V. M. POLYAKOVA

ON LEVEL LINES OF A SOLUTION OF AN ELLIPTIC EQUATION

(Presented by Academician I. G. Petrovskii on 25 V 1961)

Let us consider the equation:

\[ A(x,y)\frac{\partial^2 u}{\partial x^2} +2B(x,y)\frac{\partial^2 u}{\partial x\,\partial y} +C(x,y)\frac{\partial^2 u}{\partial y^2} +D(x,y)\frac{\partial u}{\partial x} \]
\[ +E(x,y)\frac{\partial u}{\partial y} +F(x,y)u=0, \tag{1} \]

defined in the strip \(K_h=\{0\le x\le \infty,\ -h\le y\le h\}\), \(0<h<1\).
Let \(A,B,C\) be twice continuously differentiable, \(D\) and \(E\) continuously differentiable, and, together with the indicated derivatives, bounded in absolute value by one; moreover, \(-1<F\le 0\). Suppose that the equation is uniformly elliptic:

\[ A\xi^2+2B\xi\eta+C\eta^2\ge \alpha(\xi^2+\eta^2), \qquad \alpha>0, \tag{2} \]

for arbitrary \(\xi\) and \(\eta\).

Let \(u(x,y)\) be a solution of equation (1), and let \(|u(x,y)|<1\) in the strip \(K_h\). Denote by \(\Omega\) the level set \(u(x,y)=0\). We shall call a component \(L\) of the set \(\Omega\) proper if it starts on the left side of \(K_h\) and goes to infinity without touching its upper or lower sides. Let \(L_1\) and \(L_2\) be proper components of \(\Omega\), with \(L_1\) lying above \(L_2\). Denote by \(y_1(x_0)\) and \(y_2(x_0)\), respectively, the upper point of intersection of \(L_1\) and the lower point of intersection of \(L_2\) with the line \(x=x_0\). Put \(\rho(x_0)=y_1(x_0)-y_2(x_0)\).

Theorem. Let \(L_1,L_2\) be proper components. Then

\[ \lim_{x\to\infty}\left[\ln\frac{1}{\rho(x)}:x\right]<\infty. \tag{3} \]

Proof. Suppose that (3) is false. This means that there exists a function \(g(x)\to\infty\) as \(x\to\infty\) such that

\[ \rho(x)\le e^{-xg(x)}. \tag{3a} \]

Denote by \(G\) the closed region lying between \(L_1\) and \(L_2\). We may assume that inequality (3a) holds starting from \(x=0\), and that \(g(x)\) increases monotonically. In the proof we shall use two lemmas analogous to Lemmas 1.6.1 and 1.1.1 from \((^1)\).

Let \(h\), \(0<h<1\), be an arbitrary number. Let \(\Gamma_h\) be a \(\Gamma\)-shaped strip: \(x_0-h<x<+\infty\) for \(y_0-h<y<y_0+h\); \(x_0-h<x<x_0+h\) for \(-\infty<y<y_0+h\). Suppose that in the strip \(\Gamma_h\) there are two smooth curves \(\Gamma_1\) and \(\Gamma_2\), each of which has limit points on both sides of \(\Gamma_h\). Suppose they do not intersect, and that \(\Gamma_1\) is situated above or to the right of \(\Gamma_2\); denote by \(G^{(1)}\) the region lying between them, and by \(G^{(2)}\) the part of the strip \(\Gamma_h\) not separated from infinity by the curve \(\Gamma_2\).

Lemma 1. Let \(N>2^{10}h\) be an arbitrary number and suppose that at least one of the following assumptions holds:

1) In the domain \(G^{(1)}+G^{(2)}\) there is defined a solution \(u(x,y)\) of equation (1) such that: a) \(\left.|u(x,y)|\right|_{\Gamma_1}>a>0\); b) \(\left.|u(x,y)|\right|_{G^{(2)}}>a\cdot 2^{-2N/h}\); c) \(|u(x,y)|<1\) in \(G^{(1)}\). We assume that \(\Gamma_2\) is projected one-to-one onto the \(y\)-axis.

