UDC 517.946

MATHEMATICS

E. M. LANDIS

(s)-CAPACITY AND THE BEHAVIOR OF THE SOLUTION OF A SECOND-ORDER ELLIPTIC EQUATION WITH DISCONTINUOUS COEFFICIENTS IN A NEIGHBORHOOD OF A BOUNDARY POINT

(Presented by Academician I. G. Petrovsky on 24 VI 1967)

In this note we consider a linear elliptic equation of the second order

[
Lu \equiv \sum_{i,k=1}^{n} a_{ik}(x)\frac{\partial^2 u}{\partial x_i \partial x_k}
+\sum_{i=1}^{n} b_i(x)\frac{\partial u}{\partial x_i}
+c(x)u=0 .
\tag{1}
]

It is assumed that there exist two positive constants (\alpha) and (M) such that, in the domain where the equation is considered,

[
\sum_{i,k=1}^{n} a_{ik}\xi_i\xi_k \geq \alpha |\xi|^2,
\qquad
\sum_{i=1}^{n} a_{ii}<M;
\tag{2}
]

[
|b_i|<1,\qquad -1<c\leq 0.
\tag{3}
]

By a solution of such an equation we shall mean a classical solution—a continuous function possessing the derivatives entering into the equation and satisfying the equation identically.

In the note the question of the regularity of a boundary point for the Dirichlet problem for this equation will be considered.

If in equation (1) the coefficients of the highest derivatives are smooth or even merely continuous, but with a modulus of continuity satisfying the Dini condition, then the conditions for regularity of a boundary point are the same as for the Laplace equation ((^{1-3})). The same will be true if the equation has divergence form (the coefficients in this case may be arbitrary bounded ones) ((^4)). In the general case, however, as is shown by the example given in Sec. (3^\circ), this is no longer so.

Definition 1. Let positive numbers (\alpha) and (M) be given. A point (x^0) of the boundary (\Gamma) of a domain (D) is called (\alpha, M)-regular with respect to the Dirichlet problem if there exists a continuous function (\omega(r)), (0\leq r0) for (r>0), such that, whatever the neighborhood (\sigma) of the point (x^0), the domain (D'\subset D) with boundary (\Gamma'), and the equation (1), for which inequalities (2) hold with the given constants (\alpha, M), and whatever the solution (u(x)<1) of this equation, continuous on (D'\cup\Gamma') and nonpositive on the intersection (\Gamma'\cap\sigma), there is a neighborhood (\sigma') of the point (x^0), depending only on (\sigma,\alpha), and (M), such that (u(x)<\omega(|x-x^0|)) for (x\in\sigma'\cap D).

We shall give sufficient conditions for regularity of a boundary point.

(1^\circ.) (s)-capacity. Let (s) be a positive number. Let (E) be a (B)-set of (n)-dimensional Euclidean space. Consider all possible measures (\mu), defined on subsets of the set (E), such that

[
\int_E \frac{d\mu(y)}{|x-y|^s}\leq 1
\quad \text{for } x\notin E;
\tag{4}
]

the number (C_s(E)=\sup \mu(E)), where the least upper bound is taken over all measures (\mu) satisfying inequality (4), will be called the (s)-capacity of the set (E).

Lemma 1. The (s)-capacity of a ball of radius (R) is not less than (R^s).

Lemma 2. Let (s>1) and let (\Pi_{\rho,h}) be the cylinder:
[
\sum_{i=1}^{n-1} x_i^2<\rho^2,\qquad 0<x_n<h,
]
where (\rho<h). Then (C_s(\Pi_{\rho,h}) \ge Ch\rho^{s-1}), where (C) is a positive constant depending on (s).

Lemma 3. Let (s0) is a constant depending on (s) and (n) (by (\operatorname{mes} E) is denoted the Lebesgue measure of the set (E)).

Lemma 4. For any (a>0) and (M>0) there exist (s>0) and (r_0>0) such that, for every operator (L) of the form (1) satisfying inequalities (2) and (3), the inequality
[
L\left(1/|x-x^0|^s\right)\ge 0
]
holds, where (x^0) is a fixed point of (R_n), (x) is a point of the domain where the operator (L) is defined, and (|x-x^0|<r_0).

