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MATHEMATICS
B. V. FEDOSOV
ASYMPTOTIC FORMULAS FOR THE EIGENVALUES OF THE LAPLACE OPERATOR IN THE CASE OF A POLYGONAL DOMAIN
(Presented by Academician S. L. Sobolev on 13 February 1963)
We consider the boundary-value problem for the equation
\[ \Delta u=-\nu^2 u \]
in a domain \(D\) with boundary condition \(u|_{\Gamma}=0\) or \(\left.\dfrac{\partial u}{\partial n}\right|_{\Gamma}=0\). The domain \(D\) is an arbitrary polygon (possibly nonconvex), and \(\Gamma\) is the polygonal line without self-intersections that bounds it.
Theorem. Let \(\nu_n^2\) be the \(n\)-th eigenvalue and
\[ n(\nu)=\sum_{\nu_n\leqslant 1} 1 . \]
Then
\[ n(\nu)=\frac{S}{4\pi}\nu^2+O(\nu), \tag{1} \]
\[ \int_0^\nu n(t)\,dt=\frac{S}{12\pi}\nu^3 \mp \frac{l}{8\pi}\nu^2+O(\nu), \tag{2} \]
where \(S\) is the area of \(D\), and \(l\) is the perimeter of \(\Gamma\). Here and below the minus sign corresponds to the boundary condition \(u|_{\Gamma}=0\), and the plus sign to the condition \(\left.\dfrac{\partial u}{\partial n}\right|_{\Gamma}=0\).
Formulas (1) and (2) are obtained from the expression for the trace of the Green function of the mixed boundary-value problem with the aid of V. A. Marchenko’s Tauberian theorem for Fourier integrals*.
We begin by considering an auxiliary mixed boundary-value problem for the wave equation
\[ \Delta w=\frac{\partial^2 w}{\partial t^2}\quad \text{in } D \]
with initial conditions
\[ w|_{t=0}=0,\qquad \left.\frac{\partial w}{\partial t}\right|_{t=0} =\delta\left(|\bar{x}-\bar{y}|\right); \qquad \bar{x},\bar{y}\in D, \]
and boundary condition \(w|_{\Gamma}=0\) or \(\dfrac{\partial w}{\partial n}=0\). Here \(|\bar{x}-\bar{y}|\) is the distance between the points \(\bar{x}\) and \(\bar{y}\).
Let \(D\) be a polygon with angles \(\omega_n=\pi/\alpha_n\), \(\alpha_n>1/2\). Choose \(t_0\) so small that the disk \(K_{\bar{x},t_0}\) of radius \(t_0\) centered at any point \(\bar{x}\in D\) intersects no more than two adjacent sides of the polygon. For \(t<t_0\), the function \(w(\bar{x},\bar{y},t)\) can be found easily. Indeed, if \(K_{\bar{y},t_0}\) lies entirely in \(D\), then
\[ w(\bar{x},\bar{y},t)= \frac{\sigma\left(t-|\bar{x}-\bar{y}|\right)} {2\pi\sqrt{t^2-|\bar{x}-\bar{y}|^2}}, \]
where
\[ \sigma(t)= \begin{cases} 0, & \text{for } t\leqslant 0,\\ 1, & \text{for } t>0; \end{cases} \]
\[ \text{* Such a plan was proposed by Prof. A. Ya. Povzner at I. M. Gelfand’s seminar at Moscow University in 1959.} \]
if \(K_{\bar y,t_0}\) intersects only one side, then
\[ w(\bar x,\bar y,t)= \frac{\sigma(t-|\bar x-\bar y|)}{2\pi\sqrt{t^2-|\bar x-\bar y|^2}} \mp \frac{\sigma(t-|\bar x-\bar y'|)}{2\pi\sqrt{t^2-|\bar x-\bar y'|^2}}, \]
where \(\bar y'\) is the point symmetric to \(\bar y\) with respect to this side. The remaining points are near one of the corners, and during the time \(t_0\) the influence of the other corners will not be felt. For such \(\bar y\)
\[ w(\bar x,\bar y,t)=F(\varphi-\psi)\mp F(\varphi+\psi) \]
(\(\bar x\) and \(\bar y\) have polar coordinates \(r,\varphi\) and \(s,\psi\); here \(r\) and \(s\) are the distances from the vertex of the angle, and \(\varphi\) and \(\psi\) are measured from one of the sides of the angle). The function \(F(\theta)\) is expressed in terms of a contour integral (see (¹))\(^*\)
\[ F(\theta)=\frac{\alpha_n}{4\pi i}\int_{C_t} \frac{\operatorname{ctg}\alpha_n\frac{z-\theta}{2}\,dz} {2\pi\sqrt{t^2-r^2-s^2+2rs\cos z}}. \]
The contour \(C_t\) is determined as follows. In the \(z\)-plane draw vertical cuts from the branch points of the integrand to \(\infty\) for \(t>r+s\) (Fig. 1), and for \(t<r+s\) draw cuts as shown in Fig. 2. Then the contour \(C_t\) consists of two lines joining the branch points (in the figures the contour \(C_t\) is indicated by a solid line). For \(t<r+s\), \(F(\theta)\) is equal to the sum of the residues at the poles of
\[ \operatorname{ctg}\alpha_n\frac{z-\theta}{2} \]
inside \(C_t\). For \(t>r+s\), \(F(\theta)\) is equal to the sum of the residues at the poles of
\[ \operatorname{ctg}\alpha_n\frac{z-\theta}{2}, \]
lying between \(-\pi\) and \(\pi\), plus the integral
\[ \frac{\alpha_n}{4\pi i}\int_{S_t} \frac{\operatorname{ctg}\alpha_n\frac{z-\theta}{2}\,dz} {2\pi\sqrt{t^2-r^2-s^2+2rs\cos z}}; \]
the contour \(S_t\) is indicated in Fig. 1 by a dashed line.
