Semigroup approach for identification of the unknown diffusion coefficient in a quasi-linear...

MATHEMATICAL METHODS IN THE APPLIED SCIENCESMath. Meth. Appl. Sci. 2007; 30:1283–1294Published online 22 January 2007 in Wiley InterScience(www.interscience.wiley.com) DOI: 10.1002/mma.837MOS subject classification: 35R 30; 47D 06; 39K 05

Semigroup approach for identification of the unknown diffusioncoefficient in a quasi-linear parabolic equation

Ali Demir1 and Ebru Ozbilge2,∗,†

1Applied Mathematical Sciences Research Center and Department of Mathematics, Kocaeli University,Ataturk Bulvari, Izmit-Kocaeli 41300, Turkey

2Department of Mathematics, Faculty of Science and Literature, Izmir University of Economics,Sakarya Caddesi, No. 156, Balcova-Izmir 35330, Turkey

Communicated by M. Grinfeld

SUMMARY

This article presents a semigroup approach for the mathematical analysis of the inverse coefficient prob-lems of identifying the unknown coefficient k(u(x, t)) in the quasi-linear parabolic equation ut (x, t)=(k(u(x, t))ux (x, t))x , with Dirichlet boundary conditions u(0, t)= �0, u(1, t)= �1. The main purposeof this paper is to investigate the distinguishability of the input–output mappings �[·] :K→C1[0, T ],�[·] :K→C1[0, T ] via semigroup theory. In this paper, it is shown that if the null space of the semigroupT (t) consists of only zero function, then the input–output mappings �[·] and �[·] have the distinguisha-bility property. It is also shown that the types of the boundary conditions and the region on which theproblem is defined play an important role in the distinguishability property of these mappings. More-over, under the light of measured output data (boundary observations) f (t) := k(u(0, t))ux (0, t) or/andh(t) := k(u(1, t))ux (1, t), the values k(�0) and k(�1) of the unknown diffusion coefficient k(u(x, t))at (x, t)= (0, 0) and (x, t)= (1, 0), respectively, can be determined explicitly. In addition to these, thevalues ku(�0) and ku(�1) of the unknown coefficient k(u(x, t)) at (x, t)= (0, 0) and (x, t)= (1, 0),respectively, are also determined via the input data. Furthermore, it is shown that measured output dataf (t) and h(t) can be determined analytically by an integral representation. Hence the input–outputmappings �[·] :K→C1[0, T ], �[·] :K→C1[0, T ] are given explicitly in terms of the semigroup.Copyright q 2007 John Wiley & Sons, Ltd.

KEY WORDS: semigroup approach; coefficient identification; parabolic equation

∗Correspondence to: Ebru Ozbilge, Department of Mathematics, Faculty of Science and Literature, Izmir Universityof Economics, Sakarya Caddesi, No. 156, Balcova-Izmir 35330, Turkey.

†E-mail: [email protected]

Contract/grant sponsor: Scientific and Technical Research Council (TUBITAK) of TurkeyContract/grant sponsor: Izmir University of Economics

Copyright q 2007 John Wiley & Sons, Ltd. Received 1 September 2006

1284 A. DEMIR AND E. OZBILGE

1. INTRODUCTION

Consider the following initial-boundary value problem:⎧⎪⎪⎨⎪⎪⎩

ut (x, t) = (k(u(x, t))ux (x, t))x , (x, t) ∈ �T

u(x, 0)= g(x), 0<x<1

u(0, t) =�0, u(1, t) = �1, 0<t<T

(1)

where�T ={(x, t) ∈ R2 : 0<x<1, 0<t�T }. The left and right boundary values�0, �1 are assumedto be constants. The functions c1>k(u(x, t))�c0>0 and g(x) satisfy the following conditions:

(C1) k(u(x, t)) ∈C2,1(�T );(C2) g(x)∈C2[0, 1], g(0)=�0, g(1) =�1.

Under these conditions the initial-boundary value problem (1) has the unique solution u(x, t) ∈C2,1(�T ) ∩C2,0(�T ).

