Mathematics

SHI CHZHUN-DI

ON THE ASYMPTOTIC BEHAVIOR OF EIGENFUNCTIONS OF THE STURM–LIOUVILLE PROBLEM COMPUTED BY PERTURBATION THEORY

(Presented by Academician A. A. Dorodnitsyn, 16 XII 1959)

We consider the differential operator

\[ Au=-\frac{d^2u}{dx^2}+pu,\qquad u(0)=u(\pi)=0. \]

Suppose that \(p^{(k)}(x)\) satisfies a Lipschitz condition of order 1 and that all odd derivatives vanish at the ends of the interval \((0,\pi)\) (therefore

\[ \int_0^\pi p\cos nx\,dx=O\left(\frac{1}{n^{k+1}}\right), \]

moreover, without loss of generality, one may assume that

\[ \int_0^\pi p(x)\,dx=0, \]

since if a constant is subtracted from the function \(p(x)\), then all eigenvalues are decreased by this constant, while the eigenfunctions do not change.

In the work of L. A. Dikii \({}^{(1)}\) it was shown that if the operator \(Au\) is regarded as the operator \(-d^2u/dx^2\) perturbed by the operator \(pu\), then the approximations to the eigenvalues \(\lambda_n\), computed by perturbation theory up to order \(k\), give the asymptotics for large \(n\) up to order \(2k\). Below it is shown that the corresponding fact also holds for the eigenfunctions.

Let \(\lambda_n\) be the eigenvalues of the operator \(A\), and let \(\varphi_n\) be the corresponding eigenfunctions. Then, by perturbation theory, \(\lambda_n\) and \(\varphi_n\) are computed as follows:

\[ \lambda_n=n^2+\mu_n^{(1)}+\mu_n^{(2)}+\cdots, \]

where

\[ \mu_n^{(1)}=\frac{2}{\pi}\int_0^\pi p\sin^2 nx\,dx,\qquad \mu_n^{(2)}=\frac{4}{\pi^2}\sum_{m\ne n} \frac{\left(\int_0^\pi p\sin nx\sin mx\,dx\right)^2}{n^2-m^2}; \]

\[ \varphi_n(x)=\sqrt{\frac{2}{\pi}}\sin nx+\psi_n^{(1)}+\psi_n^{(2)}+\cdots, \]

where

\[ \psi_n^{(1)}=\sqrt{\frac{2}{\pi}}\,\frac{2}{\pi} \sum_{m\ne n} \frac{\int_0^\pi p\sin nx\sin mx\,dx}{n^2-m^2}\sin mx, \]

\[ \psi_n^{(2)}=\frac{2}{\pi}\sum_{m\ne n} \frac{\int_0^\pi \left(p\psi_n^{(1)}-\mu_n^{(1)}\psi_n^{(1)}\right)\sin mx\,dx}{n^2-m^2}\sin mx -\frac{1}{2}\sqrt{\frac{2}{\pi}}\int_0^\pi \left(\psi_n^{(1)}\right)^2 dx\,\sin nx. \]

Theorem. For the normalized eigenfunctions of the operator \(A\), the following formulas hold (uniformly in \(x\)):

\[ \varphi_n(x)-\sqrt{\frac{2}{\pi}}\sin nx-\psi_n^{(1)}=O\left(\frac{1}{n^2}\right); \tag{1} \]

\[ \varphi_n(x)-\sqrt{\frac{2}{\pi}}\sin nx-\psi_n^{(1)}-\psi_n^{(2)}=O\left(\frac{1}{n^3}\right); \tag{2} \]

\[ \cdots\cdots\cdots\cdots\cdots\cdots\cdots \]

\[ \varphi_n(x)-\sqrt{\frac{2}{\pi}}\sin nx-\psi_n^{(1)}-\psi_n^{(2)}-\cdots-\psi_n^{(k)} =O\left(\frac{1}{n^{k+1}}\right). \tag{k} \]

Proof. It is easy to verify that

\[ \sum_{m\ne n} \frac{\displaystyle\int_0^\pi p\sin nx\sin mx\,dx}{n^2-m^2}\,\sin mx = \frac{1}{4n} \sum_{m\ne n} \frac{\displaystyle\int_0^\pi p\cos(n-m)x\,dx}{n-m}\,\sin mx + \]

\[ +\frac{1}{4n} \sum_{m\ne n} \frac{\displaystyle\int_0^\pi p\cos(n-m)x\,dx}{n+m}\,\sin mx + O\left(\frac{1}{n^{k+1}}\right) = S_1+S_2+O\left(\frac{1}{n^{k+1}}\right). \]

