φ k y k H ( I ) η . Thus using (3.50), the fact that H ( I ) is continuously imbedded in C ( I ) and having I = [ ˜ ξ , ˜ ξ ]in (3.46) we obtain k y ◦ φ − ^ y ◦ φ k L ([ ξ ,ξ ]) ≤ ( k y ′ k L ∞ ( J ) k φ − φ k L ([ ξ ,ξ ]) + k y k H ( J ) √ C I p C φ η ) . Hence, using (3.43) we have (3.47). (cid:3)
Let us recall that I = g ( γ ([0 , g , g ], I ε = g ε ( γ ([0 , g ε , g ε ] and for ε < ( g − g ) /
4, let˜ I ε = I ∩ I ε = [˜ g ε , ˜ g ε ] as in (3.42). By (3.26) we have k g − g ε k W , ∞ (Γ) ≤ ε and thus(3.51) k g ◦ γ − g ε ◦ γ k L ∞ ([0 , = sup s ∈ [0 , | g ◦ γ ( s ) − g ε ◦ γ ( s ) | ≤ k g − g ε k W , ∞ (Γ) ≤ ε. Now, additionally let ε ≤ C g /
2. Then, by (3.36) and (2.14) g ε and γ are bijective, and so ( g ε ◦ γ ) − is continuous. Thus ( g ε ◦ γ ) − ( ˜ I ε ) is a closed non-degenerate interval. In other words(3.52) ˜ I ε = [ ˜ g ε , ˜ g ε ] = g ε ( γ ([ t ε , t ε ])for some t ε and t ε in [0 ,
1] with t ε < t ε .Now, for ε ≤ min { ( g − g ) / , C g / } , let T ε : L ( ˜ I ε ) → L ([0 , T ǫ ( ζ )( s ) = (cid:26) ζ ( g ε ( γ ( s ))) s ∈ [ t ε , t ε ]0 g ε ( γ ( s )) ∈ [0 , \ [ t ε , t ε ] . Now, we prove some properties of T ε . Theorem 3.22.
Let T ε be as defined in (3.53). Then, for ζ ∈ W , k T ζ − T ε ζ | ˜ Iε k L ([0 , ≤ ( C g,γ,I k ζ k H ( I ) ) ε, where C g,γ,I = C I (cid:16) √ √ C g C γ (cid:17) with C I as in (3.43).Proof. Let ζ ∈ W . For any s ∈ [0 , | ( g ◦ γ ) ′ ( s ) | ≥ C g C γ . By (3.51) and (3.52), we have k g ◦ γ − g ε ◦ γ k ≤ ε and ˜ I ε = I ∩ I ε = ( g ε ◦ γ )([ t ε , t ε ]), respectively. Now ζ ∈ W ⊂ H ( I ). Then, by definition of T and T ε , we have T ( ζ ) = ζ ◦ g ◦ γ ∈ L ([0 , ζ ◦ g ε ◦ γ ) | [ tε ,tε = ( T ε ( ζ | ˜ Iε )) | [ tε ,tε ∈ L ([ t ε , t ε ]) . Hence, taking φ as g ◦ γ and φ as g ε ◦ γ in Lemma 3.21, we have k T ζ − T ε ζ ε k L ([0 , = k ζ ◦ g ◦ γ − ( ζ ◦ g ε ◦ γ ) | [ tε ,tε k L ([0 , ≤ C I (cid:16) √ p C g C γ (cid:17) k ζ k H ( I ) ε. This completes the proof. (cid:3)
ONLINEAR INVERSE PROBLEM 17
Theorem 3.23.
The map T ε : L ( ˜ I ε ) → L ([0 , , defined as in (3.53), is bounded linear and boundedbelow. In fact, for every ζ ∈ L ( ˜ I ε ) , (3.54) r C g C γ k T ε ( ζ ) k L ([0 , ≤ k ζ k L ( ˜ I ε ) ≤ q C ′ g C ′ γ k T ε ( ζ ) k L ([0 , , where C γ , C ′ γ and C g , C ′ g are as in (2.14) and (2.15), respectively.Proof. Clearly, T ε is a linear map. Since (3.36) and (2.14) hold, using Lemma 2.4, and (3.53) weobtain k T ε ( ζ ) k L ([0 , = Z | T ε ( ζ )( s ) | ds = Z t ε t ε | ζ ( g ε ( γ ( s ))) | ds ≤ C g C γ Z ˜ g ε ˜ g ε | ζ ( z ) | dz = 2 C g C γ k ζ k L ([ ˜ g ε , ˜ g ε ]) , k T ε ( ζ ) k L ([0 , = Z | T ε ( ζ )( s ) | ds = Z t ε t ε | ζ ( g ε ( γ ( s ))) | ds ≥ C ′ g C ′ γ Z ˜ g ε ˜ g ε | ζ ( z ) | dz = 12 C ′ g C ′ γ k ζ k L ([ ˜ g ε , ˜ g ε ]) . Hence we have the proof. (cid:3)
Now, by Theorem 3.23 we know that T ε is a bounded linear map which is bounded below. Thususing Lemma 3.5, the operator ( T ε ) † := (( T ε ) ∗ T ε ) − ( T ε ) ∗ is a bounded linear operator and is the generalized inverse of T ε . The following theorem, which alsofollows from Lemma 3.5, shows that the family (cid:26) ( T ε ) † : 0 < ε ≤ min { C g , g − g } (cid:27) is in fact uniformly bounded. Theorem 3.24.
For every ζ ∈ L ([0 , , (3.55) k ( T ε ) † ζ k L ( ˜ I ε ) ≤ q C ′ g C ′ γ k ζ k L ([0 , , where C ′ g and C ′ γ are as in (2.14) and (2.15). In order to obtain an approximate solution of (2.6) under the nosy data ( j δ , g ε ) satisfying (3.26) and(3.27), we adopt the following operator procedure: First we consider the following operator equation(3.56) ( T ε ) ∗ ( T ε ) ζ = ( T ε ) ∗ f j δ . Let ˜ ζ ε,δ ∈ L ( ˜ I ε ) be the unique solution of (3.56), that is, ˜ ζ ε,δ := ( T ε ) † f j δ . Then, we see that ζ ε,δ = (cid:26) ˜ ζ ε,δ on ˜ I ε , I \ ˜ I ε , belongs to L ( I ). Next, we consider the operator equation(3.57) ( T α ) ∗ ( T α )( w ) = ( T α ) ∗ ζ ε,δ . Let b α,ε,δ be the unique solution of equation (3.57). Thus by solving the operator equations (3.56)and (3.57) we obtain b α,ε,δ . Since b α,ε,δ ∈ W ⊂ R ( T ), a α,ε,δ := b ′ α,ε,δ is the solution of the equation T ( a ) = b α,ε,δ . We show that a α,ε,δ is a candidate for an approximate solution to Problem (P). Lemma 3.25.
Under the assumptions in Assumption 2.3 on ( j, g ) , let a ∈ H ( I ) be the solution of T ( a ) = f j . Assume further that a ( g ) = 0 . For ζ ∈ L ( I ) , let b α,ζ ∈ H ( I ) be such that ( T α ) ∗ ( T α )( b α,ζ ) = ( T α ) ∗ ζ, and let a α,ζ = b ′ α,ζ . Then k a − a α,ζ k H ( I ) ≤ C α + k ζ − b k L ( I ) α , (3.58) k a − a α,ζ k L ( I ) ≤ √ α k a ′ k L ( I ) + k ζ − b k L ( I ) √ α , (3.59) where C α > is such that C α → as α → . In addition, if a ∈ H ( I ) , then k a − a α,ζ k H ( I ) ≤ (1 + C L ) α k a ′ k H ( I ) + k ζ − b k L ( I ) α , (3.60) k a − a α,ζ k L ( I ) ≤ (1 + C L ) α k a ′ k H ( I ) + k ζ − b k L ( I ) √ α . (3.61) Here C L is as (3.3).Proof. Let b = T ( a ). Then, as a ( g ) = 0, we have b ∈ W . Now, by definition of a α,ζ and, H ( I )and H ( I ) norms, for r ∈ { , }k a − a α,ζ k H r ( I ) = k a − ((( T α ) ∗ T α ) − ( T α ) ∗ ζ ) ′ k H r ( I ) ≤ k b − (( T α ) ∗ T α ) − ( T α ) ∗ ζ k H r +1 ( I ) ≤ k b − (( T α ) ∗ T α ) − ( T α ) ∗ T ( b ) k H r +1 ( I ) + k (( T α ) ∗ T α ) − ( T α ) ∗ ( ζ − T ( b )) k H r +1 ( I ) Hence, for r ∈ { , } , k a − a α,ζ k H r ( I ) ≤ k b − (( T α ) ∗ T α ) − ( T α ) ∗ T ( b ) k H r +1 ( I ) (3.62) + k (( T α ) ∗ T α ) − ( T α ) ∗ ( ζ − T ( b )) k H r +1 ( I ) . By Theorem 3.9 we have(3.63) k b − (( T α ) ∗ T α ) − ( T α ) ∗ T ( b ) k H ( I ) → α → . Also, by Theorem 3.7-(iii) we have(3.64) k b − (( T α ) ∗ T α ) − ( T α ) ∗ T ( b ) k H ( I ) ≤ k b ′′ k L ( I ) √ α. Again, using (3.11) and (3.12), we have(3.65) k (( T α ) ∗ T α ) − ( T α ) ∗ ( ζ − T ( b )) k H ( I ) ≤ α k ζ − T ( b ) k L ( I ) and(3.66) k (( T α ) ∗ T α ) − ( T α ) ∗ ( ζ − T ( b )) k H ( I ) ≤ √ α k ζ − T ( b ) k L ( I ) . Thus combining (3.62), (3.63) and (3.65) we have (3.58) with C α := k b − (( T α ) ∗ T α ) − ( T α ) ∗ T ( b ) k H ( I ) , and combining (3.62), (3.64) and (3.66) we have (3.59).Next, let a ∈ H ( I ), b = T ( a ) ∈ W ∩ H ( I ). Then, using theorem 3.7-(ii) we have, for r ∈ { , } ,(3.67) k b − (( T α ) ∗ T α ) − ( T α ) ∗ T ( b ) k H r +1 ( I ) ≤ (1 + C L ) k b ′′ k H ( I ) α. Thus combining (3.62), (3.67) and (3.65) we have (3.60), and combining (3.62), (3.67) and (3.66) wehave (3.61). (cid:3)
ONLINEAR INVERSE PROBLEM 19
Now, we prove one of the main theorems of this paper.
