SOME EIGENVALUE PROBLEMS FOR INTEGRO-DIFFERENTIAL EQUATIONS WITH A SMALL PARAMETER AT THE HIGHEST DERIVATIVE

A. B. VASIL'EVA, V. F. BUTUZOV

In [1] the eigenvalues and eigenfunctions of the integro-differential equation

\[ \mu^2 \frac{d^2 y}{dx^2}+\lambda^2 Q^2(x)y=\int_0^l K(x,t)y(t)\,dt \tag{1} \]

with boundary conditions

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

were studied. It was proved that, in a neighborhood of each eigenvalue of the same boundary-value problem (2) for the differential equation obtained by omitting the integral term in (1),

\[ \mu^2 \frac{d^2 y}{dx^2}+\lambda^2 Q^2(x)y=0, \tag{3} \]

for sufficiently small \(\mu\) there exists an eigenvalue of problem (1)—(2). The asymptotics of the eigenvalues and eigenfunctions of this problem for small \(\mu\) were then constructed.

Thus, in [1] the connection was studied between the eigenvalues of the integro-differential equation (1) and the eigenvalues of the corresponding differential equation (3). However, the spectrum of eigenvalues of the integro-differential equation is not exhausted by these eigenvalues. If now in equation (1) the term \(\mu^2 \dfrac{d^2y}{dx^2}\) is omitted, then the resulting integral equation (which we shall call degenerate, in accordance with the terminology adopted for equations with a small parameter at the highest derivative)

\[ \lambda^2 Q^2(x)y=\int_0^l K(x,t)y(t)\,dt \tag{4} \]

may have its own spectrum of eigenvalues, in neighborhoods of which eigenvalues of problem (1)—(2) may also be found.

Thus there arises the problem of the connection between the eigenvalues of the integro-differential equation and the eigenvalues of the degenerate integral equation. In the present paper this question is investigated for certain simpler cases than (1). Here the parameter \(\lambda\) will be written as a factor at the integral term, and we shall consider only those equations for which the corresponding differential equation, obtained by omitting

of the integral term, does not lie on the spectrum (see below (5), (6) and (8), (9)).

In § 1 the Cauchy problem is considered for a first-order integro-differential equation with zero initial condition

\[ \mu \frac{dy}{dx}+A(x)y=\lambda \int_0^1 K(x,t)y(t)\,dt, \tag{5} \]

\[ y(0)=0 \tag{6} \]

(\(\mu\) is a small positive parameter, \(\lambda\) is a certain, generally speaking, complex parameter). Suppose that the corresponding degenerate equation

\[ A(x)y=\lambda \int_0^1 K(x,t)y(t)\,dt \tag{7} \]

has eigenvalues \(\overline{\lambda}_1,\overline{\lambda}_2,\ldots,\overline{\lambda}_n,\ldots\). The investigation shows that, for sufficiently small \(\mu\), in a neighborhood of each \(\overline{\lambda}_i\) there exists some value \(\lambda_i(\mu)\) which is an eigenvalue of problem (5), (6) in the sense that for \(\lambda=\lambda_i(\mu)\) there exists a nontrivial solution of problem (5)—(6). As \(\mu\to 0\), \(\lambda_i(\mu)\to \overline{\lambda}_i\), and the corresponding eigenfunction tends to one of the eigenfunctions of the degenerate equation (7) everywhere except at the point \(x=0\), in whose neighborhood a boundary layer arises. For \(\lambda=\lambda_i(\mu)\), generally speaking (i.e., for all \(\mu\), with the exception, perhaps, of certain values), there is no solution of equation (5) satisfying the initial condition

\[ y(0)=y^0\ne 0. \]

If, however, \(\lambda\ne \lambda_i(\mu)\), then there exists a unique solution of the Cauchy problem with initial condition \(y(0)=y^0\). If, in addition, \(\lambda\ne \overline{\lambda}_i\) for every \(i\), then, as \(\mu\to 0\), this solution tends to zero, having a boundary layer in a neighborhood of the point \(x=0\). (The case \(\lambda=\overline{\lambda}_i\) was considered by us in another paper.)

In § 2 an analogous investigation is carried out for the problem

\[ \mu^2 \frac{d^2y}{dx^2}-A^2(x)y=\lambda \int_0^1 K(x,t)y(t)\,dt, \tag{8} \]

\[ y(0)=0,\qquad y(1)=0. \tag{9} \]

§ 1. Thus, consider problem (5)—(6). We shall assume that \(A(x)>0\) for \(0\le x\le 1\). Let us pass to the integral equation

\[ y(x)=\int_0^x \frac{1}{\mu} e^{-\frac{1}{\mu}\int_{\tau}^{x} A(x)\,dx} \,d\tau\,\lambda \int_0^1 K(\tau,t)y(t)\,dt \]

or

\[ y(x)=\lambda \int_0^1 G(x,t,\mu)y(t)\,dt, \tag{10} \]

where

\[ G(x,t,\mu)=\int_0^x \frac{1}{\mu}e^{-\frac{1}{\mu}\int_\tau^x A(x)\,dx}K(\tau,t)\,d\tau . \]

