Analysis for the identification of an unknown diffusion coefficient via semigroup approach

MATHEMATICAL METHODS IN THE APPLIED SCIENCESMath. Meth. Appl. Sci. 2009; 32:2405–2415Published online 1 April 2009 in Wiley InterScience(www.interscience.wiley.com) DOI: 10.1002/mma.1141MOS subject classification: 35R 30; 47D 06; 39K 05

Analysis for the identification of an unknown diffusioncoefficient via semigroup approach

Ali Demir1 and Ebru Ozbilge2,∗,†

1Department of Mathematics, Kocaeli University, Umuttepe, 41380 Izmit-Kocaeli, Turkey2Department of Mathematics, Faculty of Science and Literature, Izmir University of Economics, Sakarya Caddesi,

No. 156, 35330 Balcova-Izmir, Turkey

Communicated by Y. S. Xu

SUMMARY

This paper presents a semigroup approach for the mathematical analysis of the inverse coefficient problemsof identifying the unknown coefficient k(ux ) in the inhomogenenous quasi-linear parabolic equationut (x, t)=(k(ux )ux (x, t))x +F(u), with the Dirichlet boundary conditions u(0, t)=�0,u(1, t)=�1 andsource function F(u). The main purpose of this paper is to investigate the distinguishability of the input–output mappings �[·] :K→C1[0,T ],�[·] :K→C1[0,T ] via semigroup theory. Copyright q 2009 JohnWiley & Sons, Ltd.

KEY WORDS: semigroup approach; coefficient identification; quasi-linear parabolic equation

1. INTRODUCTION

Consider the following initial boundary value problem:

⎧⎪⎨⎪⎩ut (x, t)=(k(ux )ux (x, t))x +F(u), (x, t)∈�T

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

u(0, t)=�0, u(1, t)=�1, 0<t<T

(1)

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

†E-mail: [email protected]

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

Copyright q 2009 John Wiley & Sons, Ltd. Received 31 October 2008

2406 A. DEMIR AND E. OZBILGE

where �T ={(x, t)∈ R2 :0<x<1, 0<t�T }. The left and right boundary values �0,�1 areassumed to be constants. The functions c1>k(ux (x, t))�c0>0 and g(x) satisfy the followingconditions:

(C1) |kux (u1x )−kux (u2x )|<d|u1x −u2x |.(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(ux ) from one or/andtwo Dirichlet and Neumann type of measured output data at the boundaries x=0 and x=1,respectively:

ux (0, t)= f (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) are assumedto be noisy free measured output data. In this context the parabolic problem (1) will be referred asa direct ( forward) problem [1, 2], with the inputs g(x) and k(ux ). It is assumed that the functionsf (t) and h(t) belong to C1[0,T ] and satisfy the consistency conditions f (0)=g′(0),h(0)=g′(1).We denote byK :={k(ux ) :c1>k(ux (x, t))�c0>0, |kux (u1x )−kux (u2x )|<d|u1x −u2x |,u(x, t)∈

C2,1(�T )∩C2,0(�T )}, the set of admissible coefficients k=k(ux ), and introduce the input–outputmappings �[·] :K→C1[0,T ],�[·] :K→C1[0,T ], where

�[k]=ux (x, t;k)|x=0, �[k]=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 [3, 4].

The purpose of this paper is to study a distinguishability of the unknown coefficient via the aboveinput–output mappings. We say that the mapping �[·] :K→C1[0,T ] (or �[·] :K→C1[0,T ]) hasthe distinguishability property, if �[k1] �=�[k2] (�[k1] �=�[k2]) implies k1(ux ) �=k2(ux ). This, inparticular, means injectivity of the inverse mappings �−1 and �−1.

The aim of this study is to use the semigroup approach to determine the unknown diffusioncoefficient in a quasi-linear parabolic equation with a source function as well as to prove thedistinguishability of the input–output mappings. It is shown that the source function contributesto the distinguishability of these mappings F(u).

Semigroup approach, which is an effective approach to the identification of coefficients, providesfurther insight into the identifiability of unknown coefficient and exhibits how much informationcan be extracted from the measured output data, in particular in the case of constant boundaryvalues.

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.A 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.

