UDC 517.944

MATHEMATICS

S. I. GRINBERG

SOME ASYMPTOTIC FORMULAS FOR THE SPECTRA OF THE THIRD BOUNDARY-VALUE PROBLEM RELATED TO VARIATION OF THE FUNCTION ENTERING THE BOUNDARY CONDITION

(Presented by Academician L. V. Kantorovich on 25 VIII 1966)

Let \(V\) be a bounded domain in three-dimensional space, and let \(S\) be its boundary. Further, let \(\{\mu_k(\sigma)\}\) \((\mu_1(\sigma)\leq \mu_2(\sigma)\leq \ldots)\) be the eigenvalues of the operator \(L u\), generated by the operation \(-\Delta u\) under the boundary condition
\[ \left[\partial u/\partial n+\sigma u\right]_S=0, \]
where \(\partial u/\partial n\) is differentiation in the direction of the outward normal.

It is known (see \((^1)\)) that
\[ \mu_n(\sigma)\sim (6\pi^2 v^{-1}n)^{2/3}\qquad (n\to\infty) \tag{1} \]
(\(v\) is the volume of the domain \(V\)); from this it is not difficult to derive the relation
\[ \sum_{\mu_k(\sigma)\leq x}\mu_k(\sigma_1)\sim \frac{v}{10\pi^2}x^{5/2}\qquad (x\to +\infty), \]
where \(\sigma(Q)\) and \(\sigma_1(Q)\) \((Q\in S)\) are two, generally speaking, different functions (the summation extends over all values of \(k\) for which \(\mu_k(\sigma)\leq x\)).

In this note the asymptotic behavior of the difference
\[ \sum_{\mu_k(\sigma)\leq x}\mu_k(\sigma_2)- \sum_{\mu_k(\sigma)\leq x}\mu_k(\sigma_1); \]
is studied; namely, the equality
\[ \sum_{\mu_k(\sigma)\leq x}\bigl[\mu_k(\sigma_2)-\mu_k(\sigma_1)\bigr] = \frac{1}{3\pi^2}\int_S [\sigma_2(Q)-\sigma_1(Q)]\,ds\cdot x^{3/2} +o(x^{3/2}) \qquad (x\to +\infty). \tag{2} \]
is established.

Some of its generalizations are also given.

Formula (2) may be regarded as a generalization to the three-dimensional case of the well-known formula of I. M. Gelfand and B. M. Levitan for the difference of the traces of two Sturm–Liouville operators (see \((^2)\)).

Let \(Q\) be an arbitrary point of the surface \(S\). We shall assume that some neighborhood of this point on the surface \(S\) can be given by the equation \(z=z_Q(x,y)\) (the \(z\)-axis is directed along the normal to \(S\) at the point \(Q\)), and that the function \(z_Q(x,y)\) everywhere in the disk \(x^2+y^2<\rho_0^2\) (where \(\rho_0\) does not depend on \(Q\)) has second partial derivatives satisfying a Lipschitz condition (with exponent equal to 1 and with a constant independent of \(Q\)).

Theorem 1. If the function \(\sigma(Q)\) satisfies a Lipschitz condition, then as \(p\to +\infty\)
\[ \sum_{k=1}^{\infty}\frac{1}{[\mu_k(\sigma)+p]^2} = \frac{v}{8\pi}\frac{1}{\sqrt p} + \frac{s}{16\pi}\frac{1}{p} + \frac{1}{24\pi}\int_S [h(Q)-3\sigma(Q)]\,ds\,\frac{1}{p^{3/2}} + O\!\left(\frac{1}{p^2}\right), \tag{3} \]
where \(v\) and \(s\) are the volume of the domain \(V\) and the area of the surface \(S\); \(h(Q)\) is the mean curvature of the surface \(S\) at the point \(Q\).

Proof. Let \(G(M, M_1, -\chi^2)=(4\pi r_{MM_1})^{-1}e^{-\chi r_{MM_1}}-g(M,M_1,\chi)\) be the Green’s function of the expression \(\Delta u-\chi^2 u\) (\(\chi\) is a sufficiently large positive constant) for the domain \(V\) under the boundary condition \([\partial u/\partial n+\sigma u]_S=0\). As \(M_1\to M\), from the equality

\[ G(M,M_1,-\chi_1^2)-G(M,M_1,-\chi^2) = (\chi^2-\chi_1^2)\sum_{k=1}^{\infty} \frac{u_k(M)u_k(M_1)} {[\mu_k(\sigma)+\chi_1^2][\mu_k(\sigma)+\chi^2]}, \]

where \(\{u_k(M)\}\) are the eigenfunctions of the operator \(Lu\), and \(\chi_1\ne \chi\), it follows that

\[ \frac{\chi_1-\chi}{4\pi} + g(M,M,\chi_1)-g(M,M,\chi) = (\chi_1^2-\chi^2)\sum_{k=1}^{\infty} \frac{u_k^2(M)} {[\mu_k(\sigma)+\chi_1^2][\mu_k(\sigma)+\chi^2]} . \]