Fig. 1
Thus:
for \(t<r+s\)
\[ w(\bar x,\bar y,t)= \sum_{|\varphi-\psi+2\omega_n k|<\tau} \frac{1}{2\pi\sqrt{t^2-r^2-s^2+2rs\cos(\varphi-\psi+2\omega_n k)}} \mp \]
\[ \mp \sum_{|\varphi+\psi+2\omega_n k|<\tau} \frac{1}{2\pi\sqrt{t^2-r^2-s^2+2rs\cos(\varphi+\psi+2\omega_n k)}}, \]
\(\tau\) is the solution of the equation \(t^2-r^2-s^2+2rs\cos\tau=0\);
for \(t>r+s\)
\[ w(\bar x,\bar y,t)= \sum_{|\varphi-\psi+2\omega_n k|<\pi} \frac{1}{2\pi\sqrt{t^2-r^2-s^2+2rs\cos(\varphi-\psi+2\omega_n k)}} \mp \]
\[ \mp \sum_{|\varphi+\psi+2\omega_n k|<\pi} \frac{1}{2\pi\sqrt{t^2-r^2-s^2+2rs\cos(\varphi+\psi+2\omega_n k)}} + \]
\[ + \frac{\alpha_n}{2\pi i}\int_{S_t} \frac{\operatorname{ctg}\alpha_n\frac{z-\varphi+\psi}{2}\,dz} {2\pi\sqrt{t^2-s^2-r^2+2rs\cos z}} \mp \frac{\alpha_n}{4\pi i}\int_{S_t} \frac{\operatorname{ctg}\alpha_n\frac{z-\varphi-\psi}{2}\,dz} {2\pi\sqrt{t^2-r^2-s^2+2rs\cos z}}; \]
\(k\) is an integer.
\(^*\) In (¹) the Green’s function of the equation \(\Delta u=k^2u\) is constructed for an angle, but with inessential changes these arguments are also applicable to the mixed problem.
Putting \(\bar x=\bar y\) and integrating over the domain, we obtain
\[ w(t)=\int_D w(\bar x,\bar y,t)\,d\bar x =\frac{S}{2\pi t}\mp\frac{l}{8}+ct \quad \text{for } t<t_0; \]
\(c=\sum c_n\), where \(c_n\) are constants depending on the magnitudes of the angles \(\omega_n\).
Starting from the Fourier method, we can obtain the following expression for \(w(\bar x,\bar y,t)\):
\[ w(\bar x,\bar y;t)=\sum_{n=1}^{\infty} \frac{u_n(\bar x)u_n(\bar y)}{\nu_n}\sin \nu_n t; \]
here \(u_n(\bar y)\) is the normalized eigenfunction of the problem corresponding to the eigenvalue \(\nu_n^2\).
For \(w(t)\) we hence obtain
\[ w(t)=\sum_{n=1}^{\infty}\frac{\sin \nu_n t}{\nu_n} =\int_0^{\infty}\frac{\sin \nu t}{\nu}\,dn(\nu). \]
The convergence of the series and the integral is understood in the “weak” sense on three-times differentiable finite functions.
Thus,
\[ \int_0^{\infty}\frac{\sin \nu t}{\nu}\,dn(\nu) =\frac{S}{2\pi t}\mp\frac{l}{8}+ct \quad \text{for } t<t_0. \tag{3} \]
The asymptotic formulas for \(n(\nu)\) are obtained from formula (3) with the help of the following Tauberian theorem for Fourier integrals (see \((^{2,3})\)).
