Open Access

Slowly Oscillating Solutions of a Parabolic Inverse Problem: Boundary Value Problems

Boundary Value Problems20102010:471491

DOI: 10.1155/2010/471491

Received: 11 October 2010

Accepted: 20 December 2010

Published: 29 December 2010


The existence and uniqueness of a slowly oscillating solution to parabolic inverse problems for a type of boundary value problem are established. Stability of the solution is discussed.

1. Introduction

It is well known that the space of almost periodic functions and some of its generalizations have many applications (e.g., [113] and references therein). However, little has been done for to inverse problems except for our work in [1416]. Sarason in [17] studied the space of slowly oscillating functions. This is a -subalgebra of , the space of bounded, continuous, complex-valued functions on with the supremum norm . Compared with , is a quite large space (see [1720]). What we are interested in is based on the belief that certainly has a variety of applications in many mathematical areas too. In [15], we studied slowly oscillating solutions of a parabolic inverse problem for Cauchy problems. In this paper, we devote such solutions for a type of boundary value problem.

Set . Let (resp., , where ) denote the -algebra of bounded continuous complex-valued functions on (resp., ) with the supremum norm. For (resp., ) and , the translate of by is the function (resp., , ).

Definition 1.1.
  1. (1)

    A function is called slowly oscillating if for every , , the space of the functions vanishing at infinity. Denote by the set of all such functions.

  2. (2)

    A function is said to be slowly oscillating in and uniform on compact subsets of if for each and is uniformly continuous on for any compact subset . Denote by the set of all such functions. For convenience, such functions are also called uniformly slowly oscillating functions.

  3. (3)

    Let be a Banach space, and let be the space of bounded continuous functions from to . If we replace in (1) by , then we get the definition of .


As in [17], we always assume that is uniformly continuous.

The following two propositions come from [15, Section 1].

Proposition 1.2.

Let be such that is uniformly continuous on . Then .

For , suppose that for all . Define by

The following proposition shows that the composite is also slowly oscillating.

Proposition 1.3.

Let . If and for all , then .

In the sequel, we will use the notations: , . means that is slowly oscillating in and uniformly for ; means that is slowly oscillating in and uniformly on .


be the fundamental solution of the heat equation [21].

2. A Type of Boundary Value Problem

We will keep the notation in Section 1 and at the same time introduce the following new notation:

In this section, we always assume the following: , , , , , , , , and , .


be Green's function for the boundary value problems [22, 23].

The following estimates are easily obtained:

where ( ) are positive and increasing for and as .

To show the main results of this section, the following lemmas are needed. The first lemma is Lemma 3.1 on page 15 in [24].

Lemma 2.1.

Let , , and be real, continuous functions on with . If

Lemma 2.2.

Let be a continuous function on . If , , and are nondecreasing and nonnegative on and


Replacing in the two integrals of (2.6) by the expression on the right hand side in (2.6), changing the integral order of the resulting inequality and making use of the monotonicity of , and , one gets

Apply Lemma 2.1 to get the conclusion.

Lemma 2.3.

Let , , and . Then the problem
has a unique solution , and is in and satisfies

where .

One sees that depends on only and is bounded near zero.


The existence and uniqueness of the solution comes from Theorem 5.3 on page 320 in [25].

As in [22, 23], the solution can be written as

By Lemma 2.1, one gets the desired inequality.

Now we show that . As in the proofs of Lemmas 2.1 and 2.3 in [15], one gets . For , with ,
Note that
where is a constant and
By Lemma 2.1, one has

where is a constant. Since and are slowly oscillating, the right-hand sides of the inequality above approaches zero as . This means that . The proof is complete.

Consider the following problem.

Problem 1.

Find functions and such that
One sees that
It follows from (2.24) that

Let , and let . We have the following two additional problems for and , respectively.

Problem 2.

Find functions and such that

Problem 3.

Find functions and such that

Lemma 2.4.

Problems 1, 2, and 3 are equivalent to each other.


The existence and uniqueness of the solution of Problem 2 can be easily obtained from that of the solution of Problem 1. Conversely, let ( ) be the solution of Problem 2. We show that Problem 1 has a unique solution ( ). The uniqueness comes from the uniqueness of (2.19)–(2.21). For the existence, let
Obviously, and satisfies (2.22). Also satisfies (2.21) because . By (2.23) and (2.27), one sees that (2.20) is true. Finally, we show that satisfies (2.19) and therefore, along with , constitutes a solution of Problem 1. In fact,
Thus, we have shown the equivalence of Problems 1 and 2. Replacing (2.34) by the function

the equivalence of Problems 2 and 3 can be proved similarly. The proof is complete.

By Lemma 2.4, to solve Problem 1, we only need to solve Problem 3. By (2.30)–(2.32), we have the integral equation about :
Rewrite (2.33) as

where is determined by (2.37).

One can directly test that Problem 3 is equivalent to (2.37)-(2.38).

Note that for a given , Lemma 2.3 shows that (2.30)–(2.32) (or equivalently, (2.37)) have a unique solution . Thus, (2.38) does define an operator . Therefore, we only need to show that the integral (2.38) has a unique solution and . That is, has a fixed point in . Let
Set , where . If , then, by Lemma 2.3, is in , and so, by (2.38), is in with
Equation (2.37) gives the estimate
Choose such that when , one has . It follows that
Choose such that when , one has

and therefore, .

Let , . By (2.38), . Note that the function is the solution of the problem
So, by Lemma 2.3, one has

Choose such that for , . Now, set . Then is a contraction from into itself, and therefore, has a unique fixed point. Thus, we have shown.

Theorem 2.5.

Let functions , , , and be as above. Then, for small , Problem 3 has a unique solution ( ) in with and .

Let be the solutions of Problem 3 in for the functions , , , and . Set , , , and . For the stability of the solution, we have the following.

Theorem 2.6.

For , one has

where depends on , , , , , , , and .


By (2.33),
Note that the function is the solution of the problem
Using a formula similar to (2.37) and Lemma 2.2 for the function , one gets
Applying Lemma 2.2 and (2.48), one gets the desired conclusion with
and is majorant of . One can specially assume that

The proof is complete.

Corollary 2.7.

Under the conditions in Theorem 2.6, the solution of Problem 3 is unique.



The research is supported by the NSF of China (no. 11071048).

Authors’ Affiliations

Department of Mathematics, Harbin Institute of Technology


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© F. Yang and C. Zhang. 2010

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