Consider the inverse problem of determining the unknown coefficient k = k(u(x, t)) from oneor/and two Neumann type of measured output data at the boundaries x = 0 and x = 1:

k(u(0, t))ux (0, t) = f (t), k(u(1, t))ux (1, t) = h(t), t ∈ (0, T ] (2)

Here u = u(x, t) is the solution of the parabolic problem (1). The functions f (t), h(t) areassumed to be noisy free measured output data. In this context, the parabolic problem (1) willbe referred as a direct ( forward) problem, with the inputs g(x) and k(u(x, t)). It is assumedthat the functions f (t) and h(t) belong to C1[0, T ] and satisfy the consistency conditionsf (0) = k(u(0, 0))g′(0)= k(�0)g

′(0), h(0)= k(u(1, 0))g′(1) = k(�1)g′(1).

We denote by K:={k(u(x, t))∈C2,1(�T ):c1>k(u(x, t))�c0>0, u(x, t)∈C2,1(�T )∩C2,0(�T )},the set of admissible coefficients k = k(u(x, t)), and introduce the input–output mappings�[·] :K→C1[0, T ], �[·] :K→C1[0, T ], where

�[k] = k(u(x, t))ux (x, t; k)|x=0, �[k] = k(u(x, t))ux (x, t; k)|x=1, k ∈K (3)

Then the inverse problem with the measured output data f (t) and h(t) can be formulated as thefollowing operator equations:

�[k] = f, �[k] = h, k ∈K, f, h ∈C1[0, T ] (4)

The monotonicity, continuity, and hence invertibility of the input–output mappings �[·] :K→C1[0, T ] and �[·] :K→C1[0, T ] are given in [1–4].

The purpose of this paper is to study the distinguishability of the unknown coefficient viathe above input–output mappings. We say that the mapping �[·] :K→C1[0, T ] (or �[·] :K→C1[0, T ]) has the distinguishability property, if �[k1] �= �[k2] (�[k1] �= �[k2]) implies k1((x, t)) �=k2((x, t)). This, in particular, means injectivity of the inverse mappings �−1 and �−1.

The results presented here are the first one, to the knowledge of the authors, from a point of viewof semigroup approach to inverse problems. This approach sheds more light on the identifiabilityof unknown coefficient and shows how much information can be extracted from the measuredoutput data, in particular in the case of constant boundary values.

Copyright q 2007 John Wiley & Sons, Ltd. Math. Meth. Appl. Sci. 2007; 30:1283–1294DOI: 10.1002/mma

SEMIGROUP APPROACH FOR IDENTIFICATION OF UNKNOWN DIFFUSION COEFFICIENT 1285

The paper is organized as follows. In Section 2, an analysis of the semigroup approach is givenfor the inverse problem with the single measured output data f (t) given at the boundary x = 0.The similar analysis is applied to the inverse problem with the single measured output data h(t)given at the point x = 1, in Section 3. The inverse problem with two Neumann measured data f (t)and h(t) is discussed in Section 4. Some concluding remarks are given in Section 5.

2. AN ANALYSIS OF THE INVERSE PROBLEM WITH GIVEN MEASURED DATA f (t)

Consider now the inverse problem with one measured output data f (t) at x = 0. In order toformulate the solution of the parabolic problem (1) in terms of semigroup, let us first arrange theparabolic equation as follows:

ut (x, t) − (k(u(0, 0))ux (x, t))x = ([k(u(x, t)) − k(u(0, 0))]ux (x, t))x , (x, t) ∈ �T

or it can be rewritten as

ut (x, t) − (k(�0)ux (x, t))x = ([k(u(x, t)) − k(�0)]ux (x, t))x , (x, t) ∈ �T

since u(0, 0)=�0. Then the initial-boundary value problem (1) can be rewritten in the followingform:

ut (x, t) − k(u(0, 0))uxx (x, t) = ((k(u(x, t)) − k(u(0, 0)))ux (x, t))x , (x, t) ∈ �T