Theorem 3.26.
Let ε < min { ( g − g ) / , C g / } . Let a , g and j be as in Lemma 3.25. Let g ε ∈ C (Γ) , j δ ∈ W − /p,p ( ∂ Ω) with p > , ζ ε,δ be the solution of (3.56), and a α,ε,δ = b ′ α,ε,δ where b α,ε,δ is the solution of (3.57). Also, let g ε and j δ satisfy (3.26) and (3.27), respectively. Then k a − a α,ε,δ k H ( I ) ≤ C α + 1 α [ q C ′ g C ′ γ ( C I,g,γ k b k H ( I ) ε + ˜ C γ δ ) + C I k b k H ( I ) ε ] , (3.68) k a − a α,ε,δ k L ( I ) ≤ √ α k a ′ k L ( I ) + 1 √ α [ q C ′ g C ′ γ ( C I,g,γ k b k H ( I ) ε + ˜ C γ δ ) + C I k b k H ( I ) ε ] , (3.69) where C α > is such that C α → as α → .In addition if a ∈ H ( I ) , then (3.70) k a − a α,ε,δ k H ( I ) ≤ (1+ C L ) k a ′ k H ( I ) α + 1 α [ q C ′ g C ′ γ ( C I,g,γ k b k H ( I ) ε + ˜ C γ δ )+ C I k b k H ( I ) ε ] , (3.71) k a − a α,ε,δ k L ( I ) ≤ (1 + C L ) k a ′ k H ( I ) α + 1 √ α [ q C ′ g C ′ γ ( C I,g,γ k b k H ( I ) ε + ˜ C γ δ ) + C I k b k H ( I ) ε ] . Here, b = T ( a ) , and C L , ˜ C γ , C I , C I,g,γ , C ′ g and C ′ γ are constants as in (3.3), Lemmas 3.13 and3.19, Theorem 3.22, (2.14) and (2.15) respectively.Proof. Since a ( g ) = 0, we have b ∈ W . Now let us note that, by Remark 3.18, we have | g − ˜ g ε | < ε and | g − ˜ g ε | < ε . Hence, taking J and J as I and ˜ I ε respectively in Lemma 3.20, and with ourchoice of ε , by Lemma 3.20-(i) we have,(3.72) k b k L ( I \ ˜ I ε ) ≤ C I k b k H ( I ) ε. Since ˜ ζ ε,δ = ( T ε ) † f j δ , T ( T ( b )) = f j , and ( T ε ) † T ε is identity, we have k ˜ ζ ε,δ − b | ˜ I ε k L ( ˜ I ε ) = k ˜ ζ ε,δ − ( T ( b )) | ˜ I ε k L ( ˜ I ε ) = k ( T ε ) † f j δ − ( T ( b )) | ˜ I ε k L (˜ I ε ) ≤ k ( T ε ) † T ( T ( b )) − ( T ( b )) | ˜ I ε k L (˜ I ε ) + k ( T ε ) † ( f j − f j δ ) k L (˜ I ε ) ≤ k ( T ε ) † ( T ( T ( b )) − T ε (( T ( b )) | T ( b ) )) k L (˜ I ε ) + k ( T ε ) † ( f j − f j δ ) k L (˜ I ε ) . Now, by (3.55) and Theorem 3.22, we obtain k ( T ε ) † ( f j − f j δ ) k L (˜ I ε ) ≤ q C ′ g C ′ γ ˜ C γ δ, k ( T ε ) † ( T ( T ( b )) − T ε (( T ( b )) | T ( b ) )) k L (˜ I ε ) ≤ q C ′ g C ′ γ C I,g,γ k b k H ( I ) ε. Therefore,(3.73) k ˜ ζ ε,δ − b | ˜ I ε k L ( ˜ I ε ) ≤ q C ′ g C ′ γ [( C I,g,γ ) k b k H ( I ) ε + ˜ C γ δ ] . Now by definition of ζ ε,δ we have k ζ ε,δ − b k L ( I ) ≤ k ˜ ζ ε,δ − b | ˜ I ε k L ( ˜ I ε ) + k b k L ( I \ ˜ I ε ) . Hence, by (3.72) and (3.73) we have, k ζ ε,δ − b k L ( I ) ≤ q C ′ g C ′ γ [( C I,g,γ ) k b k H ( I ) ε + ˜ C γ δ ] + C I k b k H ( I ) ε. Now by definition, b α,ε,δ is the unique solution of equation (3.57). Thus, with ζ ε,δ in place of ζ inLemma 3.25, we have the proof. (cid:3) Remark 3.27.
Let a and a α,ε,δ be as defined in Theorem 3.26. Then (3.68) and (3.69) take theforms k a − a α,ε,δ k H ( I ) ≤ C α + K ε + δα , k a − a α,ε,δ k L ( I ) ≤ √ α k a ′ k L ( I ) + K ε + δ √ α , respectively, where C α > C α → α →
0, and if, in addition, a ∈ H ( I ), then (3.70)and (3.71) take the forms k a − a α,ε,δ k H ( I ) ≤ (1 + C L ) k a ′ k H ( I ) α + K ε + δα , k a − a α,ε,δ k L ( I ) ≤ (1 + C L ) k a ′ k H ( I ) α + K ε + δ √ α , respectively, where K , K , K , K are positive constants independent of α, ε, δ and C L ≥ k id − L k ,where L is the bounded operator as in Proposition 3.3. .Then, choosing α = √ δ and ε = δ in (3.68) we have k a − a α,ε,δ k H ( I ) = o (1) . Thus using the new regularization method we obtain a result better than the order O (1) in [4] obtainedusing Tikhonov regularization. On choosing α = δ = ε in (3.69) we have k a − a α,ε,δ k L ( I ) = O ( √ δ ) , which is same as the estimate obtained in [4]. Next, under the source condition a ∈ H ( I ) and for α = √ δ and ε = δ , (3.70) gives the order as k a − a α,ε,δ k H ( I ) = O ( √ δ ) . This estimate is similar to a result obtained in [7] with source condition a ∈ H ( I ) and trace of a being Lipschitz which is stronger than the source condition needed in our result, whereas under thesame source condition a ∈ H ( I ), the choice of α = δ / and ε = δ in (3.71) gives the rate as k a − a α,ε,δ k L ( I ) = O ( δ / ) . This is better than the rate O ( δ / ) mentioned in [4] as the best possible estimate under L ( I ) norm(under realistic boundary condition) using Tikhonov regularization. ♦ Relaxation of assumption on perturbed data
In the previous section we have carried out our analysis assuming that the perturbed data g ε isin C (Γ), along with (3.26). This assumption can turn out to be too strong for implementation inpractical problems. Hence, here we consider a weaker and practically relevant assumption on ourperturbed data g ε , namely g ε ∈ L (Γ) with(4.1) k g − g ε k L (Γ) ≤ ε. What we essentially used in our analysis in Section 3 to derive the error estimates is that g ε ◦ γ is close to g ◦ γ in appropriate norms. Here, we consider ˜ g εγ := Π h ( g ε ◦ γ ) in place of g ε ◦ γ , whereΠ h : L ([0 , → L ([0 , W , ∞ ([0 , g εγ is close to g ◦ γ in appropriate norms, and then obtain associated error estimates. For thispurpose, we shall also assume more regularity on g ◦ γ , namely, g ◦ γ ∈ H ([0 , h : L ([0 , → L ([0 , L h which is the spaceof all continuous real valued piecewise linear functions w on [0 ,
1] defined on a uniform partition0 = t < t < · · · t N = 1 of mesh size h , that is, t i := ( i − h for i = 1 , · · · N and h = 1 /N .Thus, w ∈ L h if and only if w ∈ C [0 ,
1] such that w | [ ti − ,ti ] is a polynomial of degree at most 1. Let T h := { [ t i − , t i ] : i = 1 , · · · ( h + 1) } . ONLINEAR INVERSE PROBLEM 21
In the following, for w ∈ L ([0 , k w k H m ( τ h ) and k w k W m, ∞ ( τ h ) whenever w | τh belong to H m ( τ h ) and W m, ∞ ( τ h ), respectively. As a particular case of inverse inequality stated inLemma 4.5.3 in [1], for m ∈ { , } , we have(4.2) k Π h w k W m, ∞ ( τ h ) ≤ C ′ m h (1 / m ) k Π h w k L ( τ h ) , where C ′ m is a positive constant. Proposition 4.1.