Let \(\rho(\nu)\) satisfy the following conditions:
a) on every finite interval the function \(\rho(\nu)\) has bounded variation;
b) as \(\mu\to +\infty\),
\[ \bigvee_{\mu}^{\mu+1}\rho(\nu)=O(\mu^p),\qquad p>0; \]
c) for every even function \(g_\Lambda(t)\) having a bounded \((p+2)\)-nd derivative and equal to zero outside the interval \((-\Lambda,\Lambda)\),
\[ \int_0^\infty h(\nu)\,d\rho(\nu)=0, \]
where
\[ h(\nu)=\frac{1}{\pi}\int_0^\Lambda g_\Lambda(t)\cos \nu t\,dt. \]
Fig. 2
Then, as \(\mu\to+\infty\), the estimate
\[ \int_0^\mu\left(1-\frac{\nu^2}{\mu^2}\right)^r d\rho(\nu) =O(\mu^{p-r}),\qquad r\ge 0. \]
is valid.
By the method expounded, for example, in \((^4)\), one can verify that
\[ \bigvee_{\mu}^{\mu+a} n(\nu)<A\mu, \]
where \(a\) is fixed.
Put
\[ n_1(\nu)=n(\nu)-\frac{S}{4\pi}\nu^2\pm\frac{l}{4\pi}\nu-c; \qquad n_1(0)=0. \]
Then
\[ \int_0^\infty \frac{\sin \nu t}{\nu}\,dn_1(\nu)=0 \quad \text{for } t<t_0. \]
Take an even, three-times differentiable function \(f(t)\), equal to zero outside the interval \((-t_0,t_0)\), and the function \(h(\nu)\), equal to
\[ h(\nu)=\frac{1}{\pi}\int_0^{t_0} f(t)\cos \nu t\,dt =-\frac{1}{\pi\nu}\int_0^{t_0} f'(t)\sin \nu t\,dt . \]
Then
\[ \int_0^\infty h(\nu)\,dn_1(\nu) =\int_0^\infty \frac{1}{\pi\nu} \left(\int_0^{t_0} f'(t)\sin \nu t\,dt\right)dn_1(\nu) = \]
\[ =-\frac{1}{\pi}\int_0^{t_0} f'(t) \left(\int_0^\infty \frac{\sin \nu t}{\nu}\,dn_1(\nu)\right)dt =0 . \]
It is clear that
\[ \bigvee_\mu^{\mu+a} n_1(\nu)=O(\mu). \]
Consequently, by Tauber’s theorem,
\[ \int_0^\mu \left(1-\frac{\nu^2}{\mu^2}\right)^r dn_1(\nu) = O(\mu^{1-r}),\qquad r\geqslant 0. \]
For \(r=0\) we obtain formula (1). For \(r=1\) we have:
\[ \int_0^\mu \left(1-\frac{\nu^2}{\mu^2}\right)dn_1(\nu)=O(1). \]
Integrating twice by parts, we obtain
\[ \frac{N_1(\mu)}{\mu} -\frac{1}{\mu^2}\int_0^\mu N_1(\nu)\,d\nu =O(1); \qquad N_1(\mu)=\int_0^\mu n_1(\nu)\,d\nu, \]
i.e.
\[ \left(\frac{1}{\mu}\int_0^\mu N_1(\nu)\,d\nu\right)'=O(1). \]
Hence
\[ \int_0^\mu N_1(\nu)\,d\nu=O(\mu^2), \]
and for \(N_1(\mu)\) we have \(N_1(\mu)=O(\mu)\), which gives formula (2).
From formulas (1) and (2) one can also obtain the formula
\[ \int_0^\mu (\mu-\nu)^r\,dn_1(\nu)=O(\mu) \qquad \text{for } 0\leqslant r\leqslant 1, \]
i.e.
\[ \int_0^\mu (\mu-\nu)^r\,dn(\nu) = \frac{S}{2\pi(r+1)(r+2)}\mu^{r+2} \mp \frac{l}{4\pi(r+1)}\mu^{r+1} +O(\mu). \]
The author takes this opportunity to express his deep gratitude to Prof. V. B. Lidskii for posing the problem and for constant assistance in the work.
Moscow Institute of Physics and Technology
Received
2 II 1963
REFERENCES
- G. Herglotz, Math. Ann., 124, 219 (1952).
- V. A. Marchenko, Izv. AN SSSR, Ser. Mat., 19, 381 (1955).
- B. M. Levitan, Appendix V to the book by E. C. Titchmarsh, Eigenfunction Expansions, Part I, 1960.
- B. M. Levitan, Appendix VI to the book by E. C. Titchmarsh, Eigenfunction Expansions, Part II, 1961.