On Integral Perturbations in Differential Equations with Rapidly Oscillating Solutions

A. B. Vasil’eva

Recently many works have appeared devoted to integro-differential equations with a small parameter multiplying the highest derivative (see, for example, [1—11]). For integro-differential equations one may now regard as established an asymptotic regularity analogous to that which holds for differential equations with a small parameter multiplying the highest derivative; moreover, as in the latter case, it is valid when a certain characteristic equation has no roots lying on the imaginary axis. This regularity consists in the fact that the asymptotics of the solution has the form of a power expansion in \(\mu\) with the addition of a boundary-layer series whose terms are of the form

\[ \mu^k e^{-\frac{\alpha t}{\mu}}. \]

At the same time, one should note the following property: a boundary-layer function of order

\[ \mu^k e^{-\frac{\alpha t}{\mu}} \]

when substituted into the integral term gives a quantity of order \(\mu^{k+1}\). Therefore the integral term may, in a certain sense, be regarded as a perturbation. Since, however, a function of the form

\[ \mu^k e^{\frac{i\alpha t}{\mu}} \]

also gives, upon integration, a higher power of \(\mu\), it is natural to try to consider the integral term as a perturbation also in the case of rapidly oscillating solutions (when the characteristic equation mentioned above has purely imaginary roots).

In the present note we shall consider the simplest linear integro-differential equations which, if the integral term is omitted, describe an oscillatory process,

\[ \mu^2 y''+Q^2(x)y=0, \]

and we shall construct the asymptotics of solutions of such equations with respect to the parameter \(\mu\). We note that problems of this type occur in physics—for example, if in the equations of quantum mechanics one takes into account exchange forces, which have an integral character; at the same time, however, it is necessary to stipulate that the problems considered here have no direct application in physics and serve only as an example of a mathematical approach to the investigation of phenomena of this kind.

The asymptotics of the solution of the differential equation

\[ \mu^2 y''+Q^2(x)y=0 \]

can be obtained by means of the so-called WKB method (see, for example, [13]). Since, according to the considerations indicated above, the integral term is a perturbation, it is natural to try to apply the same formalism to the integro-differential equation as well, with certain modifications suggested by the results of the above-mentioned works.

It is precisely this idea that is implemented in the present paper. A modification of the WKB method for integro-differential equations is given. In § 1 the Cauchy problem for an equation of Volterra type is considered; in § 2, the eigenvalue problem for an equation of Fredholm type.

§ 1. APPLICATION OF THE WKB METHOD TO INTEGRO-DIFFERENTIAL EQUATIONS OF VOLTERRA TYPE

Consider the equation*:

\[ \mu^2 y'' + Q^2(x)y = \int_0^x K(x,t)y(t)\,dt + f(x) \tag{1} \]

and pose the Cauchy problem

\[ y(0)=y^0,\qquad y'(0)=y'. \tag{2} \]

We shall assume that \(Q(x)\) is different from zero everywhere in some interval of variation of \(x\), \(0 \le x \le l\)**. This same requirement was also imposed in the consideration of the differential equation [13]. Let us construct an asymptotic expansion with respect to the small parameter \(\mu\) for the solution of problem (2). First we give a description of the formal scheme for constructing the asymptotics, assuming that the differentiations arising in this process are permissible. After this, smoothness conditions will be formulated on the functions \(Q(x)\) and \(K(x,t)\) entering the equation, ensuring the required order of the remainder term.

Consider the homogeneous equation corresponding to (1):

\[ \mu^2 y'' + Q^2(x)y = \int_0^x K(x,t)y(t)\,dt. \tag{3} \]

Represent \(y\) in the form of the formal expansion

\[ y = e^{\frac{i}{\mu}\int_0^x Q(x)\,dx} \left[A_0(x)+\mu A_1(x)+\ldots\right] +\mu C_1(x)+\mu^2 C_2(x)+\ldots = \tag{4} \]

\[ = A(x,\mu)e^{\frac{i}{\mu}\int_0^x Q(x)\,dx}+C(x,\mu). \]

(When the WKB method is applied to a differential equation, the solution is sought in the form

\[ A(x,\mu)e^{\frac{i}{\mu}\int_0^x Q\,dx}, \]

without \(C(x,\mu)\). Substituting (4) into (1), we obtain on the left

\[ \mu^2\left(A''+\frac{2i}{\mu}A'Q+\frac{i}{\mu}AQ'\right) e^{\,i\int_0^x \frac{Q\,dx}{\mu}} + \]

\[ +Q^2 A e^{\,i\int_0^x \frac{Q\,dx}{\mu}} +\mu^2 C''+Q^2 C, \]

* We note that the case \(\mu^2 y''-Q^2(x)y\) (the nonoscillatory case) could also be considered by the WKB method, but we shall not do so here, since in this case other methods developed in the above-mentioned works are applicable.