Let w ∈ L ([0 , , m ∈ N ∪ { } and τ h ∈ T h . Then the following inequalities hold. k w k H m ( τ h ) ≤ h / C k w k H m +1 ( τ h ) whenever w | τh ∈ H m +1 ( τ h ) , (4.3) k w k W m, ∞ ( τ h ) ≤ C h / k w k H m +2 ( τ h ) whenever w | τh ∈ W m, ∞ ( τ h ) , (4.4) k Π h w k W m, ∞ ( τ h ) ≤ C ′ m C (2+2 m )0 h / k w k H (2 m +2) ( τ h ) whenever w | τh ∈ H m +2 ( τ h ) , (4.5) where C := 2 C [0 , with C [0 , as in (3 . and C ′ m is as in (4.2).Proof. If w ( j ) | τh ∈ H ( τ h ) for some j ∈ N ∪ { } , then using (3.43) and the fact that τ h is of length h ,we obtain k w ( j ) k L ( τ h ) ≤ h / k w ( j ) k L ∞ ( τ h ) ≤ h / C I k w ( j ) k H ( τ h ) , where I := [0 , k w k H m ( τ h ) = m X j =0 k w ( j ) k L ( τ h ) ≤ m X j =0 h / C I k w ( j ) k H ( τ h ) ≤ C I h / k w k H ( m +1) ( τ h ) . Thus, taking C = 2 C I , we have (4.3).By repeatedly using (3.43) and then by (4.3), we obtain k w k W m, ∞ ( τ h ) ≤ C I k w k H m +1 ( τ h ) ≤ C I C h / k w k H m +2 ( τ h ) . As we have taken C = 2 C I , we have the proof of (4.4).Since Π h is an orthogonal projection, from (4.2) we obtain, k Π h w k W m, ∞ ( τ h ) ≤ C ′ m h (1 / m ) k Π h w k L ( τ h ) ≤ C ′ m h (1 / m ) k w k L ( τ h ) , and, by repeatedly using (4.3) we have C ′ m h (1 / m ) k w k L ( τ h ) ≤ C (2+2 m )0 C ′ m h (1 / m ) h ((2 m +2) / k w k H (2 m +2) ( τ h ) ≤ C (2+2 m )0 C ′ m h / k w k H (2 m +2) ( τ h ) . Hence we have the proof of (4.5). (cid:3)
For simplifying the notation, we shall denote g γ := g ◦ γ, g εγ = g ε ◦ γ. By definition, Π h ( g εγ ) ∈ W , ∞ ([0 , h ( g εγ ) is close to g γ with respect toappropriate norms, we assume that(4.6) g γ ∈ H ([0 , . Theorem 4.2.
Let τ h ∈ T h and (4.6) be satisfied. Then, the following inequalities hold. (i) k Π h g εγ − g γ k L ∞ ( τ h ) ≤ ˜ C h / k g γ k H ( τ h ) + C ′ C γ εh / , (ii) k Π h g εγ − g γ k W , ∞ ( τ h ) ≤ ˜ C h / k g γ k H ( τ h ) + C ′ C γ εh / . (iii) | (Π h g εγ ) ′ ( s ) | ≤ C ′ g C ′ γ + ˜ C h / k g γ k H ( τ h ) + C ′ C γ εh / , ∀ s ∈ τ h , (iv) | (Π h g εγ ) ′ ( s ) | ≥ C g C γ − ˜ C h / k g γ k H ( τ h ) − C ′ C γ εh / ∀ s ∈ τ h . Proof.
Using triangle inequality we have k Π h g εγ − g γ k L ∞ ( τ h ) ≤ k Π h g εγ − Π h g γ k L ∞ ( τ h ) + k Π h g γ − g γ k L ∞ ( τ h ) , (4.7) k Π h g εγ − g γ k W , ∞ ( τ h ) ≤ k Π h g εγ − Π h g γ k W , ∞ ( τ h ) + k Π h g γ − g γ k W , ∞ ( τ h ) . (4.8)Assumption (2.14), Lemma 2.4 and (4.1) imply(4.9) C γ k g εγ − g γ k L ( τ h ) ≤ k g ε − g k L (Γ) ≤ ε so that, using (4.2) and the fact that Π h is an orthogonal projection, we have(4.10) k Π h g εγ − Π h g γ k L ∞ ( τ h ) ≤ C ′ h / k g εγ − g γ k L ( τ h ) ≤ C ′ C γ εh / , and(4.11) k Π h g εγ − Π h g γ k W , ∞ ( τ h ) ≤ C ′ h / k g εγ − g γ k L ( τ h ) ≤ C ′ C γ εh / , By (4.4), (4.5) and (4.3) k Π h g γ − g γ k L ∞ ( τ h ) ≤ k Π h g γ k L ∞ ( τ h ) + k g γ k L ∞ ( τ h ) ≤ C ) h / k g ◦ γ k H ( τ h ) ≤ C ) h / k g ◦ γ k H ( τ h ) . Hence, using (4.7) and (4.10), and taking ˜ C = 2( C ) , we have (i). By (4.4) and (4.5), k Π h g γ − g γ k W , ∞ ( τ h ) ≤ k Π h g γ k W , ∞ ( τ h ) + k g γ k W ∞ ( τ h ) ≤ ( C ) h / k g ◦ γ k H ( τ h ) +( C ) h / k g ◦ γ k H ( τ h ) . Hence, using (4.11) and (4.8), and taking ˜ C = ( C ) + ( C ) we have (ii).To prove (iii) and (iv), let s ∈ [0 , | ( g γ ) ′ ( s ) | − k Π h g εγ − g γ k W , ∞ ( τ h ) ≤ | (Π h g εγ ) ′ ( s ) | ≤ k ( g γ ) ′ k L ∞ ( τ h ) + k Π h g εγ − g γ k W , ∞ ( τ h ) . Using (2.14) and (2.15) the above implies C g C γ − k Π h g εγ − g γ k W , ∞ ( τ h ) ≤ | (Π h g εγ ) ′ ( s ) | ≤ C ′ g C ′ γ + k Π h g εγ − g γ k W , ∞ ( τ h ) . Hence using (ii) we have (iii) and (iv). (cid:3)
From (iii) and (iv) in Theorem 4.2 we obtain the following corollary.
Corollary 4.3.
Let h be such that (4.12) ˜ C h k g γ k H ( τ h ) + C ′ C γ ε ≤ C g C γ h / . Then, (4.13) C g C γ ≤ | (Π h g εγ ) ′ ( s ) | ≤ C ′ g C ′ γ . Since ( g γ ) ′ = 0, for any τ h ∈ T h , g ( γ ( τ h )) = [ g h , g h ] for some g h < g h . Let us denote(4.14) I h := [ g h , g h ] , I hε := Π h g εγ ([ τ h ]) . Proposition 4.4.
Let h and ε satisfy (4.12) and (4.15) ˜ C h k g γ k H ( τ h ) + C ′ C γ ε < h / . Then, for τ h ∈ T h , I h ∩ I hε is a closed interval with non-empty interior, say ˜ I hε = [ g h,ε , g h,ε ] for some g h,ε < g h,ε , and | g h − g h,ε | < ˜ C h / k g γ k H ( τ h ) + C ′ C γ εh / , (4.16) | g h − g h,ε | < ˜ C h / k g γ k H ( τ h ) + C ′ C γ εh / . (4.17) ONLINEAR INVERSE PROBLEM 23
Proof.
Since h satisfies (4.12), by Corollary 4.3, Π h g εγ satisfies (4.13). Thus I hε is a closed non-degenerate interval. So, by Lemma 3.17, taking φ = ( g γ ) | τ h and φ = (Π h g εγ ) | τh we have thefollowing. I h ∩ I hε = [ g h,ε , g h,ε ] for some g h,ε < g h,ε . Also, since (4.15) is satisfied, we have (4.16) and(4.17). (cid:3) Let us recall that, in Section 3 we have the perturbed operator T ε corresponding to the perturbeddata g ε . Here, we are working with Π h ( g εγ ). Now, let us define the corresponding operator which shallbe used in place of T ε , so that we can carry out the analysis similar to that of Section 3. In order todo that, let us first observe the following.Let h and ε satisfy (4.12) and (4.15). Then, by Corollary 4.3, Π h g εγ satisfies (4.13). Thus, Π h g εγ isbijective and, for any τ h ∈ T h , (Π h g εγ ) − is continuous on I hε . Hence, there exists t h,ε and t h,ε in τ h such that(4.18) ˜ I hε = [ g h,ε , g h,ε ] = Π h g εγ ([ t h,ε , t h,ε ]) . For y ∈ L ( ˜ I ε ), let S h,ǫ ( y )( s ) = ( y (Π h g εγ ( s )) s ∈ [ t h,ε , t h,ε ]0 s ∈ τ h \ [ t h,ε , t h,ε ]and, let(4.19) ( T h,ε y )( s ) = ( S h,ε y )( s ) for s ∈ τ h , τ h ∈ T h . We observe that T h,ε : L ( ˜ I ε ) → L ([0 , Theorem 4.5.
