- Research Article
- Open Access

# The Jump Problem for Mixed-Type Equations with Defects on the Type Change Line

- Ahmed Maher
^{1, 2}Email author

**Received:**2 January 2010**Accepted:**4 June 2010**Published:**27 June 2010

## Abstract

The jump problem and problems with defects on the type change line for model mixed-type equations in the mixed domains are investigated. The explicit solutions of the jump problem are obtained by the method of integral equations and by the Fourier transformation method. The problems with defects are reduced to singular integral equations. Some results for the solution of the equation under consideration are discussed concerning the existence and uniqueness for the solution of the suggested problem.

## Keywords

- Green Function
- Conjugation Condition
- Automorphic Function
- Cauchy Kernel
- Unknown Solution

## 1. Introduction

are the jump problems on the type change line for (1.1). Obviously, the classical Tricomi problem is the jump problem with zero jump. Two methods are used in this papre to solve the jump problem: the method of integral equations and the method of integral Fourier transformation. It is shown that explicit solutions of the jump problem can be used as potentials under researching boundary value problems with defects.

## 2. The Jump Problem for Lavrent'ev-Bitsadze Equation: The Method of Integral Equations

Let and be the elliptic and the hyperbolic parts of the mixed domain .

(here is arc abscissa of arc being measured from the point to the point ).

Assume that on the segment there is the finite number of points in which functions can have discontinuities of the first kind and functions can have singularities of integrable order. We say that such points are the exclusive points.

Denote by is the set of points of the segment which are not the exclusive ones.

(the D'Alembert formula) where . Here is a class of functions which can have singularities of integrable order at the points and , but satisfy Hoelder's condition with some index at any part of the interval . As it is known the generalized solution of the class will be regular if we assume in addition that .

In the jump problem on the type change line for Lavrent'ev-Bitsadze equation, we need to seek a function which

(1)is regular or generalized solution of (2.1) in ;

(2)satisfies the boundary conditions (2.2);

(3)has the limiting values on and

are fulfilled.

Let us obtain functional correlations at the segment which connect functions , and . The general scheme of reasoning is just the same as under solving the Tricomi problem.

Similar equation can be obtained for the function .

for analytic functions satisfying the condition (2.19). The solutions of problem (2.20) should be limited at the points and at infinity.

is simple automorphic function of group . As it is shown in [13, page 111], there is a unique opportunity to choose numbers ( being stationary points of group of the homographic transformations).

The function to be found can be obtained by differentiation, but as it will be shown later it is not obligatory.

here it is taken into account that function satisfies (2.17). Expressions of functions can be obtained from conditions (2.5).

The solution of the jump problem in the domain can be easily derived by the D'Alembert formula (2.4), and it is not necessary to seek the expression of the function for this, it is sufficient to have formula to calculate its primitive. The solution of the jump problem in the domain can be obtained by two methods: either as a solution of problem or as a solution of the Dirichlet problem.

More general statement is formulated in [2, page 30]. If the function maps the domain of the plane onto the domain of the plane and is the Green function of the Dirichlet problem for the domain , then is the Green function of the Dirichlet problem for the domain .

In the case of problem it is also possible to use the method of conformal mappings [2, Section ]. Let domains and be bounded by segment of real axis and by curves and placed in the upper half-plane. Let function map the domain onto the domain in such way that goes over into and ends of this segment remain stationary. If is the Green function of problem for the domain , then is the Green function of problem for the domain .

So the integral equation (2.11) can be transformed into equation of the form (2.17) by substitution of variables.

## 3. The Jump Problem for Lavrent'ev-Bitsadze Equation: The Method of Fourier Transformation

Let us construct the solution of the jump problem for Lavrent'ev-Bitsadze equation in the unbounded mixed domain by the method of the integral Fourier transformation.

Preliminary, we consider two auxiliary Cauchy problems in the upper and lower half-planes using some results of the works [15, 16]. We will use the following denotions: under Fourier transformation function goes over into function (distribution) .

Note that the boundary value problems in the half-space for partial differential equations have been investigated quite adequately (see, [17]). If the Cauchy problem in the half-space is overdetermined, then analysis of the algebraic equation for the Fourier transform of the unknown solution gives necessary and sufficient conditions for the boundary functions.

