UDC 517.948.32

APPROXIMATE SOLUTION OF THE WIENER–HOPF EQUATION IN AN EXCEPTIONAL CASE

Yu. I. Cherskii

1. Introduction. This paper studies the Wiener—Hopf equation

\[ \lambda f_+(x)+\frac{1}{\sqrt{2\pi}}\int_0^\infty k(x-t)f_+(t)\,dt=g_+(x),\qquad x>0, \tag{1} \]

in which \(\lambda\) is a given constant, while the kernel has the following properties: \(k(x)\in L_2(-\infty,\infty)\), and its Fourier integral\(^*\)

\[ K(x)\equiv \lim_{a\to\infty}\frac{1}{\sqrt{2\pi}}\int_{-a}^{a} k(t)e^{ixt}\,dt\equiv Vk \tag{2} \]

is such that the function

\[ \frac{\sqrt{x^2+q^2}}{|x|}\,[\lambda+K(x)]\equiv \Omega(x),\qquad q=\operatorname{const}>0 \tag{3} \]

is nonzero and satisfies the Hölder condition on the closed axis. Without loss of generality one may assume that \(\Omega(0)=1\). We shall prescribe the function \(g_+(x)\) in \(L_2(0,\infty)\); the solution \(f_+\) will be sought in the space \(\overline{M_+(1)}\), defined in Sec. 2.

Under the assumptions made, some boundary-value problems for partial differential equations reduce to equation (1).

2. Notation. Let \(n\) be an integer. Denote by \(\overline{M(n)}\) the space of functions \(\Psi(x)\) such that

\[ \Psi(x)\frac{(x+iq)^n}{x^n}\in L_2(-\infty,\infty). \]

Denote by \(\overline{M^+(n)}\) and \(\overline{M^-(n)}\) the spaces of functions \(\Psi^+(x)\) and \(\Psi^-(x)\) belonging to \(\overline{M(n)}\) and being boundary values of functions analytic respectively in the upper and lower half-planes, with

\[ \int_{-\infty}^{\infty} \left|\Psi^+(x+iy)(x+iy+iq)^n(x+iy)^{-n}\right|^2\,dx<\operatorname{const} \quad \text{for all } y\geqslant 0, \]

\[ \int_{-\infty}^{\infty} \left|\Psi^-(x+iy)(x+iy-iq)^n(x+iy)^{-n}\right|^2\,dx<\operatorname{const} \quad \text{for all } y\leqslant 0. \]

\(^*\) Fourier integrals will be denoted by the corresponding capital letters. Equality (2) defines the operator \(V\), the Fourier transform. By the known rules, this operator can also be defined on the spaces of linear functionals introduced in Sec. 2.

By the symbols \(M(n)\), \(M_{+}(n)\), and \(M_{-}(n)\) we shall denote the spaces of Fourier integrals
\(\psi(x)=V^{-1}\Psi\), \(\psi_{+}(x)=V^{-1}\Psi^{+}\), \(\psi_{-}(x)=V^{-1}\Psi^{-}\) of the functions \(\Psi(x)\), \(\Psi^{+}(x)\), and \(\Psi^{-}(x)\), belonging respectively to the spaces \(\overline{M(n)}\), \(\overline{M_{+}(n)}\), and \(\overline{M_{-}(n)}\).

For \(n \leqslant 0\) the functions \(\psi\), \(\psi_{+}\), and \(\psi_{-}\) are ordinary functions belonging to \(L_{2}(-\infty,\infty)\), with \(\psi_{+}(x)\equiv 0\) if \(x<0\), and \(\psi_{-}(x)\equiv 0\) if \(x>0\). In the case \(n>0\), the elements of the space \(M(n)\) are linear functionals \(\psi\) on the space \(M(-n)\), bounded in the following sense:

\[ |(\psi,\varphi)|^{2}\leqslant \operatorname{const}\int_{-\infty}^{\infty}\left|\left(\frac{x+iq}{x}\right)\Phi(x)\right|^{2}\,dx \]

and only such functionals. Here \(\operatorname{const}\) does not depend on \(\varphi\), and \(\Phi(x)=V\varphi \in \overline{M(-n)}\).

