Lobachevskii Journal of Mathematics http://ljm.ksu.ru Vol. 18, 2005, 151 – 175

© Ye.A. Utkina

ABSTRACT. The objective of this article is to construct in a hyper-rectangular region of the 4-dimensional Euclidean space a solution of the Goursat problem for the following equation:

Introduction. Let $D=\left\{x_{10}<x_1<x_{11},x_{20}<x_2<x_{21},x_{30}<x_3<x_{31},x_{40}<x_4<x_{41}\right\}$, where $X_1$, $X_2$, $X_3$, $X_4$ are the faces of $D$ for $x_i=x_{i0}$ (i=1,2,3,4), respectively. The objective of this article is to construct in $D$ a solution of the Goursat problem for the following equation:

where $a_{m_1{m}_2{m}_3{m}_4}\equiv 1$, while the smoothness of remaining coefficients is defined by the inclusions

Here $C^{{\alpha }_1+{\alpha }_2+{\alpha }_3+{\alpha }_4}$ stands for a class of functions continuous in $\overline{D}$ together with their partial derivatives

The Goursat problem: Determine in $D$ a regular solution of the equation (1), which satisfies the following conditions:

where

and the boundary values in (3) are coordinated on the edges of $D$ as follows:

and the coordinated values are supposed to be differentiable.

Equation (1) can be treated as an analog of the two-dimensional pseudo-parabolic equation in [1]. However, a version of the Riemann’s method to be applied here differs essentially from that used in [1]. One can note that equations considered in this paper have modifications frequently used in applications. One can consider, for example, the Boussinesq-Love equation in the theory of oscillations (see [1], formula (20)), the Aller equation in [2], p.261, applied in the mathematical simulation of moisture absorption by roots of plants.

1. First of all we consider a special case, where the derivatives with respect to the variables $x_2,x_3,x_4$ are not present:

Here and on in this section we denote by $y$ the set of coordinates $x_2$, $x_3$, $x_4$. One item of the investigation of (4) will be also used in the general case (see Section 2). In addition, it seems to be of interest to compare the investigation of (4) with the study of the following ordinary differential equation:

Obviously, (4) and (5) possess a similar structure; however, the general theory of the second case (see [3], p.145) depends essentially on the theorem stating that the Wronskian $m$ of its solutions either does not vanish or is identically equal to zero. However, this theorem cannot be generalized to equation (1): one can easily see, for example, that the equation $u_{xx}-3y{u}_x+2y^{2}u=0$ possesses the solutions $exp\left(xy\right)$, $exp\left(2xy\right)$. At the same time, its Wronskian $w=y\cdot exp\left(3xy\right)$ vanishes for $y=0$ and differs from zero in any other point of a bounded domain. Therefore there arises a problem of construction of a theory for the solutions of (4), which will not use the above property of the Wronskian. In what follows we develop a version of the Riemann method in [4], [5].

The Goursat problem can be transformed to the following well-known Cauchy problem: find in $D$ a solution of equation (1) in the class $C^{m+0}$, which will satisfy the following conditions:

We assume that the coefficients $a_i$ are continuous in $\overline{D}$ with respect to the variables $x_2$, $x_3$, $x_4$ and belong to the class $C^i\left(\overline{D}\right)$ on x${}_1$ . We define the Riemann function $V$ as a solution of the following Volterra integral equation:

which is known to exist and be unique. In addition, from (7) it follows that $V$ is a solution of the equation conjugate to (4):

To indicate that $V$ depends not only on $x_1$, $y$ but also on $\xi$, we will write this function as $R\left(x_1,y,\xi \right)$. The next identities (which will be of use later) easily follow from (7):

The following identity takes place for any function $u$ from the class $C^{m+0}\left(D\right)$:

where $a_i$ depend on ($x_1$, $y$), while $R$ and its derivatives depend on $\left(x_1,y,\xi \right)$. In this situation, $K_{ib}=\sum _{\alpha =b}^i{\left(-1\right)}^{i-\alpha }{C}_{\alpha }^b-M_{ib}$, $M_{ib}=1$ if $b=i$, and $M_{ib}=0$ otherwise. To prove (10) rewrite this identity as follows:

In the right-hand side the first and last addends cancel each other. Transfer the addend from the left-hand side to the right-hand side and add to the second term:

By using the equality

we see that identity (10) holds.

To construct the solution of the problem we represent (10) in another way:

In order to verify that (11) is valid, one can use the following equality:

The latter takes place if $b$ is odd and $i$ is even. If both $i$ and $b$ are odd, then the right-hand side of (12) can be replaced with the following expression

For $b$ even, instead of $2k+1$ one must take $2k$. Relations (12) and (13) can be verified straightforwardly. By changing the summation order in (12), we obtain the last addend in (11). Now, by taking in (11) $x_1$ equal to $\xi$ and denoting $x_1$ by $\xi$, we calculate the integral in $x_{10}<\xi <x_{11}$ and take into account identities (9). We then arrive at the following:

Obtain the required solution by integrating (14) $m-1$ times with respect to $x_1$ between the limits $x_{10}$ to $x_1$ and taking into account (6):

As in [6], p. 66, we can assume that for arbitrary ${\varphi }_i\left(y\right)$ the formula (15) supplies the general presentation of the solutions of equation (4).