Copyright q 2009 John Wiley & Sons, Ltd. Math. Meth. Appl. Sci. 2009; 32:2405–2415DOI: 10.1002/mma

IDENTIFICATION OF AN UNKNOWN DIFFUSION COEFFICIENT 2407

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 torepresent the solution of the parabolic problem (1) in terms of semigroup, the parabolic equationis arranged as follows:

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

As a result, the initial boundary value problem (1) can be rewritten in the following form:

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

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

u(0, t) = �0, u(1, t)=�1, 0<t<T

(5)

For the time being let us assume that k(ux (0,0)) were known, later this value will be determined.To formulate the solution of the parabolic problem (5) in terms of semigroup, we need to definea 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(vx (x, t)−�0+�1)−k(ux (0,0)))(vx (x, t)−�0+�1))x

+F(u(x, t)+�0(1−x)+�1x), (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

(7)

Here, A[·] :=−k(ux (0,0))d2[·]/dx2 is a second-order differential operator with domain DA={v(x)∈C2(0,1)∩C1[0,1] :v(0)=v(1)=0}. It is easy to see that g(x)∈DA, since the initial valuefunction g(x) belongs to C2[0,1].

Let T (t) denote the semigroup of linear operators generated by the operator −A [5, 6]. Thesemigroup T (t) can be easily constructed by using the eigenvalues and eigenfunctions of theinfinitesimal generator A. Hence we first consider the following eigenvalue problem:

A�(x) = ��(x)

�(0) = 0, �(1)=0(8)

This problem is called Sturm–Liouville problem. It is easy to determine that the eigenvaluesare �n =k(ux (0,0))n2�2 for all n=0,1, . . . and the corresponding eigenfunctions are �n(x)=√2sin(n�x). Hence the semigroup T (t) can be represented in the following way:

T (t)U (x,s)=∞∑n=0

〈�n(�),U (�,s)〉e−�nt�n(x) (9)

where 〈�n(�),U (�,s)〉=∫ 10 �n(�)U (�,s)d�. The null space of the semigroup T (t) of the linear

operators can be defined as follows:

N (T )={U (�, t) : 〈�n(�),U (�, t)〉=0, for all n=0,1,2,3, . . .}



It is well known that the Sturm–Liouville problem (8) generates a complete arthogonal familyof eigenfunctions hence the null space of the semigroup T (t) is trivial, i.e. N (T )={0}.

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(vx (x, t)−�0+�1)

−k(ux (0,0))(vx (x, t)−�0+�1))x +F(v(s, t)+�0(1−x)+�1x))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(ux (x,s))−k(ux (0,0)))ux (x,s))x +F(u(x,s))ds (10)

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

�(x) = (g(x)−�0(1−x)−�1x) (11)

�(x, t) = (k(ux (x, t))−k(ux (0,0)))(ux (x, t))x (12)

z(x, t) =∞∑n=0

〈�n(�),�(�)〉e−�nt�′n(x)

w(x, t,s) =∞∑n=0

〈�n(�),�(�,s)〉e−�nt�′n(x)

q(x, t,s) =∞∑n=0

〈�n(�),F(u(�,s))〉e−�nt�′n(x)

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)+F(u(x,s))ds

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

ux (0, t)=−�0+�1+z(0, t)+∫ t

0w(0, t−s,s)ds+

∫ t

0q(0, t−s,s)ds

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

f (t)=(

−�0+�1+z(0, t)+∫ t

0w(0, t−s,s)ds

)+

∫ t

0q(0, t−s,s)ds (13)

which implies that f (t) can be determined analytically. Substituting t=0 into this yields

f (0)=g′(0) (14)



which implies the consistency condition, which is given before. The right-hand side of identity (13)represents the semigroup representation of the input–output mapping �[k] on the set of admissiblesource functions K:

�[k](t) :=−�0+�1+z(0, t)+∫ t

0w(0, t−s,s)ds+

∫ t

0q(0, t−s,s)ds ∀t ∈[0,T ] (15)

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

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

−∫ t

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

∫ t

0AT (t−s)F(u(x,s))ds

Using semigroup property∫ t0 AT (t−s)u(x,s)ds=T (0)u(x, t)−T (t)u(x,0), we obtain:

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

Taking t=0 in the above identity, we get:

ut (x,0) = −T (0)g′′(x)+T (0)(kux (ux (x,0))uxx (x,0)ux (x,0))

+(k(ux (x,0))−(k(ux (0,0)))uxx (x,0))+T (0)F(u(x,0))

