UDC 517.948.34 : 513.88

ON A SINGULAR INTEGRO-DIFFERENTIAL EQUATION

Kh. M. Kogan

In this article we investigate the spectral properties of the singular integro-differential operator

\[ (A\varphi)(x)=\int_0^1 \frac{\varphi'(t)\,dt}{x-t}, \tag{1} \]

which is considered as an operator in the Hilbert space \(L_2(0,1)\), as well as the variational problem connected with this operator. The integral in (1) is understood in the sense of the principal value.

The equation

\[ 2\lambda\varphi=A\varphi \tag{2} \]

with the operator \(A\) is a special case of the (homogeneous) “airplane wing equation” (Prandtl’s equation). It has long been known, has appeared in numerous applications (aerodynamics, the plane problem of the theory of elasticity, stable processes with independent increments [3], automatic control), and a large number of works have been devoted to its study (the principal bibliographic sources of works of the 1920s and 1930s are indicated in the first edition of the monograph of N. I. Muskhelishvili [5]; see also [6]). The spectral properties of the operator \(A\) were studied by J. Elliott [7, 8] in connection with the theory of stable processes. Elliott considered the operator \(A\) as an operator from \(C(0,1)\) into \(C(0,1)\); some of its properties under such a definition do not hold or remain unclear. In particular, on the basis of her results it would have been impossible to justify the variational principle [9] for equation (2) and the possibility of applying direct methods of the calculus of variations to solve this equation.

Closely connected with equation (2) is the variational problem [1, 2, 14] of minimizing the functional

\[ C(\varphi)= \frac{ \displaystyle \int_0^1 \frac{dy}{y}\int_0^{1-y}\varphi(x+y)\varphi'(x)\,dx }{ \displaystyle \int_0^1 \varphi^2(x)\,dx }, \tag{3} \]

where \(\varphi(x)\), \(0\le x\le 1\), belongs to some class of real continuous functions satisfying the boundary conditions

\[ \varphi(0)=\varphi(1)=0. \tag{4} \]

Indeed, equation (2) is the Euler equation (more precisely, the Euler–Lagrange equation) of the variational problem and, in turn, the functional repre-

is the Rayleigh quotient for the operator \(A\), i.e., it can be transformed to the form

\[ C(\varphi)=\frac{(A\varphi,\varphi)}{2(\varphi,\varphi)}, \tag{5} \]

where \((\cdot,\cdot)\) is the usual scalar product in \(L_2(0,1)\).

Of course, in establishing the connection between the eigenvalue problem for equation (2) and the variational problem for the minimum of the functional \(C(\varphi)\), the domain of definition \(D(A)\) of the operator \(A\) must be indicated in a reasonable way, so that the eigenfunctions of equation (2) belong to \(D(A)\), and so that on \(D(A)\) the lower bound of the values of \(C(\varphi)\) and the lower bound of the spectrum of the operator \(A\) are both attained and coincide. This natural domain is the set dense in \(L_2(0,1)\), \(L_2^{0(1)}\bigl(\sqrt{x(1-x)};0,1\bigr)\), which is defined at the beginning of § 1; moreover, in § 1 some lemmas are given on the spaces \(L_p\) with weights and on singular integrals in them.

Section 2 is devoted to the proof of the fact that the functional \(C(\varphi)\) on the set \(L_2^{0(1)}\bigl(\sqrt{x(1-x)};0,1\bigr)\) can be transformed to the form (5), where \(A\) is an operator having the form (1).

In §§ 3—the main part of the paper—the spectral properties of the operator \(A\) are studied; it is proved that the operator \(A\) has a nonempty, real, discrete, positive spectrum. The smallest eigenvalue \(\lambda_1\) of the operator \(A\) exists, is simple, and it has a unique eigenfunction \(\varphi_1(x)\), nonnegative and belonging to \(L_2^{0(1)}\bigl(\sqrt{x(1-x)};0,1\bigr)\).

In §§ 4, 5 a certain lower estimate is given for \(\lambda_1\) (an approximate value of \(\lambda_1\) with a more accurate two-sided error estimate was indicated by the author in [4]) and an integral inequality connected with this estimate; also, on the basis of the results of §§ 2, 3, a variational principle is established for equation (2).

Individual facts concerning the method of inverting the operator \(A\) (here the author essentially relies on an important lemma of N. I. Akhiezer from the paper [10]), some properties of the inverse operator, the investigation of properties of eigenfunctions, etc., occurred in the above-mentioned works of Elliott, and also in Zengen, I. N. Vekua, S. G. Mikhlin [11], and possibly in some other authors. Nevertheless, the investigation of the operator \(A\) as an operator in a Hilbert space is carried out by the author, apparently, for the first time.

A brief exposition of the results of the paper is contained in the note [12].

§ 1. CLASSES \(L_2\) WITH WEIGHT AND SINGULAR INTEGRAL OPERATORS IN THEM

Let us introduce for consideration the class \(L_2^{0(1)}\bigl(\sqrt{x(1-x)};0,1\bigr)\) of absolutely continuous real functions defined on the interval \([0,1]\) and satisfying the conditions (4)

\[ \varphi(0)=\varphi(1)=0, \]

whose derivatives \(\varphi'(x)\) belong to the space \(L_2\bigl(\sqrt{x(1-x)};0,1\bigr)\), i.e., are square-integrable with weight \(p(x)=\sqrt{x(1-x)}\) on the interval \([0,1]\):

\[ \int_0^1 \sqrt{x(1-x)}\,\varphi'^2(x)\,dx<\infty. \]

Remark. To simplify some notation and arguments, henceforth in this paragraph, and also in §§ 3, 4, we shall write equation (2) in the form

\[ \lambda \varphi(x)=\int_{-1}^{1}\frac{\varphi'(t)\,dt}{x-t}, \tag{2′} \]

which is obtained by the regular change of variables

\[ x=\frac{1+\xi}{2},\qquad t=\frac{1+\tau}{2}. \]

We shall again denote the operator in (2′) by \(A\). Accordingly, we shall have in mind the space \(\overset{0}{L}{}_{2}^{(1)}(\sqrt{1-x^{2}};\,-1,1)\) instead of \(\overset{0}{L}{}_{2}^{(1)}(\sqrt{x(1-x)};\,0,1)\), the space \(L_{2}(\sqrt{1-x^{2}};\,-1,1)\) instead of \(L_{2}(\sqrt{x(1-x)};\,0,1)\), etc. In §§ 2, 5 we shall return to the former notation. In the new notation condition (4) has the form

\[ \varphi(-1)=\varphi(+1)=0. \tag{4′} \]

Lemma 1. If \(\psi(x)\in L_{2}(\sqrt{1-x^{2}};\,-1,1)\), then \(\psi(x)\in L_{p}(-1,1)\), where \((1<)p<\frac{4}{3}\).

Proof: Hölder’s inequality.

In what follows, a singular operator with Cauchy kernel will play a significant role:

\[ (S\psi)(x)\equiv \int_{-1}^{1}\frac{\psi(t)\,dt}{x-t}=G(x), \tag{6} \]

where the integral is to be understood in the sense of the principal value. It is known that the operator \(S\) has the property of “preserving” many classes of functions, i.e., of mapping these classes into themselves. Thus, for example, if \(\psi\in \mathrm{Lip}\,\alpha\), \(0<\alpha<1\), then also \(G\in \mathrm{Lip}\,\alpha\) with the same \(\alpha\); if \(\psi\in L_{2}\), then also \(G\in L_{2}\), etc. The properties of the operator \(S\) on the classes \(L_{p}\) with a weight were studied in detail by B. V. Khvedelidze [13], from whose results one can also obtain the following lemma.

Lemma 2. If \(\psi(x)\in L_{2}(\sqrt{1-x^{2}};\,-1,1)\), then also \(G(x)=S\psi(x)\in L_{2}(\sqrt{1-x^{2}};\,-1,1)\) (i.e. \(S\) “preserves” the class \(L_{2}(\sqrt{1-x^{2}};\,-1,1)\)).

We shall give a proof of this lemma without relying on the results of B. V. Khvedelidze, but proceeding from an important theorem of N. I. Akhiezer [10], which we state in a somewhat simplified formulation.

Theorem of N. I. Akhiezer. Let \(g(x)\in L_{2}(p(x);\,-1,1)\). Then the equation

\[ g(x)=\frac{1}{\pi}\int_{-1}^{1}\frac{1}{p(t)}\,\frac{f(t)\,dt}{x-t} \]

has an essentially unique solution belonging to the class \(L_{2}\left(\frac{1}{p(x)};\,-1,1\right)\), which is given by the formula

\[ f(x)=\frac{1}{\pi}\int_{-1}^{1}p(t)\,\frac{g(t)\,dt}{t-x} \tag{7} \]

almost everywhere on \([-1,1]\).

Here \(p(x)\) is a nonnegative summable function which must satisfy certain conditions indicated in [10]. The functions \(p(x)=\dfrac{1}{\sqrt{1-x^2}}\), \(\dfrac{1}{p(x)}=\sqrt{1-x^2}\) satisfy these conditions.

We shall now prove Lemma 2. Let \(p(x)=\dfrac{1}{\sqrt{1-x^2}}\) and \(\psi(x)=\dfrac{g(x)}{\sqrt{1-x^2}}\).

If \(\psi(x)\in L_2(\sqrt{1-x^2};-1,1)\), then \(g(x)\in L_2\left(\dfrac{1}{\sqrt{1-x^2}};-1,1\right)\). Indeed, \(g(x)=\sqrt{1-x^2}\,\psi(x)\) and

\[ \int_{-1}^{1}\frac{1}{\sqrt{1-x^2}}\,g^2(x)\,dx = \int_{-1}^{1}\frac{1}{\sqrt{1-x^2}}\,(\sqrt{1-x^2})^2\psi^2(x)\,dx = \]

\[ = \int_{-1}^{1}\sqrt{1-x^2}\,\psi^2(x)\,dx; \]

the last integral is finite by hypothesis. Then it follows from N. I. Akhiezer’s theorem that the function \(G(x)=\dfrac{1}{\pi}f(x)\in L_2(\sqrt{1-x^2};-1,1)\), as was required to be proved.

Lemma 3. The operator \(S\) is a bounded operator from \(L_2(\sqrt{1-x^2};-1,1)\) into \(L_2(\sqrt{1-x^2};-1,1)\).

Indeed, let \(\psi(x)\in L_2(\sqrt{1-x^2};-1,1)\). Then the finiteness of the norm

\[ \|S\|_{L_2(\sqrt{1-x^2};-1,1)} = \sup_{\|\psi\|_{L_2(\sqrt{1-x^2};-1,1)}=1} \left( \int_{-1}^{1}\sqrt{1-x^2}\,[S\psi(x)]^2\,dx \right)^{1/2} \]

follows from Lemma 2.

Lemma 4. If \(\psi\in L_2(\sqrt{1-x^2};-1,1)\), then \(S\psi\) is the limit, in the sense of convergence in the norm \(L_2(\sqrt{1-x^2};-1,1)\) (strong convergence), of the integral \(S_\varepsilon\psi\), where

\[ (S_\varepsilon\psi)(x) = \int_{-1}^{x-\varepsilon}\frac{\psi(t)\,dt}{x-t} + \int_{x+\varepsilon}^{1}\frac{\psi(t)\,dt}{x-t}. \]

Proof. Introduce the function \(\psi_\varepsilon(x)\), setting

\[ \psi_\varepsilon(x)= \begin{cases} \psi(x), & \text{for } x\in[-1,x-\varepsilon)+(x+\varepsilon,1],\\ 0, & \text{for } x\in[x-\varepsilon,x+\varepsilon]. \end{cases} \]

It is easy to see that \(\psi_\varepsilon\to\psi\) in the sense of convergence in \(L_2(\sqrt{1-x^2};-1,1)\). Indeed,

\[ \int_{-1}^{1}\sqrt{1-t^2}\,[\psi(t)-\psi_\varepsilon(t)]^2\,dt = \int_{x-\varepsilon}^{x+\varepsilon}\sqrt{1-t^2}\,\psi^2(t)\,dt \to 0, \]

since the function \(\sqrt{1-t^{2}}\,\psi^{2}(t)\in L_{1}(-1,1)\). But then, by Lemma 3,

\[ \int_{-1}^{1}\frac{\psi_{\varepsilon}(t)\,dt}{x-t} \to \int_{-1}^{1}\frac{\psi(t)\,dt}{x-t} \quad \text{as } \varepsilon\to 0 \]

in the sense of strong convergence in \(L_{2}(\sqrt{1-x^{2}};\,-1,1)\), or, if one uses the definition of the function \(\psi_{\varepsilon}(t)\),

\[ \left(\int_{-1}^{x-\varepsilon}+\int_{x+\varepsilon}^{1}\right) \frac{\psi(t)\,dt}{x-t} \to \int_{-1}^{1}\frac{\psi(t)\,dt}{x-t} \quad \text{as } \varepsilon\to 0. \]

The lemma is proved. It may be formulated as follows.

If \(\psi\in L_{2}(\sqrt{1-x^{2}};\,-1,1)\), then

\[ \int_{-1}^{1}\sqrt{1-x^{2}}\,[S\psi(x)-S_{\varepsilon}\psi_{\varepsilon}(x)]^{2}\,dx\to 0, \quad \text{as } \varepsilon\to 0 \]

or

\[ \int_{-1}^{1}\sqrt{1-x^{2}} \left\{ \int_{-1}^{1}\frac{\psi(t)\,dt}{x-t} - \left[ \int_{-1}^{x-\varepsilon}\frac{\psi(t)\,dt}{x-t} + \int_{x+\varepsilon}^{1}\frac{\psi(t)\,dt}{x-t} \right] \right\}^{2}dx\to 0 \tag{8} \]

as \(\varepsilon\to 0\).

Remark. It follows from Lemma 1 that Lemma 2 (a consequence of N. I. Akhiezer’s theorem) describes the class of solutions of the airplane-wing equation more accurately than the theorem of Zenger–Tricomi ([6], p. 229).

§ 2. FORMULATION OF THE VARIATIONAL PROBLEM. TRANSFORMATION OF FUNCTIONAL (3)

Variational problem. Determine the minimum of the functional

\[ C(\varphi)= \frac{ \displaystyle \int_{0}^{1}\frac{dy}{y}\int_{0}^{1-y}\varphi(x+y)\varphi'(x)\,dx }{ \displaystyle \int_{0}^{1}\varphi^{2}(x)\,dx } \]

on the class \(L_{2}^{0(1)}(\sqrt{x(1-x)};\,0,1)\). Prove that the minimum is attained at some function of this class. Study the properties of the extremal function.

It is difficult to compute the first variation of functional (3) in the usual way, since Fubini’s theorem cannot be applied to the repeated integral appearing in the numerator. Instead, we shall transform functional (3) to the form (5). Let us introduce the approximate functional

\[ B_{\varepsilon}(\varphi)= \int_{\varepsilon}^{1}\frac{dy}{y}\int_{0}^{1-y}\varphi(x+y)\varphi'(x)\,dx, \quad 0<\varepsilon<1. \]

Lemma 5. If \(\varphi(x)\in \overset{\circ}{L}{}_{2}^{(1)}\left(\sqrt{x(1-x)};\,0,\,1\right)\), then

\[ \lim_{\varepsilon\to 0} B_\varepsilon(\varphi)=B_0(\varphi) =\int_{0}^{1}\frac{dy}{y}\int_{0}^{1-y}\varphi(x+y)\varphi'(x)\,dx . \]

Proof. Since \(\varphi'(x)\in L_2\left(\sqrt{x(1-x)};\,0,\,1\right)\subset L_1(0,1)\) (Lemma 1), the integral mean-value theorem holds (here and below \(x,y,x+y\in[0,1]\)):

\[ \varphi(x+y)=\varphi(x)+\int_x^{x+y}\varphi'(t)\,dt . \]

Hence

\[ \left|\int_{0}^{1-y}\varphi(x+y)\varphi'(x)\,dx\right| = \left|\int_{0}^{1-y}\left[\varphi(x)+\int_x^{x+y}\varphi'(t)\,dt\right]\varphi'(x)\,dx\right| \le \]

\[ \le \frac12\,\varphi^2(1-y) + \left| \int_{0}^{1-y}\varphi'(x)\,dx \left\{ \int_x^{x+y}\frac{dt}{\sqrt{t(1-t)}}\times \right.\right. \]

\[ \left.\left. \times \int_x^{x+y}\sqrt{t(1-t)}\,\varphi'^2(t)\,dt \right\}^{1/2} \right|, \tag{9} \]

where the Bunyakovsky inequality and condition (4) were used; moreover

\[ \varphi^2(1-y)=\bigl[\varphi(1-y)-\varphi(1)\bigr]^2 = \left[\int_{1-y}^{1}\varphi'(t)\,dt\right]^2 \le \int_{1-y}^{1}\frac{dt}{\sqrt{t(1-t)}}\times \]

\[ \times \int_{1-y}^{1}\sqrt{t(1-t)}\,\varphi'^2(t)\,dt . \tag{10} \]

It is easy to verify that \(\arcsin t\in \operatorname{Lip}\frac12\). Hence the estimate follows

\[ \int_{1-y}^{1}\frac{dt}{\sqrt{t(1-t)}} = \arcsin 1-\arcsin(1-2y) \le M\sqrt{y}. \tag{11} \]

Let us continue the estimate of \(\varphi^2(1-y)\). From (10) and (11) it follows that

\[ \varphi^2(1-y) \le M\sqrt{y}\int_{0}^{1}\sqrt{t(1-t)}\,\varphi^2(t)\,dt = C_1\sqrt{y} \le C_1\sqrt[4]{y}. \tag{12} \]

Using an analogous technique, we obtain

\[ \left| \int_{0}^{1-y}\varphi'(x)\,dx \left\{ \int_x^{x+y}\frac{dt}{\sqrt{t(1-t)}} \int_x^{x+y}\sqrt{t(1-t)}\,\varphi'^2(t)\,dt \right\}^{1/2} \right| \le C_2\sqrt[4]{y}. \tag{13} \]

From (9), (12), and (13) it follows that

\[ \left|\int_{0}^{1-y}\varphi(x+y)\varphi'(x)\,dx\right| \le C\sqrt[4]{y}, \]

where \(C=\dfrac{1}{2}C_1+C_2\). Further,

\[ \left|B_0(\varphi)-B_\varepsilon(\varphi)\right| = \left| \int_0^\varepsilon \frac{dy}{y}\int_0^{1-y}\varphi(x+y)\varphi'(x)\,dx \right| \leq \]

\[ \leq C\int_0^\varepsilon \sqrt[4]{y}\,\frac{dy}{y} = 4C\sqrt[4]{\varepsilon}\to 0 \quad \text{as } \varepsilon\to 0, \]

whence the assertion of Lemma 5 follows.

From Lemma 5 it follows, in particular, that the functional \(C(\varphi)\) is bounded on the set

\[ \left\{\varphi:\varphi\in \overset{0}{L}{}_{2}^{(1)} \bigl(\sqrt{x(1-x)};\,0,1\bigr),\ \int_0^1 \varphi^2(x)\,dx=1 \right\}. \]

Now, by transforming the functional \(B_\varepsilon(\varphi)\) in two different ways (Fubini’s lemma may be applied here), we obtain

\[ B_\varepsilon(\varphi) = -\int_0^{1-\varepsilon}\varphi(x)\,dx \int_{x+\varepsilon}^{1}\frac{\varphi'(t)\,dt}{t-x} = \int_\varepsilon^1 \varphi(x)\,dx \int_0^{x-\varepsilon}\frac{\varphi'(t)\,dt}{x-t} \]

or

\[ B_\varepsilon(\varphi) = \frac12 \left\{ \int_\varepsilon^1 \varphi(x)\,dx \int_0^{x-\varepsilon}\frac{\varphi'(t)\,dt}{x-t} - \int_0^{1-\varepsilon}\varphi(x)\,dx \int_{x+\varepsilon}^{1}\frac{\varphi'(t)\,dt}{t-x} \right\}. \tag{14} \]

We shall show that the integrals

\[ I_1= \int_0^\varepsilon \varphi(x)\,dx \int_0^{x-\varepsilon}\frac{\varphi'(t)\,dt}{x-t}, \qquad I_2= -\int_{1-\varepsilon}^{1}\varphi(x)\,dx \int_{1+\varepsilon}^{1}\frac{\varphi'(t)\,dt}{t-x} \]

tend to zero as \(\varepsilon\to 0\). Take, for example, \(I_1\). To the inner integral we apply Bunyakovsky’s inequality:

\[ \left| \int_0^{x-\varepsilon}\frac{\varphi'(t)\,dt}{x-t} \right| \leq \left( \int_0^{x-\varepsilon} \frac{dt}{\sqrt{t(1-t)}(x-t)^2} \right)^{1/2} \left( \int_0^{x-\varepsilon} \sqrt{t(1-t)}\,\varphi'^2(t)\,dt \right)^{1/2}. \]

Since \(x-t\geq \varepsilon\), we have \(\dfrac{1}{(x-t)^2}\leq\dfrac{1}{\varepsilon^2}\); therefore,

\[ \left| \int_0^{x-\varepsilon} \frac{dt}{\sqrt{t(1-t)}(x-t)^2} \right| \leq \frac{1}{\varepsilon^2} \left| \int_0^{x-\varepsilon} \frac{dt}{\sqrt{t(1-t)}} \right| \leq \frac{1}{\varepsilon^2} \int_0^1 \frac{dt}{\sqrt{t(1-t)}} = \]

\[ = \frac{1}{\varepsilon^2} \left|[\arcsin(2t-1)]_{t=0}^{1}\right| = \frac{\pi}{\varepsilon^2}. \]

On the other hand,

\[ \left(\int_{0}^{x-\varepsilon}\sqrt{t(1-t)}\,\varphi'^2(t)\,dt\right)^{1/2} \le \left(\int_{0}^{1}\sqrt{t(1-t)}\,\varphi'^2(t)\,dt\right)^{1/2} = \]

\[ =\|\varphi'\|_{L_2(\sqrt{x(1-x)};\,0,\,1)}. \]

Thus,

\[ \left|\int_{0}^{x-\varepsilon}\frac{\varphi'(t)\,dt}{x-t}\right| \le \frac{\sqrt{\pi}}{\varepsilon}\, \|\varphi'\|_{L_2(\sqrt{x(1-x)};\,0,\,1)}. \]

Using the estimate obtained for the inner integral, we get

\[ |I_1| \le \frac{\sqrt{\pi}}{\varepsilon}\, \|\varphi'\|_{L_2(\sqrt{x(1-x)};\,0,\,1)} \left|\int_{0}^{\varepsilon}\varphi(x)\,dx\right|. \]

But

\[ \left|\int_{0}^{\varepsilon}\varphi(x)\,dx\right| = \left|\int_{0}^{\varepsilon}[\varphi(x)-\varphi(0)]\,dx\right| = \left|\int_{0}^{\varepsilon}dx\int_{0}^{x}\varphi'(t)\,dt\right| \le \]

\[ \le \int_{0}^{\varepsilon}dx \left(\int_{0}^{x}\frac{dt}{\sqrt{t(1-t)}}\right)^{1/2} \left(\int_{0}^{x}\sqrt{t(1-t)}\,\varphi'^2(t)\,dt\right)^{1/2}. \]

We estimate the first factor under the integral sign as above (11):

\[ \int_{0}^{x}\frac{dt}{\sqrt{t(1-t)}}\le M\sqrt{x} \quad\text{or}\quad \left(\int_{0}^{x}\frac{dt}{\sqrt{t(1-t)}}\right)^{1/2} \le \sqrt{M}\,x^{1/4}, \]

while the second factor is less than
\(\|\varphi'\|_{L_2(\sqrt{t(1-t)};\,0,\,1)}\).

Consequently,

\[ \left|\int_{0}^{\varepsilon}\varphi(x)\,dx\right| \le \|\varphi'\|_{L_2(\sqrt{x(1-x)};\,0,\,1)} \sqrt{M}\int_{0}^{\varepsilon}x^{1/4}\,dx = C_3\varepsilon^{5/4}. \]

Finally:

\[ |I_1|\le \frac{C_4}{\varepsilon}\,\varepsilon^{5/4} = C_4\varepsilon^{1/4}\to 0 \quad\text{as}\quad \varepsilon\to 0. \]

An analogous estimate also holds for \(I_2\), so that

\[ \lim_{\varepsilon\to 0} I_1 = \lim_{\varepsilon\to 0} I_2 = 0. \tag{15} \]

Now add \(\dfrac12(I_1+I_2)\) to both sides of equation (14):

\[ B_{\varepsilon}(\varphi)+\frac12(I_1+I_2) = \int_{\varepsilon}^{1}\frac{dy}{y} \int_{0}^{1-y}\varphi(x+y)\varphi'(x)\,dx + \frac12(I_1+I_2) = \]

\[ = \frac12\left\{ \int_{0}^{1}\varphi(x)\,dx \left[ \int_{0}^{x-\varepsilon}\frac{\varphi'(t)\,dt}{x-t} - \int_{x+\varepsilon}^{1}\frac{\varphi'(t)\,dt}{t-x} \right]\right\}. \tag{16} \]

and pass to the limit as \(\varepsilon \to 0\) in both parts of the equation obtained. On the left-hand side we obtain \(B_0(\varphi)\) (this follows from Lemma 5 and (15)); on the right-hand side it is necessary to carry out the limiting passage under the integral sign. This requires explanation, since Lebesgue’s theorem is not applicable here. But it follows from Lemma 4 that, if we denote

\[ F_\varepsilon(x)=\int_0^1 \frac{\varphi'(t)\,dt}{x-t}- \left[ \int_0^{x-\varepsilon}\frac{\varphi'(t)\,dt}{x-t} + \int_{x+\varepsilon}^1\frac{\varphi'(t)\,dt}{x-t} \right] = [S\varphi' - S_\varepsilon\varphi'](x), \]

then \(F_\varepsilon(x)\to 0\) in the sense of convergence in \(L_2(\sqrt{x(1-x)};\,0,1)\), i.e.

\[ \left|\int_0^1 \varphi(x)F_\varepsilon(x)\,dx\right| \le \left(\int_0^1 \frac{\varphi^2(x)\,dx}{\sqrt{x(1-x)}}\right)^{1/2} \times \]

\[ \times \left(\int_0^1 \sqrt{x(1-x)}\,F_\varepsilon^2(x)\,dx\right)^{1/2} \to 0, \]

since the first factor is bounded, while the second, according to (8), tends to zero.

Thus,

\[ \lim_{\varepsilon\to 0}\int_0^1 \varphi(x)\,dx \left[ \int_0^{x-\varepsilon}\frac{\varphi'(t)\,dt}{x-t} + \int_{x+\varepsilon}^1\frac{\varphi'(t)\,dt}{x-t} \right] = \int_0^1 \varphi(x)\,dx \int_0^1 \frac{\varphi'(t)\,dt}{x-t} \]

and from (14) we obtain

\[ \int_0^1 \frac{dy}{y}\int_0^{1-y}\varphi(x+y)\varphi'(x)\,dx = \]

\[ = \frac12\int_0^1 \varphi(x)\,dx \int_0^1 \frac{\varphi'(t)\,dt}{x-t} \tag{17} \]

for every \(\varphi(x)\in \overset{0(1)}{L}_2\bigl(1/\sqrt{x(1-x)};\,0,1\bigr)\). If we now introduce the scalar product in the real space \(L_2(0,1)\):

\[ (\varphi,\psi)=\int_0^1 \varphi(x)\psi(x)\,dx, \]

then we obtain

\[ \int_0^1 \varphi^2(x)\,dx=(\varphi,\varphi) \quad\text{and}\quad \int_0^1 \varphi(x)\,dx\int_0^1 \frac{\varphi'(t)\,dt}{x-t} = (\varphi,A\varphi), \]

where

\[ (A\varphi)(x)=\int_{0}^{1}\frac{\varphi'(t)\,dt}{x-t}. \]

The functional (3) takes the form

\[ C(\varphi)=\frac{(A\varphi,\varphi)}{2(\varphi,\varphi)} \]

and it is required to study the variational problem for the minimum of this functional. Since it is the Rayleigh quotient for the operator \(A\), in order to prove the solvability of the variational problem it is first necessary to study the spectral properties of the operator \(A\).

§ 3. STUDY OF THE OPERATOR \(A\)

Thus, we study the operator

\[ (A\varphi)(x)\equiv \int_{-1}^{1}\frac{\varphi'(t)\,dt}{x-t}=g(x) \]

from \(L_2(-1,1)\) into \(L_2(-1,1)\) with domain

\[ D(A)=\overset{0}{L}{}_{2}^{(1)}\bigl(\sqrt{1-x^2};-1,1\bigr). \]

From Lemma 2 it follows that

\[ R(A)\subseteq L_2\bigl(\sqrt{1-x^2};-1,1\bigr). \]

Lemma 6. The operator \(A\) is symmetric.

Indeed, \(A\) is the product of the operator \(S\) of singular integration with Cauchy kernel, defined on the set \(L_2(\sqrt{1-x^2};-1,1)\), and the differentiation operator \(D\), defined on the set

\[ \overset{0}{L}{}_{2}^{(1)}\bigl(\sqrt{1-x^2};-1,1\bigr) \]

of absolutely continuous functions satisfying, in particular, the boundary conditions

\[ \varphi(-1)=\varphi(+1)=0, \tag{4′} \]

\(A=SD\). Each of these operators is skew-symmetric.

We further note that the domain \(D(A)\) is dense in \(L_2(-1,1)\).

Equation (2′) is reduced to an equivalent Fredholm integral equation by inverting the operator \(A\). This method has been described repeatedly in the literature. First the operator \(S\) is inverted (on the class of functions tending to infinity at the ends of the interval; see, for example, [15])

\[ \varphi'(t)=\frac{1}{\pi^2}\frac{1}{\sqrt{1-t^2}} \int_{-1}^{1}\frac{\sqrt{1-x^2}\,g(x)\,dx}{x-t} +\frac{C_1}{\sqrt{1-t^2}}, \tag{18} \]

and then the operator \(D\):

\[ \varphi(t)=\int_{-1}^{1}K(x,t)g(x)\,dx+C_1\arcsin t+C_2, \tag{19} \]

where

\[ K(x,t)=\frac{1}{2\pi^2}\ln \frac{1-xt+\sqrt{(1-x^2)(1-t^2)}}{1-xt-\sqrt{(1-x^2)(1-t^2)}}. \tag{20} \]

For what follows it will be useful to note that the function

\[ h(t)=\frac{1}{\sqrt{1-t^2}}\int_{-1}^{1}\frac{\sqrt{1-x^2}\,g(x)\,dx}{x-t} \tag{21} \]

belongs to \(L_2\bigl(\sqrt{1-x^2};-1,1\bigr)\) if
\(g(x)\in L_2\bigl(\sqrt{1-x^2};-1,1\bigr)\). This follows from Lemma 2. Indeed, let
\(p(x)=\sqrt{1-x^2}\). Then in equation (7)
\(f(x)\in L_2\left(\dfrac{1}{\sqrt{1-x^2}};-1,1\right)\), i.e.

\[ \int_{-1}^{1}\frac{1}{\sqrt{1-t^2}}\,f^2(t)\,dt<\infty . \tag{22} \]

It is obvious (from comparison of (7) and (21)) that

\[ f(t)=\frac{1}{\pi}\sqrt{1-t^2}\,h(t). \]

From (22) we obtain

\[ \int_{-1}^{1}\frac{1}{\sqrt{1-t^2}}\cdot\frac{1}{\pi^2}(1-t^2)h^2(t)\,dt = \]

\[ =\frac{1}{\pi^2}\int_{-1}^{1}\sqrt{1-t^2}\,h^2(t)\,dt<\infty, \]

i.e. \(h(t)\in L_2\bigl(\sqrt{1-t^2};-1,1\bigr)\).

Next we study the integral operator with kernel \(K(x,t)\) (we shall have to make sure, in particular, that it is \(A^{-1}\)—the operator inverse to \(A\)).

Lemma 7. a) The kernel \(K(x,t)\) is symmetric; b) \(K(x,\pm1)=0\) for \(x\ne\pm1\); \(K(\pm1,t)=0\) for \(t\ne\pm1\); c) \(K(x,t)\ge 0\).

All the assertions of Lemma 3 follow from the form of the kernel (20).

Lemma 8. The function

\[ F(t)=\int_{-1}^{1}K^2(x,t)\,dx \]

is continuous on \([-1,1]\).

Proof. Integrating by parts, we obtain

\[ F(t)=\frac{2}{\pi^2}(1-t^2)\int_{-1}^{1}\frac{\arcsin x}{\sqrt{1-x^2}}\frac{dx}{x-t}. \tag{23} \]

We split the integral in (23) into the sum of three integrals

\[ \int_{-1}^{1} = \int_{-1}^{-1+\varepsilon} + \int_{-1+\varepsilon}^{1-\varepsilon} + \int_{1-\varepsilon}^{1}, \qquad 0<\varepsilon<\frac12 . \tag{24} \]

On the interval \([-1+\varepsilon,\,1-\varepsilon]\) the function

\[ g(x)=\frac{\arcsin x}{\sqrt{1-x^2}}\in \operatorname{Lip}1. \]

Consequently,

\[ \int_{-1+\varepsilon}^{1-\varepsilon} g(x)\frac{dx}{x-t} \in \operatorname{Lip}(1-\varepsilon_1), \]

where \(\varepsilon_1\) is arbitrarily small. On the intervals \([-1,-1+\varepsilon]\) and \([1-\varepsilon,1]\), \(\arcsin x\in \operatorname{Lip}\frac12\) and \(g(x)\in H^*\), where \(H^*\) is the class of N. I. Muskhelishvili ([16], pp. 37–38), so that the first (third) singular integral in (24) has the representation (ibid., p. 92)

\[ \int g(x)\frac{dx}{x-t}= \frac{(1\mp t)^{-\frac12}\arcsin t}{(1\pm t)^{a_0}}, \qquad a_0<\frac12 . \]

Singularities of the form \((1\pm t)^{a_0}\) are extinguished by the factor \((1-t^2)\) standing before the integral sign.

Remark. The function \(F(t)\) can be computed directly.

Theorem 1. The integral operator with kernel \(K(x,t)\) is completely continuous from \(L_2(0,1)\) into \(L_2(0,1)\) (moreover, it is a Hilbert–Schmidt operator).

