We deduce formulas for the Fréchet derivatives of cost functionals of several inverse problems for a parabolic integrodifferential equation in a weak formulation. The method consists in the application of an integrated convolutional form of the weak problem and all computations are implemented in regular Sobolev spaces.

1. Introduction

Many methods to solve inverse problems (e.g., the Landweber iteration, conjugate gradient method) use the Fréchet derivatives of the cost functionals of these problems [1]. The explicit formula for the Fréchet derivative in terms of the variation of the unknowns of the inverse problem contains the solution of an adjoint problem.

The derivation of the explicit formula for such a Fréchet derivative includes testing the direct problem with the solution of the adjoint problem and vice versa: testing the adjoint problem with the solution of the direct problem. In the case of the parabolic weak problem, such a procedure is cumbersome, because of the asymmetry of the properties of the solution and the test function. In the classical formulation of the parabolic weak problem (see, e.g., [2] and also (19) below), the test function must have higher time regularity than the weak solution. This means that in case of nonsmooth coefficients neither the solution of the direct problem nor the solution of the adjoint problem can be used as a test function. Another formulation of the weak parabolic problem consists in reducing the problem to an abstract Cauchy problem over the time variable (see, e.g., [3]). In such a case, a partial integration over the time has to be implemented within singular distributions in the derivation procedure.

In this paper, we present a new method that enables the deduction of the formulas for the Fréchet derivatives for cost functionals of inverse problems related to weak solutions of parabolic problems. The method is based on an integrated convolutional form of the weak direct problem. The requirements to the test function are weaker than in the classical case and coincide with the properties of the solution of the direct problem. All computations in the deduction procedure can be implemented within usual regular Sobolev spaces.

More precisely, we will consider inverse problems related to a parabolic integrodifferential equation that occur in heat flow with memory [46]. This equation contains a time convolution. Therefore, the convolutional form of the weak problem is especially suitable. Supposedly, the proposed method can be generalised to parabolic systems, as well.

2. Formal Direct Problem: Notation

Let be an -dimensional domain, where , and be the boundary of . Let where either or is allowed to be an empty set. In case , we assume that is sufficiently smooth, meas , and for any it holds either or meas . Denote for . Consider the problem (direct problem) to find such that where is a fixed number, , , , , , , are given functions, the subscripts ,  ,   denote the partial derivatives and denotes the time convolution. In case (), the boundary condition (4) and (5) is dropped.

The problem (2)–(5) describes the heat flow in the body with the thermal memory. Concerning the physical background, we refer the reader to [4, 6, 7]. The solution is the temperature of the body and is the heat flux relaxation (or memory) kernel. The boundary condition (5) is of the third kind where the term equals the heat flux in the direction of the conormal vector.

Let us introduce some additional notations. Let . We use the Sobolev spaces Here, , is the multiindex, and . Further, let be a Banach space. We denote by the space of abstract continuous functions from to endowed with the usual maximum norm . Moreover, let By means of these spaces, we define the following important functional spaces:

Convention. In case , the integrals , are equal to , where and is the number of points in , and is simply .

3. Weak Direct Problem and Its Convolutional Form

Let us return to the direct problem (2)–(5). Throughout the paper we assume the following basic regularity conditions on the coefficients, the kernel, and the initial and boundary functions: and the ellipticity condition (for the sake of simplicity we introduced an assumption for the extension of onto ).

The first aim is to reformulate the problem (2)–(5) in a weak form. Let us suppose that , , and (2)–(5) has a classical solution . Then, we multiply (2) with a test function from the space and integrate by parts with respect to time and space variables. We obtain the following relation: This relation makes sense also in a more general case when , satisfies only (11) and (15) and does not have regular first-order time and second-order spatial derivatives.

We call a weak solution of the problem (2)–(5) a function from the space that satisfies the relation (19) for any and in case fulfills the boundary condition (4).

Lemma 1. The following assertions are valid: (i) where if , if and if , where and are given in (12) and (14), respectively;(ii)for any it holds and , where is a constant.

Proof. Since , the assertion (i) follows from the continuous embedding of in . The assertion (ii) can be proved by means of Hölder's inequality.

Theorem 2. The problem (2)–(5) has a unique weak solution. This solution satisfies the estimate where in case , in case and is a constant depending on , , , , and .