Integrating the expression for \(G(x,t,\mu)\) by parts, we obtain

\[ \begin{aligned} G(x,t,\mu) &=\frac{K(x,t)}{A(x)} -\frac{K(0,t)}{A(0)}e^{-\frac{1}{\mu}\int_0^x A(x)\,dx} \\ &\quad -\int_0^x\left[\frac{K(\tau,t)}{A(\tau)}\right]_\tau' e^{-\frac{1}{\mu}\int_\tau^x A(x)\,dx}\,d\tau =\frac{K(x,t)}{A(x)} \\ &\quad -\frac{K(0,t)}{A(0)}e^{-\frac{1}{\mu}\int_0^x A(x)\,dx} +(\mu) =G(x,t)-G(0,t)e^{-\frac{1}{\mu}\int_0^x A(x)\,dx}+(\mu). \tag{11} \end{aligned} \]

Here, as in the preceding papers, the notation \((\mu^\alpha)\) is adopted for quantities satisfying the inequality \(|(\mu^\alpha)|<c\mu^\alpha\), where \(c\) does not depend on \(\mu\) for \(\mu\leqslant\mu_0\) and is independent of the variables \(x\) and \(t\) throughout the region under consideration in which they vary, \(0\leqslant x\leqslant 1\), \(0\leqslant t\leqslant 1\). In this notation any constant of this kind is denoted by the same letter \(c\). In what follows we shall also encounter quantities depending, in addition, on \(\lambda\); in this case \((\mu^\alpha)\) will mean that \(c\) is also independent of \(\lambda\) in the corresponding range of its variation.

In the expression (11) obtained for \(G(x,t,\mu)\), the first term

\[ G(x,t)\equiv \frac{K(x,t)}{A(x)} \]

is the kernel of the reduced equation (7). Thus the kernel \(G(x,t,\mu)\) of the integral equation (8) is a perturbed kernel with respect to the kernel of equation (7); however, this perturbation is not uniformly small, since the second term in (11) is a boundary-layer function which is not small in a neighborhood of the point \(x=0\).

Thus the question posed concerning the eigenvalues of problem (5)—(6) reduces to the investigation of the perturbed kernel \(G(x,t,\mu)\) and to determining whether the known perturbation method can be applied in the case where the perturbation has the character of a boundary-layer function. For this purpose we shall use the well-known Fredholm formulas.

Lemma 1. For the numerator \(D(x,t,\lambda,\mu)\) and denominator \(D(\lambda,\mu)\) of Fredholm for the kernel \(G(x,t,\mu)\) in the region \(0\leqslant x\leqslant 1\), \(0\leqslant t\leqslant 1\), \(|\lambda|<\lambda_0\), \(\mu\leqslant\mu_0\), where \(\mu_0\) is sufficiently small and \(\lambda_0\) is an arbitrary fixed number, the following asymptotic representation holds:

\[ D(x,t,\lambda,\mu) = D(x,t,\lambda) - D(0,t,\lambda)e^{-\frac{1}{\mu}\int_0^x A(x)\,dx} +(\mu), \tag{12} \]

\[ D(\lambda,\mu)=D(\lambda)+\mu D_1(\lambda)+(\mu^2), \tag{13} \]

where \(D(x,t,\lambda)\) and \(D(\lambda)\) are, respectively, the Fredholm numerator and denominator for the kernel \(G(x,t)\) of equation (7).

Remark. Assuming sufficient smoothness of \(A(x)\) and \(K(x,t)\), by the same method, but with more complicated calculations, one can obtain a representation for \(D(x,t,\lambda,\mu)\) and \(D(\lambda,\mu)\) with remainders of higher order. In this case the expansion for \(D(\lambda,\mu)\) has a power character in \(\mu\), while the expansion for \(D(x,t,\lambda,\mu)\) contains boundary-layer functions, as is already evident from (12).

Proof. Both (12) and (13) are not difficult to obtain from the well-known representation for the Fredholm numerator and denominator (see, for example, [2]). Let us obtain (12). We have

\[ D(x,t,\lambda,\mu) = G(x,t,\mu) + \sum_{n=1}^{\infty}(-1)^n\frac{\lambda^n}{n!}\, d_n(x,t,\mu), \]

\[ d_n(x,t,\mu) = \int_0^1\cdots\int_0^1 G_\mu \begin{pmatrix} x,t_1,\ldots,t_n\\ t,t_1,\ldots,t_n \end{pmatrix} \,dt_1\cdots dt_n, \]

\[ G_\mu \begin{pmatrix} x,t_1,\ldots,t_n\\ t,t_1,\ldots,t_n \end{pmatrix} = \left| \begin{array}{cccc} G(x,t,\mu) & G(x,t_1,\mu) & \cdots & G(x,t_n,\mu)\\ G(t_1,t,\mu) & G(t_2,t_1,\mu) & \cdots & G(t_1,t_n,\mu)\\ \cdot & \cdot & \cdot & \cdot\\ G(t_n,t,\mu) & G(t_n,t_1,\mu) & \cdots & G(t_n,t_n,\mu) \end{array} \right|. \tag{14} \]

These notations are adopted in [2]. Substituting (11) into (14), we obtain

\[ G_\mu = \left| \begin{array}{ccc} G(x,t) & \cdots & G(x,t_n)\\ G(t_1,t,\mu) & \cdots & G(t_1,t_n,\mu)\\ \cdot & \cdot & \cdot \end{array} \right| - \left| \begin{array}{ccc} G(0,t) & \cdots & G(0,t_n)\\ G(t_1,t,\mu) & \cdots & G(t_1,t_n,\mu)\\ \cdot & \cdot & \cdot \end{array} \right| e^{-\frac1\mu\int_0^x A(x)\,dx} + \]

\[ + (\mu) \left| \begin{array}{ccc} 1 & \cdots & 1\\ G(t_1,t,\mu) & \cdots & G(t,t_n,\mu)\\ \cdot & \cdot & \cdot \end{array} \right| = (1)+(2)+(3). \]