Taking x=0 in the above identity, we get:

ut (0,0)=−T (0)g′′(0)+(kux (ux (0,0))ux (0,0))uxx (0,0)+F(u(0,0))

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

0=−g′′(0)+kux (ux (0,0))�0g′′(0)+F(u(0,0))

Solving this equation for kux (ux (0,0)), we obtain the following explicit formula for the valuekux (ux (0,0)) of the first derivative kux (ux (x, t)) of the unknown coefficient:

kux (ux (0,0))=g′′(0)−F(�0)

g′′(0)�0

Now we can determine the value of k(ux (0,0)) approximately as:

k(ux (0,0))=kux (ux (0,0))ux (0,0)=f (0)(g′′(0)−F(�0))

g′′(0)�0

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

K0 :={k∈K :k(ux (0,0))= f (0)(g′′(0)−F(�0))

g′′(0)�0,kux (ux (0,0))=

g′′(0)−F(�0)

g′′(0)�0

}

The following lemma implies the relation between the coefficients k1(ux (x, t)),k2(ux (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 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(ux (x, t)),k2(ux (x, t))∈K0. Suppose thatf j (t)=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) and �q(x, t,s)=q1(x, t,s)−q2(x, t,s). Then theoutputs f j (t), j =1,2, satisfy the following integral identity:

� f ()=∫

0�w(0,−s,s)ds+

∫

0�q(0,−s,s)ds (16)

for each ∈(0,T ].ProofBy using identity (13), the measured output data f j (t) :=ux (0, t;k j ), j =1,2, can be written asfollows:

f1() = −�0+�1+z1(0,)+∫

0w1(0,−s,s)ds++

∫

0q1(0, t−s,s)ds

f2() = −�0+�1+z2(0,)+∫

0w2(0,−s,s)ds+

∫

0q2(0, t−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 1Let the conditions of Lemma 1 hold. If, in addition,

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

and

〈�n(�),F(u1(�, t))−F(u2(�, t))〉=0 ∀t ∈(0,T ], n=0,1, . . .

hold. Then f1(t)= f2(t) ∀t ∈[0,T ].Since the Sturm–Liouville problem (8) generates a complete arthogonal family of eigenfunctions,

it is obvious from Corollary 1 that the input–output mapping �[k] is distinguishable, i.e.k1(ux (x, t)) �=k2(ux (x, t)), �[k1] �=�[k2]

Theorem 1Let conditions (C1), (C2) and Fu(�0)=0 hold. Assume that �[·] :K0→C1[0,T ] be the input–output mapping defined by (3) and corresponding to the measured output f (t) :=ux (0, t). Thenthe mapping �[k] has the distinguishability property in the class of admissible coefficientsK0, i.e.

�[k1] �=�[k2] ∀k1,k2∈K0, k1(ux (x, t)) �=k2(ux (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(ux (1,0))ux (x, t))x =([k(ux (x, t))−k(ux (1,0))]ux (x, t))x +F(u), (x, t)∈�T

Hence the initial boundary value problem (1) can be rewritten in the following form:

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

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

ux (0, t) = �0, u(1, t)=�1, 0<t<T

(17)

In order to formulate the solution of the parabolic problem (17) in terms of semigroup,let us use the same function v(x, t) in identity (6), which satisfies the following parabolicproblem:

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

+F(u(x, t)+�0(1−x)+�1x), (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

(18)

Here, B[·] :=−k(ux (1,0))d2[·]/dx2 is a second-order differential operator, its domain is DB ={u∈C2(0,1)∩C1[0,1] :u(0)=u(1)=0}. It is obvious that g(x)∈DB , since the initial value functiong(x) belongs to C2[0,1].