2) In the domain \(G^{(1)}\) there is situated a subdomain \(g\), also having boundary points on both sides of \(\Gamma_h\); \(\gamma\) is the part of its boundary lying strictly inside \(\Gamma_h\). In the domain \(G^{(1)}\) there is defined a solution of equation (1), \(u(x,y)\), such that: a) \(\left.|u(x,y)|\right|_{\Gamma_1}>a>0\); b) \(u(x,y)|_\gamma=0\), \(|u(x,y)|<a\cdot 2^{-2N/h}\) for \((x,y)\in g\); c) \(|u(x,y)|<1\) in \(G^{(1)}\).

Then

\[ \mu^2 G^{(1)} > Nh/M_1, \tag{4} \]

\(M_1\) depends only on \(\alpha\) of inequality (2).

The proof of this lemma is obtained from the proof of Lemma 1.6.1 of [1] by replacing the strip \(P_h=\{-h<y<h\}\) by the \(\Gamma\)-shaped strip \(\Gamma_h\).

Lemma 2. Let a domain \(G\) be situated in the disk \(Q_R\) of radius \(R<1\), containing the center \(O\) of the disk and having boundary points on the boundary of the disk. Let \(\Gamma\) be the part of the boundary of the domain \(G\) which lies strictly inside \(Q_R\). Let equation (1) be defined in \(G\).

There exists a constant \(M_2\), depending only on the constant \(\alpha\) of inequality (2), such that if \(\mu^2G<R^2/M_2\), then for any positive solution \(u(x,y)\) of equation (1), continuous in \(G\) and vanishing on the boundary \(\Gamma\), the inequality

\[ u(0)<\frac12 \max_{P\in \bar G} u(P) \tag{5} \]

holds.

This is Lemma (1.1.1) from [1].

Proof of the theorem.

\(1^\circ\). Denote:

\[ \max_{y\in G} u(0,y)=a. \]

Suppose that, starting with \(x=x_0\), the inequality

\[ \frac{x}{2}\,g\!\left(\frac{x}{2}\right)>1 \]

holds.

Put

\[ X=\max(4M_2,x_0). \]

We shall prove that for every \(x>X\) the inequality

\[ \max_{y\in G} u(x,y) < a\cdot 2^{ \frac{-\exp\left(\frac{x}{2}g\left(\frac{x}{2}\right)\right)}{2} }. \tag{6} \]

First we shall show that

\[ \max_{y\in G} u(x_1,y)>\max_{y\in G} u(x_2,y),\quad \text{if } x_1<x_2. \tag{7} \]

Fix some number \(R\), \(0<R<1\). Since \(\rho(x)\) decreases as \(x\) increases, there will be an \(x'\) such that for every \(x''>x'\) the area of the domain \(G\) situated in the disk \(Q_R\) of radius \(R\) with center at \((x'',y)\), \(y\in G\), will be less than \(R^2/M_2\). Then by Lemma 2

\[ \max_{(x,y)\in \bar G\cap \bar Q_R} u(x,y) > 2\max_{y\in G} u(x'',y). \]

By the maximum principle, \(\max\limits_{(x,y)\in \bar G\cap \bar Q_R} u(x,y)\) is attained at a point \((x_1,y_1)\) on the boundary of the domain, and moreover on that part of it which lies on the boundary of \(Q_R\). Hence either \(x_1>x''\) or \(x_1<x''\). Suppose \(x_1>x''\). Taking the point \((x_1,y_1)\) as the new center of a disk of radius \(R\), we obtain that the maximum of \(u(x,y)\) in the intersection

the domain \(G\) with the newly obtained circle, by the maximum principle, must also lie to the right of \(x=x_1\) and, by Lemma 2, be at least twice as large. Hence the function must increase without bound as \(x\to\infty\). Consequently, \(x_1<x''\), and (7) follows from the maximum principle.

Let us now prove the required inequality (6). Take on \([0,x]\), \(x>X\), the following points:

\[ x^{(1)}=2M_2 e^{-g\cdot 0}, \]

\[ \ldots\ldots\ldots\ldots\ldots\ldots \]

\[ x^{(n)}=x^{(n-1)}+2M_2 e^{-x^{(n-1)}g(x^{(n-1)})}. \]

Then their number \(n\) satisfies the inequality

\[ n>\left[\frac{x}{2\cdot 2M_2 e^{-\frac{x}{2}g\left(\frac{x}{2}\right)}}\right]> \left[e^{\frac{x}{2}g\left(\frac{x}{2}\right)}\right]> \frac12 e^{\frac{x}{2}g\left(\frac{x}{2}\right)}, \tag{8} \]

since \(x>X\).