Hence follows the formula

\[ \sum_{k=1}^{\infty}\frac{1}{[\mu_k(\sigma)+\chi^2]^2} = \frac{v}{8\pi}\frac{1}{\chi} + \frac{1}{2\chi}\int_V g'_\chi(M,M,\chi)\,dv_M . \]

By means of a method close to that used in \((3)\), it can be proved that

\[ \int_V g'_\chi(M,M,\chi)\,dv_M = \]

\[ = \frac{s}{8\pi}\frac{1}{\chi} + \frac{1}{12\pi}\int_S [h(Q)-3\sigma(Q)]\,ds\cdot \frac{1}{\chi^2} + O\!\left(\frac{1}{\chi^3}\right) \qquad (\chi\to+\infty). \]

Combining the last two equalities and putting \(\chi^2=p\), we obtain (3).

Theorem 2. Let \(\sigma(Q)\), \(\sigma_1(Q)\), and \(\sigma_2(Q)\) \((Q\in S)\) be any three functions satisfying the Lipschitz condition. Then

\[ \sum_{\mu_k(\sigma_2)\le x} [\mu_k(\sigma_2)-\mu_k(\sigma_1)] = \frac{I}{3\pi^2}x^{3/2} + O(x^{3/2}) \qquad (x\to+\infty), \tag{4} \]

where

\[ I=\int_S[\sigma_2(Q)-\sigma_1(Q)]\,ds. \]

Proof. Suppose first that \(\sigma_2(Q)\ge \sigma_1(Q)\). Introduce for consideration the nonnegative and nondecreasing function

\[ \Phi(x)= \sum_{\mu_k(\sigma_2)\le x} [\mu_k(\sigma_2)-\mu_k(\sigma_1)]; \]

moreover, set

\[ \psi(p)= \sum_{k=1}^{\infty} \frac{[\mu_k(\sigma_2)-\mu_k(\sigma_1)]^2[\mu_k(\sigma_2)+2\mu_k(\sigma_1)+3p]} {[\mu_k(\sigma_2)+p]^3[\mu_k(\sigma_1)+p]^2}. \]

We now use Theorem 1: let us write formula (4) first for the function \(\sigma_1(Q)\), and then for the function \(\sigma_2(Q)\); subtracting the second from the first equality, we find that for any \(\alpha<\mu_1(\sigma_2)\)

\[ 2\int_{\alpha}^{+\infty}\frac{d\Phi(x)}{(x+p)^3} + \psi(p) = \frac{I}{8\pi}\frac{1}{p^{3/2}} + O\!\left(\frac{1}{p^2}\right) \qquad (p\to+\infty). \]

Taking further into account that

\[ \psi(p)=o\!\left[\int_{\alpha}^{+\infty}(x+p)^{-3}\,d\Phi(x)\right] \qquad \text{as } p\to+\infty, \]

we have

\[ \int_{\alpha}^{+\infty}\frac{d\Phi(x)}{(x+p)^3} \sim \frac{I}{16\pi}\frac{1}{p^{3/2}} \qquad (p\to+\infty). \]

Applying the Hardy–Littlewood Tauberian theorem (see, for example, (⁴)), we find:
\[ \Phi(x)=\sum_{\mu_k(\sigma_2)\le x}\left[\mu_k(\sigma_2)-\mu_k(\sigma_1)\right]\sim \frac{I}{3\pi^2}x^{3/2}\qquad (x\to+\infty). \]

Similarly, the asymptotic equality is established
\[ \sum_{\mu_k(\sigma_1)\le x}\left[\mu_k(\sigma_2)-\mu_k(\sigma_1)\right]\sim \frac{I}{3\pi^2}x^{3/2}\qquad (x\to+\infty). \]

Taking further into account that
\[ \sum_{\mu_k(\bar\sigma)\le x}\left[\mu_k(\sigma_2)-\mu_k(\sigma_1)\right]\le \sum_{\mu_k(\sigma)\le x}\le \sum_{\mu_k(\underline\sigma)\le x}, \]
where \(\bar\sigma(Q)=\max\{\sigma(Q),\sigma_2(Q)\}\), \(\underline\sigma(Q)=\min\{\sigma(Q),\sigma_1(Q)\}\), and that as \(x\to+\infty\) the functions
\[ \sum_{\mu_k(\bar\sigma)\le x}\left[\mu_k(\sigma_2)-\mu_k(\sigma_1)\right]= \]
\[ =\sum_{\mu_k(\bar\sigma)\le x}\left[\mu_k(\bar\sigma)-\mu_k(\sigma_1)\right]- \sum_{\mu_k(\bar\sigma)\le x}\left[\mu_k(\bar\sigma)-\mu_k(\sigma_2)\right], \]
\[ \sum_{\mu_k(\underline\sigma)\le x}\left[\mu_k(\sigma_2)-\mu_k(\sigma_1)\right] \]
are asymptotically equal to \(\dfrac{I}{3\pi^2}x^{3/2}\), we obtain (4) for the special case \(\sigma_2(Q)\ge\sigma_1(Q)\).