With $u=u\left(x\right)$ and $a_i=a_i\left(x\right)$ equation (4) turns into an ordinary differential equation (5). Obviously, formula (15) represents a solution for (5), moreover, here ${\varphi }_i\left(y\right)$ are constants and both the coefficients $a_i$ and the Riemann function do not depend on $y$. Thus, formula (15) contains the classical result; moreover, it is written there in a closed form which means that the classical theory indicates a method for solving, while such a solution is represented here by a formula.

Now assume $a_i=a_i\left(y\right)$, i.e., here we deal with a certain generalization of ordinary differential equations with constant coefficients. We will consider only a particular case with $m=2$. The formula (15) then takes the form

Find the Riemann function whose equation in this case is as follows:

where

and besides $H\left(x,t,y\right)=a_1\left(y\right)-\left(x-t\right)a_0\left(y\right)$. We thus have

Put $a_1\left(y\right)={\lambda }_1\left(y\right)+{\lambda }_2\left(y\right)$, $a_0\left(y\right)={\lambda }_1\left(y\right)\cdot {\lambda }_2\left(y\right)$. For the case with ${\lambda }_1\left(y\right)\ne {\lambda }_2\left(y\right)$, we obtain

where ${\lambda }_{1;2}\left(y\right)=\frac{a_1\left(y\right)\mp \sqrt{a_1^2\left(y\right)-4a_0\left(y\right)}}{2}$. By interchanging $x$ and $x_0$, we obtain

Substitute the result into (16). One can easily see that

where

Now set ${\lambda }_1\left(y\right)={\lambda }_2\left(y\right)$. Then the Riemann function is

We again interchange $x$ and $x_0$ and calculate the respective terms. We thus obtain

where

Let us write the solution in the case of conjugate complex roots ${\lambda }_1\left(y\right)$, ${\lambda }_2\left(y\right)$ of the characteristic equation. The roots can be represented as follows

where $\beta \cdot i=\sqrt{a_1^2\left(y\right)-4a_0\left(y\right)}$, $\alpha =a_1\left(y\right)$.

In this case, by the straightforward calculation yields:

Consequently, in contrast to the classical technique for solving ordinary differential equations with constant coefficients, the formula obtained contains simultaneously both the case of a non-multiple root of the characteristic equation and that of a multiple root, as well as the case of conjugate roots.

Consider also a non-homogeneous equation

Its solution has the form

Denote by $K\left(t,y,x\right)$ $\frac{{\left(x-t\right)}^{m-1}}{\left(m-1\right)!}R\left(t,y,x\right)$. One can then easily verify that we get a particular solution obtained in ([7], p. 121), by the Cauchy method. Since $K\left(t,y,t\right)=K_x^{\prime }\left(t,y,t\right)=...=K_x^{\left(m-2\right)}\left(t,y,t\right)=0$, $K_x^{\left(m-1\right)}\left(t,y,t\right)=1$ and $u_{\ast }\left(x,y\right)=\underset{x_0}{\overset{x}{\int }}K\left(t,y,x\right)f\left(t,y\right)dt$ is a particular solution of the equation, satisfying the zero initial conditions ${\varphi }_0\left(y\right)\equiv {\varphi }_1\left(y\right)\equiv ...\equiv {\varphi }_{n-1}\left(y\right)\equiv 0$. In addition, $K\left(t,y,x\right)$ is written in the explicit form.

2. In this section us consider a general case. For the sake of simplicity we use following notation introduced in [1]: $D_t^k\varphi \equiv {\partial }^k\varphi ∕\partial t^k$ for $k=1,2,..$ and $D_{t_0}^k\varphi \equiv \underset{t_0}{\overset{t}{\int }}\frac{{\left(t-\tau \right)}^{-k-1}\varphi \left(\tau \right)}{{\left(-k-1\right)}^{}!}d\tau$ if $k=-1,-2,..$ and $D_t^0$ being operator of identical transform.

By the Riemann function $R\left(x_1,x_2,x_3,x_4;{\xi }_1,{\xi }_2,{\xi }_3,{\xi }_4\right)$ we call any solution of the integral equation

which both exists and is unique (see [8], p.180). From (17) it follows that, if we consider the first four arguments of the equation, then $R$ is a solution of the following equation conjugate to (1):

In what follows we need the next proposition.

For any function $u$ in $C^{\sum _{\alpha =1}^4{m}_{\alpha }}\left(D\right)$, the following identity holds: