Mathematics

S. P. Pul’kin

THE TRICOMI PROBLEM FOR THE GENERAL LAVRENT’EV–BITSADZE EQUATION

(Presented by Academician I. G. Petrovskii, 19 VI 1957)

1°. Consider the equation

[
L(u)\equiv u_{xx}+\operatorname{sgn} y\,u_{yy}+A(x,y)u_x+B(x,y)u_y+C(x,y)u=0
\tag{L}
]

[
(C(x,y)\le 0 \text{ for } y>0).
]

Let the domain (D) be bounded by a smooth line (\Gamma), lying in the half-plane (y>0) and resting on the segment (-1\le x\le 1) of the axis (OX) (the segment (AB)), and by two characteristics (AC) and (BC) in the half-plane (y<0). The set of points of the domain (D) for which (y>0) ((y<0)) will be denoted by (D_1) ((D_2)).

Problem T. Find a function (u(x,y)) having the following properties:
1) (u(x,y)) is a solution of equation (L) in each of the domains (D_1) and (D_2);
2) (u(x,y)) is continuous in (\overline D);
3) the partial derivatives (\partial u/\partial x), (\partial u/\partial y) are continuous in (D);
4) on the line (\Gamma) and on one of the characteristics, for example (AC), (u(x,y)) assumes prescribed values: (u=\varphi(s)) on (\Gamma); (u=\psi(x)) on (AC); (\varphi(l)=\psi(-1)), where (l) is the length of the curve (\Gamma). The length of the arc (s) is measured from the point (B). The coefficients of equation (L) will be assumed to be twice continuously differentiable functions in each of the domains (D_1, D_2), up to their boundaries.

2°. In the domain (D_2) we transform equation (L) to characteristic coordinates, setting (\xi=x+y), (\eta=x-y). We obtain the equation

[
L_0(u)\equiv u_{\xi\eta}+a(\xi,\eta)u_\xi+b(\xi,\eta)u_\eta+c(\xi,\eta)u=0.
\tag{(L_0)}
]

The domain (D_2) is transformed into a domain (D_0) of the ((\xi,\eta))-plane, which is the triangle (A_0B_0C_0) with side equations (\xi=\eta), (\xi=-1), (\eta=1).

We shall assume that the coefficients of equation ((L_0)) satisfy the following conditions:

Conditions (A_1): (a\ge 0), (a_\xi+ab-c\ge 0), (c\ge 0) in (D_0).

Denote by (D^{\varepsilon}_{hk}) the triangular domain bounded by: 1) the segment (A'_0B'_0) of the line (\eta-\xi=\varepsilon); 2) the segment (A'_0C'_0) of the line (\xi=h); 3) the segment (B'_0C'_0) of the line (\eta=k) ((\varepsilon\ge 0,\ -1\le h<0<k\le 1)).

We shall call a solution (u(\xi,\eta)) of equation ((L_0)) strictly regular in (D^{\varepsilon}{hk}) if: 1) (u(\xi,\eta)) is twice continuously differentiable in (D^{\varepsilon}) together with its first-order derivatives.}); 2) (u(\xi,\eta)) is continuous in (\overline{D}^{\varepsilon}_{hk

A generalized solution of class (S) of equation ((L_0)) in (D_0) will mean a function (u(\xi,\eta)), defined in (D_0), which can be represented as the limit in (D_0) of a sequence ({u_n(\xi,\eta)}) of solutions strictly regular in (\overline{D}0), the convergence being uniform in every closed domain (\overline{D}^{\varepsilon}) ((\varepsilon>0)), and (u_n(-1,\eta)=u(-1,\eta)).

Lemma 1. Let (u(\xi,\eta)) be a generalized solution of equation ((L_0)) of class (S), continuous in (\overline{D}_0), and suppose that (u(-1,\eta)=0). If the conditions (A_1) are satisfied, then (\max u), if it is (>0), is attained on (A_0B_0).

Proof. We note that the lemma is valid for strictly regular solutions ((^{1})). Suppose, in our case, that (\max u=M>0) is attained not on (A_0B_0). Then (u=M) on some closed set (\Phi) not containing points of the segment (A_0B_0). There will be found a closed domain (\overline{D}_{-1,1-\alpha}^{\alpha}), (\alpha>0), and (n) such that (u_n(\xi,\eta)) will attain a positive maximum not on (A'_0B'_0), which contradicts the property of strictly regular solutions.

For what follows we shall also need the following known lemma ((^{2})).

Lemma 2. Let the function (u(x,y)), continuous in (\overline{D}_1), satisfy in (\overline{D}_1) the inequality (L(u)\geq 0) ((L(u)\leq 0)), with (C\leq 0). Suppose (u(x,y)) assumes its greatest positive (least negative) value in (\overline{D}_1) at the point ((x_0,0)), (-1<x_0<1). If the values of (u) on (\Gamma) are less (greater) than (u(x_0,0)), then

[
\lim_{y\to 0}\frac{\partial u(x_0,y)}{\partial y}<0\ (>0),
]

provided this limit exists.

