MATHEMATICS

Corresponding Member of the Academy of Sciences of the USSR A. V. BITSADZE

ILL-POSEDNESS OF THE DIRICHLET PROBLEM FOR EQUATIONS OF MIXED TYPE IN MIXED DOMAINS

As a model equation of mixed type we take the equation of M. A. Lavrent’ev

[
U_{xx}+\operatorname{sgn} y\cdot U_{yy}=0.
\tag{1}
]

Denote by (D) a simply connected mixed domain bounded by a smooth Jordan arc (\sigma), lying in the upper half-plane and with endpoints at the points (A(0,0)), (B(1,0)), and by curves of continuous curvature issuing from these points,
[
L:\ y=-\gamma(x),\qquad L_1:\ y=-\gamma_1(x),
]
satisfying the conditions

[
\gamma>0,\qquad \gamma_1>0,\qquad 0<\gamma'(x)<1,\qquad 0<-\gamma'_1(x)<1.
\tag{2}
]

Let (C[x_1,-\gamma(x_1)]) be the point of intersection of the curves (L) and (L_1). From the point (E(x_0,0)), where (x_1-\gamma(x_1)\le x_0\le x_1+\gamma(x_1)), draw the characteristics (EB_2:\ y=x-x_0) and (EB_1:\ y=x_0-x), where (B_2) and (B_1) are the points of intersection of the indicated characteristics with the curves (L) and (L_1). Denote by (L_2) and (L_3) the arcs (AB_2) and (BB_1) of the curves (L) and (L_1), respectively.

For equation (1) in the domain (D) the Dirichlet problem is ill-posed independently of the size and form of the hyperbolic part of the domain (D).

This follows from the fact that the following mixed problem always has, and moreover has uniquely, a solution.

It is required to find a function (U(x,y)) with the following properties: 1) (U(x,y)) is a solution of equation (1) in the domain (D) for (y\ne0,\ y\ne x-x_0,\ y\ne x_0-x); 2) it is continuous in the closed domain (\overline D); 3) the partial derivatives (U_x) and (U_y) are continuous in the closed domain (D) everywhere except, possibly, at the points (A,B) and on the segments (EB_2, EB_1), near which they may become infinite of order less than (1/2); 4) on the lines (\sigma, L_2), and (L_3) the function (U(x,y)) assumes the prescribed values

[
U\big|{\sigma}=\psi_1,\qquad
U\big|=\psi_2,\qquad
U\big|_{L_3}=\psi_3,
\tag{3}
]

where (\psi_1) is a continuous function, and (\psi_2) and (\psi_3) are twice continuously differentiable functions.

Without loss of generality one may assume that

[
\psi_1\equiv0,\qquad \psi_2(0)=\psi'_2(0)=\psi_3(1)=\psi'_3(1)=0.
]

For simplicity of computation suppose that (\gamma=\alpha x,\ \gamma_1=-\beta x+\beta), where (\alpha) and (\beta) are constants satisfying (0<\alpha<1,\ 0<\beta<1). In this case

[
C=C\left(\frac{\beta}{\alpha+\beta},-\frac{\alpha\beta}{\alpha+\beta}\right),\qquad
B_2=B_2\left(\frac{x_0}{1+\alpha},-\frac{\alpha x_0}{1+\alpha}\right),\qquad
B_1=
B_1\left[\frac{x_0+\beta}{1+\beta},\frac{\beta(x_0-1)}{1+\beta}\right].
]

The result obtained below remains valid also under general assumptions concerning (L) and (L_1), provided conditions (2) are satisfied.

The general solution of equation (1), satisfying conditions (3), in the triangles (AB_2E) and (EB_1B), respectively, has the form

[
U(x,y)=f(x+y)-f[\lambda(x-y)]+\psi_2\left[\frac{1+\lambda}{2}(x-y)\right],
\tag{4}
]

[
U(x,y)=\varphi(x-y)-\varphi[\mu(x+y)+1-\mu]+\psi_3\left[\frac{(x+y)(1+\mu)+1-\mu}{2}\right],
\tag{5}
]

where (f(t)), (0<t<x_0), (\varphi(t)), (x_0<t<1), are arbitrary twice continuously differentiable functions, and