Proof. The integral
\[ \int_{-1}^{1}\int_{-1}^{1} K^2(x,t)\,dx\,dt = \int_{-1}^{1} F(x)\,dx, \]
obviously, is finite.

Let us now substitute into (19), in place of \(g(x)\), the expression \(\lambda\varphi(x)\). Equation (19) takes the form
\[ \varphi(t)=\lambda\int_{-1}^{1}K(x,t)\varphi(x)\,dx+C_1\arcsin t+C_2 . \tag{25} \]

Nontrivial solutions of equation (25) are continuous on the interval \([-1,1]\). To prove this it is sufficient to refer to the book of S. G. Mikhlin [11], pp. 102–105, since the kernel \(K(x,t)\) satisfies the notion of continuity as a whole set forth in that book. Now in both parts of equation (25) we pass to the limit as \(t\to\pm1\); taking condition \((4')\) into account, we obtain
\[ C_1=C_2=0. \tag{26} \]

Thus, (25) takes the form
\[ \varphi(t)=\lambda\int_{-1}^{1}K(x,t)\varphi(x)\,dx. \tag{27} \]

We have obtained a homogeneous Fredholm integral equation with symmetric kernel (20). It has at least one real eigenvalue, and the corresponding eigenfunction is not identically equal to zero.

Let us now note that the integral operator with kernel \(K(x,t)\) was obtained by us as a result of inverting the operator \(A\), and we defined it in the space \(L_2(-1,1)\). In order to prove that it is indeed the inverse operator to the operator \(A\), it is necessary to verify that it maps \(R(A)\) into \(D(A)\), i.e., that its range belongs to the space
\[ \overset{\circ}{L}{}_{2}^{(1)}\left(\sqrt{1-x^2};\, -1,1\right). \]
But this is indeed so. In fact,
\[ D(A^{-1})=R(A)\subseteq L_2\left(\sqrt{1-x^2};\, -1,1\right), \]
let \(g(x)\in L_2\left(\sqrt{1-x^2};\, -1,1\right)\). Then from (26), (19), and Lemma 7 it follows that \(\varphi(\pm1)=0\), and from (18) and the following remark it further follows that
\[ \varphi'(t)\in L_2\left(\sqrt{1-x^2};\, -1,1\right). \]

Next we shall apply Jentzsch’s theorem ([17], theorem \(\beta'\)):

*Let the kernel \(K(x,t)\) be positive almost everywhere in the square \([-1,1]\times[-1,1]\) (with the measure of the set on which \(K(x,t)=0\) equal to zero) and generate a completely continuous operator from \(L_2\) into \(L_2\). Then equation (27) has a unique positive (almost everywhere) eigenfunction from \(L_2\) [which may vanish only at those points where \(K(x,t)\) or its continuous iteration \(K^n(x,t)\) vanishes (i.e., in our case, at the points \(\pm1\))]*, corresponding to its characteristic—

* The remark in square brackets follows from the second paragraph on p. 60 of [17].

characteristic number is positive, simple, and smaller than the modulus of any other characteristic number.

The kernel \(K(x,t)\) of equation (27) satisfies all the conditions of Jentzsch’s theorem. We denote by \(\varphi_1(x)\) the nonnegative eigenfunction of equation (27) corresponding to the characteristic value of equation (27), \(\lambda_1\).

We shall now prove that the operator \(A\) is positive definite. So far it is known only that all its eigenvalues are real (since the operator \(A\) is symmetric) and that the eigenvalue of smallest absolute value \(\lambda_1>0\).

But perhaps among the subsequent eigenvalues there are negative ones? The following shows that this is not so.

Theorem 2. The operator \(A\) is positive, i.e. \((A\varphi,\varphi)\geqslant 0\).

Proof. Extend \(\varphi(x)\) and \(\varphi'(x)\) to the whole real axis by setting \(\varphi(x)=0\) for \(1\leq |x|<\infty\). Then
\[ \int_{-1}^{1}\frac{\varphi'(t)\,dt}{x-t} = \int_{-\infty}^{\infty}\frac{\varphi'(t)\,dt}{x-t} \]
is the convolution of the functions \(\varphi'(x)\) and \(1/x\).

Let \(\Phi(s)\) be the Fourier transform of the function \(\varphi(x)\). Applying the convolution theorem and Parseval’s formula, we obtain (the bar over \(\Phi(s)\) denotes complex conjugation)
\[ (A\varphi,\varphi) = \int_{-1}^{1}\varphi(x)\,dx \int_{-1}^{1}\frac{\varphi'(t)\,dt}{x-t} = \]
\[ = \int_{-\infty}^{\infty}\varphi(x)\,dx \int_{-\infty}^{\infty}\frac{\varphi'(t)\,dt}{x-t} = \]
\[ = \int_{-\infty}^{\infty}\overline{\Phi}(s)(-is)\Phi(s)\pi i\,\operatorname{sign}s\,ds = \pi\int_{-\infty}^{\infty}|s|\,|\Phi(s)|^2\,ds\geqslant 0, \]
since \(s\cdot\operatorname{sign}s=|s|\).

The theorem is proved.

From Theorem 2 and the fact that \(\lambda_1>0\), it follows that the operator \(A\) is positive definite, i.e. its entire spectrum lies on the positive half-axis and \(\lambda_1\) is its smallest eigenvalue. From the complete continuity of the inverse operator \(A^{-1}\) there follows the discreteness of the spectrum. Moreover, it was proved earlier that all its eigenfunctions belong to
\[ \overset{0}{L}{}_{2}^{(1)}\left(\sqrt{1-x^2};\, -1,\,1\right), \]
and hence also
\[ \varphi_1(x)\in \overset{0}{L}{}_{2}^{(1)}\left(\sqrt{1-x^2};\, -1,\,1\right). \]

§ 4. A PRELIMINARY LOWER ESTIMATE FOR \(\lambda_1\) AND ONE INTEGRAL INEQUALITY

Lemma 9. The function \(\varphi_1(x)\) is even.