** The simplest case is meant, when the so-called turning points are absent.

and on the right

\[ \int_0^x K(x,t)A(t,\mu)e^{\frac{i}{\mu}\int_0^t Q\,dx}\,dt + \int_0^x K(x,t)C(t,\mu)\,dt . \]

Integration by parts in the first term gives

\[ \begin{aligned} \int_0^x K(x,t)A(t,\mu)e^{\frac{i}{\mu}\int_0^t Qdt}\,dt &= \mu\,\frac{K(x,t)A(t,\mu)}{iQ(t)} e^{\frac{i}{\mu}\int_0^t Qdt}\bigg|_0^x \\ &\quad -\mu\int_0^x \left[\frac{K(x,t)A(t,\mu)}{iQ(t)}\right]_t e^{\frac{i}{\mu}\int_0^t Qdt}\,dt \\ &= \left\{ \mu\,\frac{K(x,t)A(t,\mu)}{iQ(t)} -\mu^2 \left[\frac{K(x,t)A(t,\mu)}{iQ(t)}\right]_t \frac{1}{iQ(t)} +\right. \\ &\quad\left. + \cdots +(-1)^{k-1}\frac{\mu^k}{i^k} \left[ \cdots \left[ \left[\frac{K(x,t)A(t,\mu)}{Q(t)}\right]_t \frac{1}{Q(t)} \right]_t \cdots \frac{1}{Q(t)} \right]_t \frac{1}{Q} \right\} e^{\frac{i}{\mu}\int_0^t Qdt}\bigg|_0^x +\cdots \end{aligned} \]

where the indicated operation is performed \(k-1\) times.

Equating on the left and on the right the coefficients of
\(e^{\frac{i}{\mu}\int_0^x Qdx}\) and the free terms, and then the coefficients at equal powers of \(\mu\), we obtain equations for determining \(A_0,A_1,\ldots\) and \(C_1,C_2,\ldots\). In particular,

\[ 2QA_0' + A_0\left[Q' + \frac{K(x,x)}{Q(x)}\right]=0. \]

Hence (we set the constant of integration equal to 1)

\[ A_0=\frac{1}{\sqrt Q}\, e^{-\frac12\int_0^x \frac{K(x,x)}{Q^2(x)}\,dx}. \tag{5} \]

Thus, the presence of the integral term does not affect the phase, but changes the amplitude of the principal term of the expansion. Further, for \(A_1\) and \(C_1\) the following relations are obtained:

\[ Q^2C_1(x)=\int_0^x K(x,t)C_1(t)\,dt +i\,\frac{K(x,0)}{Q^{3/2}(0)}, \tag{6} \]

\[ 2QA_1' + A_1\left[Q' + \frac{K(x,x)}{Q(x)}\right] = \frac{1}{i} \left\{ -A_0'' + \left[\frac{A_0K(x,t)}{Q(t)}\right]_t\bigg|_{t=x} \frac{1}{Q(x)} \right\}. \tag{7} \]

The left-hand side of the equations determining \(A_k\) is the same as in (7); the right-hand side will be a certain known function depending on the coefficients of the preceding indices. Put \(A_k(0)=0\) \((k>0)\). The \(C_k\) will be determined by integral equations differing from (6) only in the nonhomogeneity, which depends on the coefficients of the preceding indices. Thus one can determine successively all \(A_k\) and \(C_k\) and construct a formal solution, which we shall denote by \(y_1\):

\[ y_1=(A_0+A_1\mu+\ldots)e^{\frac{i}{\mu}\int_0^x Qdx}+C_1\mu+C_2\mu^2+\ldots \tag{8} \]

It will correspond to the complex-conjugate solution

\[ y_2=(\overline{A}_0+\overline{A}_1\mu+\ldots)e^{-\frac{i}{\mu}\int_0^x Qdx} +\overline{C}_1\mu+\overline{C}_2\mu^2+\ldots \tag{9} \]

Let us note that the structure of these formal solutions resembles the structure of the solutions for the case of nonoscillatory systems mentioned at the beginning of the article: \(C(x,\mu)\) is a power series in \(\mu\), while

\[ A(x,\mu)e^{-\frac{i}{\mu}\int_0^x Qdx} \]

is an analogue of a boundary-layer series, with the difference, however, that now the terms of this series are noticeable not only near the boundary point.

We shall now seek a formal solution of the nonhomogeneous equation (1) in the form of a power series \(B_0+B_1\mu+\ldots\). Substituting in (1) and comparing the coefficients of like powers of \(\mu\), we shall have

\[ Q^2B_0=\int_0^x K(x,t)B_0(t)dt+f(x), \]

\[ \cdot\quad \cdot\quad \cdot\quad \cdot\quad \cdot\quad \cdot\quad \cdot \]

\[ B''_{2m-2}+Q^2B_{2m}=\int_0^x K(x,t)B_{2m}(t)dt. \]

In order to construct the asymptotics of the solution of problem (1), (2), we proceed as follows. Denote

\[ A_k(x,\mu)=\sum_{i=0}^{k} A_k(x)\mu^k \]