We shall use the well-known Hadamard estimate for a determinant \(\Delta\) of order \(n\): if \(|a_{ik}|<c\), then

\[ |\Delta|<n^{\frac n2}c^n=\gamma_n. \]

Then we obtain \((3)=(\mu\gamma_{n+1})\), i.e., according to the notation adopted above,

\[ |(3)|<c\,\mu c^{n+1}(n+1)^{\frac{n+1}{2}} \]

(\(c\) does not depend on \(\mu,x,t,t_1,\ldots,t_n\), or \(n\)). In turn,

\[ (2) = - \left| \begin{array}{ccc} G(0,t) & \cdots & G(0,t_n)\\ G(t_1,t) & \cdots & G(t_1,t_n)\\ G(t_2,t,\mu) & \cdots & G(t_2,t_n,\mu)\\ \cdot & \cdot & \cdot \end{array} \right| e^{-\frac1\mu\int_0^x A(x)\,dx} + \]

\[ + \left| \begin{array}{ccc} G(0,t) & \cdots & G(0,t_n)\\ G(0,t) & \cdots & G(0,t_n)\\ G(t_2,t,\mu) & \cdots & G(t_2,t_n,\mu)\\ \cdot & \cdot & \cdot \end{array} \right| e^{-\frac1\mu\int_0^x A(x)\,dx-\frac1\mu\int_0^{t_1} A(x)\,dx} + (\mu\gamma_{n+1}). \]

The second term here tends to zero. Continuing this process, we shall have

\[ (2)=-\left| \begin{array}{cccc} G(0,t)&\ldots&G(0,t_n)\\ G(t_1,t)&\ldots&G(t_1,t_n)\\ \cdot&\cdot&\cdot&\cdot\\ G(t_n,t)&\ldots&G(t_n,t_n) \end{array} \right| e^{-\frac{1}{\mu}\int_0^x A(x)\,dx} +n(\mu\gamma_{n+1})= \]

\[ =-G\left( \begin{array}{cccc} 0,&t_1,&\ldots,&t_n\\ t,&t_1,&\ldots,&t_n \end{array} \right) e^{-\frac{1}{\mu}\int_0^x A(x)\,dx} +n(\mu\gamma_{n+1}). \]

At the same time

\[ (1)= \left| \begin{array}{ccc} G(x,t)&\ldots&G(x,t_n)\\ G(t_1,t)&\ldots&G(t_1,t_n)\\ G(t_2,t,\mu)&\ldots&G(t_2,t_n,\mu)\\ \cdot&\cdot&\cdot&\cdot \end{array} \right| - \]

\[ -\left| \begin{array}{ccc} G(x,t)&\ldots&G(x,t_n)\\ G(0,t)&\ldots&G(0,t_n)\\ G(t_2,t,\mu)&\ldots&G(t_2,t_n,\mu)\\ \cdot&\cdot&\cdot&\cdot \end{array} \right| e^{-\frac{1}{\mu}\int_0^{t_1} A(x)\,dx} +(\mu\gamma_{n+1})= \]

\[ = \left| \begin{array}{ccc} G(x,t)&\ldots&G(x,t_n)\\ G(t_1,t)&\ldots&G(t_1,t_n)\\ G(t_2,t,\mu)&\ldots&G(t_2,t_n,\mu)\\ \cdot&\cdot&\cdot&\cdot \end{array} \right| +(\gamma_{n+1})e^{-\frac{1}{\mu}\int_0^{t_1} A(x)\,dx} +(\mu\gamma_{n+1})= \]

\[ = \left| \begin{array}{ccc} G(x,t)&\ldots&G(x,t_n)\\ G(t_1,t)&\ldots&G(t_1,t_n)\\ \cdot&\cdot&\cdot&\cdot\\ G(t_n,t)&\ldots&G(t_n,t_n) \end{array} \right| + \]

\[ +(\gamma_{n+1})e^{-\frac{1}{\mu}\int_0^{t_1} A(x)\,dx} +\ldots+ (\gamma_{n+1})e^{-\frac{1}{\mu}\int_0^{t_n} A(x)\,dx} +n(\mu\gamma_{n+1}). \]

Finally,

\[ G_\mu\left( \begin{array}{cccc} x,&t_1,&\ldots,&t_n\\ t,&t_1,&\ldots,&t_n \end{array} \right) = G\left( \begin{array}{cccc} x,&t_1,&\ldots,&t_n\\ t,&t_1,&\ldots,&t_n \end{array} \right) - \]

\[ -G\left( \begin{array}{cccc} 0,&t_1,&\ldots,&t_n\\ t,&t_1,&\ldots,&t_n \end{array} \right) e^{-\frac{1}{\mu}\int_0^x A(x)\,dx} + \sum_{i=1}^n (\gamma_{n+1})e^{-\frac{1}{\mu}\int_0^{t_i} A(x)\,dx} +(2n+1)(\mu\gamma_{n+1}), \]

Integrating, we obtain

\[ d_n(x,t,\mu)=d_n(x,t)-d_n(0,t)e^{-\frac1\mu\int_0^x A(x)\,dx}+(3n+1)(\mu\gamma_{n+1}). \]

Summation with respect to \(n\) from \(0\) to \(\infty\) gives (the series with terms
\(\dfrac{\lambda^n}{n!}(3n+1)(\gamma_{n+1}\) converges)

\[ D(x,t,\lambda,\mu)=D(x,t,\lambda)-D(0,t,\lambda)e^{-\frac1\mu\int_0^x A(x)\,dx}+(\mu), \]

i.e. (12).