u(x, 0) = g(x), 0<x<1 (5)

u(0, t) = �0, u(1, t) = �1, 0<t<T

For the time being we assume that k(u(0, 0))= k(�0) were known, later this value will be deter-mined. In order to formulate the solution of the parabolic problem (5) in terms of semigroup, weneed to define a new function:

v(x, t) = u(x, t) − �0(1 − x) − �1x, x ∈ [0, 1] (6)

which satisfies the following parabolic problem:

vt (x, t) + A[v(x, t)] = ((k(v(x, t) + �0(1 − x) + �1x)

− k(�0))(vx (x, t) + �1 − �0))x , (x, t) ∈ �T

v(x, 0) = g(x) − �0(1 − x) − �1x, 0<x<1 (7)

v(0, t) = 0, v(1, t) = 0, 0<t<T

Here A[v(x, t)] := −k(�0)d2v(x, t)/dx2 is a second-order differential operator, its domain is DA =

{u ∈C2(0, 1) ∩C1[0, 1] : u(0)= u(1)= 0}. It is obvious that g(x)∈ DA, since the initial valuefunction g(x) belongs to C2[0, 1].

Denote by T (t) the semigroup of linear operators generated by the operator −A [5, 6]. Notethat we can easily find the eigenvalues and eigenfunctions of the differential operator A. Moreover,the semigroup T (t) can be easily constructed by using the eigenvalues and eigenfunctions of the



infinitesimal generator A. Hence we first consider the following eigenvalue problem:

A�(x) = ��(x)

�(0) = 0, �(1) = 0(8)

We can easily determine that the eigenvalues are �n = k(u(0, 0))n2�2 for all n = 0, 1, . . . and thecorresponding eigenfunctions are �n(x)= √

2 sin(n�x). In this case, the semigroup T (t) can berepresented in the following way:

T (t)U (x, s) =∞∑n=0

〈�n(x),U (x, s)〉 e−�nt�n(x) (9)

where 〈�n(x),U (x, s)〉= ∫ 10 �n(x)U (x, s) dx . Under this representation, the null space of the

semigroup T (t) of the linear operators can be defined as follows:

N (T ) ={U (x, s) : 〈�n(x),U (x, s)〉= 0 for all n = 0, 1, 2, 3, . . .}From the definition of the semigroup T (t) we can say that the null space of it consists of only zerofunction, i.e. N (T ) ={0}. As we will see later, this result is very important for the uniqueness ofthe unknown coefficient k(u(x, t)).

The unique solution of the initial-boundary value problem (7) in terms of semigroup T (t) canbe represented in the following form:

v(x, t) = T (t)v(x, 0) +∫ t

0T (t − s)((k(v(x, t) + �0(1 − x) + �1x)

− k(�0))(vx (x, t) + �1 − �0))x ds

Hence by using the identity (6) and taking the initial value u(x, 0) = g(x) into account, the solutionu(x, t) of the parabolic problem (5) in terms of semigroup can be written in the following form:

u(x, t) = �0(1 − x) + �1x + T (t)(g(x) − �0(1 − x) − �1x)

+∫ t

0T (t − s)((k(u(x, t)) − k(u(0, 0)))ux (x, s))x ds

In order to arrange the above solution representation, let us define the following:

�(x) = (g(x) − �0(1 − x) − �1x)

�(x, t) = ((k(u(x, t)) − k(u(0, 0)))ux (x, s))x

z(x, t) =∞∑n=0

〈�n(.), �(.)〉 e−�nt�′n(x) (10)

w(x, t, s) =∞∑n=0

〈�n(.), �(., s)〉 e−�nt�′n(x) (11)

Then we can rewrite the solution representation in terms of �(x) and �(x, s) in the following form:

u(x, t) =�0(1 − x) + �1x + T (t)�(x) +∫ t

0T (t − s)�(x, s) ds (12)