(the notations \(B_k(x,\mu)\), \(C_k(x,\mu)\) are analogous). Put

\[ Y_k=\alpha\left[A_k(x,\mu)e^{\frac{i}{\mu}\int_0^x Qdx}+C_k(x,\mu)\right]+ \]

\[ +\beta\left[\overline{A}_k(x,\mu)e^{-\frac{i}{\mu}\int_0^x Qdx}+\overline{C}_k(x,\mu)\right]+B_k(x,\mu) \tag{10} \]

and require that \(Y_k\) satisfy the initial conditions (2):

\[ y^0=\alpha\left[A_k(0,\mu)+C_k(0,\mu)\right]+ \]

\[ +\beta\left[\overline{A}_k(0,\mu)+\overline{C}_k(0,\mu)\right]+B_k(0,\mu), \]

\[ y^1=\alpha\left[A'_k(0,\mu)+C'_k(0,\mu)+\frac{i}{\mu}A_k(0,\mu)Q(0)\right]+ \tag{11} \]

\[ +\beta\left[\overline{A}'_k(0,\mu)+\overline{C}'_k(0,\mu)-\frac{i}{\mu}\overline{A}_k(0,\mu)Q(0)\right]+B'_k(0,\mu). \]

This system is solvable with respect to \(\alpha\) and \(\beta\), since its determinant at \(\mu=0\) is
\(-2iQ(0)A_0^2(0)=-2i\ne0\).

We shall assume that the function \(Q(x)\) is continuously differentiable \(2k+1\) times on the interval \(0\le x\le l\), and that \(K(x,t)\) is continuously differentiable with respect to both arguments \(2k\) times. Then the following is valid.

Theorem. The expression (10), in which \(\alpha\) and \(\beta\) are determined from (11), represents the solution \(y(x,\mu)\) of the Cauchy problem (2) for equation (1) with asymptotic accuracy \(\mu^{k+1}\), i.e.

\[ |y(x,\mu)-Y_k|<C\mu^{k+1}, \tag{12} \]

where \(C\) does not depend on \(x\) and \(\mu\) for \(\mu\le\mu_0\) and \(0\le x\le l\).

Remark. When substituting \(\alpha\) and \(\beta\) into (10) and multiplying by \(A_k\), \(C_k\), etc., terms up to order \((\mu^k)\) inclusive should be retained;* the asymptotic accuracy of the simpler expression obtained in this way, compared with \(Y_k\), will be the same as that for \(Y_k\).

For the proof, set \(\Delta=y-Y_k\). Using the relations defining \(y_1,y_2\), it is not difficult to obtain that

\[ \mu^2\Delta''+Q^2\Delta=\int_0^x K(x,t)\Delta(t)\,dt+R(x,\mu), \tag{13} \]

\[ \Delta(0)=0,\qquad \Delta'(0)=0, \]

where \(R(x,\mu)=\alpha P+\beta\overline{P}-\mu^{k+2}B''_k(x)\),

\[ P=-\mu^{k+2}A''_k(x)e^{\frac{i}{\mu}\int_0^x Q\,dx}+ \]

\[ +\left\{-\mu^2\left[\frac{K(x,t)A_k(t)\mu^k}{iQ(t)}\right]_t\frac{1}{iQ(t)}\bigg|_{t=x}+\cdots+\right. \]

\[ \left. +(-1)^k\frac{\mu^{k+1}}{(i)^{k+1}} \left[\cdots\left[\frac{K(x,t)(A_k\mu^k+\cdots+A_1\mu)}{Q}\right]_t \right. \right. \]

\[ \left. \left. \times\cdots\frac{1}{Q}\right]_t \frac{1}{Q}\bigg|_{t=x} \right\} e^{\frac{i}{\mu}\int_0^x Q\,dx} - \]

\[ -\left\{ \mu^{k+1}\frac{K(x,0)A_k(0)}{iQ(0)}+\cdots+ (-1)^k\frac{\mu^{k+1}}{(i)^{k+1}}\times \right. \]

* Here and below the notation \((\mu^\alpha)\) is used for quantities satisfying the inequality \(|(\mu^\alpha)|\le C\mu^\alpha\), where \(C\) is a constant independent of \(\mu\).

\[ \times \left\{\ldots \left[\left[\frac{K(x,t)(A_k\mu^k+\ldots+A_0)}{Q}\right]_t \ldots \frac{1}{Q}\right]_t \frac{1}{Q}\bigg|_{t=0}\right\} -\mu^{k+2}C_k''(x). \]

Under the assumptions indicated above concerning \(Q(x)\) and \(K(x,t)\), as is seen from this expression, \(R(x,\mu)\) has a continuous derivative with respect to \(x\), and \(R=(\mu^{k+1})\) and \(R_x'=(\mu^{k+1})\).

To estimate \(\Delta\), we pass from (13) to the integral equation

\[ \Delta = \int_0^x Y(x,\xi)\frac{d\xi}{\mu^2} \int_0^\xi K(\xi,t)\Delta(t)\,dt + \int_0^x \frac{R(\xi,\mu)}{\mu^2}Y(x,\xi)\,d\xi, \tag{14} \]

where \(\mu^2Y_{xx}''+Q^2Y=0,\quad Y|_{x=\xi}=0,\quad Y_x'|_{x=\xi}=1\). Equation (14) may also be written in the form

\[ \Delta = \int_0^x \Delta(t)\,dt \int_t^x \frac{Y(x,\xi)K(\xi,t)}{\mu^2}\,d\xi + \int_0^x \frac{R(\xi,\mu)}{\mu^2}Y(x,\xi)\,d\xi \]

or, briefly,

\[ \Delta=\int_0^x \Delta(t)G(x,t,\mu)\,dt+H(x,\mu). \tag{15} \]