3. Exact solution of equation (1). We shall give the notation of equation (1) a more precise form:

\[ \left(f_{+},\ \lambda\varphi(x)+\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} k(t-x)\varphi(t)\,dt\right) = \int_{-\infty}^{\infty}\bigl[g_{+}(x)+f_{-}(x)\bigr]\varphi(x)\,dx . \tag{4} \]

Here \(\varphi(x)\) ranges over the space \(L_{2}(-\infty,\infty)\). The function \(g_{+}(x)\) may be regarded as given in \(M_{+}(0)\), while the function \(f_{-}(x)\) belongs to \(M_{-}(0)\) and, like \(f_{+}\), is unknown.

We shall solve the posed problem (4) by the usual method for convolution-type equations—by passing to Fourier integrals:

\[ [\lambda+K(x)]F^{+}(x)=F^{-}(x)+G^{+}(x),\qquad -\infty<x<\infty . \tag{5} \]

Here \(K(x)\) and \(G^{+}(x)\) are known functions, \(G^{+}(x)\in \overline{M_{+}(0)}\). The functions \(F^{+}(x)\) and \(F^{-}(x)\) are unknown, with

\[ F^{+}(x)\in \overline{M_{+}(1)},\qquad F^{-}(x)\in \overline{M_{-}(0)} . \tag{6} \]

The set of conditions (5), (6) is known as the Riemann problem (conjugation problem). Problem (5), (6) is equivalent to equation (1): they are either simultaneously solvable or unsolvable; their solutions \(F^{+}(x)\) and \(f_{+}\) are connected by the equality

\[ f_{+}=V^{-1}F^{+}(x). \tag{7} \]

Using the known methods for solving the Riemann problem, we arrive at the following conclusions. Let the index of the function (3) be equal to zero.* Then

1) For the solvability of problem (5), (6), it is necessary and sufficient that there exist a constant \(c\) such that**

\[ \frac{1}{[\sqrt{x}]^{+}} \left\{ \left[ \frac{G^{+}(x)[\sqrt{x}]^{-}} {X^{-}(x)[\sqrt{x-iq}]^{-}} \right]^{+} + \frac{c}{x-iq} \right\} \in L_{2}(-\infty,\infty). \tag{8} \]

* That is, the increment along the \(x\)-axis of the argument of the function \(\Omega(x)\) is zero. This case, which occurs in applications, is the only one considered in the present section for reasons of economy of space.

** The symbol \([F(x)]^{+}\), where \(F(x)\in L_{2}(-\infty,\infty)\), means the following:

\[ 2[F(x)]^{+}=V\{(1+\operatorname{sgn}x)V^{-1}F\}. \]

By \([\sqrt{z}]^+\) and \([\sqrt{z}]^-\) are denoted the square roots, analytic respectively in the upper and lower half-planes; their values have nonnegative imaginary part. The function \(X^-\) (and, analogously, \(X^+\)) is defined by the equality

\[ X^{\pm}(x)=\exp\left[\mp \frac{1}{2}\ln\Omega(x)-\frac{x}{2\pi i}\int_{-\infty}^{\infty}\ln\Omega(t)\,\frac{dt}{t(t-x)}\right]. \tag{9} \]

Here \(\ln\Omega(x)\) is a continuous function which vanishes at \(x=0\), and the integral is understood in the sense of the Cauchy principal value.

2) If the constant \(c\) exists, then the limit indicated below exists and the equality holds

\[ c=\lim_{\varepsilon\to 0}\frac{iq}{\varepsilon}\int_{0}^{\varepsilon} \left[ \frac{G^+(x)[\sqrt{x}]^-}{X^-(x)[\sqrt{x-iq}]^-} \right]^+ dx. \]

3) If the solution \(F^+(x)\) of problem (5), (6) exists, then it is unique and is determined by the formula

\[ F^+(x)=\frac{X^+(x)[\sqrt{x+iq}]^+}{x[\sqrt{x}]^+} \left\{(x-iq) \left[ \frac{G^+(x)[\sqrt{x}]^-}{X^-(x)[\sqrt{x-iq}]^-} \right]^+ +c\right\}. \tag{10} \]

