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AN ASYMPTOTIC FORMULA FOR THE EIGENVALUES OF A CIRCULAR MEMBRANE
N. V. KUZNETSOV, B. V. FEDOSOV
Let a boundary-value problem be posed in a bounded plane domain \(D\) with piecewise smooth boundary \(\Gamma\):
\[ -\Delta u=k^2 u,\quad u|_{\Gamma}=0. \tag{1} \]
Arrange the eigenvalues \(k_n^2\) of problem (1) in nondecreasing order, and let
\[ N(k)=\sum_{k_n<k} 1 . \]
As is well known (see [1]), for large \(k\)
\[ N(k)=\frac{S}{4\pi}k^2+O(k\ln k), \tag{2} \]
where \(S\) is the area of the domain \(D\).
Until now it has been unknown whether, in formula (2), one can single out the second term of the asymptotics. It has been established, however, that if a second power term exists, then formula (2) necessarily has the form
\[ N(k)=\frac{S}{4\pi}k^2-\frac{L}{4\pi}k+o(k), \tag{3} \]
where \(L\) is the length of \(\Gamma\) (see [2, 3]).
As far as we know, formula (3) has so far been proved only for the case when \(D\) is a rectangle (see, for example, [3], where the estimate for the remainder in the number of integer points in a circle, known in number theory, is used).
In the present paper formula (3) is proved for the case of a circular membrane. Moreover, it is possible to prove that the remainder term in (3) is \(O(k^{2/3})\). Thus the following is true.
Theorem. If \(D\) is a circle of radius \(r\), then
\[ N(k)=\frac{S}{4\pi}k^2-\frac{L}{4\pi}k+O(k^{2/3}), \tag{4} \]
where \(S=\pi r^2,\ L=2\pi r\).
We first set out the general scheme of the proof of the theorem\(^*\).
Without loss of generality, the radius of the circle may be assumed equal to 1.
It is known that the square roots of the eigenvalues of the circular membrane of unit radius are the positive zeros of the Bessel functions \(J_n(x)\), \(n=0,1,2,\ldots\) (for \(n\ge 1\) each zero is counted twice).
Using the asymptotic representation of Bessel functions of large orders, we reduce the problem of the number of positive zeros of the functions \(J_n(x)\) not exceeding \(k\) to the counting of the number of integer points in a certain domain. The counting of integer points is carried out with the help of a result in number theory—the van der Corput theorem (see [5]).
We now give the detailed proof.
Denote by \(N_n(k)\) the number of zeros of \(J_n(x)\) in the interval \(n<x<k\). Then
\[ N(k)=N_0(k)+2\sum_{n=1}^{[k]} N_n(k), \]
where \([k]\) denotes the integer part of the number \(k\).
\(^*\) The method we use belongs to E. C. Titchmarsh (see [4], where the asymptotics of the eigenvalues of the Schrödinger equation with a spherically symmetric potential is investigated).
Let \(\nu_0=[k^{1/3}]+\dfrac{1}{2}\), \(\nu_1=[k-k^{4/9}]+\dfrac{1}{2}\). We split the sum \(\displaystyle \sum_{n=1}^{[k]} N_n(k)\) into three terms
\[ \sum_{n=1}^{[k]} N_n(k) = \sum_{n=1}^{\nu_0-1/2} + \sum_{n=\nu_0+1/2}^{\nu_1-1/2} + \sum_{n=\nu_1+1/2}^{[k]} = \Sigma_1+\Sigma_2+\Sigma_3 \]
and estimate each of them.
For \(n\) sufficiently large, the following asymptotic representation is valid for Bessel functions (see [6]):
\[ J_n(k)= \left(\frac{4}{k^2-n^2}\right)^{1/4} \varphi^{1/4} \left\{ Ai(-\varphi) \left[1+O\left(\frac{1}{n^2}\right)\right] + Ai'(-\varphi)O\left(\frac{1}{n^{4/3}}\right) \right\}, \]
where
\[ Ai(-z)=\frac{z^{1/2}}{3} \left[ J_{-1/3}\left(\frac{2}{3}z^{3/2}\right) + J_{1/3}\left(\frac{2}{3}z^{3/2}\right) \right] \]
is the Airy function, and \(\varphi(k,n)\) is defined by the equality
\[ \frac{2}{3}\varphi^{3/2} = \eta(k,n) = \sqrt{k^2-n^2} - n\arccos\frac{n}{k}, \]
and all \(O\)-symbols are uniform in \(k\).