Denote by (u_0(\xi',\eta';\xi,\eta)) the Riemann function of equation ((L_0)). By a solution of equation ((L_0)) of class (R) in the characteristic triangle (D_0) we shall mean a function (u(\xi,\eta)) which in (D_0) can be represented by the Riemann formula

[
u(\xi,\eta)=\frac12\bigl[u_0(\xi,\xi;\xi,\eta)\tau(\xi)+u_0(\eta,\eta;\xi,\eta)\tau(\eta)\bigr]+
]
[
+\int_{\xi}^{\eta} M(t,\xi,\eta)\tau(t)\,dt+\int_{\xi}^{\eta} N(t,\xi,\eta)\nu(t)\,dt,
\tag{1}
]

where (\tau(x)) is continuous on (-1\leq x\leq 1), differentiable on (-1<x<1), and (\nu(x)) is continuous on (-1<x<1), absolutely integrable on (-1\leq x\leq 1). Here (M) and (N) are known functions entering into the Riemann formula.

Lemma 3. Every solution of class (R) belongs to class (S).

From Lemmas 1, 2, and 3 follows the uniqueness of the solution of the problem in class (R).

(3^\circ). We note that a continuously differentiable in (D_0) solution of the equation
(L_0(u)=f(\xi,\eta)), satisfying the conditions (u(-1,\eta)=0), (u_\xi(\xi,\xi)-u_\eta(\xi,\xi)=0), can be represented by the formula

[
u(\xi,\eta)=\int_{-1}^{\xi} d\xi'\int_{\xi'}^{\xi} f(\xi',\eta')S(\xi',\eta';\xi,\eta)\,d\eta' +
]
[
+\int_{-1}^{\xi} d\xi'\int_{\xi'}^{\eta} f(\xi',\eta')T(\xi',\eta';\xi,\eta)\,d\eta',
\tag{2}
]

where (S,T) are continuous in (\overline{D}_0\times\overline{D}_0). Formula (2) and the corresponding expressions for (S) and (T) can be obtained, for example, by applying to the solution of the problem the method of successive approximations.

Lemma 4. Every solution of equation ((L_0)) in (D_0) of class (R) can be represented in the form

[
u(\xi,\eta)=g(\xi,\eta)+\int_{-1}^{\xi} T_0(\xi,\eta,t)\nu(t)\,dt,
\tag{3}
]

where (g(\xi,\eta)) is expressed in terms of (\psi,a,b,c,S,T); (T_0(\xi,\eta,t)) is expressed in terms of (a,b,c,S,T); (\nu(\xi)=\left(\partial u/\partial \xi-\partial u/\partial \eta\right)_{\xi=\eta}).

From (3), putting (g(x)=g(x,x)), (T_0(x,t)=T_0(x,x,t)), (\tau(x)=u(x,x)), we find:

[
\tau(x)=g(x)+\int_{-1}^{x} T_0(x,t)\nu(t)\,dt.
\tag{4}
]

4°. Problem N for the domain (D_1). Find in (D_1) a solution of equation (L) satisfying the conditions:

[
u\big|{\Gamma}=\varphi(s),\qquad
\left.\frac{\partial u}{\partial y}\right|=\nu(x).
]

Conditions B. Let the curve (\Gamma) have continuous curvature, and let the tangents to the curve (\Gamma) at the points (A) and (B) be parallel to the (OY)-axis.

Let (G(x',y';x,y)) be the Green’s function of the Laplace equation for problem N, whose existence is not hard to prove. Following Lichtenstein’s method, we seek the solution of problem N in the form

[
u(x,y)=w(x,y)-\frac{1}{2\pi}\iint_{D_1}G(x',y';x,y)V(x',y')\,dx'\,dy',
]

where (w(x,y)) is the solution of problem N for the Laplace equation; it is expressed by the formula

[
w(x,y)=-\frac{1}{2\pi}\int_{-1}^{1}G(t,0;x,y)\nu(t)\,dt+
]

[
+\frac{1}{2\pi}\int_{-1}^{1}
\frac{\partial}{\partial n}G(x(s),y(s);x,y)\varphi(s)\,ds.
]

To find the function (V(x,y)), we form an integral equation (similarly to (3), p. 267), whose kernel (K_2(x',y';x,y)) is the first iterate of the kernel

[
K(x',y';x,y)=\frac{1}{2\pi}
\left[
A(x,y)\frac{\partial G}{\partial x}
+B(x,y)\frac{\partial G}{\partial y}
+C(x,y)G
\right].
]

If (1) is not an eigenvalue of the kernel (K_2(x',y';x,y)), then, solving the corresponding integral equation, we shall find (V(x,y)) and obtain the solution of problem N. Putting in it (y=0), we arrive at the formula

[
\tau(x)=-\int_{-1}^{1}
\left[G(t,x)+K_0(t,x)+K_{00}(t,x)\right]\nu(t)\,dt+
]