Let h and ε satisfy (4.12) and (4.15). Then, the operator T h,ε : L ( ˜ I ε ) → L ([0 , is bounded linear and bounded below. Further, we have the following. (i) For ζ ∈ L ( ˜ I ε ) , (4.20) k T h,ε ( ζ ) k L ([0 , ≤ s C g C γ k ζ k L ( ˜ I ε ) , (4.21) q C ′ g C ′ γ k T h,ε ( ζ ) k L ([0 , ≥ k ζ k L ( ˜ I ε ) , (4.22) k (( T h,ε ) ∗ T h,ε ) − ( T h,ε ) ∗ k ≤ q C ′ g C ′ γ . (ii) For ζ ∈ W , (4.23) k T h,ε ( ζ | ˜ Iε ) − T ζ k L ([0 , ≤ D g,ε,h k ζ ′ k H ( I ) + k ζ k H ( I ) D g,ε,h . (iii) If ζ ∈ W ∩ H ( I ) , then (4.24) k T h,ε ( ζ | ˜ Iε ) − T ζ k L ([0 , ≤ D g,ε,h k ζ ′ k H ( I ) + k ζ k H ( I ) D g,ε,h , where D g,ε,h = C I ( ˜ C k g ◦ γ k H ([0 , h + εC γ ) ,D g,ε,h = 4( C I ) s C g C γ ( ˜ C h / k g γ k H ([0 , + C ′ C γ εh / ) ,D g,ε,h = 8( C I ) s C g C γ ( ˜ C h / k g γ k H ([0 , + C ′ C γ εh / ) / . Here C , ˜ C , C ′ , C I , C g , C ′ g , C ′ γ , and C γ are constants as defined in (4.3), Theorem 4.2-(ii), (3.43),(2.15) and (2.14) respectively. Proof.
Clearly, T h,ε is a linear map. Now,(4.25) k T h,ε ( ζ ) k L ([0 , = X τ h ∈ T h Z τ h | T h,ε ( ζ )( s ) | ds = X τ h ∈ T h Z t h,ε t h,ε | y (Π h g εγ ( s )) | ds. Since, h satisfies (4.12), by Corollary 4.3, Π h g εγ satisfies (4.13). Hence, using Lemma 2.4, we have X τ h ∈ T h Z t h,ε t h,ε | y (Π h ( g εγ )( s ))) | ds ≤ C g C γ X τ h ∈ T h Z ˜ I hε | ζ ( z ) | dz = 2 C g C γ Z ˜ I ε | ζ ( z ) | dz, (4.26) X τ h ∈ T h Z t h,ε t h,ε | ζ (Π h ( g εγ )( s ))) | ds ≥ C ′ g C ′ γ X τ h ∈ T h Z ˜ I hε | ζ ( z ) | dz = 12 C ′ g C ′ γ Z ˜ I ε | ζ ( z ) | dz. (4.27)Hence, combining (4.25) and (4.26) we have (4.20), and combining (4.25) and (4.27) we have (4.21).Hence, T h,ε is bounded linear and bounded below. Since, T h,ε satisfies (4.20) and (4.21), from Lemma3.5, we obtain (4.22).Using the fact that Π h is a projection, and Lemma 2.4, (2.14) and (4.1), we obtain,(4.28) k Π h g γ − Π h ( g εγ ) k L ([0 , ≤ k g γ − g ε ◦ γ k L ([0 , ≤ εC γ , and, using the fact that Π h is an orthogonal projection, and (4.5), k g γ − Π h g γ k L ([0 , = X τ h ∈ T h k g γ − Π h g γ k L ( τ h ) ≤ X τ h ∈ T h k g γ k L ( τ h ) ≤ h ( C ) X τ h ∈ T h k g γ k H ( τ h ) ≤ C ) h k g γ k H ([0 , (4.29)Taking ˜ C = 2( C ) , (4.28) and (4.29) imply(4.30) k g γ − Π h g εγ k L ([0 , ≤ h ˜ C k g γ k H ([0 , + εC γ . Now, ζ ∈ W implies ζ | Ih ∈ H ( I h ). Hence, taking φ and φ as g ◦ γ | τ h and Π h g εγ | τ h respectively, inthe first part of Lemma 3.21, (4.30) and (3.43), we have, k T h,ε ( ζ | ˜ Ihε ) − T ζ k L ( ∪ τh ∈ T h [ t h,ε ,t h,ε ]) = X τ h ∈ T h k ζ ◦ g γ − ζ ◦ Π h g εγ k L ([ t h,ε ,t h,ε ]) ≤ (cid:16) h ˜ C k g γ k H ([0 , + εC γ (cid:17) X τ h ∈ T h k ζ ′ k L ∞ g ( γ ( τ h )) ≤ ( C I ) (cid:16) h ˜ C k g γ k H ([0 , + εC γ (cid:17) X τ h ∈ T h k ζ ′ k H ( g ( γ ( τ h ))) = ( C I ) (cid:16) h ˜ C k g γ k H ([0 , + εC γ (cid:17) k ζ ′ k H ( I ) Hence,(4.31) k T h,ε ( ζ | ˜ Ihε ) − T ζ k L ( ∪ τh ∈ T h [ t h,ε ,t h,ε ]) ≤ C I (cid:16) h ˜ C k g ◦ γ k H ([0 , + εC γ (cid:17) k ζ ′ k H ( I ) . Since g ′ γ >
0, we have g ( γ ([ t h,ε , t h,ε ])) = [ ˜ g h,ε , ˜ g h,ε ] ⊂ I h for some ˜ g h,ε < ˜ g h,ε . As h and ε satisfy(4.12) and (4.15), taking φ = ( g ◦ γ ) | [ t h,ε ,t h,ε ] and φ = Π h g εγ | [ t h,ε ,t h,ε ] in Lemma 3.17, we have, | g h,ε − ˜ g h,ε | < ˜ C h / k g γ k H ( τ h ) + C ′ C γ εh / , ONLINEAR INVERSE PROBLEM 25 | g h,ε − ˜ g h,ε | < ˜ C h / k g γ k H ( τ h ) + C ′ C γ εh / . Hence by (4.16) and (4.17),(4.32) | g h − ˜ g h,ε | < C h / k g γ k H ( τ h ) + 2 C ′ C γ εh / , (4.33) | g h − ˜ g h,ε | < C h / k g γ k H ( τ h ) + 2 C ′ C γ εh / . Since (2.14) and (2.15) hold, by Lemma 2.4, k ζ ◦ g γ k L ([0 , \ ( ∪ τh ∈ T h [ t h,ε ,t h,ε ])) = Z ([0 , \ ( ∪ τh ∈ T h [ t h,ε ,t h,ε ])) | ζ ( g ( γ ( s ))) | ds ≤ C g C γ Z I \∪ τh ∈ T h g ( γ ([ t h,ε ,t h,ε ])) | ζ ( z ) | dz ≤ C g C γ X τ h ∈ T h Z I h \ g ( γ ([ t h,ε ,t h,ε ])) | ζ ( z ) | dz ≤ C g C γ X τ h ∈ T h "Z ˜ g h,ε g h | ζ ( z ) | dz + Z g h ˜ g h,ε | ζ ( z ) | dz . Hence,(4.34) k ζ ◦ g γ k L ([0 , \ ( ∪ τh ∈ T h [ t h,ε ,t h,ε ])) ≤ C g C γ X τ h ∈ T h "Z ˜ g h,ε g h | ζ ( z ) | dz + Z g h ˜ g h,ε | ζ ( z ) | dz . Now, by (3.43), ζ ∈ W implies ζ ∈ W , ∞ ( I ). Hence, as (4.32) and (4.33) hold, by Lemma 3.20-(i)and then by (3.43), we have Z ˜ g h,ε g h | ζ ( z ) | dz ≤ C I ) (cid:18) ˜ C h / k g γ k H ( τ h ) + C ′ C γ εh / (cid:19) k ζ k W , ∞ ([ g h , ˜ g h,ε ]) ≤ C I ) (cid:18) ˜ C h / k g γ k H ( τ h ) + C ′ C γ εh / (cid:19) k ζ k H ([ g h , ˜ g h,ε ]) , and, similarly, Z ˜ g h,ε g h | ζ ( z ) | dz ≤ C I ) (cid:18) ˜ C h / k g γ k H ( τ h ) + C ′ C γ εh / (cid:19) k ζ k H ([ g h , ˜ g h,ε ]) Thus, from (4.34) we have (4.23).Next, let ζ ∈ H ( I ). Since (4.32) and (4.33) hold, by Lemma 3.20-(ii) and then by (3.43), we obtain Z ˜ g h,ε g h | ζ ( z ) | dz ≤ C I ) (cid:18) ˜ C h / k g γ k H ( τ h ) + C ′ C γ εh / (cid:19) k ζ k W , ∞ ([ g h , ˜ g h,ε ]) ≤ C I ) (cid:18) ˜ C h / k g γ k H ( τ h ) + C ′ C γ εh / (cid:19) k ζ k H ([ g h , ˜ g h,ε ]) , and, similarly, Z ˜ g h,ε g h | ζ ( z ) | dz ≤ C I ) (cid:18) ˜ C h / k g γ k H ( τ h ) + C ′ C γ εh / (cid:19) k ζ k H ([ g h , ˜ g h,ε ]) . Thus, from (4.34) we have (4.24). (cid:3)
Let ˜ ζ ε,δ,h ∈ L ( ˜ I ε ) be the unique solution of the equation(4.35) T h,ε ∗ T h,ε ( ζ ) = T h,ε ∗ f j δ , that is ˜ ζ ε,δ,h := ( T h,ε ) † f j δ . Now, let ζ ε,δ,h = (cid:26) ˜ ζ ε,δ,h on ˜ I ε I \ ˜ I ε . Then, ζ ε,δ,h ∈ L ( I ). Let b α,ε,δ,h be the solution of the equation(4.36) ( T α ) ∗ ( T α )( w ) = ( T α ) ∗ ζ ε,δ,h . We show that a α,ε,δ,h := b ′ α,ε,δ,h is an approximate solution of (2.6). For this purpose, we shall makeuse of the following proposition. Proposition 4.6.
