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Reports of the Academy of Sciences of the USSR
1962, Volume 143, No. 1
MATHEMATICS
I. S. ARSHON and M. A. EVGRAFOV
AN EXAMPLE OF A FUNCTION HARMONIC IN ALL SPACE AND BOUNDED OUTSIDE A CIRCULAR CYLINDER
(Presented by Academician M. V. Keldysh on 23 X 1961)
The present note, like the notes \((^{1-3})\), is devoted to a problem that may be formulated as follows.
Let \(D\) be some domain in the plane \((x_1,x_2)\), and let \(u(x,x_1,x_2)\) be a function harmonic in the half-cylinder
\[ (x_1,x_2)\in D,\qquad x>0. \tag{1} \]
Find \(\sigma(D)\), the lower bound of those values \(\sigma\) for which it follows from conditions A and B that \(u(x,x_1,x_2)\) is bounded also inside the half-cylinder (1):
A.
\[
\max_{(x_1,x_2)\in D}|u(x,x_1,x_2)|<M\exp\exp\frac{\pi x}{\sigma}\quad (x>0).
\]
B. \(u(x,x_1,x_2)\) and \(\operatorname{grad} u(x,x_1,x_2)\) are bounded on the surface of the half-cylinder (1).
The analogous problem for functions harmonic in the plane is easily reduced to the Phragmén–Lindelöf theorem on functions analytic in a strip.
In the notes \((^{1,2})\) the constant \(\sigma(D)\) was found for the case when \(D\) is a circle or a rectangle, and in the note \((^3)\) an upper estimate for \(\sigma(D)\) was obtained for the case of an arbitrary domain. To formulate the results, we introduce two geometric characteristics of the domain. By the external width \(H(D)\) of the domain \(D\) we shall mean the width of the narrowest strip into which this domain can be placed. By the internal width \(h(D)\) of the domain \(D\) we shall mean the diameter of the largest circle that can be placed in this domain.
The result of the note \((^3)\) consists, roughly speaking, in the fact that
\[ \sigma(D)\leq H(D), \]
and in the present note it will be proved that
\[ \sigma(D)\geq h(D). \]
Thus, for domains for which \(H(D)\) and \(h(D)\) coincide (in particular, for a circle, a rectangle, and a semicircle), the constant \(\sigma(D)\) is equal to their common value.
To prove the inequality \(\sigma(D)\geq h(D)\), it is sufficient to construct a function harmonic in all space, bounded outside a circular half-cylinder with base radius \(\pi\lambda/2\), and inside it growing, but more slowly than \(\exp\exp(x/\lambda_1)\) with any \(\lambda_1<\lambda\) (as \(x\to+\infty\)). The boundedness of the gradient of such a function outside our half-cylinder will be obtained from the boundedness of the function itself by virtue of the Poisson formula.
The construction of an example of such a function constitutes the content of the note.
Theorem. The function
\[ u_\lambda(x,x_1,x_2)=\int_0^\infty J_0\!\left(s\sqrt{x_1^2+x_2^2}\right)e^{xs-\lambda s\ln\lambda s}\,ds, \tag{2} \]
is harmonic in all space, bounded outside the half-cylinder
\[ x_1^2+x_2^2\leq \left(\frac{\pi\lambda}{2}\right)^2,\qquad x\geq 0, \]
not being bounded inside it, and satisfies the inequality
\[ \max_{x_1,x_2} |u_\lambda(x,x_1,x_2)|<M_\varepsilon \exp \exp \frac{x}{\lambda-\varepsilon}, \qquad x>0, \]
with any \(\varepsilon>0\).
Proof. The integral (2) and all its derivatives converge uniformly for all \(x,x_1,x_2\), and from the differential equation for the Bessel function \(J_0(z)\) it follows that \(u_\lambda(x,x_1,x_2)\) is harmonic in the whole space.
It is obvious that \(u_\lambda(x,0,0)\) is unbounded as \(x\to+\infty\) and that, for any \(x_1,x_2\), the inequality
\[ |u_\lambda(x,x_1,x_2)|\leq u_\lambda(x,0,0)<M_\varepsilon \exp \exp \frac{x}{\lambda-\varepsilon}\qquad (x>0) \]
holds with any \(\varepsilon>0\).
It remains to show that \(u_\lambda(x,x_1,x_2)\) is bounded when \(x_1^2+x_2^2>(\pi\lambda/2)^2\). We shall use the formula
\[ J_0(\xi)=\frac{1}{\pi i}\int_1^\infty \frac{e^{i t\xi}\,dt}{\sqrt{t^2-1}} -\frac{1}{\pi i}\int_1^\infty \frac{e^{-i t\xi}\,dt}{\sqrt{t^2-1}} \qquad (\xi>0), \tag{3} \]
which is easily obtained from the well-known formula
\[ J_0(\xi)=\frac{1}{2\pi i}\int_{|z|=R}\frac{e^{\xi z}\,dz}{\sqrt{z^2+1}}\qquad (R>1) \]
by deforming the contour of integration.
Substituting formula (3) into formula (2) for \(u_\lambda\), we obtain
\[ u_\lambda(x,\rho\sin\varphi,\rho\cos\varphi) = \frac{1}{\pi i}\int_0^\infty\int_1^\infty \frac{e^{i\rho s t+xs-\lambda s\ln \lambda s}}{\sqrt{t^2-1}}\,dt\,ds - \frac{1}{\pi i}\int_0^\infty\int_1^\infty \frac{e^{-i\rho s t+xs-\lambda s\ln \lambda s}}{\sqrt{t^2-1}}\,dt\,ds . \]
Rotating the ray of integration \((0,+\infty)\) in the first term by the angle \(+\pi/2\), and in the second term by the angle \(-\pi/2\), we find
\[ u_\lambda= \frac{1}{\pi}\int_0^\infty\int_1^\infty \frac{e^{-\rho t y+i x y-i\lambda \ln(i\lambda y)}}{\sqrt{t^2-1}}\,dt\,dy - \frac{1}{\pi}\int_0^\infty\int_1^\infty \frac{e^{-\rho t y-i x y+i\lambda y\ln(-i\lambda y)}}{\sqrt{t^2-1}}\,dt\,dy, \]
whence
\[ |u_\lambda(x,\rho\sin\varphi,\rho\cos\varphi)| \leq \frac{2}{\pi}\int_0^\infty\int_1^\infty e^{-\rho t y+\frac12\pi\lambda y}\, \frac{dt}{\sqrt{t^2-1}}\,dy . \]
For \(\rho>\pi\lambda/2\) the integral converges, and we obtain
\[ |u_\lambda(x,x_1,x_2)|\leq \frac{M}{\sqrt{x_1^2+x_2^2-\pi\lambda/2}} . \]
The theorem is proved.
Received
12 X 1961
References cited
- M. A. Evgrafov, I. A. Chegis, DAN, 134, No. 2 (1960).
- I. A. Chegis, DAN, 136, No. 3 (1961).
- I. S. Arshon, M. A. Evgrafov, DAN, 142, No. 4 (1962).