For \(n<\nu_1\) one may use the asymptotics of the Airy function and its derivative. Then, for \(\nu_0<n<\nu_1\), we obtain
\[ J_n(k)= \sqrt{\frac{2}{\pi}}\,(k^2-n^2)^{-1/4} \left\{ \cos\left[\eta(k,n)-\frac{\pi}{4}\right] + O\left(\frac{1}{\eta}\right) + O\left(\frac{1}{n}\right) \right\}, \tag{5} \]
where the absolute value of \(O\left(\dfrac{1}{\eta}\right)\) does not exceed
\[ \frac{1}{2\pi\eta}+\frac{1}{8\pi\eta^2}. \]
It follows from (5) that the number of zeros of the function \(J_n(x)\) in the interval \(n<x<k\) coincides with the number of zeros of \(\cos\left(x-\dfrac{\pi}{4}\right)\) in the interval \(0<x<\eta(k,n)+O\left(\dfrac{1}{\eta}\right)+O\left(\dfrac{1}{n}\right)\), and consequently,
\[ N_n(k)= \left[ \frac{1}{\pi}\eta(k,n) + \frac{1}{4} + O\left(\frac{1}{\eta}\right) + O\left(\frac{1}{n}\right) \right]. \tag{6} \]
Hence it is clear that
\[ \Sigma_2= \sum_{n=\nu_0+1/2}^{\nu_1-1/2} N_n(k) \]
coincides with the number of integer points in the region
\[ \nu_0<\nu<\nu_1;\qquad \frac{1}{2}<x< f(\nu)= \frac{1}{\pi}\eta(k,\nu) + \frac{1}{4} + O\left(\frac{1}{\eta}\right) + O\left(\frac{1}{\nu}\right). \tag{7} \]
Let \(\delta\) be the maximum of \(\left|O\left(\dfrac{1}{\nu}\right)\right|\) in the interval \((\nu_0,\nu_1)\), and let \(a\) be such that
\[ \left|O\left(\frac{1}{\eta}\right)\right|<\frac{a}{\pi\eta}. \]
Since
\[ f_1(\nu)= \frac{1}{\pi}\eta+\frac{1}{4}-\frac{a}{\pi\eta}-\delta < f(\nu) < f_2(\nu)= \frac{1}{\pi}\eta+\frac{1}{4}+\frac{a}{\pi\eta}+\delta, \]
we have \(N_1<\Sigma_2<N_2\), where \(N_1\) and \(N_2\) are the numbers of integer points in the region of the form (7), where \(f(\nu)\) is replaced by \(f_1(\nu)\) and \(f_2(\nu)\), respectively.
To count \(N_1\) and \(N_2\) we shall use the following theorem, which is a particular case of van der Corput’s theorem.
Suppose \(\nu_0-\dfrac{1}{2}\), \(\nu_1-\dfrac{1}{2}\), \(x_0-\dfrac{1}{2}\) are integers, \(\nu_0<\nu_1\). Let in the interval \(\nu_0<\nu<\nu_1\) the real function \(f(\nu)\) be twice continuously differentiable and
\[ 0<\sigma<f'(\nu)<\tau,\qquad f''(\nu)>\frac{1}{\rho},\qquad \rho>1,\qquad \rho>\sigma^{-3}. \]
Let \(N\) be the number of integer points in the region \(\nu_0<\nu<\nu_1\), \(x_0<x<f(\nu)\), and let \(A\) be the area of this region. Then
\[ N-A=O\left(\rho^{2/3}\tau\right). \]
Let us verify that \(f_1\) and \(f_2\) satisfy the conditions of the theorem. We have
\[ f_1'(\nu)=-\frac{\arccos \frac{\nu}{k}}{\pi}\left(1+\frac{a}{\eta^2}\right), \]
\[ f_1''(x)=\frac{1}{\pi\sqrt{k^2-\nu^2}}\left(1+\frac{a}{\eta^2}\right) -\frac{2a\left(\arccos \frac{\nu}{k}\right)^2}{\pi\eta^3}. \tag{8} \]
Since in the interval \(\nu_0<\nu<\nu_1\)
\[ \frac{a}{\eta^2}=O\left(k^{-1/3}\right), \]
it follows that
\[ 0<Ck^{-5/18}<-f_1'(\nu)<\frac{1}{2}+O\left(k^{-1/3}\right). \]
To estimate the second derivative from below, we first show that
\[ \frac{\left(\arccos \frac{\nu}{k}\right)^2}{\eta^3} = o\left(\frac{1}{\sqrt{k^2-\nu^2}}\right) \tag{9} \]
uniformly in \(\nu\) on the interval \((\nu_0,\nu_1)\) as \(k\to\infty\). Divide the interval \((\nu_0,\nu_1)\) into two: \((\nu_0,\nu')\) and \((\nu',\nu_1)\), where \(\nu'=k-k^{3/5}\).
In the interval \((\nu_0,\nu')\),
\[ \frac{\left(\arccos \frac{\nu}{k}\right)^2}{\eta^3} \ll \frac{1}{\eta^3(\nu')} = O\left(k^{-6/5}\right), \]
and in the interval \((\nu',\nu_1)\)
\[ \frac{\left(\arccos \frac{\nu}{k}\right)^2}{\eta^3} \ll \frac{\left(\arccos \frac{\nu'}{k}\right)^2}{\eta^3(\nu_1)} = O\left(k^{-6/5}\right). \]
At the same time,
\[ \frac{1}{\sqrt{k^2-\nu^2}}>\frac{1}{k} \]
in the interval \((\nu_0,\nu')\), and
\[ \frac{1}{\sqrt{k^2-\nu^2}}>\frac{1}{\sqrt{2}\,k^{4/5}} \]
in the interval \((\nu',\nu_1)\), whence (9) follows.