Analogous calculations lead to the representation (13), except that for this it is necessary to use a more accurate representation for \(G(x,t,\mu)\) than (11), which can be obtained by further integration by parts:

\[ \begin{aligned} G(x,t,\mu) &=\frac{K(x,t)}{A(x)}-\frac{K(0,t)}{A(0)}e^{-\frac1\mu\int_0^x A(x)\,dx} -\mu\left(\frac{K(x,t)}{A(x)}\right)'_x\frac1{A(x)} \\ &\quad+\mu\left(\frac{K(x,t)}{A(x)}\right)'_x\Bigg|_{x=0}\cdot\frac1{A(0)} e^{-\frac1\mu\int_0^x A(x)\,dx}+(\mu^2) \\ &=G(x,t)-G(0,t)e^{-\frac1\mu\int_0^x A(x)\,dx} -\mu G_1(x,t)+\mu G_1(0,t)e^{-\frac1\mu\int_0^x A(x)\,dx} \\ &\quad+(\mu^2). \end{aligned} \tag{15} \]

We shall not dwell in detail on obtaining the representation (13). For what follows we shall also need a representation for \(D(x,t,\lambda,\mu)\) with a remainder term of higher order than in (12), of the form

\[ D(x,t,\lambda,\mu)=D(x,t,\lambda)-D(0,t,\lambda)e^{-\frac1\mu\int_0^x A(x)\,dx} + \]

\[ +\mu D_1(x,t,\lambda)-\mu D_1(0,t,\lambda)e^{-\frac1\mu\int_0^x A(x)\,dx} +(\mu^2). \tag{16} \]

Concerning the representations (13) and (16), the following should be noted. The method is inconvenient for the actual calculation of the coefficients \(D_1(x,t,\lambda)\), \(D_1(0,t,\lambda)\), \(D_1(\lambda)\). Therefore, using (13) and (16), where \(D_1(x,t,\lambda)\), \(D_1(0,t,\lambda)\), \(D_1(\lambda)\) have not yet been found explicitly, we shall first obtain representations for the eigenvalues \(\lambda(\mu)\) and eigenfunctions \(\varphi(x,\mu)\) of problem (5)—(6), again with undetermined coefficients (see below (17) and (18)); and using (17) and (18), one can justify for equation (10) the perturbation method, by means of which it will already be possible actually to determine the coefficients in the representation for \(\lambda(\mu)\), as well as in the representation for the eigenfunctions.

Theorem 1. Let the eigenvalue \(\bar\lambda\) of the degenerate equation (7) be a simple root of the corresponding Fred-

of the polynomial \(D(\lambda)\). Then, in a sufficiently small neighborhood of \(\bar\lambda\), for sufficiently small \(\mu\) there exists a unique eigenvalue \(\lambda(\mu)\) of problem (5)—(6), or, equivalently, of equation (10), for which the asymptotic representation

\[ \lambda(\mu)=\bar\lambda+\mu\bar\lambda_{1}+(\mu^{2}). \tag{17} \]

holds.

The corresponding eigenfunction \(\varphi(x,\mu)\) of problem (5)—(6), as \(\mu\to 0\), tends to a certain eigenfunction \(\varphi(x)\) of the degenerate equation (7), corresponding to \(\bar\lambda\), for all \(x>0\), and the following asymptotic representation holds:

\[ \begin{aligned} \varphi(x,\mu)=&\,\varphi(x)-\varphi(0)e^{-\frac{1}{\mu}\int_{0}^{x}A(x)\,dx} \\ &+\mu\varphi_{1}(x)-\mu\varphi_{1}(0)e^{-\frac{1}{\mu}\int_{0}^{x}A(x)\,dx} +(\mu^{2})\qquad (0\leq x\leq 1). \end{aligned} \tag{18} \]

Remark. In principle, by the same method one can obtain representations of the type (17) and (18) with remainder terms of higher order of smallness,

\[ \lambda(\mu)=\bar\lambda+\mu\bar\lambda_{1}+\cdots+\mu^{n}\bar\lambda_{n}+(\mu^{n+1}), \tag{17'} \]

\[ \begin{aligned} \varphi(x,\mu)=&\,\varphi(x)-\varphi(0)e^{-\frac{1}{\mu}\int_{0}^{x}A(x)\,dx} \\ &+\cdots+\mu^{n}\varphi_{n}(x)-\mu^{n}\varphi_{n}(0)e^{-\frac{1}{\mu}\int_{0}^{x}A(x)\,dx} +(\mu^{n+1}). \end{aligned} \tag{18'} \]