To prove the theorem in the general case, it is now enough to apply the equality
\[ \sum_{\mu_k(\sigma)\le x}\left[\mu_k(\sigma_2)-\mu_k(\sigma_1)\right]= \sum_{\mu_k(\sigma)\le x}\left[\mu_k(\sigma_0)-\mu_k(\sigma_1)\right]- \sum_{\mu_k(\sigma)\le x}\left[\mu_k(\sigma_0)-\mu_k(\sigma_2)\right], \]
where \(\sigma_0(Q)=\max\{\sigma_1(Q),\sigma_2(Q)\}\), and to use the special case of this theorem considered above.

Taking (1) into account, we obtain

Corollary 1. As \(n\to\infty\),
\[ \sum_{k=1}^{n}\left[\mu_k(\sigma_2)-\mu_k(\sigma_1)\right]=\frac{2I}{v}\,n+o(n). \]

Remark. From Theorem 2 (or Corollary 1) it follows that if the integral
\[ I=\int_S[\sigma_2(Q)-\sigma_1(Q)]\,ds\ne0, \]
then for infinitely many values of the index \(k\) the difference \(\mu_k(\sigma_2)-\mu_k(\sigma_1)\) has the same sign as the integral \(I\).

With the aid of Theorem 2, the following can easily be proved.

Theorem 3. Let the functions \(\sigma(Q)\), \(\sigma_1(Q)\), \(\sigma_2(Q)\) and the quantity \(I\) be the same as in Theorem 2. Suppose, further, that on the half-axis \([a,+\infty)\) there is given a sign-definite function \(f(x)\), absolutely continuous on every interval \([a,b]\) \((b<+\infty)\); in addition, assume that the expression \(xf'(x)[f(x)]^{-1}\) is bounded almost everywhere and
\[ \int^{+\infty} x^{1/2}f(x)\,dx=\infty . \]
Then, as \(x\to+\infty\),
\[ \sum_{a<\mu_k(\sigma)\le x} f[\mu_k(\sigma)]\,[\mu_k(\sigma_2)-\mu_k(\sigma_1)] =\left(\frac{I}{2\pi^2}+o(1)\right)\int_a^x |t|^{1/2}f(t)\,dt . \]

Assuming that \(f(x)=x^m\) \((m\ge -3/2)\), we obtain

Corollary 2. As \(x\to+\infty\),

\[ \sum_{0<\mu_k(\sigma)\le x} \mu_k^m(\sigma)\,[\mu_k(\sigma_2)-\mu_k(\sigma_1)] = \begin{cases} \dfrac{I}{(2m+3)\pi^2}\,x^{m+3/2}+o\!\left(x^{m+3/2}\right), & (m>-3/2),\\[6pt] \dfrac{I}{2\pi^2}\ln x+o(\ln x), & (m=-3/2). \end{cases} \]

From Corollary 2 it is not difficult to derive the following assertions:

A. If \(\mu_k(\sigma_1)\ne 0\), then as \(n\to\infty\)

\[ \sum_{k=1}^{n}\frac{\mu_k(\sigma_2)}{\mu_k(\sigma_1)} = n+\left(\frac{6}{\pi^2\nu}\right)^{1/3} I n^{1/3} +o\!\left(n^{1/3}\right). \]

B. If \(\mu_1(\sigma_1)>0\) and \(\mu_1(\sigma_2)>0\), then as \(n\to\infty\)

\[ \sum_{k=1}^{n}\left[\mu_k^m(\sigma_2)-\mu_k^m(\sigma_1)\right] = \]

\[ = \begin{cases} \dfrac{m}{(2m+1)\pi^2} \left(\dfrac{6\pi^2}{2}\right)^{(2m+1)/3} I n^{(2m+1)/3} +o\!\left(n^{(2m+1)/3}\right), & (m>-1/2),\\[8pt] -\dfrac{I}{6\pi^2}\ln n+o(\ln n), & (m=-1/2). \end{cases} \]

Received
22 VII 1966

CITED LITERATURE

¹ R. Courant, D. Hilbert, Methods of Mathematical Physics, 1, Moscow, 1951.
² I. M. Gel'fand, B. M. Levitan, DAN, 88, 4, 593 (1953).
³ A. Pleijel, 12, Skand. Mat. Kongr., 1954, p. 222.
⁴ E. C. Titchmarsh, Eigenfunction Expansions Associated with Second-Order Differential Equations, 2, Moscow, 1961.