Differentiating both sides of the above identity with respect to x and using semigroup propertiesat x = 0 yield

ux (0, t) =−�0 + �1 + z(0, t) +∫ t

0w(0, t − s, s) ds

Taking into account the over-measured data k(u(0, t))ux (0, t) = k(�0)ux (0, t) = f (t) we get

f (t) = k(u(0, t))

(−�0 + �1 + z(0, t) +

∫ t

0w(0, t − s, s) ds

)(13)

Using the over measured data k(u(0, t))ux (0, t) = f (t), we can write k(u(0, t))= f (t)/ux (0, t)for all t�0 which can be rewritten in terms of semigroup in the following form:

k(u(0, t))= f (t)

/(−�0 + �1 + z(0, t) +

∫ t

0w(0, t − s, s) ds

)

Taking limit as t → 0 in the above identity yields

k(u(0, 0))= f (0)/(−�0 + �1 + z(0, 0))

Since u(0, t)=u(0, 0)=�0, we have k(u(0, t))=k(u(0, 0)). Therefore, we can replace k(u(0, t))by k(u(0, 0)) at (13).

The right-hand side of identity (13) defines the semigroup representation of the input–outputmapping �[F] on the set of admissible source functions F:

�[F](t) := k(�0)

(−�0 + �1 + z(0, t) +

∫ t

0w(0, t − s, s) ds

)∀t ∈ [0, T ] (14)

Now differentiating both sides of identity (12) with respect to t and using the definitions (10)and (11) yield

ut (x, t) = T (t)A(u(x, 0) − �0(1 − x) − �1x) + ([k(u(x, t)) − k(u(0, 0))]ux (x, t))x

+∫ t

0AT (t − s)([k(u(x, t)) − k(u(0, 0))]ux (x, s))x ds

Using semigroup property∫ t0 AT (s)g(s) ds = T (t)g(t)−T (0)g(0) in the above equation, we obtain

ut (x, t) = T (t)g′′(x) + 2T (0)([k(u(x, t)) − k(u(0, 0))]ux (x, t))x− T (t)([k(u(x, t)) − k(u(0, 0))]ux (x, 0))x

Taking x = 0 in the above identity, we get

ut (0, t) = T (t)g′′(0) + 2T (0)([k(u(0, t)) − k(u(0, 0))]ux (0, t))x− T (t)([k(u(0, t)) − k(u(0, 0))]ux (0, t))x

Since u(0, t) =�0 we have ut (0, t) = 0. Taking into account this and substituting t = 0 yield

0= g′′(0) + ku(u(0, 0))u2x (0, 0)



Solving this equation for k′(u(0, 0)) and substituting ux (0, 0) = f (0)/k(u(0, 0)), we obtain thefollowing explicit formula for the value k′(u(0, 0)) of the first derivative k′(u(x, t)) of the unknowncoefficient:

ku(u(0, 0))=− (k(u(0, 0)))2g′′(0)( f (0))2

Here again since u(0, t) = u(0, 0) =�0, we have k(u(0, t))= k(u(0, 0)) and ku(u(0, t))=ku(u(0, 0)). Therefore we have

ku(u(0, t)) = − (k(u(0, 0)))2g′′(0)( f (0))2

Then we can redefine the admissible set of diffusion coefficients as follows:

K0 := {k ∈K : k(u(0, t))= f (0)/(−�0 + �1 + z(0, 0)),

ku(u(0, t))=−((k(u(0, 0)))2g′′(0))/( f (0))2}The following lemma implies the relation between the coefficients k1(u(x, t)), k2(u(x, t))∈K0

at x = 0 and the corresponding outputs f j (t) := k j (0)ux (0, t; k j ), j = 1, 2.