For \(Y(x,\xi)\) one can obtain the representation, using the usual WKB method,

\[ Y(x,\xi)= \mu\,\frac{1}{\sqrt{Q(x)Q(\xi)}}\, \sin \frac{\displaystyle\int_\xi^x Q\,dt}{\mu} + (\mu^2). \]

We then have

\[ H(x,\mu)= \int_0^x \frac{R(\xi,\mu)}{\mu^2}Y(x,\xi)\,d\xi = (\mu^{k+1})+ \]

\[ + \frac{1}{\mu} \int_0^x \frac{R(\xi,\mu)}{\sqrt{Q(x)Q(\xi)}}\, \sin \frac{\displaystyle\int_\xi^x Q\,dt}{\mu}\,d\xi. \]

The integral term on the right is also \((\mu^{k+1})\), as is not difficult to verify by integrating by parts and using the properties of \(R(x,\mu)\) noted above. Thus, \(H(x,\mu)=(\mu^{k+1})\). In exactly the same way, by integration by parts we obtain \(G(x,t,\mu)=(\mu^0)\). In other words, \(|G|<C_1,\ |H|<C_2\mu^{k+1}\).

Now from (15) it is already not difficult to obtain (12), for example by the method of a majorant equation, introducing \(\bar{\Delta}\) satisfying the integral equation

\[ \bar{\Delta}=\int_0^x C_1\bar{\Delta}\,dt+\mu^{k+1}C_2 \]

and related to \(\Delta\) by the inequality \(|\Delta|<\bar{\Delta}\). It is easy to find \(\bar{\Delta}\) by direct calculation and obtain

\[ \bar{\Delta}=\mu^{k+1}C_2e^{C_1x}. \]

Consequently,

\[ |\Delta|<\mu^{k+1}C_2e^{C_1x},\qquad 0\le x\le l, \]

i.e. \(\Delta=(\mu^{k+1})\).

§ 2. AN EQUATION OF FREDHOLM TYPE.

THE EIGENVALUE PROBLEM

Consider the equation

\[ \mu^2 y''+\Lambda^2 Q^2(x)y=\int_0^l K(x,t)y(t)\,dt \tag{16} \]

under the additional conditions \(y(0)=0,\ y(l)=0\).

As before, it is assumed that \(Q(x)\ne 0\) for \(0\le x\le l\). With regard to equation (16), the following questions may be posed.

1. Consider the differential equation

\[ y''+\lambda^2 Q^2(x)y=0 \tag{17} \]

under the same additional conditions. As is known, this problem has eigenvalues of the following form (the asymptotics are given for large indices):

\[ \lambda_n=\frac{\pi n}{\displaystyle\int_0^l Q\,dx} \left[1+O\left(\frac{1}{n^2}\right)\right]. \tag{18} \]

Denoting \(\displaystyle \int_0^l Q\,dx/\pi n=\mu\), we rewrite (17) in the form

\[ \mu^2 y''+\Lambda^2 Q^2(x)y=0,\qquad y(0)=y(l)=0. \tag{19} \]

This problem has an eigenvalue close to unity: \(\Lambda=1+O(\mu^2)\). Let us now turn to equation (16), which differs from (19) by an integral term which, for the reason indicated above, may be regarded as a certain perturbation. It is natural to ask whether problem (16) has an eigenvalue in a neighborhood of \(\Lambda=1\), and if so, what its asymptotics with respect to \(\mu\) are.

1. Let the corresponding integral equation for \(\mu=0\) have spectrum \(\Lambda_1,\Lambda_2,\ldots\). It is asked whether, in a neighborhood of \(\Lambda_p\), for sufficiently small \(\mu\), there is an eigenvalue of problem (16), and what its asymptotics with respect to \(\mu\) are. One may expect that the principal term of this asymptotics will be \(\Lambda_p\).

To answer both questions posed, it is necessary to construct the asymptotics of the general solution of equation (16), in the first case for \(\Lambda\) close to \(1\) and, generally speaking, far from \(\Lambda_k\) \((k=1,2,\ldots)\), and in the second case for \(\Lambda\) close to some \(\Lambda_p\).

In the present article, problem 1 will be investigated in detail. With regard to problem 2, we shall confine ourselves only to some remarks at the end of the article.