[
+\int_{0}^{l}
\left[G^*(s,x)+\overline{K}0(s,x)+\overline{K}(s,x)\right]\varphi(s)\,ds,
\tag{5}
]

where (\tau(x)=u(x,0)); (G(t,x)=\dfrac{1}{2\pi}G(t,0;x,0)); (G^*(s,x)=\dfrac{1}{2\pi}\dfrac{\partial G(x(s),y(s);x,0)}{\partial n}); (K_0,\overline{K}0) are expressed in terms of (G,A,B,C), while (K), in addition, are also expressed in terms of the resolvent of the kernel (K_2(x',y';x,y)).},\overline{K}_{00

5°. Solution of problem T. Comparing (4) and (5), we obtain an integral equation of the first kind with the unknown function (\nu(x)). In order to obtain an equation of the second kind, we differentiate the resulting equation of the first kind with respect to (x). As a result we obtain the equation

[
\nu(x)+\frac{1}{\pi}\int_{-1}^{1}
\left(\frac{1}{t-x}-\frac{t}{1-xt}\right)\nu(t)\,dt
=
\int_{-1}^{1}\mathcal{K}(t,x)\nu(t)\,dt+g_0(x).
\tag{6}
]

Making estimates of (\partial T_0/\partial x), (\partial K_0/\partial x), (\partial K_{00}/\partial x), and also (\partial G(t,0;x,0)/\partial x-\partial G_0(t,0;x,0)/\partial x), where (G_0) is the Green’s function for the case of a circle, we see that the function (\mathscr K(t,x)) has singularities no higher than of logarithmic order. We shall seek the function (\nu(x)) in the class of functions satisfying the Hölder condition on (-1<x<1), absolutely integrable with some power (\alpha>1). Then the first term of the right-hand side will be a function satisfying the Hölder condition on (-1\le x\le 1). Assuming the function (\varphi(s)) to have continuous second derivatives, at least in neighborhoods of the points (A) and (B), and that (\psi'(x)) satisfies the Hölder condition, we arrive at the conclusion that (g_0(x)) satisfies the Hölder condition on (-1\le x\le 1). Applying the known methods of regularization (see, for example, (5)), we arrive at the integral equation

[
\nu(x)=\int_{-1}^{1}\left{\mathscr K(t,x)-\frac{1}{2\pi}\sqrt{\frac{1+x}{1-x}}\int_{-1}^{1}\sqrt{\frac{1-u}{1+u}}\left(\frac{1}{u-x}-\frac{u}{1-xu}\right)\mathscr K(t,u)\,du\right}\nu(t)\,dt+
]

[
+g_0(x)-\frac{1}{2\pi}\sqrt{\frac{1+x}{1-x}}\int_{-1}^{1}\sqrt{\frac{1-t}{1+t}}\left(\frac{1}{t+x}-\frac{t}{1-xt}\right)g_0(t)\,dt .
\tag{7}
]

This is a Fredholm equation, and the solvability of equation (7) follows from the uniqueness of the solution of problem T. Having found (\nu(x)) from (7), we find (\tau(x)) from (5), and then find the desired solution in (D_1) and (D_2). This will be the solution of problem T of class (R). Thus, we have:

Theorem 1. Let the contour (\Gamma) satisfy condition B, and let the coefficients of equation (L) be such that, for (y<0), conditions (A_1) are fulfilled and (1) is not an eigenvalue of the kernel (K_2(x',y';x,y)). If the function (\varphi(s)) is continuous on (0\le s\le l) together with its second-order derivatives, and (\psi(x)) has a first derivative satisfying the Hölder condition on (-1\le x\le 0), then in the class (R) there exists, moreover a unique, solution of problem T.

6°. Particular case. Consider the equation

[
u_{xx}+\operatorname{sgn} y\cdot u_{yy}+C(x,y)u=0
\qquad (C\le 0 \text{ for } y>0).
\tag{F}
]

If (C(x,y)) depends only on (x) or only on (y), then problem T for equation (F) always has a solution, provided the above-stated conditions imposed on the contour (\Gamma) and on the functions (\varphi(s)) and (\psi(x)) are fulfilled.

Kuibyshev State Pedagogical Institute
named after V. V. Kuibyshev

Received
17 VI 1957

CITED LITERATURE

({}^{1}) S. Agmon, L. Nirenberg, M. Protter, Comm. Pure and Appl. Math., 6, No. 4 (1954).
({}^{2}) K. I. Babenko, On the theory of equations of mixed type, Dissertation, 1951.
({}^{3}) R. Courant, D. Hilbert, Methods of Mathematical Physics, 2, 1951.
({}^{4}) S. Gellerstedt, Sur un problème aux limites pour une équation linéaire aux dérivées partielles du second ordre de type mixte, Thèse, Uppsala, 1935.
({}^{5}) S. G. Mikhlin, Uspekhi Mat. Nauk, 2, issue 3 (1948).