Let a and g be as defined in Lemma 3.25. Let h and ε satisfy the relations in(4.12) and (4.15). Let g ε ∈ L ( I ) be such that (4.1) is satisfied. Then, b = T ( a ) satisfies, (4.37) k b k L ( I \ ˜ I ε ) ≤ k b k H ( I ) ( C I ) (cid:16) ˜ C h / k g γ k H ([0 , + C ′ C γ εh / (cid:17) , and, in addition, if a ∈ H ( I ) , then, (4.38) k b k L ( I \ ˜ I ε ) ≤ k b k H ( I ) C I ) (cid:16) ˜ C h / k g γ k H ([0 , + C ′ C γ εh / (cid:17) / , Proof.
Since, h and ε satisfy (4.12), for any τ h ∈ T h , as (4.16) holds, by Lemma 3.20-(i) and then by(3.43), we have k b k L ( I h \ ˜ I hε ) ≤ C I k b k W , ∞ ( I h ) (cid:18) ˜ C h / k g γ k H ( τ h ) + C ′ C γ εh / (cid:19) ≤ ( C I ) k b k H ( I h ) (cid:18) ˜ C h / k g γ k H ( τ h ) + C ′ C γ εh / (cid:19) , and, if a ∈ H ( I ), b ∈ H ( I ) and so, by Lemma 3.20-(ii) and then by (3.43), k b k L ( I h \ ˜ I hε ) ≤ C I ) k b k W , ∞ ( I h ) (cid:18) ˜ C h / k g γ k H ( τ h ) + C ′ C γ εh / (cid:19) / ≤ C I ) k b k H ( I h ) (cid:18) ˜ C h / k g γ k H ( τ h ) + C ′ C γ εh / (cid:19) / . Since k b k L ( I \ ˜ I ε ) = P τ h ∈ T h k b k L ( I h \ ˜ I hε ) , the required inequalities follow. (cid:3) Theorem 4.7.
Let a , g and j be as in Lemma 3.25. Let g ε ∈ L ( I ) , j δ ∈ W − /p,p ( ∂ Ω) with p > .Also, let g ε and j δ satisfy (4.1) and (3.27), respectively, and h and ε satisfy the relations in (4.15)and (4.12), and a α,ε,δ,h = b ′ α,ε,δ,h . Then the following results hold. (i) With the original assumption that a ∈ H ( I ) , k a − a α,ε,δ,h k H ( I ) ≤ C α + 2 α k b k H ( I ) ( C I ) C g,ε,h (4.39) + 2 α q C ′ g C ′ γ [ D g,ε,h k b ′ k H ( I ) + D g,ε,h k b k H ( I ) + ˜ C γ δ ] , k a − a α,ε,δ,h k L ( I ) ≤ √ α k a ′ k L ( I ) + 2 √ α k b k H ( I ) ( C I ) C g,ε,h (4.40) + 2 √ α q C ′ g C ′ γ [ D g,ε,h k b ′ k H ( I ) + D g,ε,h k b k H ( I ) + ˜ C γ δ ] . ONLINEAR INVERSE PROBLEM 27 (ii) If a ∈ H ( I ) , then, k a − a α,ε,δ,h k L ( I ) ≤ √ α k a ′ k L ( I ) + 2 √ α k b k H ( I ) ( C I ) ( C g,ε,h ) / (4.41) + 2 √ α q C ′ g C ′ γ [ D g,ε,h k b ′ k H ( I ) + D g,ε,h k b k H ( I ) + ˜ C γ δ ] . (iii) If a ∈ H ( I ) , then k a − a α,ε,δ,h k H ( I ) ≤ (1 + C L ) k a ′ k H ( I ) α + 2 α k b k H ( I ) ( C I ) ( C g,ε,h ) / (4.42) + 2 α q C ′ g C ′ γ [ D g,ε,h k b ′ k H ( I ) + D g,ε,h k b k H ( I ) + ˜ C γ δ ] , k a − a α,ε,δ,h k L ( I ) ≤ (1 + C L ) k a ′ k H ( I ) α + 2 √ α k b k H ( I ) ( C I ) ( C g,ε,h ) / (4.43) + 2 √ α q C ′ g C ′ γ [ D g,ε,h k b ′ k H ( I ) + D g,ε,h k b k H ( I ) + ˜ C γ δ ] . In the above C α > is such that C α → as α → , b = T ( a ) , C g,ε,h = ( ˜ C h / k g γ k H ([0 , + C ′ C γ εh / ) D g,ε,h = C I ( ˜ C k g ◦ γ k H ([0 , h + εC γ ) ,D g,ε,h = 4( C I ) s C g C γ ( ˜ C h / k g γ k H ([0 , + C ′ C γ εh / ) ,D g,ε,h = 8( C I ) s C g C γ ( ˜ C h / k g γ k H ([0 , + C ′ C γ εh / ) / , and C , C L , ˜ C , C ′ , C I C g , C ′ g , C ′ γ , C γ are constants as defined in (4.3), Proposition 3.3, Theorem4.2-(ii), (3.43), (2.15) and (2.14) respectively.Proof. By definition of ζ ε,δ,h ,(4.44) k ζ ε,δ,h − b k L ( I ) ≤ k ˜ ζ ε,δ,h − b | ˜ I ε k L ( ˜ I ε ) + k b k L ( I \ ˜ I ε ) . We use the notation ( T h,ε ) † := (( T h,ε ) ∗ T h,ε ) − ( T h,ε ) ∗ . Then, by (4.22), and using the fact that( T h,ε ) † ( T h,ε ) ∗ is identity, we have k ( T h,ε ) † T ( T ( b )) − ( T ( b )) | ˜ I ε k L ( ˜ I ε ) ≤ q C ′ g C ′ γ k T ( T ( b )) − T h,ε (( T ( b )) | ˜ I ε ) k L ([0 , , and, in addition, using (3.34),(4.45) k ( T h,ε ) † ( f j − f j δ ) k L ( I ) ≤ q C ′ g C ′ γ k f j δ − f j k L ([0 , ≤ q C ′ g C ′ γ ˜ C γ δ. Also, since a ( g ) = 0 we have b = T ( a ) ∈ W , so that, by (4.23) and (4.24),(4.46) k ( T h,ε ) † T ( T ( b )) − ( T ( b )) | ˜ I ε k L ( ˜ I ε ) ≤ q C ′ g C ′ γ [ D g,ε,h k b ′ k H ( I ) + D g,ε,h,b ] , where, D g,ε,h,b := (cid:26) D g,ε,h k b k H ( I ) if b ∈ W ,D g,ε,h k b k H ( I ) if b ∈ H ( I ) ∩ W . Now, by the definition of ˜ ζ ε,δ,h and the fact that T ( T ( b )) = f j , we have k ˜ ζ ε,δ,h − b | ˜ I ε k L ( ˜ I ε ) ≤ k ( T h,ε ) † f j δ − ( T ( b )) | ˜ I ε k L ( ˜ I ε ) ≤ k ( T h,ε ) † T ( T ( b )) − ( T ( b )) | ˜ I ε k L ( ˜ I ε ) + k ( T h,ε ) † ( f j − f j δ ) k L ( I ) Hence, from (4.46) and (4.45) we have(4.47) k ˜ ζ ε,δ,h − b | ˜ I ε k L ( ˜ I ε ) ≤ q C ′ g C ′ γ [ D g,ε,h k b ′ k H ( I ) + D g,ε,h,b + ˜ C γ δ ] . Thus, from (4.44), (4.37) and (4.47) we have k ζ ε,δ,h − b k L ( I ) ≤ k b k H ( I ) ( C I ) (cid:16) ˜ C k g ◦ γ k H ([0 , h + εC γ (cid:17) (4.48) + q C ′ g C ′ γ [ D g,ε,h k b ′ k H ( I ) + D g,ε,h k b k H ( I ) + ˜ C γ δ ] . If a ∈ H ( I ) then b ∈ H ( I ), and thus from (4.44), (4.38) and (4.47) we have, k ζ ε,δ,h − b k L ( I ) ≤ k b k H ( I ) ( C I ) (cid:18) ˜ C k g ◦ γ k H ([0 , h + εC γ (cid:19) / (4.49) + q C ′ g C ′ γ [ D g,ε,h k b ′ k H ( I ) + D g,ε,h k b k H ( I ) + ˜ C γ δ ] . Our aim is to find an estimate for the error term ( a − a α,ε,δ,h ) in L ( I ) and H ( I ) norms. Now b α,ε,δ,h is the unique solution of equation (4.36). Thus, according to Lemma 3.25 we need an estimateof k ζ ε,δ,h − b k L ( I ) in order to find our required estimates. Inequalities (4.48) and (5.19) give usestimates of k ζ ε,δ,h − b k L ( I ) under different conditions on b . Hence, taking ζ ε,δ,h in place of ζ inLemma 3.25 we have the proof. (cid:3) Remark 4.8.