We shall seek a fundamental system of solutions of equation (16) in the form \(y_1=u_1+\Delta,\ y_2=y_1\), where \(u_1\) is the usual WKB representation in the absence of the integral term, i.e.

\[ y_1=e^{\frac{i\Lambda}{\mu}\int_0^x Q\,dx}\,(A_0+\mu iB_1)+\Delta \tag{20} \]

\[ \left(A_0=\frac{1}{\sqrt{Q(x)}},\quad 2B_1'Q\Lambda+2B_1Q'\Lambda=A_0'',\quad B_1(0)=0\right). \]

By direct substitution of (20) into (16), it is not difficult to obtain the following equation for \(\Delta\):

\[ \mu^2\Delta''+\Lambda^2Q^2\Delta=\int_0^l K(x,t)\Delta(t)\,dt+ \tag{21} \]

\[ +\mu\left[a_1(x)e^{\frac{i\Lambda}{\mu}\int_0^l Q\,dx}+b_1(x)\right]+R_1(x,\mu), \]

where

\[ a_1(x)=\frac{A_0(l)K(x,l)}{i\Lambda Q(l)},\qquad b_1(x)=-\frac{A_0(0)K(x,0)}{i\Lambda Q(0)}. \]

If \(K(x,t)\) and \(Q(t)\) have continuous derivatives with respect to \(t\) up to and including the third order, then one may assert that \(R_1(x,\mu)=(\mu^2)\), \(R'_{1x}(x,\mu)=(\mu^2)\). Put \(\Delta=\mu\Delta_1+\eta\). Then

\[ \Lambda^2Q^2\Delta_1=\int_0^l K(x,t)\Delta_1\,dt+ \left[a_1(x)e^{\frac{i\Lambda}{\mu}\int_0^l Q\,dx}+b_1(x)\right], \tag{22} \]

\[ \mu^2\eta''+\Lambda^2Q^2\eta=\int_0^l K(x,t)\eta\,dt+R_2,\qquad R_2=R_1-\mu^2\Delta_1''. \tag{23} \]

Consider the integral equation (22). We shall assume \(K(x,t)\) symmetric. Introducing \(\delta_1=\Delta_1Q(x)\) and \(\widetilde K(x,t)=K(x,t)/Q(x)Q(t)\), one may pass to the integral equation

\[ \Lambda^2\delta_1=\int_0^l \widetilde K(x,t)\delta_1\,dt+ \frac{1}{Q}\left[a_1e^{\frac{i\Lambda}{\mu}\int_0^l Q\,dx}+b_1\right], \tag{24} \]

whose solution can be written in the form

\[ \delta_1=\frac{1}{Q(x)\Lambda^2} \left[a_1(x)e^{\frac{i\Lambda}{\mu}\int_0^l Q\,dx}+b_1(x)\right] +\frac{1}{\Lambda^2}\sum_{k=1}^{\infty} \frac{\Lambda_k^2 f_k\varphi_k(x)}{\Lambda^2-\Lambda_k^2}, \tag{25} \]

\[ f_k=\int_0^l \frac{1}{Q(x)} \left[a_1(x)e^{\frac{i\Lambda}{\mu}\int_0^l Q\,dx}+b_1(x)\right]\varphi_k(x)\,dx, \]

where \(\varphi_k(x)\) are the normalized eigenfunctions of the kernel \(\widetilde K(x,t)\).

In view of problem 1, we shall suppose that \(\Lambda\) lies in a neighborhood of the value \(1\) and at a distance \((\mu^0)\) (uniformly in \(k\)) from any \(\Lambda_k\).

We shall not set ourselves the goal of obtaining a complete asymptotic expansion of the fundamental system of solutions of equation (16), but shall find only the principal terms. Form the expression \(Y_1=u_1+\mu\Delta_1\). As the investigation shows, this expression differs from the exact solution \(y_1\), satisfying the same initial conditions

\[ y_1(0)=A_0(0)+\mu\Delta_1(0), \]

\[ y_1'(0)=\frac{i\Lambda\sqrt{Q(0)}}{\mu}+A_0'(0)+\mu iB_1'(0)+\mu\Delta_1'(0), \]

by a quantity \((\mu^2)\). The consideration of \(y_2\) is analogous. The proof can be carried out by passing

from the integro-differential equation (23) with zero initial conditions, which is satisfied by the remainder term \(\eta\), to the integral equation

\[ \eta=\int_{0}^{l}G(x,t,\Lambda,\mu)\eta(t)\,dt+H(x,\Lambda,\mu), \tag{26} \]

where

\[ G=\int_{0}^{x}\frac{Y(x,\xi)K(\xi,t)}{\mu^{2}}\,d\xi,\qquad H=\int_{0}^{x}\frac{R_{2}}{\mu^{2}}Y(x,\xi)\,d\xi, \]

\(Y(x,\xi)\) is the Cauchy function already encountered above, in § 1. We shall not give the proof in full here, but note only its main points. By integration by parts it is not difficult to obtain that \(H=(\mu^{2})\) (analogously to what was done in § 1) and that

\[ G=-\frac{1}{\Lambda^{2}Q^{1/2}(x)} \left[ \frac{K(x,t)}{Q^{3/2}(x)} -\frac{K(0,t)}{Q^{3/2}(0)} \cos\frac{\Lambda}{\mu}\int_{0}^{x}Q\,dx \right]+(\mu). \tag{27} \]

Using this representation of \(G\), we see that, for all sufficiently small \(\mu\), in some sufficiently small \(\omega\)-neighborhood of \(\Lambda=1\), the homogeneous equation corresponding to (26) has only the trivial solution. Suppose the contrary. Denote \(\bar{\eta}=\max \eta\), where \(\eta\) is the corresponding eigenfunction. Write the homogeneous equation (26) in the form

\[ \eta=-\frac{1}{\Lambda^{2}}\int_{0}^{l}\frac{K(x,t)}{Q^{2}(x)}\eta\,dt+F(x,\Lambda,\mu), \]

where

\[ F=-\frac{1}{\Lambda^{2}Q^{1/2}(x)Q^{3/2}(0)} \cos\frac{\Lambda}{\mu}\int_{0}^{x}Q\,dx \times \]

\[ \times\int_{0}^{l}K(0,t)\eta\,dt+\int_{0}^{l}(\mu)\eta\,dt. \]