Proof. The assertion of the theorem in case is well known from the theory of parabolic equations (see, e.g., [2]). Let be the operator that assigns to the data vector the weak solution of the problem (2)–(5) in case . Then it holds where is the right-hand side of (20).
Further, let us formulate the problem for the difference . Introducing the linear operator by the formula the weak problem (2)–(5) for the solution equivalent to the following operator equation for the quantity : We have to estimate . For this purpose, we firstly prove the following auxiliary inequality: for any and .
Denoting , , making use of the following property of the Bochner integral: for functions and the Cauchy's inequality, the relation (24) can be deduced by means of the following computations:
Next, let and introduce the operator Due to the causality we have for any . Using these relations, the continuity of the linear operator , the inequality (24), and the boundedness of , we compute the following: with some constants and depending on , , , , . Using Lemma 1, we obtain Using this relation in (27), we arrive at the following basic estimate for : where is a constant depending on , , , , . Let us define the weighted norms in : where . The estimate (29) implies the further estimate Since as , there exists , depending on and , such that . Thus, . The operator is a contraction in . This implies the existence and uniqueness assertions of the theorem.
To prove the estimate (20), we firstly deduce from (23) the inequality . This implies . Using the equivalence relations , where and (21), we reach (20).

We note the upper integration bound in (19) can be released to be any number from the interval . Indeed, (19) is equivalent to the following problem: for any . This assertion can be proved using the standard technique defining the test function as follows: and letting the parameter to approach .

Next we transform the weak direct problem (31) to a form that does not contain a time derivative of the test function . This form enables the extension of the test space. This is useful for treatment of problems for adjoint states of quasisolutions of inverse problems in next sections.

Theorem 3. The function satisfies the relation (19) for any if and only if it satisfies the following relation: for any .

Here, according to the definition of the time convolution in the previous section, .

Proof. It is sufficient to prove that satisfies (31) for any if and only if it satisfies (33) for any . Suppose that satisfies (31) and choose an arbitrary . Let be an arbitrary number on the interval and choose some function such that the relation is valid. For instance, it is possible to define as a periodic function with respect to , that is, for , for , for and so on. Using the relation (31) with replaced by and setting there we obtain the equality where Note that the time derivative of can be removed from by integration. Indeed, let . Then Changing the order of the integrals over and in the last term, we easily obtain Integrating now the whole equality (35) over from to , observing (37) and (39), and finally redenoting by , we reach the desired relation (33). Summing up, we have proved that (33) holds for any . But all terms in the right-hand side of (33) are well defined for , too. Since is densely embedded in , we conclude that (33) holds for any .
It remains to show that (33) implies (31). Suppose that satisfies (33) and choose an arbitrary and . Again, let be a function from such (34) is valid. Inserting instead of into (33), differentiating with respect to and setting we come to the relation (31). Theorem is proved.

Corollary 4. A function is a weak solution of (2)–(5) if and only if it satisfies the relation (33) for any and in case fulfills the boundary condition (4).

4. Inverse Problems and Quasisolutions

In the sequel, let us pose some inverse problems for the weak solution of (2)–(5). These problems are selected in order to demonstrate the wide possibilities of the method that we will introduce in Section 5.

Firstly, we suppose that (2)–(5) has the following specific form: where is unknown. The coefficients and other given functions , , , , are assumed to satisfy (11)–(17). Moreover, is prescribed.

IP1. Find the vector such that the weak solution of (40) satisfies the following instant additional conditions: where and , are given functions (observations of ).

Since for , the weak solution of (40) exists in ; hence, it has traces , . In practice, the term may represent an approximation of a more general function , where , form a basis in .

Further, let also be unknown.

IP2. Find the vector and such that the weak solution of (40) satisfies the following integral additional conditions: where , are given observation functions and , are given weights that satisfy the following condition: Note that the integral in (42) belongs to for any and . Indeed, for such and it holds , which implies In practice, the weights are usually concentrated in neighborhoods of some fixed values of time .

Finally, let us pose a nonlinear inverse problem for the coefficient and the kernel . Assume that . Then any coefficient that belongs to satisfies (12). Moreover, let us set if and . The other coefficients and the given functions , , , , are assumed to satisfy (11)–(17).

IP3. Find and such that the weak solution of (2)–(5) satisfies the following integral additional conditions: where ,    are given observation functions and is a given weight function such that .

As in IP1, we can show that the trace belongs to . Moreover, using the property , the embedding of in and Hölder’s inequality, one can immediately check that the term in (45) belongs to .

Available existence, uniqueness, and stability results for IP1–IP3 require stronger smoothness of the data than imposed in the present paper. Let us cite some of these results.

In case , the well posedness of IP1 was proved in [8]. Partial results were deduced earlier in [9]. A more general problem involving both IP1 and IP2 without the unknown in case was studied in [10] by means of different techniques. IP1 and IP2 in case and were treated in many papers, for example, [1114]. The case is open even if . Inverse problems to determine with given were studied in a number of papers, for example, [7, 1523]. The problem for with given was treated in [8].

In the present paper, we will deal with quasisolutions of IP1–IP3 and related cost functionals. Denote . Let . The quasi-solution of IP1 in the set is an element , where is the following cost functional and is the solution of (40) that corresponds to a fixed element .

Similarly, let . The quasi-solution of IP2 in the set is , where is the cost functional and is the weak solution of (40) that corresponds to a given vector .