[
\lambda=\frac{1-\alpha}{1+\alpha}, \qquad \mu=\frac{1-\beta}{1+\beta}.
]

From (4) and (5) we have

[
U_x(x,0)+U_y(x,0)=2f_x(x), \qquad 0<x<x_0;
]

[
U_x(x,0)-U_y(x,0)=2\varphi_x(x), \qquad x_0<x<1;
]

[
U_x(x,0)-U_y(x,0)=-2f_x(\lambda x)+2\psi_{2x}\left(\frac{1+\lambda}{2}x\right), \qquad 0<x<x_0;
\tag{6}
]

[
U_x(x,0)+U_y(x,0)=-2\varphi_x(\mu x+1-\mu)+
]

[
+2\psi_{3x}\left[\frac{x(1+\mu)+1-\mu}{2}\right], \qquad x_0<x<1.
\tag{7}
]

Denote by (F_0(z)=U_0(x,y)+iV_0(x,y)) the function analytic in the elliptic part of the domain (D), whose real part is the solution of the homogeneous mixed problem.

Under the additional requirement imposed on the arc (\sigma):

[
\omega(s)=\operatorname{Im}\left{z(1-z)(x_0-z)\left(\frac{dy}{dx}+i\right)^2\,dz\right}_{\sigma}\le 0,
]

where (s) is the length of the arc (\sigma) (measured from the point (B)), the uniqueness of the solution of the problem posed above follows from the fact that

[
I=\int_0^1 x(1-x)(x_0-x)U_{0x}U_{0y}\,dx
=\int_{\sigma}\omega(s)\left(\frac{\partial U_0}{\partial y}\right)^2\,ds=0.
]

In particular, when (\sigma) coincides with the semicircle (\sigma_0: 2z=1+e^{i\varphi}), (\operatorname{Im} z\ge 0), the function (\omega(s)=-1/8).

Denoting the right-hand sides in formulas (6) and (7), respectively, by (\omega_1(x)) and (\omega_2(x)), we can rewrite these formulas in the form

[
\operatorname{Re}(1-i)F'(x)=\omega_1(x), \qquad 0<x<x_0;
]

[
\operatorname{Im}(1-i)F'(x)=-\omega_2(x), \qquad x_0<x<1,
\tag{8}
]

where (F(z)=U(x,y)+iV(x,y)) is a holomorphic function in the elliptic part of the domain (D), whose real part is the solution of the problem posed above.

We shall restrict ourselves to considering the case where (\sigma) coincides with the semicircle (\sigma_0), and (x_0>1/2). A holomorphic function (F(z)) satisfying conditions (8), whose real part is zero on (\sigma_0), is determined by the formula

[
\begin{aligned}
(1-i)F'(z)={}&\frac{1}{\pi i}\sqrt{\frac{z(1-z)}{(x_0-z)(\xi_0-z)}}\left{\int_0^{x_0}\sqrt{\frac{(x_0-t)(\xi_0-t)}{t(1-t)}}\left(\frac{1}{t-z}+\frac{1-2t}{t+z-2tz}\right)\right.\
&\left.\times \omega_1(t)\,dt-\int_{x_0}^1\sqrt{\frac{(t-x_0)(\xi_0-t)}{t(1-t)}}\left(\frac{1}{t-z}+\frac{1-2t}{t+z-2tz}\right)\omega_2(t)\,dt\right},
\end{aligned}
\tag{9}
]

where ((2x_0-1)\xi_0=x_0), and by the root is meant its single-valued branch in the plane cut along the segments ([0,x_0]), ([1,\xi_0]), positive for (0<z<x_0).