Introducing the notation

\[ a(x,\Lambda,\mu)=-\frac{1}{\Lambda^{2}Q^{1/2}(x)Q^{3/2}(0)} \cos\frac{\Lambda}{\mu}\int_{0}^{x}Q\,dx, \]

\[ B=\int_{0}^{l}K(0,t)\eta\,dt,\qquad \beta=\int_{0}^{l}(\mu)\eta\,dt, \]

we write

\[ \tilde{\eta}=\int_{0}^{l}\overline{R}(aB+\beta)\,dt+aB+\beta, \]

where \(\overline{R}\) is the resolvent of the integral equation obtained from (16) for \(\mu=0\). By assumption, \(\Lambda=1\) is not an eigenvalue of this equation, hence

therefore, \(\overline R\) is continuous in \(x,t,\Lambda\) for \(0 \leq x \leq l\), \(0 \leq t \leq l\), \(\Lambda \in \omega\), if the \(\omega\)-neighborhood of the value \(\Lambda=1\) is sufficiently small, and then \(|\overline R|<C\). The estimates \(|\alpha|<C\), \(\beta=(\mu)\overline\eta\), are also valid. We further have

\[ \int_0^l \overline R(\alpha B+\beta)\,dt=(\mu)B+(\mu)\overline\eta. \]

Hence

\[ \eta=\alpha B+(\mu)\overline\eta+(\mu)B \]

and, consequently,

\[ B=B\int_0^l K(0,t)\,dt+(\mu)\overline\eta+(\mu)B=(\mu)B+(\mu)\overline\eta. \]

From this relation, for sufficiently small \(\mu\), we obtain \(B=(\mu)\overline\eta\), and then

\[ \overline\eta=\alpha(\mu)\overline\eta+(\mu)\overline\eta+(\mu)(\mu)\overline\eta=(\mu)\overline\eta, \]

i.e.

\[ \overline\eta \leq C\mu\,\overline\eta . \]

It follows that \(\overline\eta=0\), provided only that \(\mu\) is sufficiently small, and thus the homogeneous equation corresponding to (26) has only the trivial solution.

It follows from this that the Fredholm resolvent for (26) is continuous, and that its derivatives with respect to \(\Lambda\) are continuous for \(0 \leq x \leq l\), \(0 \leq t \leq l\), all \(\mu \leq \mu_0\) and \(\Lambda \in \omega\). To prove this one may use Fredholm formulas (see, for example, [12]). On the basis of these properties of the resolvent, it is not difficult to verify that \(\eta=(\mu^2)\), and also to establish the continuity of \(y_1\) and \(y_2\) in \(x\) and \(\Lambda \in \omega\), and the continuity of the derivatives with respect to \(\Lambda\) of these solutions.

Thus,

\[ \begin{aligned} y_1&=u_1+\mu\Delta_1+(\mu^2),\\ y_2&=\overline u_1+\mu\overline\Delta_1+(\mu^2). \end{aligned} \tag{28} \]

Similarly one can obtain the asymptotics for the derivatives with respect to \(\Lambda\) of \(y_1\) and \(y_2\), namely:

\[ \begin{aligned} \frac{\partial y_1}{\partial \Lambda} &= \frac{iA_0}{\mu} \int_0^x Q\,dx\, e^{\frac{i\Lambda}{\mu}\int_0^x Q\,dx} +(\mu^0), \\[6pt] \frac{\partial y_2}{\partial \Lambda} &= -\frac{iA_0}{\mu} \int_0^x Q\,dx\, e^{-\frac{i\Lambda}{\mu}\int_0^x Q\,dx} +(\mu^0). \end{aligned} \tag{29} \]

(We note that these expressions may be obtained by formal differentiation of (28).)

Using the asymptotics (28), (29), we shall prove that in a neighborhood of \(\Lambda=1\) there is an eigenvalue of problem (16), and we shall find its asymptotics with accuracy \((\mu^2)\). We shall seek the solution of problem (16) in the form \(y=\alpha y_1+\beta y_2\). Requiring \(y(0)=0\), \(y(l)=0\), and equating to zero the determinant \(\Delta(\mu,\Lambda)\) of the resulting system with respect to \(\alpha\) and \(\beta\), we arrive at an equation for determining the eigenvalues \(\Delta(\mu,\Lambda)=0\). With accuracy \((\mu^2)\), \(\Delta\) has the form

\[ \begin{aligned} &- e^{-\frac{i\Lambda}{\mu}\int_0^l Q\,dx} \left\{ \frac{1}{\sqrt{Q(0)Q(l)}}-\mu\,\frac{iB_1(l)}{\sqrt{Q(0)}}+\mu\,\frac{\Delta_1(0)}{\sqrt{Q(l)}} \right\} +\mu\,\frac{\overline{\Delta}_1(l)}{\sqrt{Q(0)}} \\ &\quad - e^{\frac{i\Lambda}{\mu}\int_0^l Q\,dx} \left\{ \frac{1}{\sqrt{Q(0)Q(l)}}+\mu\,\frac{iB_1(l)}{\sqrt{Q(0)}}+\mu\,\frac{\overline{\Delta}_1(0)}{\sqrt{Q(l)}} \right\} -\mu\,\frac{\Delta_1(l)}{\sqrt{Q(0)}} . \end{aligned} \tag{30} \]

