/ / Article

Review Article | Open Access

Volume 2009 |Article ID 182371 | 15 pages | https://doi.org/10.1155/2009/182371

# Well-Posedness of the Cauchy Problem for Hyperbolic Equations with Non-Lipschitz Coefficients

Accepted16 May 2009
Published26 Aug 2009

#### Abstract

We consider hyperbolic equations with anisotropic elliptic part and some non-Lipschitz coefficients. We prove well-posedness of the corresponding Cauchy problem in some functional spaces. These functional spaces have finite smoothness with respect to variables corresponding to regular coefficients and infinite smoothness with respect to variables corresponding to singular coefficients.

#### 1. Introduction

Let us consider the Cauchy problem for a second-order hyperbolic equation: where the matrix is real and symmetric for all , .

Suppose that (1.1) is strictly hyperbolic, that is, there exists such that for all .

It is known that if satisfies the Lipschitz condition and , then for any the problem (1.1), (1.2) has a unique solution where (see [1, Chapter ] and [2, Chapter ]).

If we reject the Lipschitz condition, this result, generally speaking, stops to be valid (see ).

In the paper  it is proved that if that is, if satisfies the logarithmic Lipschitz condition: where monotonically decreasing tends to zero, and tends to infinity, then there exists such that, for all , the problem (1.1), (1.2) has a unique solution (this behavior goes under the name of loss of derivatives).

In the paper  it is considered the case when , a part of coefficients belongs to the class and another part of coefficients satisfies the Lipschitz condition. It is proved that the loss of derivatives occurs in those variables for which appropriate coefficient belongs to the class .

It is interesting to investigate the Cauchy problem for (1.1), with singular coefficients. Many interesting results have been obtained in this direction. For example, in the paper  it is supposed that for each and where , . It is proved that if , the problem (1.1), (1.2) is well-posed in . If and where , then the problem (1.1), (1.2) is well-posed in the Geverey class , (see ). If the coefficients satisfy only Holder conditions of order then in  it is established that the problem (1.1), (1.2) is well-posed for all . In this direction see also the results obtained in the papers .

In this paper we consider the Cauchy problem for a higher-order hyperbolic equation with anisotropic elliptic part: where .

Here the coefficients satisfy different conditions of type (1.6) and (1.7), so that and corresponding to different are different. The smoothness of the solution depending on smoothness on initial data with respect to each variable depends not only on but also on and .

#### 2. Statement of the Problem and Results

We considered the Cauchy problem (1.8). Suppose that and satisfy the following conditions:

In order to formulate the basic results we introduce some denotation. Let be some Hilbert space. By we will denote a functional space with the norm where , and is a Fourier transformation with respect to variable .

For by we will denote a functional space with the norm

Denote ,

If then , , and , where is the Geverey space of order (see [12, 13]). If then is Hilbert-valued anisotropic Sobolev space . For the read valued functions the anisotropic Sobolev spaces are stated in . The basic results led in  are also valid for abstract-valued functions.

We introduce also the following denotation:

The main results are the following theorems.

Theorem 2.1. Let the conditions (2.1)–(2.3) be satisfied, where
Then for any the energy estimates hold, where and are some constants indepent of ,

Theorem 2.2. Let the conditions (2.1)–(2.3) be satisfied, where Additionally, let the conditions be satisfied, where . Then for any , and the energy estimates, hold, where and are some constants independent of ,

Remark 2.3. It is clear by our notation that and we can write

Remark 2.4. It is possible to replace the conditions and (2.8) or (2.12) by Lipschitz conditions.

The following theorems are obtained from Theorems 2.1 and 2.2.

Theorem 2.5. Let condition (2.1)–(2.9) be satisfied. Then for any , the problem (1.1), (1.2) admits a unique solution

Theorem 2.6. Let conditions (2.1)–(2.3) and (2.12)–(2.14) be satisfied. Then for any , the problem (1.1), (1.2) admits a unique solution

In particular it follows from Theorem 2.1 that if the conditions (2.1)–(2.3) are satisfied, then the problem (1.1), (1.2) is well-posed in , and if the conditions (2.1)–(2.3) and (2.12)–(2.14) are satisfied then the problem (1.1), (1.2) is well-posed in the Geverey class .

#### 3. Proof of Theorems

At first we reduce some auxiliary statements.

We denote and define the weighted energetic function in the following way: where

The following auxiliary lemmas are proved similar to the paper . The proofs of the lemmas are in appendix.

Lemma 3.1. If , then there exits such , that
If , then there exits such , that

Lemma 3.2. If , then there exits such constant , that .
If then there exits such , that

By the definition of we have

On the other hand from (1.8) we have where .

From (3.6) and (3.7) we obtain

If then by definition of and we have

By our supposition for. Therefore we can easily see that with some constant .

Using the Cauchy inequality, definition of , , and we have

where .

From (3.10)–(3.13) we get that when , then there exists such a constant , that If then by definition of and from (3.9) we have that On the other hand From (3.13), (3.15), and (3.16) we again get inequality (3.14).

It follows from (3.14) that where .

Proof of Theorem 2.1. Let , then . From Lemmas 3.1 and 3.2 we have where , .
Thus,
On the other hand from the definition of and we have It follows from (3.17)–(3.20) that

Proof of Theorem 2.2. Let , then . Taking into account Lemmas 3.1 and 3.2 and Theorem 2.5 we have
Further using the inequality we obtain where .
On the other hand from the definition of and we have
From inequalities (3.17), (3.23), and (3.24) it follows that

Proof of Theorem 2.5. For any the problem (3.7), (3.8) has a unique solution (see [15, Chapter I]).
Let , , then for any , where is some constant dependent on and .
Taking into account Theorem 2.1 it follows from (3.20) that that is, It follows from (3.28) that
By the expression of it follows that the function is the solution of problem (1.8).
The uniqueness of the solution is proved by standard method.

The proof of Theorem 2.6 is carried out in the similar way.

#### A. Proof of Lemmas

Proof of Lemma 3.1. Let . Then from (2.2) we have It follows from (2.1) and (2.14) that
By definition of for we have If , and , then from (A.1) we have If , then using (A.1) we get Consequently if , the statement of the lemma follows from (A.2)–(A.5).
Let . By definition of for we have
If and then
If then
Thus if then the statement of the lemma follows from (A.2), (A.6), and (A.8).
The lemma is proved.

Proof of Lemma 3.2. At first we consider the case when . If then where .
If then
Now let us consider the case In this case . If then If then
As and it follows that and . Then according to the Young inequality there exists such that Thus, by (A.9)–(A.13) the following inequality is valid: where .

#### B. Example

Let us consider the Cauchy problem in : where .

It follows from Theorem 2.5 that the problem (B.1) has a unique solution

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