Denote by S(t) the semigroup of linear operators generated by the operator −B [5, 6]. As it isbefore, the semigroup S(t) can be easily constructed by using the eigenvalues and eigenfunctionsof the infinitesimal generator B

S(t)U (x,s)=∞∑n=0

〈�n(�),U (�,s)〉e−�nt�n(x) (19)

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

group S(t) of the linear operators can be defined as follows:

N (S)={U (�, t) : 〈�n(�),U (�, t)〉=0 for all n=0,1,2,3, . . .}As in the previous section, we can say that the null space of it is trivial, i.e. N (S)={0}.The unique solution of the inhomogeneous initial boundary value problem (18) can be

represented in the following form:

v(x, t) = S(t)v(x,0)+∫ t

0S(t−s)((k(vx (x, t)−�0+�1)

−k(ux (1,0))(vx (x, t)−�0+�1))x +F(v(s, t)+�0(1−x)+�1x))ds



Hence by using the identity (6) and taking the initial value u(x,0)=g(x) into account, thesolution u(x, t) of the parabolic problem (17) in terms of semigroup can be written in the followingform:

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

+∫ t

0S(t−s)(((k(ux (x,s))−k(ux (1,0)))ux (x,s))x +F(u(x,s)))ds (20)

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

�(x) = (g(x)−�0(1−x)−�1x) (21)

�(x, t) = (k(ux (x, t))−k(ux (1,0)))(ux (x,s))x (22)

z(x, t) =∞∑n=0

〈�n(�),�(�)〉e−�nt�′n(x)

w(x, t,s) =∞∑n=0

〈�n(�),�(�,s)〉e−�nt�′n(x)

q(x, t,s) =∞∑n=0

〈�n(�),F(u(�,s))〉e−�nt�′n(x)

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+S(t)�(x)+∫ t

0S(t−s)�(x,s)+F(u(x,s))ds

Differentiating both sides of the above identity with respect to x and using semigroup propertiesat x=1 yields:

ux (1, t)=−�0+�1+z(1, t)+∫ t

0w(1, t−s,s)ds+

∫ t

0q(1, t−s,s)ds

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

h(t)=(

−�0+�1+z(1, t)+∫ t

0w(1, t−s,s)ds

)+

∫ t

0q(1, t−s,s)ds (23)

which implies that h(t) can be determined analytically. Substituting t=0 into this yields

h(0)=g′(1) (24)

Hence we have the consistency condition, which is given before. The right-hand side of identity (23)defines the semigroup representation of the input–output mapping �[k] on the set of admissiblesource functions K:

�[k](t) :=−�0+�1+z(1, t)+∫ t

0w(1, t−s,s)ds+

∫ t

0q(1, t−s,s)ds ∀t ∈[0,T ] (25)



Now differentiating both sides of identity (20) with respect to t and using the definitions (21) and(22) yield:

ut (x, t) = −S(t)B(u(x,0)−�1(1−x)−�1x)+([k(ux (x, t))−k(ux (1,0))]ux (x, t))x +F(u(x, t))

−∫ t

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

∫ t

0BS(t−s)F(u(x,s))ds

Using semigroup property∫ t0 BS(t−s)u(x,s)ds= S(0)u(x, t)−S(t)u(x,0), we obtain:

ut (x, t)=−S(t)g′′(x)+S(t)((k(ux (x, t))−k(ux (1,0)))ux (x,0))x +S(t)(F(u(x,0)))

Taking t=0 in the above identity, we get:

ut (x,0) = −S(0)g′′(x)+S(0)(kux (ux (x,0))uxx (x,0)ux (x,0))

+(k(ux (x,0))−(k(ux (1,0)))uxx (x,0))+S(0)F(u(x,0))

Taking x=1 in the above identity, we get:

ut (1,0)=−S(0)g′′(1)+(kux (ux (1,0))ux (1,0))uxx (1,0)+F(u(1,0))

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

0=−g′′(1)+kux (ux (1,0))�1g′′(1)+F(u(1,0))

Solving this equation for kux (ux (1,0)), we obtain the following explicit formula for the valuekux (ux (1,0)) of the first derivative kux (ux (x, t)) of the unknown coefficient:

kux (ux (1,0))=g′′(1)−F(�1)

g′′(1)�1

Now we can determine the value of k(ux (1,0)) approximately as:

k(ux (1,0))=kux (ux (1,0))ux (1,0)=h(0)(g′′(1)−F(�1))

g′′(1)�1

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

K1 :={k∈K :k(ux (1,0))= h(0)(g′′(1)−F(�1))

g′′(1)�1,kux (ux (1,0))=

g′′(1)−F(�1)

g′′(1)�1

}

The following lemma implies the relation between the coefficients k1(ux (x, t)),k2(ux (x, t))∈K1 at x=1 and the corresponding outputs h j (t) :=k j (1)ux (1, t;k j ), j =1,2.