Hence, in the zeroth approximation we have

\[ \Delta_0 = -\frac{2i}{\sqrt{Q(0)Q(l)}}\, \sin \frac{\Lambda}{\mu}\int_0^l Q\,dx = 0, \]

\(\Lambda=1\) is one of the roots of this equation. Consider the interval
\[ |\Lambda-1|=\frac{1}{2n}. \]
At the boundary of this interval
\[ \sin \frac{\Lambda}{\mu}\int_0^l Q\,dx = \sin \Lambda\pi n = \sin\left(1\pm\frac{1}{2n}\right)\pi n = \pm(-1)^n, \]
i.e. \(\Delta_0/i\) assumes values of opposite sign. And since \(\Delta=\Delta_0+(\mu)\), for sufficiently small \(\mu\) the same property is possessed by \(\Delta/i\). By continuity of \(y_1\) and \(y_2\), \(\Delta/i\) is a continuous function of \(\Lambda\). Then in the interval
\[ |\Lambda-1|\le \frac{1}{2n} \]
the equation \(\Delta=0\) has a root. If, moreover, (33) is taken into account, then it is not difficult to obtain that \(\Delta/i\) is a monotone function of \(\Lambda\), since the principal term of the derivative of \(\Delta/i\) with respect to \(\Lambda\) has the form

\[ -\frac{2}{\mu}\, \frac{\displaystyle\int_0^l Q\,dx}{\sqrt{Q(l)Q(0)}}\, \cos \frac{\Lambda}{\mu}\int_0^l Q\,dx \]

and preserves its sign in the indicated interval. By virtue of the monotonicity of \(\Delta/i\), the root of the equation \(\Delta=0\) in the interval
\[ |\Lambda-1|\le \frac{1}{2n} \]
will be unique.

Let us find the asymptotics of this root. Equating (30) to zero, we obtain (for convenience we indicate the dependence of \(\Delta_1\) not only on \(x\), but also on \(\Lambda\))

\[ -\frac{1}{\sqrt{Q(0)Q(l)}}\, \sin \frac{\Lambda}{\mu}\int_0^l Q\,dx + \frac{\mu B_1}{\sqrt{Q(0)}}\, \cos \frac{\Lambda}{\mu}\int_0^l Q\,dx - \]

\[ -\frac{\mu}{\sqrt{Q(0)}}\,\operatorname{Im}\Delta_1(l,\Lambda) + \frac{\mu}{\sqrt{Q(l)}}\, \operatorname{Im} \left[ \Delta_1(0,\Lambda)\, e^{-\frac{i\Lambda}{\mu}\int_0^l Q\,dx} \right] =0. \tag{31} \]

Put
\[ \Lambda=1+\frac{\delta}{\pi n}. \]
Equation (31) shows that \(\delta\) must be small. Then

\[ \sin \frac{\Lambda}{\mu}\int_0^l Q\,dx = \sin \left(1+\frac{\delta}{\pi n}\right)\pi n = (-1)^n \sin \delta \sim (-1)^n\delta. \]

Next, \(\Delta_1(x,1)\) satisfies the equation

\[ Q^2\Delta_1=\int_0^l K(x,t)\Delta_1\,dt -i\left[\frac{K(x,l)}{Q(l)^{3/2}}e^{\pi n i} -\frac{K(x,0)}{Q(0)^{3/2}}\right], \]

whence it is seen that \(\Delta_1\) is purely imaginary, i.e. \(\Delta_1(x,1)=iD(x)\). Taking all this into account, we obtain from (31) the following expression for \(\delta\):

\[ \delta=\mu\left[B_1(l,1)\sqrt{Q(l)}+(-1)^{n+1}D_1(l)\sqrt{Q(l)}+D_1(0)\sqrt{Q(0)}\right]=\mu\Lambda_1. \tag{32} \]

Thus, the asymptotic formula for \(\Lambda\) has the form

\[ \Lambda=1+\Lambda_1\frac{\displaystyle\int_0^l Q\,dx}{(\pi n)^2}. \tag{33} \]

For \(D_1=0\), from (33) we obtain (18) with an explicit expression for \(O\!\left(\frac{1}{n^2}\right)\).

Let us formulate the result obtained as a theorem. We list once more the conditions under which the derivation was carried out, and call the collection of these conditions conditions A: the kernel \(K(x,t)\) is symmetric; \(K(x,t)\) and \(Q(x)\) possess continuous derivatives up to the third order inclusive; the integral equation obtained from (16) for \(\mu=0\) has no eigenvalue equal to \(1\); \(Q(x)\ne 0\) for \(0\le x\le l\).