In this case we shall say that the (s)-capacity is upper for the operator (L) in radius (r_0).

Remark 1. The number (s) can be made to depend only on the ratio (M/a).

The proof of these lemmas is elementary.

2°. Sufficient conditions for regularity of a boundary point. In what follows, by (Q_R^{x^0}) we shall denote the ball of radius (R) with center at the point (x^0).

Lemma 5 (main). Let a domain (D) be situated in the ball (Q_{4R}^{x^0}), have limit points on the surface of the ball (Q_R^{x^0}), and intersect the ball (Q_R^{x^0}). Let (H) be the intersection of the complement of the domain (D) with the ball (Q_R^{x^0}). Let (\Gamma) be that part of the boundary of the domain (D) which is situated strictly inside (Q_{4R}^{x^0}). Let equation (1) be defined in (D). Let (u(x)) be its solution, positive in (D), continuous in (\overline D), and vanishing on (\Gamma). Let the numbers (s>0) and (r_0>0) be such that the (s)-capacity is upper for the operator (L) in radius (r_0) and (8R<r_0).

Then
[
\max_{\overline D} u
\ge
\left[
1+\frac{\xi}{R^s}C_s(H)
\right]
\max_{D\cap Q_R^{x^0}} u,
]
where (\xi>0) is a constant depending on (s).

Proof. Choose an arbitrary (\varepsilon>0), and let the measure (\mu) be such that
[
U(x)=\int_H \frac{d\mu(y)}{|x-y|^s}\le 1
\quad \text{for } x\notin H,
\qquad
\mu(H)>C_s(H)-\varepsilon.
]
Let
[
M=\max_{\overline D}u.
]
Put
[
v(x)=M\left[
1-U+\frac{C_s(H)}{(3R)^s}
\right].
]
We have (Lv\le 0) in (D), and (v) is not less than (u) on the boundary of the domain (D). Therefore
[
\max_{D\cap Q_R^{x^0}} u
\le
\max_{D\cap Q_R^{x^0}} v
\le
M\left[
1-\frac{C_s(H)}{R^s}\left(\frac1{2^s}-\frac1{3^s}\right)
+\frac{\varepsilon}{(3R)^s}
\right],
]
which, by the arbitrariness of (\varepsilon), proves the lemma.

With the help of Lemma 5 one proves

Lemma 6. Let, in the spherical layer
[
R<|x-x^0|<4R,
]
there be situated a domain (D), having limit points on both spheres (|x-x^0|=R) and (|x-x^0|=4R). Let (\Gamma) be that part of the boundary of the domain (D) which is situated strictly inside the layer. Let equation (1) be defined in (D), and let (u(x)) be its solution, positive in (D), continuous in (\overline D), and vanishing on (\Gamma). Let the numbers (s>0) and (r_0>0) be such that the (s)-capacity is upper for the operator (L) in radius (r_0) and (8R<r_0).

Then

[
\max_{x\in \overline D} u(x) >
\left[1+\frac{\zeta C_s(H)}{R^s}\right]
\max_{x\in D,\ |x-x^0|=2R} u(x),
]

where (\zeta) is a constant depending on (s) and (r_0), and
(H=(Q_{3R}^{x^0}\setminus Q_{2R}^{x^0})\setminus D).

With the aid of Lemma 5 one proves

Theorem 1. Let there be a domain (D) with boundary (\Gamma), and let (x^0\in\Gamma). Denote by (H) the complement of the domain (D). Let (\alpha>0) and (M>0) be given numbers, and let (s>0) be such that (s)-capacity is an upper one for every operator (L) satisfying inequality (2).

Set (\gamma_m=C_s(H\cap Q_{2^{-m}}^{x^0})), and suppose that

[
\sum_{m=1}^{\infty} 2^{ms}\gamma_m=\infty .
\tag{5}
]

Then the point (x^0) is (\alpha,M)-regular, and, if we denote
[
S(t)=\sum_{m=1}^{[t]}2^{ms}\gamma_m,
]
then for the function (\omega(r)), for sufficiently small (r), the estimate
[
\omega(r)< e^{-aS(|\ln r|)}
]
will hold, where the constant (a>0) depends on (\alpha) and (M).