Lemma 2Let u1(x, t)=u(x, t;k1) and u2(x, t)=u(x, t;k2) be the solutions of the direct problem (17)corresponding to the admissible coefficients k1(ux (x, t)),k2(ux (x, t))∈K0. Suppose that h j (t)=ux (1, t;k j ), j=1,2, are the corresponding outputs, and denote by �h(t)=h1(t)−h2(t),



�w(x, t,s)=w1(x, t,s)−w2(x, t,s) and �q(x, t,s)=q1(x, t,s)−q2(x, t,s). Then the outputsh j (t), j =1,2, satisfy the following integral identity:

�h()=∫

0�w(1,−s,s)ds+

∫

0�q(1,−s,s)ds (26)

for each ∈(0,T ].ProofBy using identity (23), the measured output data h j (t) :=ux (1, t;k j ), j=1,2 can be written asfollows:

h1() = −�0+�1+z1(1,)+∫

0w1(1,−s,s)ds++

∫

0q1(1,−s,s)ds

h2() = −�0+�1+z2(1,)+∫

0w2(1,−s,s)ds+

∫

0q2(1,−s,s)ds

respectively. From identity (20) it is obvious that z1(1,)= z2(1,) for each ∈(0,T ]. Hence thedifference of these formulas implies the desired result. �

Corollary 2Let the conditions of Lemma 1 hold. If, in addition,

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

and

〈�n(�),F(u1(�, t))−F(u2(�, t))〉=0 ∀t ∈(0,T ], n=0,1, . . .

As mentioned above the Sturm–Liouville problem generates a complete arthogonal family ofeigenfunctions. Therefore, it is obvious from Corollary 2 that the input–output mapping �[k] isdistinguishable, i.e.

k1(ux (x, t)) �=k2(ux (x, t)), �[k1] �=�[k2]Theorem 2Let conditions (C1), (C2) and Fu(�0)=0 hold. Assume that �[·] :K1→C1[0,T ] be the input–output mapping defined by (3) and corresponding to the measured output h(t) :=ux (1, t). Then themapping �[k] has the distinguishability property in the class of admissible coefficients K0, i.e.

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

4. THE INVERSE PROBLEM WITH DIRICHLET–NEUMANNMEASURED OUTPUT 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(ux (0,0)) as well as k(ux (1,0)) can be defined by the



above explicit formulae. Based on this result, let us define now the set of admissible coefficientsK2 as an intersection

K2 :=K0∩K1={k∈K :k(ux (0,0))= f (0)(g′′(0)−F(�0))

g′′(0)�0,kux (ux (0,0))=

g′′(0)−F(�0)

g′′(0)�0,

k(ux (1,0))= h(0)(g′′(1)−F(�1))

g′′(1)�1,kux (ux (1,0))=

g′′(1)−F(�1)

g′′(1)�1

}

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

Corollary 3The 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(ux (x, t)),k2(ux (x, t))∈K2, k1(ux (x, t)) �=k2(ux (x, t)), �[k1] �=�[k2], �[k1] �=�[k2]

5. CONCLUSION

The aim of this study was to analyze 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 study, we reach the result that having Sturm–Liouville problem, which generatesa complete arthogonal family of eigenfunctions, plays a central role in the distinguishability of theinput–output mappings.

ACKNOWLEDGEMENTS

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

REFERENCES

1. Cannon JR. The One-Dimensional Heat Equation, Encyclopedia of Mathematics and Its Applications, vol. 23.Addison Wesley: Massachusets, 1984.

2. DuChateau P. Introduction to inverse problems in partial differential equations for engineers, physicistsand mathematicians. In Parameter Identification and Inverse Problems in Hydrology, Geology and Ecology,Gottlieb J, DuChateau P (eds). Kluwer Academic Publishers: The Netherland, 1996; 3–38.

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

4. DuChateau P, Thelwell R, Butters G. Analysis of an adjoint problem approach to the identification of an unknowndiffusion coefficient. Inverse Problems 2004; 20:601–625.

5. Renardy M, Rogers RC. An Introduction to Partial Differential Equations. Springer: New York, 2004.6. Showalter RE. Monotone Operators in Banach Spaces and Nonlinear Partial Differential Equations. American

Mathematical Society: Providence, RI, 1997.


Analysis for the identification of an unknown diffusion coefficient via semigroup approach

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