Lemma 2.1Let u1(x, t) = u(x, t; k1) and u2(x, t) = u(x, t; k2) be the solutions of the direct problem (5)corresponding to the admissible coefficients k1(u(x, t)), k2(u(x, t))∈K0. Suppose that f j (t) =k j (0)ux (0, t; k j ), j=1, 2, are the corresponding outputs, and denote by � f (t)= f1(t) − f2(t),�w(x, t, s) =w1(x, t, s)−w2(x, t, s). Then the outputs f j (t), j = 1, 2, satisfy the following integralidentity:

� f (�) = k

(�0

∫ �

0�w(0, � − s, s) ds

)ds (15)

for each � ∈ (0, T ].ProofBy using identity (13), the measured output data f j (t) := k j (u(0, 0))ux (0, t; k j ), j = 1, 2, can bewritten as follows:

f1(�) = k(�0)

(−�0 + �1 + z1(0, �) +

∫ t

0w1(0, � − s, s) ds

)

f2(�) = k(�0)

(−�0 + �1 + z2(0, �) +

∫ t

0w2(0, � − s, s) ds

)

respectively. From identity (10) it is obvious that z1(0, �) = z2(0, �) for each �∈ (0, T ]. Hence thedifference of these formulas implies the desired result. �

Corollary 2.1Let the conditions of Lemma 2.1 hold. If, in addition,

〈�n(x), �1(x, t) − �2(x, t)〉 = 0 ∀t ∈ (0, T ], n = 0, 1, . . .

hold. Then f1(t) = f2(t), ∀t ∈ [0, T ].



Since the null space of semigroup contains only zero function, i.e. N (T ) = {0}, Corollary 2.1states that f1 = f2 if and only if �1(x, t) − �2(x, t) = 0 for all (x, t) ∈ �T . From the definition of�(x, t), it implies that k1(u(x, t)) = k2(u(x, t)) for all (x, t) ∈ �T . Furthermore, this implies thatthe input–output mapping �[k] is distinguishable, i.e.

k1(u(x, t)) �= k2(u(x, t)), �[k1] �= �[k2]Theorem 2.1Let conditions (C1) and (C2) hold. Assume that �[·] :K0 →C1[0, T ] be the input–output mappingdefined by (3) and corresponding to the measured output f (t) := k(u(0, t))ux (0, t). Then themapping �[k] has the distinguishability property in the class of admissible coefficients K0, i.e.

�[k1] �=�[k2] ∀k1, k2 ∈K0, k1(u(x, t)) �= k2(u(x, t))

3. THE INVERSE PROBLEM WITH THE MEASURED DATA h(t)

Consider now the inverse problem with one measured output data h(t) at x = 1. As in the previoussection, let us arrange the parabolic equation in parabolic problem (1) as follows:

ut (x, t) − k(u(1, 0))uxx (x, t) = ([k(u(x, t)) − k(u(1, 0))]ux (x, t))x , (x, t) ∈ �T

or it can be rewritten as

ut (x, t) − (k(�1)ux (x, t))x = ([k(u(x, t)) − k(�1)]ux (x, t))x , (x, t) ∈ �T

since u(1, 0)=�1. Then the initial-boundary value problem (1) can be rewritten in the followingform:

ut (x, t) − k(u(1, 0))uxx (x, t) = ((k(u(x, t)) − k(u(1, 0)))ux (x, t))x , (x, t) ∈ �T

u(x, 0) = g(x), 0<x<1 (16)

ux (0, t) = �0, ux (1, t) = �1, 0<t<T

In order to formulate the solution of the above parabolic problem in terms of semigroup, let ususe the same function v(x, t) in identity (6) which satisfies the following parabolic problem:

vt (x, t) + B[v(x, t)] = (k(v(x, t) + �0(1 − x) + �1x) − k(�1))ux (x, t))x , (x, t) ∈ �T

v(x, 0) = g(x) − �0(1 − x) − �1x, 0<x<1

v(0, t) = 0, v(1, t) = 0, 0<t<T

(17)

Here B[v(x, t)]:=−k(�1)d2v(x, t)/dx2 is a second-order differential operator, its domain is