Example. Let, in equation (1), \(2g_+(x)=(1+\operatorname{sgn}x)e^{-\alpha x}\), \(\operatorname{Re}\alpha>0\). Then \(\sqrt{2\pi}G^+(x)= i(x+i\alpha)^{-1}\),

\[ \left[ \frac{G^+(x)[\sqrt{x}]^-}{X^-(x)[\sqrt{x-iq}]^-} \right]^+ = \frac{i[\sqrt{-i\alpha}]^-} {\sqrt{2\pi}\,X^-(-i\alpha)[\sqrt{-i\alpha-iq}]^-(x+i\alpha)} \]

and the solvability condition (8) will be satisfied for

\[ c=\frac{q} {\sqrt{2\pi}\,X^-(-i\alpha)[\sqrt{-i\alpha-iq}]^-[\sqrt{-i\alpha}]^-}. \]

Formula (10) takes the form

\[ F^+(x)= \frac{i[\sqrt{-i\alpha-iq}]^- X^+(x)[\sqrt{x+iq}]^+} {\sqrt{2\pi}\,X^-(-i\alpha)[\sqrt{-i\alpha}]^-[\sqrt{x}]^+(x+i\alpha)}. \tag{11} \]

The solution \(f_+(x)\) of equation (1), determined by formulas (7) and (11), for large positive \(x\) has the form

\[ f_+(x)=\frac{D}{\sqrt{x}}+f_0(x), \tag{12} \]

where \(D\) is a constant, and \(f_0(x)\in L_2(N,\infty)\).

4. Approximate solution.

Let, as before, the index of the function (3) be equal to zero. We pose the problem of finding in the space \(M_+(1)\) an approximate solution \(\tilde f_+\), so close to the exact solution \(f_+\) that the difference of these solutions is an element of the narrower space \(M_+(0)\) (i.e., a function of class \(L_2(0,\infty)\)). We also pose the problem of estimating the mean-square error.

The solution of these problems will be based on Theorem 2 of [1]. Let us consider all conditions 1)—8) of this theorem (we use the notation adopted in [1]).

1) Put \(A=M^+(1)\), \(B=M^+(2)\). In this case the elements \(H_1^+(x)\) and \(H_2^+(x)\) of the space \(B\) are regarded as equal if these functions differ by no more

than by the function \(c_0x^{-2}+c_1x^{-1}\), where \(c_0\) and \(c_1\) are constants: \(H_1^+=H_2^+\) in \(B\), if

\[ H_1^+(x)\equiv H_2^+(x)+\frac{c_0}{x^2}+\frac{c_1}{x}. \]

The operators \(K\) and \(\widetilde K\) may be taken in the form

\[ KF^+=\frac{x^2+q^2}{x^2}\,[(\lambda+K(x))F^+(x)]^+ \quad \text{in } B, \]

\[ \widetilde K F^+=\frac{x^2+q^2}{x^2}\,[(\widetilde\lambda+\widetilde K(x))F^+(x)]^+ \quad \text{in } B. \]

Here \(\widetilde K(x)\) and \(\widetilde\lambda\) are such that the function

\[ \widetilde\Omega(x)=\frac{\sqrt{x^2+q^2}}{|x|}\,(\widetilde\lambda+\widetilde K(x)) \tag{13} \]

satisfies the Hölder condition on the closed \(x\)-axis.

2) The equation \(Kf=g\) in the case under consideration has the form

\[ \frac{x^2+q^2}{x^2}\,[(\lambda+K(x))F^+(x)]^+=H^+(x) \quad \text{in } B. \tag{14} \]

Here \(F^+(x)\) is the unknown function belonging to \(\overline{M^+(1)}\), and

\[ H^+(x)\equiv \frac{x^2+q^2}{x^2}\,G^+(x)\in B. \tag{15} \]

It can be shown that equations (14) and (1) are solvable or unsolvable simultaneously; assuming equation (1) solvable, we obtain that equation (14) is solvable.