Proof. We have
\[ D(\lambda)=D(\bar\lambda)+D'(\bar\lambda)(\lambda-\bar\lambda)+\cdots =A(\lambda,\bar\lambda)(\lambda-\bar\lambda), \]
since \(D(\bar\lambda)=0\), and since \(D'(\bar\lambda)\ne 0\) (a simple root), there are constants \(a\) and \(A\) such that
\[ a<|A(\lambda,\bar\lambda)|<A. \]
Construct about \(\bar\lambda\) a circle \(S\) of radius \(\alpha\mu\), where \(\alpha\) is some constant independent of \(\mu\), to be chosen appropriately. We have
\[ |D(\lambda)|_{S}>a|\lambda-\bar\lambda|=a\alpha\mu, \]
while
\[ |D_{1}(\lambda)\mu+(\mu^{2})|_{S}<B\mu, \]
and if \(\alpha a>B\) (i.e. \(\alpha>B/a\)), then
\[ |D_{1}(\lambda)\mu+(\mu^{2})|_{S}<|D(\lambda)|_{S}, \]
and therefore
\[ D(\lambda,\mu)=D(\lambda)+\mu D_{1}(\lambda)+(\mu^{2}) \]
by Rouché’s theorem has inside \(S\) a root \(\lambda(\mu)\), and moreover only one. Hence the representation
\[ \lambda(\mu)=\bar\lambda+(\mu) \]
already follows. By analogous reasoning the representation (17) is proved, in which
\[ \bar\lambda_{1}=-\frac{D_{1}(\bar\lambda)}{D'(\bar\lambda)}. \]
Indeed, denote
\[ \lambda(\mu)=\bar\lambda+\mu\bar\lambda_{1}+\Delta \left(\bar\lambda_{1}=-\frac{D_{1}(\bar\lambda)}{D'(\bar\lambda)}\right). \]
We have
\[ \begin{aligned} D(\lambda(\mu),\mu) &=D(\bar\lambda+\mu\bar\lambda_{1}+\Delta) +\mu D_{1}\bigl(\bar\lambda+(\mu)\bigr)+(\mu^{2}) \\ &=D(\bar\lambda+\mu\bar\lambda_{1})+A_{1}\Delta +\mu D_{1}(\bar\lambda)+(\mu^{2}) \\ &=D(\bar\lambda)+\mu\bar\lambda_{1}D'(\bar\lambda)+(\mu^{2}) +A_{1}\Delta+\mu D_{1}(\bar\lambda)+(\mu^{2}) \\ &=A_{1}\Delta+(\mu^{2}), \end{aligned} \]
where \(A_{1}\ne 0\) by virtue of the condition \(D'(\lambda)\ne 0\). Hence we obtain
\[ \Delta=(\mu^{2}), \]
which proves formula (17).

Further, from (16) and (17) follows the representation

\[ D(x,t,\lambda,\mu)=D(x,t,\bar\lambda)-D(0,t,\bar\lambda)e^{-\frac1\mu\int_0^x A(x)\,dx} +\mu D_1(x,t)-\mu D_1(0,t)e^{-\frac1\mu\int_0^x A(x)\,dx}+(\mu^2). \tag{19} \]

Since \(D(x,t,\lambda(\mu),\mu)\) is, with respect to the variable \(x\), the eigenfunction \(\varphi(x,\mu)\) of equation (10) corresponding to the eigenvalue \(\lambda(\mu)\), while \(D(x,t,\bar\lambda)\) is the eigenfunction \(\varphi(x)\) of equation (7) corresponding to the eigenvalue \(\bar\lambda\), expression (19) can be rewritten in the form (18), and thus Theorem 1 is proved.

Remark. From (18) it is seen that the norm of \(\varphi(x,\mu)\), as \(\mu\to0\), differs infinitesimally from the norm of \(\varphi(x)\). Therefore one may say that the eigenfunction \(\varphi(x,\mu)\), normalized to unity

\[ \left(\int_0^1|\varphi(x,\mu)|^2dx=1\right), \]

as \(\mu\to0\) tends precisely to that \(\varphi(x)\) which is also normalized to unity.

Using now (17) and (18), one can, as was already said above, develop for equation (10) the perturbation method for the purpose of actually determining the coefficients in the representation for the eigenvalues and eigenfunctions.

Let us find, for example, \(\bar\lambda_1\). We shall proceed directly from equation (10), which is satisfied by \(\varphi(x,\mu)\):

\[ \varphi(x,\mu)=\lambda(\mu)\int_0^1G(x,t,\mu)\varphi(t,\mu)\,dt. \]

Substituting here (15), (17), and (18), we obtain

\[ \begin{aligned} &\varphi(x)-\varphi(0)e^{-\frac1\mu\int_0^x A(x)\,dx} +\mu\varphi_1(x)-\mu\varphi_1(0)e^{-\frac1\mu\int_0^x A(x)\,dx} +(\mu^2) \\ &=[\bar\lambda+\mu\bar\lambda_1+(\mu^2)] \int_0^1 \left[ G(x,t)-G(0,t)e^{-\frac1\mu\int_0^x A(x)\,dx} -\mu G_1(x,t)+\mu G_1(0,t)e^{-\frac1\mu\int_0^x A(x)\,dx} +(\mu^2) \right] \\ &\quad\times \left[ \varphi(t)-\varphi(0)e^{-\frac1\mu\int_0^t A(t)\,dt} +\mu\varphi_1(t)-\mu\varphi_1(0)e^{-\frac1\mu\int_0^t A(t)\,dt} +(\mu^2) \right]dt. \end{aligned} \]

Equating terms of the same order on both sides of the equality, we shall have:

\[ \mu^0)\qquad \varphi(x)=\bar\lambda\int_0^1G(x,t)\varphi(t)\,dt, \]

\[ \mu^0 e^{-\frac{1}{\mu}\int_0^x A(x)\,dx}\right) \]

\[ -\varphi(0)=-\bar{\lambda}\int_0^1 G(0,t)\varphi(t)\,dt \]