DB=C2(�T ).Denote by S(t) the semigroup of linear operators generated by the operator B. As mentioned

above, in order to construct semigroup S(t) we need to know the eigenvalues and eigenfunctions



of the infinitesimal generator B. Therefore, we first consider the following eigenvalue problem:

B�(x) = ��(x)

�(0) = 0, �(1)= 0

We can easily determine that the eigenvalues are �n = k(u(1, 0))n2�2 for all n = 0, 1, . . . and thecorresponding eigenfunctions are �n(x)= √

2 sin(n�x). Then the semigroup S(t) can be repre-sented in the following form:

S(t)U (x, s) =∞∑n=0

〈�n(x),U (x, s)〉 e−�nt�n(x)

Then by using above representation, we can define the null space of the semigroup S(t) of thelinear operators as follows:

N (S)={U (x, s) : 〈�n(x),U (x, s)〉= 0 for all n = 1, 2, 3, . . .}From the definition of the semigroup S(t) we can say that the null space of it consists of onlyzero function, i.e. N (S)={0}. As we mentioned in the previous section, this result plays a veryimportant role in the uniqueness of the unknown coefficient k(u(x, t)). Also, as mentioned beforeit is easy to see that the region where the variable x ranges plays a very important role in the nullspace of the semigroup S(t). For instance, if we let x range between −�/2 and �/2 then we cansee that even functions defined on this interval will be in the null space of the semigroup S(t).

The unique solution of the initial value problem (17) in terms of semigroup S(t) can berepresented in the following form:

v(x, t) = S(t)v(x, 0)+∫ t

0S(t−s)((k(v(x, t)

+�0(1−x)+�1x)−k(�1))(vx (x, t)+�1 − �0))x ds

Hence by using the identity (6), the solution u(x, t) of the parabolic problem (16) in terms ofsemigroup can be written in the following form:

u(x, t) = �0(1 − x) + �1x + S(t)(u(x, 0) − �0(1 − x) − �1x)

+∫ t

0S(t − s)((k(u(x, t)) − k(u(1, 0)))ux (x, s))x ds (18)

Defining the following:

�(x, s) = ((k(u(x, t)) − k(u(1, 0)))(ux (x, s) + 2�1 − 2�0))x

z1(x, t) =∞∑n=0

〈�n(x), �(x)〉 e−�nt�′n(x) (19)

w1(x, t, s) =∞∑n=0

〈�n(x), �(x, s)〉 e−�nt�′n(x) (20)



The solution representation of the parabolic problem (16) can be rewritten in the following form:

u(x, t) =�0(1 − x) + �1x + S(t)� +∫ t

0S(t − s)�(x, s) ds

Now differentiating both sides of the above identity with respect to x and substituting x=1 yield

ux (1, t) =−�0 + �1 + z1(1, t) +∫ t

0w1(1, t − s, s) ds

Taking into account the over-measured data k(u(1, 0))ux (1, t) = h(t), we get

h(t) = k(u(1, 0))

(−�0 + �1 + z1(1, t) +

∫ t

0w1(1, t − s, s) ds

)(21)

Now we can determine the value k(u(1, 0)). From the over-measured data k(u(1, 0))ux (1, t) =h(t), we can write k(u(1, 0))= h(t)/ux (1, t) for all t�0, which can be rewritten in terms ofsemigroup in the following form:

k(u(1, 0))= h(t)

/(−�0 + �1 + z(1, t) +

∫ t

0w(1, t − s, s) ds

)

Taking limit as t → 0 in the above identity yields

k(u(1, 0))= h(0)/(−�0 + �1 + z(1, 0))

Since u(1, t)=u(1, 0)=�1, we have k(u(1, t))=k(u(1, 0)). Therefore, we can replace k(u(1, t))by k(u(1, 0)) at (21).