3) As \(B_0\) we take the linear space of functions \(\Psi^+(x)\) of the form

\[ \Psi^+(x)\equiv \Psi_0^+(x)+\frac{c_0}{x^2}+\frac{c_1}{x}, \qquad \Psi_0^+(x)\in \overline{M^+(0)}. \tag{16} \]

As in the space \(B\), \(\Psi_1^+=\Psi_2^+\) in \(B_0\), if

\[ \Psi_1^+(x)\equiv \Psi_2^+(x)+\frac{c_0}{x^2}+\frac{c_1}{x} \]

(\(c_0\) and \(c_1\) are constants). We introduce the norm:

\[ \|\Psi^+\|_{B_0}= \left(\int_{-\infty}^{\infty}|\Psi_0^+(x)|^2\,dx\right)^{1/2}. \]

We require that the function \(\widetilde K(x)\) and the constant \(\widetilde\lambda\) satisfy the condition

\[ \max_{-\infty<x<\infty} \left| \frac{(x^2+q^2)^{3/2}}{x^3}\,[\lambda+K(x)-\widetilde\lambda-\widetilde K(x)] \right|<\infty. \tag{17} \]

It is not difficult to verify that condition (17) is sufficient for \((K-\widetilde K)f_1\in B_0\) for any element \(f_1\) of \(A\). From condition (17), in particular, it follows that \(\widetilde\Omega(0)=1\).

4) Let \(A_1\) be the linear space of functions \(F^+(x)\) belonging to \(\overline{M^+(1)}\) and having as their Fourier integrals \(V^{-1}F^+\) functions \(f_+\), representable for large positive \(x\) in the form (12). We define the functional \(T\) by the equality \((T,F^+)=D\). Suppose that the function (13) differs

from zero on the closed axis and its index is equal to zero. Then the equation \(\tilde K \tilde f_0 = 0\), or, equivalently,

\[ \frac{x^2+q^2}{x^2}\left[(\tilde\lambda+\tilde K(x))\tilde F_0^+(x)\right]^+=0 \quad \text{in } B \]

has, both in the space \(A\) and in the space \(A_1\), only one linearly independent solution

\[ \tilde F_0^+(x)= \frac{\tilde X^+(x)\,[\sqrt{x+iq}]^+}{[\sqrt{x}]^+(x+iq)} . \tag{18} \]

Here

\[ \tilde X^\pm(x)= \exp\left[ \mp \frac{1}{2}\ln \tilde\Omega(x) -\frac{x}{2\pi i}\int_{-\infty}^{\infty} \ln \tilde\Omega(t)\,\frac{dt}{t(t-x)} \right]. \tag{19} \]

The solution (18) has the property

\[ (T,\tilde F_0^+)=-i\sqrt{\frac{2}{q}}\ne 0. \]

Next, in the space \(B\), take an element \(\tilde H^+(x)\) close to the element (15) in the sense of the condition \(g-\tilde g\in B_0\). In other words, put

\[ \tilde H^+(x)=\frac{x^2+q^2}{x^2}\,\tilde G^+(x), \]

where the function \(\tilde G^+(x)\) belongs to \(\overline{M^+(0)}\) and satisfies the condition

\[ \frac{x^2+q^2}{x^2}\,[G^+(x)-\tilde G^+(x)]\in \overline{M^+(0)}. \tag{20} \]

Now write the equation \(\tilde K\tilde f=\tilde g\):

\[ \frac{x^2+q^2}{x^2}\left[(\tilde\lambda+\tilde K(x))\tilde F^+(x)\right]^+ =\tilde H^+(x) \quad \text{in } B. \tag{21} \]

This equation is equivalent to the Riemann problem (cf. problem (5), (6)):

\[ [\tilde\lambda+\tilde K(x)]\,\tilde F^+(x) = \tilde F^-(x)+\tilde G^+(x)+\frac{\mathrm{const}}{x+iq}, \tag{22} \]

\[ \tilde F^+(x)\in \overline{M^+(1)},\qquad \tilde F^-(x)\in \overline{M^-(0)}, \tag{23} \]

where \(\mathrm{const}\) is an arbitrary constant. A necessary and sufficient condition for the solvability of problem (22), (23) has the form

\[ \frac{1}{[\sqrt{x}]^+} \left\{ \left[ \frac{\tilde G^+(x)\,[\sqrt{x}]^-} {\tilde X^-(x)\,[\sqrt{x-iq}]^-} \right]^+ +\frac{\tilde c}{x-iq} \right\} \in L_2(-\infty,\infty). \tag{24} \]