Thus, for the second derivative we obtain the estimate
\[ f_1''(\nu)=\frac{1}{\pi\sqrt{k^2-\nu^2}} +o\left(\frac{1}{\sqrt{k^2-\nu^2}}\right)>\frac{B}{k}. \tag{10} \]
From estimates (8) and (10) it follows that \(f_1(\nu)\) satisfies the conditions of the Van der Corput theorem with
\(\sigma=Ck^{-5/18}\), \(\tau=\frac{1}{2}+O\left(k^{-1/3}\right)\), \(\mu=\frac{k}{B}\).
Thus,
\[ N_1=\int_{\nu_0}^{\nu_1}\left(f_1(\nu)-\frac{1}{2}\right)d\nu+O\left(k^{2/3}\right). \]
Further, by virtue of the estimate \(\eta'>Ck^{-5/18}\) for \(\nu_0<\nu<\nu_1\),
\[ \int_{\nu_0}^{\nu_1}\frac{d\nu}{\eta} = \int_{\eta(\nu_0)}^{\eta(\nu_1)} \frac{d\eta}{\eta'\eta} = O\left(k^{-5/18}\ln k\right). \]
Moreover,
\[ \int_{\nu_0}^{\nu_1}\delta\,d\nu=O\left(k^{2/3}\right) \]
and, consequently,
\[ N_1= \int_{\nu_0}^{\nu_1} \left\{ \frac{1}{\pi}\tau_1(k,\nu)-\frac{1}{4} \right\}d\nu + O\left(k^{2/3}\right). \]
For \(N_2\) the same formula is obtained. Thus,
\[ \Sigma_2=\int_{\nu_0}^{\nu_1}\left\{\frac{1}{\pi}\eta(k,\nu)-\frac{1}{4}\right\}\,d\nu+O(k^{2/3}). \tag{11} \]
To estimate \(\Sigma_3\), note that \(N_n(k)<N_{\nu_1-1/2}(k)\) for \(n>\nu_1-1/2\), and \(N_{\nu_1-1/2}(k)=O(k^{1/6})\), as follows from formula (6). Hence it follows that
\[ \Sigma_3=O(k^{11/18}). \tag{12} \]
We now compute \(\Sigma_1\). We have
\[ \sum_{n=1}^{\nu_0-1/2} N_n(k) = N_{\nu_0-1/2}\left(\nu_0-\frac{1}{2}\right) + \sum_{n=1}^{\nu_0-1/2}\left[N_n(k)-N_{\nu_0-1/2}(k)\right]. \]
From formula (6),
\[ N_{\nu_0-1/2}(k)=\frac{k}{\pi}+O(k^{1/3}). \]
Since
\[ N_n(k)-N_{\nu_0-1/2}(k)<N_0(k)-N_{\nu_0-1/2}(k), \]
and since
\[ N_0(k)=\frac{k}{\pi}+O(1), \tag{13} \]
we have \(N_n(k)-N_{\nu_0-1/2}(k)=O(k^{1/3})\), and, consequently,
\[ \Sigma_1=\frac{k}{\pi}\left(\nu_0-\frac{1}{2}\right)+O(k^{2/3}). \tag{14} \]
From equalities (11), (12), (13), (14) we obtain
\[ N(k)=2\frac{k}{\pi}\nu_0 + 2\int_{\nu_0}^{\nu_1}\left\{\frac{1}{\pi}\eta(k,\nu)-\frac{1}{4}\right\}\,d\nu + O(k^{2/3}). \]
Since
\[ k\nu_0=\int_0^{\nu_0}\eta\,d\nu+O(k^{1/3}) \quad\text{and}\quad \int_{\nu_1}^{k}\eta\,d\nu=O(k^{11/18}), \]
it follows that
\[ N(k)=\frac{2}{\pi}\int_0^k \eta(k,\nu)\,d\nu-\frac{k}{2}+O(k^{2/3}), \]
which coincides with (4) if we put \(r=1\). The theorem is proved.*)
The same method, with an inessential modification, is applicable to the problem of the number of eigenvalues smaller than \(k^2\), if the domain \(D\) in which the boundary-value problem is posed is a sector or part of a sector or of an annulus bounded by two concentric circles. Formula (4) also holds in this case.
In conclusion, the authors take the opportunity to express their sincere gratitude to Professor V. B. Lidskii for valuable advice and comments.
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Received by UMN October 30, 1964
Moscow Institute of Physics and Technology
*) As one of the authors has succeeded in showing, this method permits one to prove the existence of the second term for all cases in which the boundary-value problem in the plane admits separation of variables.