Passing to the limit in (9) as (z\to x) and taking the imaginary part for (0<x<x_0), and the real part for (x_0<x<1), we obtain

[
f_x(x)+\frac{1}{\pi}\int_0^{x_0}\sqrt{\frac{x(1-t)(x_0-t)(\xi_0-t)}{t(1-x)(x_0-x)(\xi_0-x)}}\left(\frac{1}{t-x}+\frac{1}{t+x-2tx}\right)f_t(\lambda t)\,dt=\rho_1(x),
\tag{10}
]

[
\varphi_x(x)-\frac{1}{\pi}\int_{x_0}^{1}\sqrt{\frac{t(1-x)(t-x_0)(\xi_0-t)}{x(1-t)(x-x_0)(\xi_0-x)}}\left(\frac{1}{t-x}-\frac{1}{t+x-2tx}\right)
]
[
\times \varphi_t(\mu t+1-\mu)\,dt=\rho_2(x),
\tag{11}
]

[
\rho_1(x)=\psi(x)+\frac{1}{\pi}\int_{x_0}^{1}\sqrt{\frac{x(1-x)(t-x_0)(\xi_0-t)}{t(1-t)(x_0-x)(\xi_0-x)}}\left(\frac{1}{t-x}+\frac{1-2t}{t+x-2tx}\right)
]
[
\times \varphi_t(\mu t+1-\mu)\,dt,
]

[
\rho_2(x)=\psi(x)-\frac{1}{\pi}\int_0^{x_0}\sqrt{\frac{x(1-x)(x_0-t)(\xi_0-t)}{t(1-t)(x-x_0)(\xi_0-x)}}\left(\frac{1}{t-x}+\frac{1-2t}{t+x-2tx}\right)f_t(\lambda t)\,dt,
]

[
\begin{aligned}
\psi(x)={}&\frac{1}{\pi}\sqrt{\frac{x(1-x)}{(x_0-x)(\xi_0-x)}}\left{\int_0^{x_0}\sqrt{\frac{(x_0-t)(\xi_0-t)}{t(1-t)}}\right.\
&\times \left(\frac{1}{t-x}+\frac{1-2t}{t+x-2tx}\right)\psi_{2t}!\left(\frac{1+\lambda}{2}\,t\right)\,dt\
&\left.-\int_{x_0}^{1}\sqrt{\frac{(t-x_0)(\xi_0-t)}{t(1-t)}}\right.\
&\left.\times \left(\frac{1}{t-x}+\frac{1-2t}{t+x-2tx}\right)\psi_{3t}!\left[\frac{t(1+\mu)+1-\mu}{2}\right]\,dt\right},\qquad 0<x<x_0;
\end{aligned}
]

[
\begin{aligned}
\psi(x)={}&\frac{1}{\pi}\sqrt{\frac{x(1-x)}{(x-x_0)(\xi_0-x)}}\left{\int_0^{x_0}\sqrt{\frac{(x_0-t)(\xi_0-t)}{t(1-t)}}\right.\
&\times \left(\frac{1}{t-x}+\frac{1-2t}{t+x-2tx}\right)\psi_{2t}!\left(\frac{1+\lambda}{2}\,t\right)\,dt\
&\left.-\int_{x_0}^{1}\sqrt{\frac{(t-x_0)(\xi_0-t)}{t(1-t)}}\right.\
&\left.\times \left(\frac{1}{t-x}+\frac{1-2t}{t+x-2tx}\right)\psi_{3t}!\left[\frac{t(1+\mu)+1-\mu}{2}\right]\,dt\right},\qquad x_0<x<1.
\end{aligned}
]

By a simple change of variables, equations (10) and (11), in the notation

[
\sqrt{x}\,\mu_1(x)=f_x(x)\sqrt{(1-x)(x_0-x)(\xi_0-x)},
]

[
\sqrt{x}\,h_1(x)=\rho_1(x)\sqrt{(1-x)(x_0-x)(\xi_0-x)},
]

[
\sqrt{1-x}\,\mu_2(x)=\varphi_x(x)\sqrt{x(x-x_0)(\xi_0-x)},
]

[
\sqrt{1-x}\,h_2(x)=\rho_2(x)\sqrt{x(x-x_0)(\xi_0-x)}
]

can be rewritten in the following form:

[
\mu_1(x)+\frac{1}{\pi}\int_0^{\lambda x_0}
\sqrt{\frac{(\lambda-t)(\lambda x_0-t)(\lambda \xi_0-t)}
{(1-t)(x_0-t)(\xi_0-t)}}
\left(\frac{1}{t-\lambda x}+\frac{1}{t+\lambda x-2tx}\right)\mu_1(t)\,dt
=
]

[
= h_1(x), \qquad 0