The proof is obvious.

We now note that, by virtue of condition \((4')\) (see [5], p. 384),
\[ \int_{-1}^{1}\frac{\varphi'(t)\,dt}{x-t} = \frac{d}{dx}\int_{-1}^{1}\frac{\varphi(t)\,dt}{x-t}. \]

Substitute \(\varphi_1\) in place of \(\varphi\) and \(\lambda_1\) in place of \(\lambda\) into equation \((2')\), and integrate the resulting identity from \(0\) to \(x\):

\[ \lambda_1 \int_0^x \varphi_1(t)\,dt = \int_{-1}^1 \frac{\varphi_1(t)\,dt}{x-t} - \int_{-1}^1 \frac{\varphi_1(t)\,dt}{-t}; \]

the last integral is equal to zero by Lemma 9. Put \(x=1\):

\[ \lambda_1 \int_0^1 \varphi_1(t)\,dt = \int_{-1}^1 \frac{\varphi_1(t)\,dt}{1-t} = \left( \int_{-1}^0 + \int_0^{+1} \right) \frac{\varphi_1(t)\,dt}{1-t}. \]

In the first integral we put \(t=-\tau\), use the oddness of the function \(\varphi_1(t)\), and then again write \(t\) instead of \(\tau\):

\[ \lambda_1 \int_0^1 \varphi_1(t)\,dt = \int_0^1 \left( \frac{1}{1+t}+\frac{1}{1-t} \right) \varphi_1(t)\,dt = 2\int_0^1 \frac{\varphi_1(t)\,dt}{1-t^2}, \]

whence

\[ \int_0^1 \left( \lambda_1-\frac{2}{1-t^2} \right) \varphi_1(t)\,dt=0. \]

The values of the function \(\dfrac{2}{1-t^2}\) fill the interval \([2,+\infty)\). Since \(\varphi_1(t)\) preserves its sign on the interval \([0,1]\), the function

\[ \psi(t)=\lambda_1-\frac{2}{1-t^2} \]

must change sign; this is possible only for \(\lambda_1>2\). Applying this estimate to the functional (3), we obtain

\[ \int_0^1 \frac{dy}{y}\int_0^{1-y}\varphi(x+y)\varphi'(x)\,dx \geq 2\int_0^1 \varphi^2(x)\,dx . \tag{28} \]

Applying more precise estimates for \(\lambda_1\), one can strengthen the coefficient in (28). The most accurate, of course, is the inequality

\[ \int_0^1 \frac{dy}{y}\int_0^{1-y}\varphi(x+y)\varphi'(x)\,dx \geq \lambda_1 \int_0^1 \varphi^2(x)\,dx \]

with the value \(\lambda_1=3.637\pm10^{-3}\), contained (in implicit form—through the number \(\mu_1=\dfrac{\pi^2}{2\lambda_1}\)) in [4].

§ 5. VARIATIONAL PRINCIPLE

In § 3 it was proved that the operator \(A\) is positive definite and has a nonempty spectrum. It follows from this that the variational principle can be applied to it, which consists in the fact that the least proper value \(\lambda_1>0\) of the equation

\[ 2\lambda\varphi=A\varphi \]

is the lower bound of the values of the Rayleigh quotient \(\dfrac{(A\varphi,\varphi)}{2(\varphi,\varphi)}\), attained on some function \(\varphi_1(x)\in \overset{0}{L}{}_{2}^{(1)}(\sqrt{1-x^2};-1,1)\), and

\[ \lambda_1=\frac{(A\varphi_1,\varphi_1)}{2(\varphi_1,\varphi_1)}. \]

Next we return to the limits of integration \((0,1)\) and shall again deal with equation (2) (instead of \((2')\)). Our next task is to recover the functional (3) from the Rayleigh quotient (5). We multiply both sides of equation (2) by \(\varphi(x)\) and integrate from 0 to 1:

\[ 2\lambda\int_0^1 \varphi^2(x)\,dx = \int_0^1 \varphi(x)\,dx\int_0^1 \frac{\varphi'(t)\,dt}{x-t} = \]

\[ = \lim_{\varepsilon\to 0} \left\{ \int_0^1 \varphi(x)\,dx \left[ \int_0^{x-\varepsilon}\frac{\varphi'(t)\,dt}{x-t} - \int_{x+\varepsilon}^1\frac{\varphi'(t)\,dt}{t-x} \right] \right\} = \]

\[ = \lim_{\varepsilon\to 0} \left\{ \int_0^1 \varphi(x)\,dx \int_0^{x-\varepsilon}\frac{\varphi'(t)\,dt}{x-t} - \int_0^1 \varphi(x)\,dx \int_{x+\varepsilon}^1\frac{\varphi'(t)\,dt}{t-x} \right\}. \tag{29} \]

The validity of passing to the limit under the integral sign was proved in § 2.

Transforming each of the iterated integrals in (29) separately, we obtain that each of them is equal to

\[ \int_\varepsilon^1 \frac{dy}{y}\int_0^{1-y}\varphi(x+y)\varphi'(x)\,dx. \]

Thus,

\[ 2\lambda\int_0^1 \varphi^2(x)\,dx = \lim_{\varepsilon\to 0} 2\int_\varepsilon^1 \frac{dy}{y}\int_0^{1-y}\varphi(x+y)\varphi'(x)\,dx, \]

whence

\[ \lambda_1 = \inf_{\varphi\in \overset{0}{L}{}_{2}^{(1)}(\sqrt{x(1-x)};\,0,1)} C(\varphi) = \]

\[ = \frac{ \displaystyle \int_0^1 \frac{dy}{y}\int_0^{1-y}\varphi_1(x+y)\varphi_1'(x)\,dx }{ \displaystyle \int_0^1 \varphi_1^2(x)\,dx }. \]

The passage to the limit here is possible (see Lemma 5).

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Received by the editors
July 16, 1966

VINITI, Moscow