On the other hand,

\[ \max_{y\in G} u\left(x^{(k)},y\right)> 2\max_{y\in G} u\left(x^{(k+1)},y\right). \]

Indeed, for each \(x^{(k)}\) we find a point \(P^{(k)}\) at which the maximum of \(u(x^{(k)},y)\), \(y\in G\), is attained. Construct circles \(Q^{(k)}\) with radii

\[ r_k=2M_2 e^{-x^{(k-1)}g(x^{(k-1)})} \]

and centers at \(P^{(k)}\). Then

\[ \mu_2 Q^{(k)}\cap G<4e^{-2x^{(k-1)}g(x^{(k-1)})}M_2=\frac{r_k^2}{M_2}; \]

hence Lemma 2 is applicable to each \(Q^{(k)}\). Therefore,

\[ \max_{\overline{Q}^{(k)}\cap \overline{G}} u(x,y)>2u\left(P^{(k)}\right). \]

Hence, all the more,

\[ u\left(P^{(k-1)}\right)>2u\left(P^{(k)}\right), \]

but since \(n\) satisfies condition (8), it follows that

\[ \max u(x,y)<a\cdot 2^{-\frac{\exp\left[\frac{x}{2}g\left(\frac{x}{2}\right)\right]}{2}}, \]

which was required to be proved.

\(2^\circ\). Let \((0,y_0)\) be the point at which the maximum of \(u(0,y)\) for \(y\in G\) is attained; \(u(0,y_0)=a_0\). By the continuity of \(u(x,y)\) there exists a number \(\delta>0\) such that

\[ \min_{y_0-\delta<y<y_0+\delta} u(0,y)>\frac{a_0}{2}. \]

Denote \(h_1=h-|y_0|-\delta\). Since \(u(x,y)=0\) on \(L_1\) and \(L_2\) (the boundary of the domain \(G\)), we have \(h_1>0\). We show that for all \(x>X+2h_1\) in the strip \(K_\delta=\{y_0-\delta<y<y_0+\delta\}\) the inequality

\[ |u(x,y)|\le a_1(x)\,2^{2N/h_1}=a_2, \tag{9} \]

holds, where

\[ a_1(x)=a_0\cdot 2^{-\frac{\exp\{(x/2-h_1)g(x/2-h_1)\}}{2}},\qquad N=\max(2^{10}h_1,\,8M_1h). \]

Suppose that this is not so. Let there exist a point \((\hat x,\hat y)\in K_\delta\), \(\hat x>X+2h_1\), such that \(|u(\hat x,\hat y)|>a_1(\hat x)2^{2N/h_1}\). Let \(u(\hat x,\hat y)>0\) (the case \(u(x,y)<0\) is symmetric). By the Kronrod–Landis theorem \((^2)\), for almost all \(c\) from the range of values of the function \(u(x,y)\), the level line \(u(x,y)=c\) contains no points with zero gradient of the function \(u(x,y)\). Therefore one can find a point \((\hat x,\hat y)\) such that \(u(\hat x,\hat y)>a_2\) and at the same time, on the level line of the function \(u(x,y)\) passing through this point, \(\operatorname{grad} u(x,y)\ne 0\). Then the component \(l\) of the indicated level line passing through the point \((\hat x,\hat y)\) behaves as follows: a) either it intersects the left side of the strip \(\hat x-2h_1<x<\hat x\) (the right side of the strip \(\hat x<x<\hat x+2h_1\)), b) or it exits onto the boundary of the strip \(K_h\). In any of these cas-

than it separates the \(\Gamma\)-shaped strip \(\Gamma_{h_1}=\{\hat x-2h_1<x<\hat x\) for \(y<\tilde y;\ x>\hat x-2h_1\) for \(\tilde y+2h_1>y>\tilde y\}\), or the strip \(\Gamma_{h_1}=\{-\infty<x<\hat x+2h_1\) for \(\tilde y-2h_1<y<\tilde y;\ \hat x<x<\hat x+2h_1\) for \(y>\tilde y\}\). From \(1^\circ\) it follows that, for all \(x>X\), the inequality