In the case where the coefficients of the equation satisfy the Hölder condition, an analogous result was obtained in [5].

With the aid of Lemma 6 one proves

Theorem 2. Let (D,\Gamma), and (x^0) have the same meaning as in Theorem 1. Let equation (1) be defined in (D), and let (s>0) be such that (s)-capacity is upper for the operator (L). Let (\gamma_m) have the same meaning as above, and let condition (5) be fulfilled. Then a bounded solution of equation (1), continuous on (\Gamma) in a neighborhood of (x^0), is also continuous at (x^0).

Such a point will be called (O)-regular.

From Lemma 1 and Theorems 1 and 2 it follows that

Theorem 3. If a boundary point of a domain can be touched by a vertex cone from outside the domain, then such a point is regular (in both senses) for any equation (1), and as (\omega(r)) one may take (r^\beta) for sufficiently small (\beta>0) ((\beta) depends on (\alpha) and (M), more precisely on their ratio).

Lemma 2 makes it possible to give a milder sufficient condition for regularity:

Theorem 4. Let the point (x^0) of the boundary (\Gamma) of the domain (D) possess the following property. One can introduce an orthogonal coordinate system (y_1,\ldots,y_n) with origin at this point such that the points belonging to the cusp

[
\left(\sum_{i=1}^{n-1} y_i^2\right)^{1/2}
<
\frac{y_n}{(\ln 1/y_n)^{1/(s-1)}},
\qquad
0<y_n<h
\tag{6}
]

for some (h), (0<h<1), do not belong to the domain (D). Then the point (x^0) is (O)-regular for every equation (1) for which (s)-capacity is upper. For any (\alpha) and (M) to which this (s) corresponds, the point (x^0) is (\alpha,M)-regular.

This condition corresponds to the Uryson condition [6] for the Laplace equation when (n\ge 4). It is all the more interesting that when (n=3), for a general equation (1), in contrast to the Laplace equation, this condition cannot be substantially weakened, as is shown by the example given in the following section.

If (M/\alpha<n+2), then one can find (s<n) for which (s)-capacity is upper for the corresponding equations. Lemma 3 permits, in this case, replacing in Theorems 1 and 2 the capacity of the intersection (H\cap Q_{2^{-m}}^{x^0}) by its Lebesgue measure.

3°. One example of an irregular point. For any (\delta>0) one can find a homogeneous elliptic equation with three independent

variables

[
\sum_{i,k=1}^{3} a_{ik}(x_1,x_2,x_3)\frac{\partial^2 u}{\partial x_i\partial x_k}=0,
]

whose coefficients differ from the coefficients of Laplace’s equation by less than (\delta), defined in a domain (D), the boundary (\Gamma) of which has a point (x^0) of the same type as in Theorem 4, but with another exponent of the logarithm in inequality (6):

[
\left(\sum_{i=1}^{n-1} y_i^2\right)^{1/2}
<
\frac{y_n}{(\ln 1/y_n)^a},
]

where (a>0) is a number depending on (\delta), and find a solution (u(x)) of this equation in the domain (D) such that: (u(x)) is bounded, (u(x)) is continuous at all boundary points except (x^0), in a neighborhood of (x^0) it assumes on the boundary the value equal to one, and, at the same time,

[
\lim_{x\to x^0} u(x)<1.
]

Let (\varepsilon>0) be arbitrary and (a>(1+\varepsilon)/\varepsilon \ln 2). Put (s=1+\varepsilon). Put

[
u(x_1,x_2,x_3)=
\sum_{m=N}^{\infty}\frac{2^{-m(s-1)}}{m^s}
\int_{2^{-(m+1)}}^{2^{-m}}
\frac{dt}{\left[x_1^2+x_2^2+(x_3-t)^2\right]^{s/2}}.
]

The function (u) is defined outside the positive half-axis (x_3). Choose the number (N) so that (u(0,0,0)<1/2), which is possible, since at this point the series converges. The set of points where (u(x_1,x_2,x_3)>1) forms a funnel (B) around the positive half-axis (x_3), containing the funnel

[
\left(\sum_{i=1}^{2} x_i^2\right)^{1/2}
<
\frac{x_3}{(\ln 1/x_3)^a},
\qquad
0