Finally, defining , the quasi-solution of IP3 in is an element , where is the cost functional and is the weak solution of the direct problem (2)–(5) corresponding to given . Here, we restricted the space for the unknown to , because we will seek for the Fréchet derivative of in a Hilbert space. Moreover, the kernel of the second addend corresponding to in the representation formula of (90) is an element of and in general does not belong to the adjoint space .

According to the above-mentioned arguments, the functionals , , are well-defined in , , and , respectively.

5. The Fréchet Derivatives of Cost Functionals of Inverse Problems

5.1. General Procedure

Suppose that the solution of the direct problem depends on a vector of parameters that has to be determined in an inverse problem making use of certain measurements of . Let the quasi-solution of the inverse problem be sought by a method involving the Fréchet derivative of the cost functional (i.e., some gradient-type method). Usually in practice, a solution of a proper adjoint problem is used to represent the Fréchet derivative.

We introduce a general procedure to deduce such adjoint problems. Assume that is the difference of solutions of the direct problem corresponding to a difference of the vector of the parameters . More precisely, we suppose that is a solution of the following problem: with some data , , , depending on . We restrict ourselves to the case when the Dirichlet boundary condition of the state is independent of . Therefore, the condition (51) for is homogeneous.

In practice, the adjoint parabolic problems are usually formulated as backward problems. In our context, it is better to pose adjoint problems in the forward form. The involved memory term with is defined via a forward convolution and from the practical viewpoint, it is preferable to have the direct and adjoint problems in a similar form (e.g., to simplify parallelisation of computations).

More precisely, let the adjoint state be a solution of the following problem: where , , , and are some data depending on and the cost functional under consideration.

Assume that the quadruplets , , , , and , , , satisfy the conditions (14)–(16). Then, due to Theorem 2, the problems (49)–(52) and (53) have unique weak solutions in the space . Actually, we have because of the homogeneous boundary conditions on .

Let us write the relation (33) for and use the test function . Then we obtain for any Secondly, let us write this relation for and use the test function . Then we have for any Subtracting (54) from (55), using the commutativity of the convolution, the symmetricity relations and differentiating with respect to , we arrive at the following basic equality that can be used in various inverse problems:

5.2. Derivative of

Theorem 5. The functional is the Fréchet differentiable in and where , , are the unique -dependent weak solutions of the following problems: .

Proof. Let us fix some . One can immediately check that it holds where is the weak solution of the following problem: Applying the estimate (20) to the solution of this problem we deduce the following estimate for the second term in the right-hand side of (59): with some constant . This implies that is the Fréchet differentiable and the first term in the right-hand side of (59) represents the Fréchet derivative, that is,
Further, let us use the method presented in Section 5.1 to deduce the proper adjoint problems. Comparing (60) with (49)–(52) we see that , . Therefore, the relation (56) has the form In order to deduce a formula for the component in the quantity , we set , and in (63). Then we immediately have where according to (53) and the definition of , the function is the weak solution of the problem (58) in the domain instead of . Due to Theorem 2, this problem has a unique solution. From (62) and (64) we obtain (57). The latter formula contains the values of in . Therefore, we can restrict the problem (58) from to .

To use the formula (57) one has to solve weak problems for the functions in domains . In the following theorem, we will show that computational work related to the evaluation of the Fréchet derivative can be considerably reduced. Actually, it is sufficient to solve weak problems in the smaller domains , . Here, .

Theorem 6. The Fréchet derivative of the functional can also be written in the form where are the unique -dependent weak solutions of the following sequence of recursive problems in the domains : where . Here, and the function and the vector are defined via as follows: , and , for .

Proof. Firstly, let us check that (66) indeed have unique weak solutions in . To this end we can use Theorem 2. For the problem this is immediate, because the initial condition of the problem for belongs to and other equations in this problem are homogeneous. Further, we use the induction. Choose some in the range and suppose that for all such that . The aim is to us to show that then the problem for has a unique weak solution in . Let us represent the th addend in (68) in the form For any in the range we have where for , for and . Since and , we have and . Due to the Young’s theorem for convolutions, we get . Therefore, . This implies that belongs to . From the latter relation and we immediately have . Using the embedding theorem and Lemma 1 we see that satisfies the property (14). Finally, the initial condition belongs to , because ,  . All assumptions of Theorem 2 are satisfied for the problem for . Consequently, it possesses a unique weak solution in .
Secondly, let us define the functions where and are the solutions of (58). We are going to show that , . From the definition of using the value of and simple computations, we immediately get Let us fix and choose some . We continue by the formulae for and for . Further, let us define where . By the definition, it holds .
Let us write down the weak form (31) for the problem for (58) with the test function . We fix some and compute the difference of this weak problem with replaced by and replaced by and take the sum over . This results in the following expression: where Using the definitions of and and the formula (72), we have Similarly, using the definitions of and and changing the variable of integration in , we deduce By the change of variable, the quantity is transformed to Let us consider the term in the latter formula. We compute