Suppose2 ε / < min C γ C g ˜ C k g ◦ γ k H ([0 , + C ′ C ′ γ / , C k g ◦ γ k H ([0 , + C ′ C ′ γ . Then, for ε = δ and h = δ / , (4.15) and (4.12) are satisfied. Hence, by Theorem 4.7, we have thefollowing:(1) Choosing α = √ δ , we have k a − a α,ε,δ,h k H ( I ) = o (1) . (2) If a ∈ H ( I ) and α = δ / , then k a − a α,ε,δ,h k H ( I ) = O ( √ δ ) , k a − a α,ε,δ,h k L ( I ) = O ( δ / ) . (3) Choosing α = δ , we have k a − a α,ε,δ,h k L ( I ) = O ( δ / ) . (4) If a ∈ H ( I ), then k a − a α,ε,δ,h k L ( I ) = O ( δ / ) . Results in (1) and (2) above are analogous to the corresponding results for a − a α,ε,δ in Remark 3.27.The estimate in (4) is same as the corresponding estimate in Remark 3.27, except for the fact thathere we need an additional condition that a ∈ H ( I ). ♦ ONLINEAR INVERSE PROBLEM 29 With exact solution having non-zero value at g In the previous two sections we have considered the exact solution with assumption that a ( g ) = 0.Here we consider the case when a ( g ) = 0 but is assumed to be known. Let a ( g ) = c . Since a isthe solution to Problem (P), by (2.6) we have f j = T ( a ) which implies(5.1) f j = T ( a − c + c ) = T ( a − c ) + cT (1)Now by definition of T we have(5.2) T (1)( s ) = Z g ◦ γ ( s ) g dt = g ◦ γ ( s ) − g , s ∈ [0 , . Thus, combining (5.1) and (5.2) we have(5.3) T ( a − c ) = f j − c ( g γ − g )Hence a − c is the solution of the following operator equation,(5.4) T ( a ) = f j − c ( g γ − g ) , where clearly f j − c ( g γ − g ) ∈ L ([0 , a − c )( g ) = 0. Now, let us define b ,c ( x ) = Z xg ( a ( t ) − c ) dt, x ∈ I. Then b ,c ∈ W . Thus, the analysis of the previous two sections can be applied here to obtain a stableapproximate solution of equation (5.4). Let a c,α := b ′ c,α , where b c,α is the solution to the followingequation.(5.5) ( T α ) ∗ ( T α )( w ) = ( T α ) ∗ ζ c , where ζ c is the solution of the equation(5.6) ( T ) ∗ ( T ) ζ = ( T ) ∗ ( f j − c ( g γ − g )) . Now, let g ε and j δ be the perturbed data as defined in Theorem 4.7. Also, let g be such that g ◦ γ ∈ H ([0 , ζ c,ε,δ,h be the solution of the equation(5.7) T h,ε ∗ T h,ε ( ζ ) = T h,ε ∗ ( f j δ − c (Π h ( g εγ ) − g )) , where Π h ( g εγ ) is as defined in Section 4. Now, ζ c,ε,δ,h defined as, ˜ ζ c,ε,δ,h on ˜ I ε and 0 on I \ ˜ I ε , is in L ( I ). Let b c,ε,δ,h be the solution of the equation(5.8) ( T α ) ∗ ( T α )( w ) = ( T α ) ∗ ζ c,ε,δ,h Then we have the following theorem.
Theorem 5.1.
Let a , c and b ,c be as defined in the beginning of the section. Let g and j be asdefined in Lemma 3.25, and g ◦ γ ∈ H ([0 , . Let h and ε satisfy (4.12) and (4.15), respectively. Also,let g ε ∈ L (Γ) , j δ ∈ W − /p,p ( ∂ Ω) with p > , and g ε and j δ satisfy (4.1) and (3.27) respectively. Let a c,α,ε,δ,h := b ′ c,α,ε,δ,h , and let C g,ε,h := ˜ C h / k g γ k H ( τ h ) + C ′ C γ εh / ,D g,ε,h = C I ( ˜ C k g ◦ γ k H ([0 , h + εC γ ) ,D g,ε,h,b ,c = C I ) q C g C γ ( ˜ C h / k g γ k H ( τ h ) + C ′ C γ εh / ) if b ,c ∈ W , C I ) q C g C γ (cid:16) ˜ C h / k g γ k H ( τ h ) + C ′ C γ εh / (cid:17) / if b ,c ∈ H ( I ) ∩ W . H ( c, ε, δ, h ) := q C ′ g C ′ γ (cid:16) D g,ε,h k b ′ ,c k H ( I ) + D g,ε,h,b ,c + ˜ C γ δ + cD g,ε,h (cid:17) . Then k a − ( c + a c,α,ε,δ,h ) k H ( I ) ≤ C α + 2 α (cid:0) k b ,c k H ( I ) C I C g,ε,h + H ( c, ε, δ, h ) (cid:1) , (5.9) k a − ( c + a c,α,ε,δ,h ) k L ( I ) ≤ √ α k a ′ k L ( I ) + 2 √ α (cid:0) k b ,c k H ( I ) C I C g,ε,h + H ( c, ε, δ, h ) (cid:1) , (5.10) where C α > is such that C α → as α → . Further, we have the following. (i) If a ′ ∈ L ∞ ( I ) , then, (5.11) k a − ( c + a α,ε,δ,h ) k L ( I ) ≤ √ α k a ′ k L ( I ) + 2 √ α (cid:16) k b ,c k H ( I ) ( C I ) ( C g,ε,h ) / + H ( c, ε, δ, h ) (cid:17) . (ii) If a ∈ H ( I ) , then (5.12) k a − ( c + a c,α,ε,δ,h ) k H ( I ) ≤ (1 + C L ) k a ′ k H ( I ) α + 2 α (cid:16) k b ,c k H ( I ) ( C I ) ( C g,ε,h ) / + H ( c, ε, δ, h ) (cid:17) , (5.13) k ( a − c ) − a c,α,ε,δ,h k L ( I ) ≤ (1 + C L ) k a ′ k H ( I ) α + 2 √ α (cid:16) k b ,c k H ( I ) ( C I ) ( C g,ε,h ) / + H ( c, ε, δ, h ) (cid:17) , where C L , ˜ C , C ′ , C I C g , C ′ g , C ′ γ , and C γ are constants as in Proposition 3.3, Theorem 4.2, (3.43),(2.15) and (2.14) respectively.Proof. By definition of ζ c,ε,δ,h ,(5.14) k ζ c,ε,δ,h − b ,c k L ( I ) ≤ k ˜ ζ c,ε,δ,h − b ,c | ˜ I ε k L ( ˜ I ε ) + k b ,c k L ( I \ ˜ I ε ) . Here also we use the notation ( T h,ε ) † := (( T h,ε ) ∗ T h,ε ) − ( T h,ε ) ∗ . By (4.22), k ( T h,ε ) † T ( T ( b ,c )) − ( T ( b ,c )) | ˜ I ε k L ( ˜ I ε ) ≤ q C ′ g C ′ γ k T ( T ( b ,c )) − T h,ε (( T ( b ,c )) | ˜ I ε ) k L ([0 , , k ( T h,ε ) † ( f j − c ( g γ − g ) − f j δ − c (Π h g εγ − g )) k L ( I ) ≤ q C ′ g C ′ γ k [ f j − c ( g γ − g ) − f j δ − c (Π h g εγ − g )] k L ([0 , ≤ q C ′ g C ′ γ k [ f j − f j δ − c ( g ◦ γ − Π h g εγ )] k L ( ˜ I ε ) . Hence, by (3.34) and (4.30),(5.15) k ( T h,ε ) † ( f j − c ( g γ − g ) − f j δ − c (Π h g εγ − g )) k L ( I ) ≤ q C ′ g C ′ γ ( ˜ C γ δ + cD g,ε,h ) , and, by definition b ,c ∈ W and so, by (4.23) and (4.24)(5.16) k ( T h,ε ) † T ( T ( b ,c )) − ( T ( b ,c )) | ˜ I ε k L ( ˜ I ε ) ≤ q C ′ g C ′ γ [ D g,ε,h k b ′ ,c k H ( I ) + D g,ε,h,b ,c ] . Now by definition of ˜ ζ c,ε,δ,h and the fact that T ( T ( b ,c )) = f j − c ( g γ − g ), we have k ˜ ζ c,ε,δ,h − b ,c | ˜ I ε k L ( ˜ I ε ) ≤ k ( T h,ε ) † ( f j δ − c (Π h g εγ − ( T ( b ,c )) | ˜ I ε k L ( ˜ I ε ) ≤ k ( T h,ε ) † ( T ( T ( b ,c )) − ( T ( b ,c )) | ˜ I ε ) k L ( ˜ I ε ) + k ( T h,ε ) † ( f j − c ( g γ − g ) − f j δ − c (Π h g εγ − g )) k L ( I ) . Hence, from (5.16) and (5.15) we have(5.17) k ˜ ζ ε,δ,h − b ,c | ˜ I ε k L ( ˜ I ε ) ≤ q C ′ g C ′ γ [ D g,ε,h k b ′ ,c k H ( I ) + D g,ε,h,b ,c + ˜ C γ δ + cD g,ε,h ] . ONLINEAR INVERSE PROBLEM 31
Thus, from (5.14), (4.37) and (5.17) we have(5.18) k ζ ε,δ,h − b ,c k L ( I ) ≤ k b ,c k H ( I ) ( C I ) (cid:18) ˜ C k g ◦ γ k H ([0 , h + εC γ (cid:19) + H ( c, ε, δ, h ) . If a ,c ∈ H ( I ), from (5.14), (4.38) and (5.17) we have,(5.19) k ζ ε,δ,h − b ,c k L ( I ) ≤ k b ,c k H ( I ) ( C I ) (cid:18) ˜ C k g ◦ γ k H ([0 , h + εC γ (cid:19) / + H ( c, ε, δ, h ) . By definition, b c,α,ε,δ,h is the unique solution of equation (5.8). Thus, putting ζ c,ε,δ,h in place of ζ inLemma 3.25, we have the proof using (4.48) and (5.19). (cid:3) From Theorem 5.1, we see that c + a c,α,ε,δ,h is a stable approximate solution of Problem (P), witherror estimates obtained from Theorem 5.1. Remark 5.2.