(the equalities hold, since \(\varphi(x)\) is an eigenfunction of the degenerate equation),

\[ \begin{aligned} \varphi_1(x)={}&\frac{1}{\lambda}\int_0^1 G(x,t)\varphi_1(t)\,dt -\bar{\lambda}\int_0^1 G_1(x,t)\varphi(t)\,dt +\bar{\lambda}_1\int_0^1 G(x,t)\varphi(t)\,dt \\ &-\bar{\lambda}\,\frac{G(x,0)}{A(0)}\,\varphi(0). \end{aligned} \tag{20} \]

The last term in (20) is obtained by extracting the term of order \(\mu^1\) from the term

\[ -\bar{\lambda}\int_0^1 G(x,t)\varphi(0) e^{-\frac{1}{\mu}\int_0^t A(t)\,dt}\,dt. \]

From (20) \(\bar{\lambda}_1\) is found if one uses the fact that, for the solvability of equation (20), the orthogonality condition must be satisfied:

\[ \int_0^1\left[ -\bar{\lambda}\int_0^1 G_1(x,t)\varphi(t)\,dt +\bar{\lambda}_1\int_0^1 G(x,t)\varphi(t)\,dt -\bar{\lambda}\frac{G(x,0)}{A(0)}\varphi(0) \right]\psi(x)\,dx=0 \]

(\(\psi(x)\) is an eigenfunction of the equation adjoint to equation (7),
\(\psi(x)=\bar{\lambda}\int_0^1 K(t,x)\psi(t)\,dt\)). Hence we obtain (under the condition
\(\int_0^1 \varphi(x)\psi(x)\,dx\ne 0\))

\[ \bar{\lambda}_1= \frac{ \frac{1}{\bar{\lambda}}\int_0^1\left[ \int_0^1 G_1(x,t)\varphi(t)\,dt +\frac{G(x,0)}{A(0)}\varphi(0) \right]\psi(x)\,dx }{ \int_0^1\left[\int_0^1 G(x,t)\varphi(t)\,dt\right]\psi(x)\,dx } = \]

\[ = \frac{ \bar{\lambda}^2\int_0^1\left[ \int_0^1 G_1(x,t)\varphi(t)\,dt +\frac{G(x,0)}{A(0)}\varphi(0) \right]\psi(x)\,dx }{ \int_0^1 \varphi(x)\psi(x)\,dx }. \]

After this, \(\varphi_1(x)\) is determined from (20), and it suffices to take any particular solution.

The process can be continued if, instead of (17) and (18), one uses (17′) and (18′) and an asymptotic representation for \(G(x,t,\mu)\) more accurate than (15), which can be obtained by further integration by parts, as was already indicated above. The equation for \(\lambda_k\) will each time be obtained from the solvability condition for the equation for \(\varphi_k(x)\). In this way one can determine all coefficients in (17′), (18′). The representation for the entire one-parameter family of eigenfunctions can be obtained by multiplying (18′) by an arbitrary constant \(A\). If \(\varphi(x,\mu)\) is normalized \(\left(\int_0^1 |\varphi(x,\mu)|^2 dx = 1\right)\), then \(A\) is determined in the form of an asymptotic power series in \(\mu\), whose principal term is

\[ A_0=\frac{1}{\sqrt{\int_0^1 |\varphi(x)|^2 dx}}. \]

Let now, instead of (6), the initial condition \(y(0)=y^0\) be prescribed. Then, instead of (10), we shall have the equation

\[ y(x)=y^0 e^{-\frac{1}{\mu}\int_0^x A(x)\,dx} +\lambda\int_0^1 G(x,t,\mu)y(t)\,dt. \tag{21} \]

Let \(\lambda \ne \overline{\lambda_i}\) for all \(i\). Then, for sufficiently small \(\mu\), \(\lambda\) is not an eigenvalue of the homogeneous equation corresponding to equation (21), and, consequently, equation (21) has a unique solution expressible by Fredholm’s formula

\[ \begin{aligned} y(x) &=y^0 e^{-\frac{1}{\mu}\int_0^x A(x)\,dx} +\lambda\int_0^1 \frac{D(x,t,\lambda,\mu)}{D(\lambda,\mu)} \,y^0 e^{-\frac{1}{\mu}\int_0^t A(t)\,dt}\,dt \\ &=y^0 e^{-\frac{1}{\mu}\int_0^x A(x)\,dx}+(\mu). \end{aligned} \]

It follows from this that \(y(x)\to 0\) as \(\mu\to 0\) everywhere except at the point \(x=0\), in a neighborhood of which there is a boundary layer \(\left(y^0 e^{-\frac{1}{\mu}\int_0^x A(x)\,dx}\right)\). If, however, \(\lambda=\lambda(\mu)\), then equation (21), generally speaking, has no solution, since

\[ e^{-\frac{1}{\mu}\int_0^x A(x)\,dx}, \]

generally speaking, is not orthogonal to \(\psi(x)\), and hence also not to the function \(\psi(x,\mu)\)—the eigenfunction of the equation adjoint to equation (10).

§ 2. Let us now consider problem (8)—(9) on eigenvalues for a linear integro-differential equation of the second order. As in § 1, we assume \(A(x)\) and \(K(x,t)\) to be sufficiently smooth functions and \(A(x)\ne 0\) for \(0\le x\le 1\). Without loss of generality, one may assume \(A(x)>0\).