The right-hand side of identity (21) defines the semigroup representation of the input–outputmapping �[k] on the set of admissible source functions F:

�[k](t) := k(�1)

(−�0 + �1 + z1(1, t) +

∫ t

0w1(1, t − s, s) ds

)∀t ∈ [0, T ] (22)

Differentiating both sides of identity (18) with respect to t yields

ut (x, t) = T (t)A(u(x, 0) − �0(1 − x) − �1x) + ([k(u(x, t)) − k(u(1, 0))]ux (x, t))x

+∫ t

0AT (t − s)([k(u(x, t)) − k(u(1, 0))]ux (x, s))x ds

Using semigroup properties, we obtain

ut (x, t) = T (t)g′′(x) + 2T (0)([k(u(x, t)) − k(u(1, 0))]ux (x, t))x− T (t)([k(u(x, t)) − k(u(1, 0))]ux (x, 0))x

Taking x = 1 in the above identity, we get

ut (1, t) = T (t)g′′(1) + 2T (0)([k(u(1, t)) − k(u(1, 0))]ux (1, t))x− T (t)([k(u(1, t)) − k(u(1, 0))]ux (1, t))x



Since u(1, t) =�1 we have ut (1, t) = 0. Taking into account this and substituting t = 0 yield

0= g′′(1) + ku(u(1, 0))u2x (1, 0)

Solving this equation for ku(u(1, 0)) and substituting ux (1, 0) = h(0)/k(u(1, 0)), we obtain thefollowing explicit formula for the value k′(u(1, 0)) of the first derivative ku(u(x, t)) of the unknowncoefficient:

ku(u(1, 0))=− (k(u(1, 0)))2g′′(1)(h(0))2

As before, since u(1, t) = u(1, 0) =�1, we have k(u(1, t))= k(u(1, 0)) and ku(u(1, t))=ku(u(1, 0)). Therefore, we have

ku(u(1, t)) = − (k(u(1, 0)))2g′′(1)(h(0))2

Then we can define the admissible set of diffusion coefficients as follows:

K1 := {k ∈K : k(u(1, t))= h(0)/(−�0 + �1 + z(1, 0)),

ku(u(1, t))=−((k(u(1, 0)))2g′′(1))/(h(0))2}

The following lemma implies the relation between the coefficients k1(u(x, t)), k2(u(x, t))∈K1at x = 1 and the corresponding outputs h j (t) := k j (1)ux (1, t; k j ), j = 1, 2.

Lemma 3.1Let u1(x, t) = u(x, t; k1) and u2(x, t) = u(x, t; k2) be solutions of the direct problem (16) cor-responding to the admissible coefficients k1(u(x, t)), k2(u(x, t))∈K1. Suppose that h j (t) =u(1, t; k j ), j=1, 2, are the corresponding outputs and denote by�h(t)=h1(t)−h2(t),�w1(x, t, s)=w11(x, t, s) − w2

1(x, t, s). Then the outputs h j (t), j = 1, 2, satisfy the following integralidentity:

�h(�) = k

(�1

∫ �

0�w1(1, � − s, s) ds

)ds (23)

for each � ∈ (0, T ].ProofBy using identity (21), the measured output data h j (t) := k j (u(1, 0))ux (1, t; k j ), j = 1, 2, can bewritten as follows:

h1(�) = k(�1)

(−�0 + �1 + z11(1, �) +

∫ t

0w11(1, � − s, s) ds

)

h2(�) = k(�1)

(−�0 + �1 + z21(1, �) +

∫ t

0w21(1, � − s, s) ds

)

respectively. From identity (19) it is obvious that z11(1, �) = z21(1, �) for each �∈ (0, T ]. Hence thedifference of these formulas implies the desired result. �



Corollary 3.1Let conditions of Lemma 3.1 hold. If, in addition,

〈�n(x), �1(x, t) − �2(x, t)〉 = 0 ∀t ∈ (0, T ], n = 0, 1, . . .

hold. Then h1(t) = h2(t), ∀t ∈ [0, T ].Since the null space of it consists of only zero function, i.e. N (S)= {0}, Corollary 3.1 states that

f1 = f2 if and only if �1(x, t) − �2(x, t) = 0 for all (x, t) ∈ �T . From the definition of �(x, t), itimplies that k1(u(x, t))= k2(u(x, t)) for all (x, t) ∈ �T . Furthermore, this implies that the input–output mapping �[k] is distinguishable, i.e.

k1(u(x, t)) �= k2(u(x, t)), �[k1] �= �[k2]Theorem 3.1Let conditions (C1) and (C2) hold. Assume that �[·] :K0 →C1[0, T ] be the input–output mappingdefined by (3) and corresponding to the measured output f (t) := k(u(0, t))ux (0, t). Then themapping �[k] has the distinguishability property in the class of admissible coefficients K0, i.e.