Theorem. Let in equation (16) \(\mu=\dfrac{1}{\pi n}\displaystyle\int_0^l Q\,dx\). Then, under conditions A, in some neighborhood of \(\Lambda=1\) (of order \(\dfrac{1}{n}\)) problem (16) has a unique eigenvalue, whose asymptotic form is (33), where \(\Lambda_1\) is given by formula (32). The principal term of this eigenvalue, equal to \(1\), coincides with the principal term of the corresponding eigenvalue of problem (19).

Thus, the presence of the integral term in the equation affects not the principal term of \(\Lambda\), but the correction of order \(1/n^2\).

Remark concerning case 2. In order to study case 2, it is necessary to obtain a representation for \(y_1\) and \(y_2\) in the region \(\Lambda-\Lambda_p=O(\mu)\). In expression (25), the principal term in this case will be

\[ \frac{\Lambda_p^2}{\Lambda^2}\, \frac{f_p\varphi_p(x)}{\Lambda^2-\Lambda_p^2} = \frac{\Lambda_p^2 f_p\varphi_p(x)}{\Lambda^2\varkappa\mu} \qquad (\Lambda^2-\Lambda_p^2=\varkappa\mu,\ \varkappa\sim 1) \tag{34} \]

and, therefore, the principal term \(y_1\) has the form \((y_1-Y_1=(\mu))\)

\[ Y_1=\frac{1}{\sqrt{Q(x)}}\, e^{\frac{i\Lambda}{\mu}\int_0^x Q\,dx} + \frac{\Lambda_p^2 f_p\varphi_p(x)}{\Lambda^2\varkappa Q(x)}. \tag{35} \]

Similarly, the principal term \(y_2\)

\[ Y_2=-\frac{1}{\sqrt{Q(x)}}e^{-\frac{i\Lambda}{\mu}\int_0^x Q\,dx} +\frac{\Lambda_p^2\overline{f}_p\,\varphi_p(x)}{\Lambda^2\varkappa Q(x)}. \tag{36} \]

Representing now the solution in the form \(y=\alpha y_1+\beta y_2\) and satisfying the boundary conditions, we obtain, by setting the determinant equal to zero and retaining terms of zero order,

\[ -2iA_0(0)A_0(l)\sin\frac{\Lambda}{\mu}\int_0^l Q\,dx +\frac{1}{\varkappa}\left[ A_0(0)(\overline{B}_0(l)-B_0(l))+\right. \]

\[ \left. + A_0(l)\left( B_0(0)e^{-\frac{i\Lambda}{\mu}\int_0^l Q\,dx} -\overline{B}_0(0)e^{\frac{i\Lambda}{\mu}\int_0^l Q\,dx} \right)\right]=0, \tag{37} \]

where

\[ A_0(x)=\frac{1}{\sqrt{Q(x)}}, \qquad B_0(x)=\frac{f_p\varphi_p(x)}{Q(x)}. \]

Hence, after some transformations, we shall have

\[ \varkappa=\Lambda_p\left\{ \cos\frac{\Lambda}{\mu}\int_0^l Q\,dx \left[\sqrt{Q(l)}\,\psi_p^2(l)+\sqrt{Q(0)}\,\psi_p^2(0)\right] -\psi_p(0)\psi_p(l)\left[\sqrt{Q(l)}+\right.\right. \]

\[ \left.\left. +\sqrt{Q(0)}\right]\right\} \left\{A_0(0)A_0(l)\sin\frac{\Lambda}{\mu}\int_0^l Q\,dx\right\}^{-1}, \tag{38} \]

where

\[ \Lambda=\Lambda_p+\frac{\varkappa}{2\Lambda_p}\mu . \]

This equation with respect to \(\varkappa\) has, generally speaking, an infinite number of roots, and therefore one may expect that in the neighborhood of each eigenvalue \(\Lambda_p\) of the equation

\[ \Lambda^2 Q^2(x)y=\int_0^l K(x,t)y\,dt \]

there is, generally speaking, an infinite number of eigenvalues of problem (16).

References

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4. Bykov Ya. V. and Imanaliev M. In: Investigations on Integro-Differential Equations in Kirghizia. Publishing House of the Academy of Sciences of the Kirghiz SSR, issue 2, 1962, pp. 3–20.
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1. K. I. Kvantaliani. Reports of the Academy of Sciences of the Georgian SSR, 27, No. 2, 1961, Tbilisi.

2. K. I. Kvantaliani. Candidate’s dissertation, Tbilisi, 1963.

3. A. B. Vasil’eva and V. F. Butuzov. In: Numerical Methods for Solving Differential and Integral Equations and Boundary Problems (supplement to Journal of Computational Mathematics and Mathematical Physics, 1964). Nauka, Moscow, pp. 183–191.

4. R. Courant and D. Hilbert. Methods of Mathematical Physics, I, Chap. III, § 7, GITTL, 1951.

5. E. Kamke. Handbook of Ordinary Differential Equations, IL, Moscow, 1950, p. 209.

6. M. Imanaliev. Doctoral dissertation, Frunze, 1964.

Received by the editors
January 18, 1965

Moscow State University
named after M. V. Lomonosov