Using the Green’s function \(Q_\mu(x,s)\) for the corresponding differential equation

\[ \mu^2 \frac{d^2 y}{dx^2} - A^2(x)y = 0 \tag{22} \]

with boundary conditions (9), one can pass from equation (8) to the integral equation

\[ y(x)=\int_0^1 Q_\mu(x,s)\left[-\lambda \int_0^1 K(s,t)y(t)\,dt\right]ds \]

or

\[ y(x)=\lambda \int_0^1 G(x,t,\mu)y(t)\,dt, \tag{23} \]

where

\[ G(x,t,\mu)=-\int_0^1 Q_\mu(x,s)K(s,t)\,ds. \tag{24} \]

We shall obtain an asymptotic representation for the kernel \(G(x,t,\mu)\). For this it is necessary to construct an asymptotic representation of the Green’s function; to this end we use the asymptotics of two linearly independent solutions of equation (22),

\[ y_1(x)=\left[\frac{1}{\sqrt{A(x)}}+(\mu)\right] e^{-\frac{1}{\mu}\int_0^x A(x)\,dx}, \]

\[ y_2(x)=\left[\frac{1}{\sqrt{A(x)}}+(\mu)\right] e^{\frac{1}{\mu}\int_0^x A(x)\,dx} \tag{25} \]

and of their derivatives

\[ y_1'(x)=\left[-\frac{1}{\mu}\sqrt{A(x)}+(\mu^0)\right] e^{-\frac{1}{\mu}\int_0^x A(x)\,dx}, \]

\[ y_2'(x)=\left[\frac{1}{\mu}\sqrt{A(x)}+(\mu^0)\right] e^{\frac{1}{\mu}\int_0^x A(x)\,dx}. \tag{25'} \]

These formulas follow, for example, from paper [3]. Constructing now the Green’s function according to the known rules (see, for example, [4]),

\[ Q_\mu(x,s)= \begin{cases} -\dfrac{1}{\mu^2}\dfrac{U(x)V(s)}{U(s)V'(s)-V(s)U'(s)}, & (0\le x\le s),\\[1.2em] -\dfrac{1}{\mu^2}\dfrac{V(x)U(s)}{U(s)V'(s)-V(s)U'(s)}, & (s\le x\le 1), \end{cases} \]

where

\[ U(x)=y_1(x)y_2(0)-y_2(x)y_1(0),\qquad V(x)=y_1(x)y_2(1)-y_2(x)y_1(1), \]

And substituting here expressions (25) and (25′), we obtain

\[ Q_\mu(x,s)= \]

\[ = \left\{ \begin{array}{ll} \displaystyle -\frac{1}{2\mu\sqrt{A(x)A(s)}}\, e^{-\frac{1}{\mu}\int_x^s A(x)\,dx} \left[ \left(1-e^{-\frac{2}{\mu}\int_s^1 A(s)\,ds}\right) \left(1-e^{-\frac{2}{\mu}\int_0^x A(x)\,dx}\right)+(\mu) \right], & (0\le x\le s),\\[2.2em] \displaystyle -\frac{1}{2\mu\sqrt{A(x)A(s)}}\, e^{-\frac{1}{\mu}\int_s^x A(x)\,dx} \left[ \left(1-e^{-\frac{2}{\mu}\int_x^1 A(x)\,dx}\right) \left(1-e^{-\frac{2}{\mu}\int_0^s A(s)\,ds}\right)+(\mu) \right], & (s\le x\le 1). \end{array} \right. \]

Using this formula and integrating by parts in (24), we obtain an asymptotic representation for the kernel \(G(x,t,\mu)\)

\[ G(x,t,\mu)=\frac{K(x,t)}{A^2(x)} - \]

\[ -\frac{K(0,t)}{\sqrt{A(x)A^3(0)}}\, e^{-\frac{1}{\mu}\int_0^x A(x)\,dx} - \frac{K(1,t)}{\sqrt{A(x)A^3(1)}}\, e^{-\frac{1}{\mu}\int_x^1 A(x)\,dx} +(\mu). \tag{26} \]

The first term in (26) is the kernel of the degenerate equation

\[ A^2(x)y(x)=\lambda\int_0^1 K(x,t)y(t)\,dt, \tag{27} \]

and the next two summands are boundary-layer functions. Thus, as in § 1, the question of the eigenvalues of problem (8)—(9) reduces to the study of the eigenvalues of the perturbed kernel \(G(x,t,\mu)\), in which the perturbation has the character of boundary-layer functions. True, in contrast to § 1, the boundary layer appears here in neighborhoods of both boundary points (\(x=0\) and \(x=1\)).

As in § 1, one can now prove an analogue of the lemma on the asymptotic representation of the numerator \(D(x,t,\lambda,\mu)\) and the denominator \(D(\lambda,\mu)\) of Fredholm for the kernel \(G(x,t,\mu)\).

\[ D(x,t,\lambda,\mu)=D(x,t,\lambda) - D(0,t,\lambda)\sqrt{\frac{A(0)}{A(x)}}\, e^{-\frac{1}{\mu}\int_0^x A(x)\,dx} - \]

\[ - D(1,t,\lambda)\sqrt{\frac{A(1)}{A(x)}}\, e^{-\frac{1}{\mu}\int_x^1 A(x)\,dx} +(\mu), \]

\[ D(\lambda,\mu)=D(\lambda)+\mu D_1(\lambda)+(\mu^2), \]

where \(D(x,t,\lambda)\) and \(D(\lambda)\) are respectively the Fredholm numerator and denominator for the kernel

\[ \frac{K(x,t)}{A^2(x)} \]

of equation (27). These representations make it possible to prove a theorem analogous to Theorem 1.