Let us relax the assumption on the exact solution a even more. Let us assume that a ( g ) is not equal to the known number c but is known to be “close” to it, i.e,(5.20) | a ( g ) − c | < η, for some η >
0. Let c := a ( g ). Define b ,c ( x ) = R xg ( a ( t ) − c ) dt for x ∈ I . Then b ,c ∈ W . Also,let g , j , g ε , j δ , h , ζ c,ε,δ,h , b c,α,ε,δ,h and a c,α,ε,δ,h be as defined in Theorem 5.1. Since (5.20) holds, k ( f j − f j δ − ( c − c )( g γ + g ) − c ( g ◦ γ − Π h g εγ ) k L ( ˜ I ε ) k L ([0 , ≤ k f j δ − f j k L ([0 , + c k g ◦ γ − Π h g εγ k L ([0 , + ( k g ◦ γ k L ([0 , + | g | ) η and, by (3.34) and (4.30), we have(5.21) k ( f j − f j δ − ( c − c )( g γ + g ) − c ( g ◦ γ − Π h g εγ ) k L ( ˜ I ε ) k L ([0 , ≤ ˜ C γ δ + a ( g ) D g,ε,h +( k g ◦ γ k L ([0 , + | g | ) η, with D g,ε,h as in Theorem 5.1. Now, as T ( T ( b ,c )) = f j − c ( g γ − g ), k ˜ ζ c,ε,δ,h − b ,c | L ( ˜ I ε ) k L ( ˜ I ε ) = k ˜ ζ c,ε,δ,h − T ( b ,c ) | L ( ˜ I ε ) k L ( ˜ I ε ) ≤ k ( T h,ε ) † ( f j δ − c (Π h g εγ − g )) − T ( b ,c ) | L ( ˜ I ε ) k L ( ˜ I ε ) ≤ k ( T h,ε ) † T ( T ( b ,c )) − T ( b ,c ) | L ( ˜ I ε ) k L ( ˜ I ε ) + k ( T h,ε ) † ( f j − c ( g γ − g ) − f j δ + c (Π h g εγ − g )) k L ( ˜ I ε ) ≤ k ( T h,ε ) † ( T T ( b ,c ) − T h,ε ( T ( b ,c ) | L ( ˜ I ε ) ) k L ( ˜ I ε ) + k ( T h,ε ) † ( f j − f j δ − ( c − c )( g γ + g ) − c ( g ◦ γ − Π h g εγ )) k L ( ˜ I ε ) , and, by (4.22), (4.23) and (4.24) k ( T h,ε ) † T ( T ( b ,c )) − ( T ( b ,c )) | ˜ I ε k L ( ˜ I ε ) ≤ q C ′ g C ′ γ [ D g,ε,h k b ′ ,c k H ( I ) + D g,ε,h,b ,c + k f j − f j δ k L ([0 , + k ( c − c )( g γ + g ) + c ( g ◦ γ − Π h g εγ ) k L ([0 , ]with D g,ε,h,b ,c as in Theorem 5.1. Hence, by (5.21) k ˜ ζ c,ε,δ,h − b ,c | L ( ˜ I ε ) k L ( ˜ I ε ) ≤ q C ′ g C ′ γ [ D c,g,ε,h k b ′ ,c k H ( I ) + D g,ε,h,b ,c + ˜ C γ δ + cD g,ε,h + ( k g ◦ γ k L ([0 , + | g | ) η ] . Thus, using similar arguments as in the proof of Theorem 5.1, we obtain estimates for k ( a − c ) − a c,α,ε,δ,h k H ( I ) and k ( a − c ) − a c,α,ε,δ,h k L ( I ) . Using the fact that ( a − ( a c,α,ε,δ,h + c )) = (( a − c ) − a c,α,ε,δ,h ) + ( c − c ) , we obtain ( a c,α,ε,δ,h + c ) as a stable approximate solution to Problem (P), and obtain the correspondingerror estimates. ♦ Illustration of the procedure
In order to find a stable approximate solution of Problem (P) using the new regularization methodwe have to undertake the following.Let j δ ∈ W − /p,p ( ∂ Ω) with p > g ε ∈ L ( ∂ Ω) be the perturbed data satisfying (3.27) and (4.1)respectively, and let f j δ = v j δ ◦ γ . Also let us assume g ◦ γ ∈ H ([0 , a α,ε,δ .Step (i): (a) Suppose g ε ∈ W , ∞ (Γ) and it satisfies (3.26). Let ˜ ζ ε,δ be the unique element in L ([0 , T ε ) ∗ ( T ε )˜ ζ ε,δ = ( T ε ) ∗ f j δ with T ε defined as in (3.53). Define ζ ε,δ to be equal to ˜ ζ ε,δ on ˜ I ε , and equal to 0 on I \ ˜ I ε .(b) Suppose g ε ∈ L (Γ) \ W , ∞ (Γ). Then under the assumption g ◦ γ ∈ H ([0 , ζ ε,δ,h ∈ L ([0 , T h,ε ) ∗ ( T h,ε )˜ ζ ε,δ,h = ( T h,ε ) ∗ f j δ with T h,ε defined as in (4.19). Define ζ ε,δ,h to be equal to ˜ ζ ε,δ,h on ˜ I ε , and equal to 0 on I \ ˜ I ε .We denote the solution obtained in this step by ζ ε,δ .Step (ii): Let ζ ε,δ be as in Step (i). Let b α,ε,δ be the unique element in H ( I ) such that(6.3) ( T α ) ∗ T α ( b α,ε,δ ) = ( T α ) ∗ ζ ε,δ with T α defined as in (3.1).Step (iii): Define a α,ε,δ := b ′ α,ε,δ , the derivative of b α,ε,δ .We now explain how to solve (6.1) and (6.2) and obtain ζ ε,δ . Let us observe that, for g ε ∈ W , ∞ (Γ)and for f ∈ L ([0 , T ǫ ) ∗ ( f )( z ) = ( f (( g ε ◦ γ ) − ( z ))( g ε ◦ γ ) ′ ( γ − (( g ε ) − ( z ))) z ∈ ˜ I ε z ∈ I \ ˜ I ε . Hence, it can be seen that, ζ ε,δ ( z ) = (cid:26) f j δ (( g ε ◦ γ ) − ( z )) z ∈ ˜ I ε z ∈ I \ ˜ I ε . For g ε ∈ L (Γ) \ W , ∞ (Γ), for any f ∈ L ([0 , T h,ε ) ∗ ( f )( z ) := ( S h,ε,δ ) ∗ ( f )( z ) , z ∈ ˜ I hε , where ( S h,ǫ,δ ) ∗ ( f )( z ) = ( f ((Π h g ε ◦ γ ) − ( z ))(Π h g ε ◦ γ ) ′ ( γ − ((Π h g ε ) − ( z ))) z ∈ ˜ I hε z ∈ I h \ ˜ I hε , Hence, it can be seen that, ζ h,ε,δ ( z ) = χ h,ε,δ ( z ) , z ∈ ˜ I hε , ONLINEAR INVERSE PROBLEM 33 where, χ h,ε,δ ( z ) = (cid:26) f j δ ((Π h g ε ◦ γ ) − ( z )) z ∈ ˜ I hε z ∈ I h \ ˜ I hε . Thus we have ζ ε,δ . Next let us consider Step (ii). Let us consider the case when ζ ε,δ ∈ C ( I ). If ζ ε,δ ∈ R ( T α ) then the solution of(6.4) T α ( b ) = ζ ε,δ is the solution of (6.3). Now let us note that, finding a solution of (6.4) is same as solving the ODE(6.5) − αb ′′ + b = ζ ε,δ with boundary condition(6.6) b ( g ) = 0and(6.7) b ′ ( g ) = 0 . Hence, if j δ and g ε are such that ζ ε,δ ∈ R ( T α ) ∩ C ( I ) then the solution of the ODE (6.5)-(6.7) givesus our desired b α,ε,δ . Also, by Step (iii) a α,ε,δ = b ′ α,ε,δ is our desired regularized solution. Now let usnote that, if ζ ε,δ ∈ L ( I ) \ C ( I ) then there exists ζ ε,δn ∈ C ( I ) such that k ζ ε,δ − ζ ε,δn k L ( I ) = O ( 1 n )for n ∈ N . Since by (3.4) we have k (( T α ) ∗ T α ) − ( T α ) ∗ ( ζ ε,δ − ζ ε,δn ) k H ( I ) ≤ α k ζ ε,δ − ζ ε,δn k L ( I ) , if ζ ε,δn ∈ R ( T α ) then the solution b α,ε,δ,n of (6.5)-(6.7) with ζ ε,δn in place of ζ ε,δ is an approximationof b α,ε,δ . Again, as k b ′ α,ε,δ,n − b ′ α,ε,δ k H ( I ) ≤ k b α,ε,δ,n − b α,ε,δ k H ( I ) , executing Step (iii) b ′ α,ε,δ,n is our desired approximate regularized solution. Hence, if j δ and g ε aresuch that either ζ ε,δ or ζ ε,δn is in R ( T α ) ∩ C ( I ), then we have a stable approximate solution. Thus inthis case we obtain a stable approximate solution to Problem (P) using steps among which the mostcritical one turns out to be that of solving an ODE.7. Appendix
Lemma 7.1.