The first term

\[ S_1=\frac{1}{4n} \sum_{m\ne n} \frac{\displaystyle\int_0^\pi p\cos(n-m)x\,dx}{n-m}\,\sin mx = \]

\[ = -\frac{1}{2n} \sum_{m=1}^{n-1} \frac{\displaystyle\int_0^\pi p\cos mx\,dx}{m}\,\sin mx\cos nx + O\left(\frac{1}{n^{k+2}}\right) = \]

\[ = -\frac{1}{2n} \sum_{m=1}^{\infty} \left(\int_0^\pi\int_0^x p(t)\,dt\,\sin mx\,dx\right) \sin mx\cos nx + O\left(\frac{1}{n^{k+2}}\right) = \]

\[ = -\frac{\pi}{4n}\int_0^x p(t)\,dt\,\cos nx + O\left(\frac{1}{n^{k+2}}\right). \]

Now let us find the asymptotics of the second term \(S_2\). For \(k=1\), obviously,
\(S_2=O\left(\frac{1}{n^2}\right)\). For \(k>1\),

\[ S_2= \frac{1}{4n} \sum_{m\ne n} \frac{\displaystyle\int_0^\pi p\cos(n-m)x\,dx}{n+m}\,\sin mx = \sum_{m=1}^{n-1} \frac{\displaystyle\int_0^\pi p\cos mx\,dx}{4n^2-m^2}\,\cos mx\sin nx - \]

\[ -\frac{1}{2n} \sum_{m=1}^{n-1} \frac{m\displaystyle\int_0^\pi p\cos mx\,dx}{4n^2-m^2}\,\sin mx\cos nx + O\left(\frac{1}{n^{k+2}}\right) = \]

\[ = \frac{\pi}{8n^2} \left[ p(x)-\frac{1}{4n^2}p''(x)+\cdots+ \left(-\frac{1}{4n^2}\right)^{\left[\frac{k-1}{2}\right]} p^{\,2\left[\frac{k-1}{2}\right]}(x) \right]\sin nx + \]

\[ + \frac{\pi}{16n^3} \left[ p'(x)-\frac{1}{4n^2}p'''(x)+\cdots+ \left(-\frac{1}{4n^2}\right)^{\left[\frac{k-1}{2}\right]} p^{\,2\left[\frac{k-1}{2}\right]+1}(x) \right]\cos nx + O\left(\frac{1}{n^{k+1}}\right). \]

We have, therefore,

\[ \psi_n^{(1)}=-\sqrt{\frac{2}{\pi}}\,\frac{1}{2n}\int_0^x p(t)\,dt\cos nx+O\!\left(\frac{1}{n^2}\right) \qquad \text{for } k=1; \]

\[ \psi_n^{(1)}=-\sqrt{\frac{2}{\pi}}\,\frac{1}{2n}\int_0^x p(t)\,dt\cos nx +\sqrt{\frac{2}{\pi}}\,\frac{1}{4n^2}p(x)\sin nx +O\!\left(\frac{1}{n^3}\right) \qquad \text{for } k=2. \]

Hence, using the known asymptotic formula for the eigenfunctions

\[ \varphi_n(x)=\sqrt{\frac{2}{\pi}}\sin nx+ \sqrt{\frac{2}{\pi}}\,\frac{1}{n}a_1(x)\cos nx+ O\!\left(\frac{1}{n^2}\right), \]

where

\[ a_1(x)=-\frac12\int_0^x p(t)\,dt, \]

we obtain (1).

Let us proceed to the proof of (2). If one takes into account that

\[ \mu_n^{(1)}=\frac{2}{\pi}\int_0^\pi p\sin^2 nx\,dx =O\!\left(\frac{1}{n^3}\right), \]

then the first term of \(\psi_n^{(2)}\) is simplified:

\[ \frac{2}{\pi}\sum_{m\ne n} \frac{\displaystyle\int_0^\pi\bigl(p\psi_n^{(1)}-\mu_n^{(1)}\psi_n^{(1)}\bigr)\sin mx\,dx} {n^2-m^2}\sin mx = \]

\[ =\sqrt{\frac{2}{\pi}}\,\frac{2}{\pi}\,\frac{1}{n} \sum_{m\ne n} \frac{\displaystyle\int_0^\pi p a_1\cos nx\sin mx\,dx} {n^2-m^2}\sin mx +O\!\left(\frac{1}{n^3}\right). \]