Theorem 2. Suppose that the eigenvalue \(\bar{\lambda}\) of the degenerate equation (27) is a simple root of the corresponding Fredholm denominator \(D(\lambda)\). Then, in a sufficiently small neighborhood of \(\bar{\lambda}\), for sufficiently small \(\mu\) there exists a unique eigenvalue \(\lambda(\mu)\) of problem (8)—(9) (or, equivalently, of equation (23)), for which the asymptotic representation

\[ \lambda(\mu)=\bar{\lambda}+\mu \bar{\lambda}_1+(\mu^2) \tag{28} \]

holds.

The corresponding eigenfunction \(\varphi(x,\mu)\), as \(\mu\to 0\), tends to some eigenfunction \(\varphi(x)\) of the degenerate equation (27) on the interval \(0<x<1\), and the asymptotic representation

\[ \begin{aligned} \varphi(x,\mu)=\varphi(x) &-\varphi(0)\sqrt{\frac{A(0)}{A(x)}}\, e^{-\frac{1}{\mu}\int_0^x A(x)\,dx} \\ &-\varphi(1)\sqrt{\frac{A(1)}{A(x)}}\, e^{-\frac{1}{\mu}\int_x^1 A(x)\,dx} +(\mu), \qquad (0\le x\le 1). \end{aligned} \tag{29} \]

By the same methods one can obtain representations of the type (28) and (29) with remainders of a higher order of smallness.

As in § 1, the actual value of the coefficients of the expansions (28), (29) can be found by the perturbation method. For \(\bar{\lambda}_1\), for example, we obtain

\[ \bar{\lambda}_1= \left\{ \bar{\lambda}^2 \left[ \int_0^1\int_0^1 G_1(x,t)\varphi(t)\,dt +\frac{G(x,0)\varphi(0)}{A(0)} +\frac{G(x,1)\varphi(1)}{A(1)} \right]\psi(x)\,dx \right\} \left[ \int_0^1 \varphi(x)\psi(x)\,dx \right]^{-1}, \]

where \(\psi(x)\) is the eigenfunction of the equation adjoint to equation (27).

Suppose now that, instead of (9), the boundary conditions

\[ y(0)=y^0,\qquad y(1)=y^1 \tag{30} \]

are given.

Using formulas (25), we construct a solution of the differential equation (22) satisfying the boundary conditions (30). It has the form

\[ \tilde{y} = y^0\sqrt{\frac{A(0)}{A(x)}}\, e^{-\frac{1}{\mu}\int_0^x A(x)\,dx} + y^1\sqrt{\frac{A(1)}{A(x)}}\, e^{-\frac{1}{\mu}\int_x^1 A(x)\,dx} +(\mu). \]

Making now the substitution \(y=\tilde{y}+z\), we obtain for the function \(z\) the equation

\[ \mu^2\frac{d^2 z}{dx^2}-A^2(x)z = \lambda\int_0^1 K(x,t)z(t)\,dt + \lambda\int_0^1 K(x,t)\tilde{y}(t)\,dt \]

with zero boundary conditions \(z(0)=0,\ z(1)=0\). With the help of the Green’s function we pass to the integral equation

\[ z(x)=\lambda\int_{0}^{1}G(x,t,\mu)z(t)\,dt+ \int_{0}^{1}Q_{\mu}(x,s)\left[\lambda\int_{0}^{1}K(s,t)\tilde y(t)\,dt\right]ds. \tag{31} \]

It is not difficult to see that the nonhomogeneity in this equation has the estimate

\[ \int_{0}^{1}Q_{\mu}(x,s)\left[\lambda\int_{0}^{1}K(s,t)\tilde y(t)\,dt\right]ds \equiv f(x,\mu)=(\mu). \]

Therefore, if \(\lambda\ne \overline{\lambda_i}\) for every \(i\), then, for sufficiently small \(\mu\), \(\lambda\) will not be an eigenvalue of the homogeneous equation corresponding to equation (31), and, consequently, equation (31) will have a unique solution and the estimate \(z(x)=(\mu)\) will be valid. Thus, in this case the solution of problem (8), (30) has the form

\[ y(x,\mu)=y^{0}\sqrt{\frac{A(0)}{A(x)}}\, e^{-\frac{1}{\mu}\int_{0}^{x}A(x)\,dx} + y^{1}\sqrt{\frac{A(1)}{A(x)}}\, e^{-\frac{1}{\mu}\int_{x}^{1}A(x)\,dx} +(\mu), \]

i.e. \(y(x,\mu)\to 0\) as \(\mu\to 0\) in the interval \(0<x<1\), while in a neighborhood of the boundary points there is a boundary layer.

If, however, \(\lambda=\lambda(\mu)\), then equation (31), generally speaking, has no solutions, since \(f(x,\mu)\), generally speaking, is not orthogonal to \(\psi(x,\mu)\), the eigenfunction of the equation adjoint to (23).

References

1. Vasil’eva A. B. Differential Equations, 1, No. 6, 717–730, 1965.
2. Smirnov V. I. A Course of Higher Mathematics, 4, 1957, p. 32.
3. Birkhoff G. D. Trans. Amer. Math. Soc., 9, 219–231, 1908.
4. Tricomi F. Differential Equations, 1962, p. 171.

Received by the editors
April 13, 1965.

Moscow State University
named after M. V. Lomonosov,
Faculty of Physics