Let J be a closed interval in R . Then, (7.1) k y k L ∞ ( J ) ≤ C J k y k H ( J ) , where C J = C max { , (2 | J | + 1) } . In particular, for any interval J ′ contained in J , (7.2) k y k L ∞ ( J ′ ) ≤ C J k y k H ( J ′ ) . Proof.
Let J = [ c, d ] for some c < d . Let ˜ c, ˜ d ∈ R be such that ˜ c < c , d < ˜ d and(7.3) max { ( c − ˜ c ) , ( ˜ d − d ) } < ( d − c ) . Then, let us define the function ˜ y ( t ) = t ∈ R \ [˜ c, ˜ d ] y ( c ) (cid:16) t − ˜ cc − ˜ c (cid:17) t ∈ [˜ c, c ] y ( t ) t ∈ Jy ( d ) (cid:16) ˜ d − t ˜ d − d (cid:17) t ∈ [ d, ˜ d ] Then, it can be seen that ˜ y ∈ H ( R ) and(7.4) k ˜ y k L ([˜ c, ˜ d ]) = ( y ( c )) ( c − ˜ c ) Z c ˜ c ( t − ˜ c ) dt + k y k L ([ c,d ]) + ( y ( d )) ( ˜ d − d ) Z ˜ dd ( ˜ d − t ) dt. Now,(7.5) ( y ( c )) ( c − ˜ c ) Z c ˜ c ( t − ˜ c ) dt = ( y ( c )) c − ˜ c ) ( c − ˜ c ) = ( y ( c )) c − ˜ c )and(7.6) ( y ( d )) ( ˜ d − d ) Z ˜ dd ( ˜ d − t ) dt = ( y ( d ))
3( ˜ d − d ) ( ˜ d − d ) = ( y ( d )) d − d ) . By the fundamental theorem of calculus, for any t ∈ [ c, d ], y ( c ) = − R tc y ′ ( s ) ds + y ( t ) , which implies, | y ( c ) | = | y ( t ) − Z tc y ′ ( s ) ds | ≤ | y ( t ) | + | Z tc y ′ ( s ) ds | ) . Hence, using Schwartz inequality as we have | Z tc y ′ ( s ) ds | ≤ (cid:18)Z tc | y ′ ( s ) | ds (cid:19) ≤ ( t − c ) k y ′ k L ([ c,d ]) , | y ( c ) | ≤ | y ( t ) | + ( t − c ) k y ′ k L ([ c,d ]) )holds. This implies | y ( c ) | ( d − c ) = Z dc | y ( c ) | dt ≤ Z dc | y ( t ) | dt + k y ′ k L ([ c,d ]) Z dc ( t − c ) dt ! . Thus,(7.7) | y ( c ) | ( d − c ) ≤ k y k L ([ c,d ]) + ( d − c ) k y ′ k L ([ c,d ]) ) . Again, by the fundamental theorem of calculus, for any t ∈ [ c, d ], y ( d ) = R dt y ′ ( s ) ds + y ( t ) , whichimplies, | y ( d ) | = | y ( t ) + Z dt y ′ ( s ) ds | ≤ | y ( t ) | + | Z dt y ′ ( s ) ds | ) . Hence, using Schwartz inequality as we have | Z dt y ′ ( s ) ds | ≤ Z dt | y ′ ( s ) | ds ! ≤ ( d − t ) k y ′ k L ([ c,d ]) , | y ( d ) | ≤ | y ( t ) | + ( d − t ) k y ′ k L ([ c,d ]) )holds. This implies | y ( d ) | ( d − c ) = Z dc | y ( d ) | dt ≤ Z dc | y ( t ) | dt + k y ′ k L ([ c,d ]) Z dc ( d − t ) dt ! . Thus,(7.8) | y ( d ) | ( d − c ) ≤ k y k L ([ c,d ]) + ( d − c ) k y ′ k L ([ c,d ]) ) . Hence, combining (7.4), (7.5), (7.6), (7.7) and (7.8), we obtain k ˜ y k L ([˜ c, ˜ d ]) ≤
43 ( k y k L ([ c,d ]) + ( d − c ) k y ′ k L ([ c,d ]) ) + k y k L ([ c,d ]) (7.9) ≤ k y k L ([ c,d ]) + 43 ( d − c ) k y ′ k L ([ c,d ]) . ONLINEAR INVERSE PROBLEM 35
Now, ˜ y ′ ( t ) = t ∈ R \ [˜ c, ˜ d ] y ( c ) t ∈ [˜ c, c ] y ′ ( t ) t ∈ J − y ( d ) t ∈ [ d, ˜ d ]Hence, k ˜ y ′ k L ([˜ c, ˜ d ]) ≤ Z c ˜ c ( y ( c )) dt + k y ′ k L ([ c,d ]) + Z ˜ dd ( y ( d )) dt (7.10) ≤ ( y ( c )) ( c − ˜ c ) + k y ′ k L ([ c,d ]) + ( y ( d )) ( ˜ d − d ) . Thus, from (7.7) and (7.8), we obtain k ˜ y ′ k L ([˜ c, ˜ d ]) ≤ k y k L ([ c,d ]) + ( d − c ) k y ′ k L ([ c,d ]) ) + k y k L ([ c,d ]) + k y ′ k L ([ c,d ]) (7.11) ≤ k y k L ([ c,d ]) + (2( d − c ) + 1) k y ′ k L ([ c,d ]) and from (7.9) and (7.11), we obtain k ˜ y k H ([˜ c, ˜ d ]) ≤ k y k L ([ c,d ]) + r
103 ( d − c ) + 1 ! k y ′ k L ([ c,d ]) ≤ k y k L ([ c,d ]) + (2( d − c ) + 1) k y ′ k L ([ c,d ]) . Hence,(7.12) k ˜ y k H ([˜ c, ˜ d ]) ≤ max { , (2( d − c ) + 1) }k y k H ([ c,d ]) . Since, H ( R ) is continuously imbedded in C ( R ) ∩ L ∞ ( R ) (cf. [6]), there exists C > k ˜ y k L ∞ ( R ) ≤ C k ˜ y k H ( R ) , so that k y k L ∞ ([ c,d ]) ≤ k ˜ y k L ∞ ( R ) ≤ C k ˜ y k H ( R ) = C k ˜ y k H ([˜ c, ˜ d ]) . Hence, by (7.12) k y k L ∞ ([ c,d ]) ≤ C max { , (2( d − c ) + 1) }k y k H ([ c,d ]) . (cid:3) References [1] S. C. Brenner and L. R. Scott,
The Mathematical Theory of Finite Element Methods , Springer, 2008.[2] J.R. Cannon, Determination of the unknown coefficient k ( u ) in the equation ∇ .k ( u ) ∇ u = 0 from overspecifiedboundary data, J. Math. Anal. Appl. , Vol. 18, (1967), 112-4.[3] P. Grisvard,
Elliptic Problems in Nonsmooth Domains , Pitman Advanced Publishing Program, 1985.[4] M.F. Herbert Egger, Jan-Frederik Pietschmann and Matthias Schlottbom , Numerical diffusion law via regulariza-tion in Hilbert scales,
Inverse Problems , Vol. 30, (2014), 025004 (14pp).[5] D B Ingham and Y Yuan, The solution of a non-linear inverse problem in heat transfer,
IMA J. Appl. Math. , Vol.50, (1993), 113-32.[6] S. Kesavan,
Topics in Applicable Functional Analysis and Applications , New Age International Limited, 1989.[7] P. Kugler, Identification of a temperature dependent heat conductivity from single boundary measurements,
SIAMJ. Numer. Anal. , Vol. 41, (2003), 1543-63.[8] M.T. Nair,
Functional Analysis: A First Course , PHI Learning, New Delhi, 2002 (Fourth Print: 2014).[9] M.T. Nair,
Linear Operator Equations: Approximation and Regularization , World Scientific, 2009.[10] R.E.Showalter,
Monotone Operators in Banach Spaces and Nonlinear Partial Differential Equations , AMS, Prov-idence, RI, 1997.
Department of Mathemtics, Indian Institute of Technology Madras, Chennai 600036, India.
E-mail address ::