Further, it is not difficult to verify the validity of the following transformations:

\[ \sum_{m\ne n} \frac{\displaystyle\int_0^\pi p a_1\cos nx\sin mx\,dx} {n^2-m^2}\sin mx = \]

\[ =-\frac{1}{2n}\sum_{m=1}^{n-1} \frac{\displaystyle\int_0^\pi p a_1\sin mx\,dx}{m} \cos mx\sin nx +O\!\left(\frac{1}{n^2}\right) = \]

\[ =-\frac{1}{2n}\sum_{m=1}^{\infty} \frac{\displaystyle\int_0^\pi p a_1\sin mx\,dx}{m} \cos mx\sin nx +O\!\left(\frac{1}{n^2}\right) = \]

\[ =-\frac{1}{8n}\sum_{m=1}^{\infty} \left[ \int_0^\pi \left(\int_0^x p(t)\,dt\right)^2 \cos mx\,dx \right] \cos mx\sin nx +O\!\left(\frac{1}{n^2}\right). \]

The second term of \(\psi_n^{(2)}\) is easily computed:

\[ -\frac12\sqrt{\frac{2}{\pi}}\int_0^\pi \psi_n^{(1)^2}\,dx\,\sin nx = -\frac{1}{8\pi}\sqrt{\frac{2}{\pi}}\,\frac{1}{n^2} \int_0^\pi\left(\int_0^x p(t)\,dt\right)^2 dx\,\sin nx +O\!\left(\frac{1}{n^3}\right). \]

Thus, the second perturbation term \(\psi_n^{(2)}\) has the asymptotic form

\[ \begin{aligned} \psi_n^{(2)} &= -\frac{1}{4\pi}\sqrt{\frac{2}{\pi}}\,\frac{1}{n^2} \sum_{m=1}^{\infty} \left[ \int_{0}^{\pi} \left(\int_{0}^{x} p(t)\,dt\right)^2 \cos mx\,dx \right]\cos mx\sin nx \\ &\quad -\frac{1}{8\pi}\sqrt{\frac{2}{\pi}}\,\frac{1}{n^2} \int_{0}^{\pi} \left(\int_{0}^{x} p(t)\,dt\right)^2 dx\,\sin nx +O\!\left(\frac{1}{n^3}\right) \\ &= -\sqrt{\frac{2}{\pi}}\,\frac{1}{8n^2} \left(\int_{0}^{x} p(t)\,dt\right)^2 \sin nx +O\!\left(\frac{1}{n^3}\right). \end{aligned} \]

Thus, we obtain

\[ \psi_n^{(1)}+\psi_n^{(2)} = \sqrt{\frac{2}{\pi}}\,\frac{1}{n}a_1(x)\cos nx + \sqrt{\frac{2}{\pi}}\,\frac{1}{n^2} \left[ \frac{1}{4}p(x) -\frac{1}{8} \left(\int_{0}^{x} p(t)\,dt\right)^2 \right]\sin nx +O\!\left(\frac{1}{n^3}\right). \]

According to the result of L. A. Dikii \((^2)\),

\[ \varphi_n(x) = \sqrt{\frac{2}{\pi}}\sin nx \left(1+\frac{a_2(x)}{n^2}\right) + \sqrt{\frac{2}{\pi}}\,\frac{1}{n}a_1(x)\cos nx + O\!\left(\frac{1}{n^3}\right), \]

where

\[ a_2(x)=\frac{1}{4}p(x)-\frac{1}{8} \left(\int_{0}^{x} p(t)\,dt\right)^2, \]

we conclude that

\[ \varphi_n-\psi_n^{(1)}-\psi_n^{(2)} = O\!\left(\frac{1}{n^3}\right). \]

Approximations of higher orders are considered analogously.

I express my deep gratitude to my scientific advisers A. A. Abramov and L. A. Dikii for their attention to this work and for valuable advice.

Computing Center
Academy of Sciences of the USSR

Received
14 XII 1959

CITED LITERATURE

1. L. A. Dikii, UMN, 13, no. 3(81), 111 (1958).
2. L. A. Dikii, Izv. AN SSSR, ser. matem., 19, 4, 187 (1955).