Equality (3.3) is the main correlation between boundary functions and .

and boundary functions can be given arbitrary.

Note that, if we pass in this formula from the Fourier transforms of the boundary functions to their prototypes we obtain the D'Alembert formula (2.4).

In the particular case under , the jump problem coincides with the Tricomi problem in the unbounded mixed domain. Without loss of generality we can assume that .

and the condition (3.3). Here are the Fourier transforms of functions being completed by zero up to the whole axis.

where is the Fourier transform of the boundary function being completed by zero up to the whole axis.

we obtain (3.17).

Equality (3.21) is the condition of the Riemann boundary value problem with discontinuous coefficient given on the real axis (this equality is being understood as the equality of distributions).

Note that solution of the jump problem in the whole plane without condition on the characteristic (3.11) is not unique but is determined within the arbitrary function.

where is a single-valued branch of the power function which is chosen in the plane with a cut along positive semiaxis of real axis and takes on the real values on the upper side of the cut.

Now we can easily obtain the expressions of the other auxiliary functions and consequently, the solution of the jump problem in the domains and .

The technique of the integral Fourier transformation can be used also in the cases when the mixed domain in the jump problem has another form.

If the elliptic part of the mixed domain is, for example, a semidisc then the Fourier transformation method can be modificated in the following way. Assume that the unknown solution of the jump problem on the semidisc is equal to zero. Continue the function to the whole upper half-plane symmetrically about , that is, in such way that values of function are equal at the points symmetrical about semidisc. Besides the solution and its normal derivative should be continuous on the semidisc. Then all formulas obtained at the beginning of the section remain valid but after substitution of variable integrals on infinite intervals can be transformed into integrals on segment. This method can be used in more general case when the elliptic part of the mixed domain is a half of the symmetrical about real axis fundamental domain of group of homographic transformations [12, Chapter III].

## 4. The Boundary Value Problems with Defect on the Line of Type Change for Lavrent'ev-Bitsadze Equation

Let the mixed domain be bounded by the line with the ends at the points and of the real axis and by characteristics and of Lavrent'ev-Bitsadze equation (2.1). Let be a set of disjoint segments placed inside the segment and let be a complement of with respect to .

We should seek the function with the following properties:

(1) satisfies (2.1) in under (classical or generalized solution);

(4) satisfies on the conjugation condition (1.4).

If the set is empty and (there are no defects), then the problem under consideration coincides with the classical Tricomi problem. If , then we have two independent boundary value problems: the Dirichlet problem for the Laplace equation in and the Goursat problem in .

Later on for simplicity we will assume that in the set there is only one segment and .

If in the problem with defect the values of the unknown solution are given on the type change line, then we say that such defect is the defect of the first kind. If on the values of the derivative of the unknown solution are given (the defect of the 2nd kind), then by the main correlation (2.9) nothing changes in fact. By the same reason the problem with defect of the 3d kind (when on the linear combination of the solution and its derivative are given) can be reduced to the problem with defect of the 1st kind. Note that defect can be considered as a cut and independent boundary conditions can be given on every side of the cut.

We will seek a solution of the problem with defect on the line of type change as a solution of the jump problem (see Section 1). Let the elliptic part of the mixed domain be a semidisc. Without loss of generality we can assume that .

is supplementary unknown constant.

is fulfilled. From the equality (4.13) the constant will be determined and so the Riemann problem (4.12) will have the unique solution.

Further operations are evident. The difference of the limiting values of the solution of the Riemann problem gives the unknown function on , by this the function will be determined and the function if it is necessary. But as it was mentioned above, it is sufficient to have only the expression of primitive of the function but not of this function itself.

If the mixed domain is unbounded it is convenient to use under solving the problem with defect on the type change line the results of Section 2 obtained by the Fourier transformation method. Depending on the kind of defect one of the auxiliary functions and will be identically equal to zero and the values of another function on the defect will remain unknown. Immediately from the formula (4.4) it is easy to get the integral equation equivalent to the problem with defect.

## Declarations

### Acknowledgment

The author wish to thank Professor N. B. Pleshchinskii at Kazan University (Russian) for his critical reading of the manuscript and his valuable comments.

## Authorsâ€™ Affiliations

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