�[k1] �=�[k2] ∀k1, k2 ∈K0, k1(u(x, t)) �= k2(u(x, t))

4. THE INVERSE PROBLEM WITH TWO NEUMANN MEASURED DATA

Consider now the inverse problem (1)–(2) with two measured output data f (t) and h(t). As shownbefore, having these two data, the values k(u(0, 0)) as well as k(u(1, 0)) can be defined by theabove explicit formulae. Based on this result, let us define now the set of admissible coefficientsK2 as an intersection:

K2 :=K0 ∩K1

={k ∈K : k(u(0, t))= f (0)/(−�0 + �1 + z(0, 0)),

k(u(1, t)) = h(0)/(−�0 + �1 + z(1, 0)),

ku(u(0, t)) = −k(u(0, 0))g′′(0)/( f (0) + k(u(0, 0))(�0 − �1)),

ku(u(1, t)) = −k(u(1, 0))g′′(1)/(h(0) + k(u(1, 0))(�0 − �1))}

On this set both input–output mapping �[k] and �[k] have distinguishability property.

Corollary 4.1The input–output mappings �[·] :K2 →C1[0, T ] and �[·] :K2 →C1[0, T ] distinguish any twofunctions k1(u(x, t)) �= k2(u(x, t)) from the set K2, i.e.

∀k1(u(x, t)), k2(u(x, t))∈K2, k1(u(x, t)) �= k2(u(x, t)), �[k1] �= �[k2]

�[k1] �= �[k2]



5. CONCLUSION

The aim of this study was to analyse distinguishability properties of the input–output mappings�[·] :K2 →C1[0, T ] and �[·] :K2 →C1[0, T ], which are naturally determined by the measuredoutput data. In this paper we show that if the null spaces of the semigroups T (t) and S(t) includeonly zero function then the corresponding input–output mappings �[·] and �[·] are distinguishable.

This study shows that boundary conditions and the region on which the problem is defined playan important role in the distinguishability of the input–output mappings �[·] and �[·] since thesekey elements determine the structure of the semigroup T (t) of linear operators and its null space.

ACKNOWLEDGEMENTS

The research was supported in part by the Scientific and Technical Research Council (TUBITAK) ofTurkey and Izmir University of Economics.

REFERENCES

1. DuChateau P. Monotonicity and invertibility of coefficient-to-data mappings for parabolic inverse problems. SIAMJournal on Mathematical Analysis 1995; 26:1473–1487.

2. Renardy M, Rogers RC. An Introduction to Partial Differential Equations. Springer: New York, 2004.3. Showalter RE. Monotone Operators in Banach Spaces and Nonlinear Partial Differential Equations. American

Mathematical Society: Providence, RI, 1997.4. Cannon JR. The One-Dimensional Heat Equation. Encyclopedia of Mathematics and its Applications, vol. 23.

Addison-Wesley: Reading, MA, 1984.5. DuChateau P, Thelwell R, Butters G. Analysis of an adjoint problem approach to the identification of an unknown

diffusion coefficient. Inverse Problems 2004; 20:601–625.6. DuChateau P. Introduction to inverse problems in partial differential equations for engineers, physicists and

mathematicians. In Parameter Identification and Inverse Problems in Hydrology, Geology and Ecology, Gottlieb J,DuChateau P (eds). Kluwer Academic Publishers: Dordrecht, The Netherlands, 1996; 3–38.


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