\[ \max_{y\in G} u(x,y)<a_1(x) \]

holds. Hence Lemma 1 (case 2) is applicable to the strip \(\Gamma_{h_1}\); here the part of \(G\) in the strip \(\Gamma_{h_1}\) plays the role of \(g\). Consequently, the area \(\sigma\) between the component under consideration of the level line \(l\) and the part of \(G\) in the strip \(\Gamma_{h_1}\) satisfies inequality (4), \(\sigma>Nh_1/M_1\). On the other hand, \(\sigma<8hh_1\). But, recalling that \(N=\max(2^{10}h_1,8M_1h)\), we arrive at a contradiction. This proves the required inequality (9).

Let us bring (9) to a somewhat different form:

\[ a_1 2^{2N/h_1} = a_0\cdot 2^{-\frac{\exp[(x/2-h_1)g(x/2-h_1)]}{2}}\cdot 2^{2N/h_1} < a_0\cdot 2^{-2^{\frac{x}{4}g\left(\frac{x}{4}\right)}}, \]

starting from some \(X_1\). Hence,

\[ u(x,y)_{\,y_0-\delta<y<y_0+\delta} < a_0\cdot 2^{-2^{\frac{x}{4}g\left(\frac{x}{4}\right)}}, \qquad x>\max(X,X_1). \tag{10} \]

\(3^\circ\). Let us prove the theorem. We have:

\[ u(0,y)_{\,y_0-\delta<y<y_0+\delta} > \frac{a_0}{2}=a', \]

\[ |u(x_0,y)|_{\,y_0-\delta<y<y_0+\delta} < a_0\cdot 2^{-2^{\frac{x}{4}g\left(\frac{x}{4}\right)}} < a'\cdot 2^{-2^{\frac{x_0}{8}g\left(\frac{x_0}{4}\right)}}, \qquad x_0>\max(X_1,X). \]

Then, by Lemma 1, the area \(S\) between the straight lines \(x=0\) and \(x=x_0\), \(x_0>\max(X,X_1)\), in the strip \(K_\delta\) must be greater than

\[ \frac{x_0}{8}\,g\!\left(\frac{x_0}{4}\right)\frac{\delta^2}{M_1}. \]

But \(S<2\delta x_0\), which is impossible for \(x_0\) satisfying the inequality \(g(x_0/4)>16M_1/\delta\). We arrive at a contradiction, which proves the theorem.

Corollary. Let the equation \(\Delta u=0\) be given. Let \(|u|<1\) in an angle \(\varphi\) (or in a part of an angle \(\varphi: r>r_0\)). The greatest possible rate of decrease of the distance between a pair of lines of one and the same level set \(u=\mathrm{const}\), lying entirely in the angle \(\varphi\), is

\[ \rho(r)=\frac{M}{r^\alpha},\qquad \alpha>0. \]

Proof. It is clear that for the Laplace equation the theorem is valid without the restriction \(h<1\), since the equation is invariant under similarity transformations, and also, instead of the zero level set, one may consider the level set \(u(x,y)=\mathrm{const}\). Make the transformation:

\[ x_1=\operatorname{Re}\ln(x+iy),\qquad y_1=\operatorname{Im}\ln(x+iy). \]

The equation \(\Delta u=0\) is invariant under it, i.e. \(\Delta u(x_1,y_1)=0\). Then the region inside the angle \(\varphi\) passes into a half-strip of width \(\varphi\), and

\[ \rho(x,y)=\tilde\rho(x_1,y_1)=Me^{-\alpha u}. \]

But this is the greatest possible rate of decrease of the distance between level lines \(u(x,y)=\mathrm{const}\) lying entirely in the strip \((+\infty>x>x_0,\ y_2<y<y_1)\).

Received
14 III 1961

REFERENCES

1. E. M. Landis, UMN, 14, no. 1 (85) (1959).
2. A. S. Kronrod, E. M. Landis, DAN, 58, No. 7 (1947).