Suppose that, in a neighborhood of the point \(x=0\), the functions (3) and (13) satisfy the Hölder condition with exponent not less than \(1/2\). Then it can be shown that condition (24) is fulfilled simultaneously with condition (8). Suppose also that the functions \(\tilde\lambda+\tilde K(x)\) and \(\tilde G^+(x)\) can be chosen so that \(\tilde F^+(x)\in A_1\), where \(\tilde F^+(x)\) is the solution of problem (22), (23).

5) As \(A_0\) we take the linear space \(\overline{M^+(0)}\), in which the norm is introduced

\[ \|\Phi^+(x)\|_{A_0} = \left( \int_{-\infty}^{\infty} |\Phi^+(x)|^2\,dx \right)^{1/2}. \]

It is obvious that \(A_0 \subset A_1\) and \((T,\Phi^+)=0\) if \(\Phi^+ \in A_0\). The equation \(\widetilde K\varphi=\psi\):

\[ \widetilde K\Phi^+ \equiv \frac{x^2+q^2}{x^2} \bigl[(\tilde\lambda+\widetilde K(x))\Phi^+(x)\bigr]^+ = \Psi^+(x) \quad \text{in } B_0 \]

has in the space \(A_0\) a unique and unconditional solution

\[ \Phi^+(x)=\widetilde K^{-1}\Psi^+(x) \equiv \frac{\widetilde X^+(x)[\sqrt{x}]^+}{[\sqrt{x+iq}]^+} \left[ \frac{\Psi_0^+(x)[\sqrt{x}]^-}{\widetilde X^-(x)[\sqrt{x-iq}]^-} \right]^+ . \]

Here the functions \(\Psi^+(x)\) and \(\Psi_0^+(x)\) are connected by the identity (16).

6) For any function \(\Phi^+(x)\) belonging to \(A_0\), we have

\[ \|(K-\widetilde K)\Phi^+\|_{B_0}^2 = \int_{-\infty}^{\infty} \left\| \left[ \frac{x^2+q^2}{x^2} (\lambda+K(x)-\tilde\lambda-\widetilde K(x))\Phi^+(x) \right]^+ \right\|^2 dx \le \]

\[ \le \int_{-\infty}^{\infty} \left| \frac{x^2+q^2}{x^2} (\lambda+K(x)-\tilde\lambda-\widetilde K(x))\Phi^+(x) \right|^2 dx \le \]

\[ \le \max_{-\infty<x<\infty} \left| \frac{x^2+q^2}{x^2} (\lambda+K(x)-\tilde\lambda-\widetilde K(x)) \right|^2 \|\Phi^+\|_{A_0}^2 . \]

7) Similarly we estimate the norm of the operator \(\widetilde K^{-1}\):

\[ \|\widetilde K^{-1}\| \le \max \left| \frac{\widetilde X^+(x)\sqrt{x}}{(x^2+q^2)^{1/4}} \right| \cdot \max \left| \frac{\sqrt{x}}{\widetilde X^-(x)(x^2+q^2)^{1/4}} \right| . \]

8) The values of the function \(\tilde\lambda+\widetilde K(x)\) must in any case be sufficiently close to \(\lambda+K(x)\) that

\[ \frac{ \max\left|\widetilde X^+(x)x^{1/2}(x^2+q^2)^{-1/4}\right| }{ \min\left|\widetilde X^-(x)x^{-1/2}(x^2+q^2)^{1/4}\right| } \cdot \max \left| \frac{x^2+q^2}{x^2} (\lambda+K(x)-\tilde\lambda-\widetilde K(x)) \right| \le \sigma<1 . \]

5. Continuation. Use of the assertions of the theorem.

The theorem mentioned in § 4 asserts that any solution of the equation \(Kf=g\) (that is, of equation (14)) belongs to \(A_1\). But equation (14) is equivalent to the Riemann problem

\[ [\lambda+K(x)]F^+(x) = F^-(x)+G^+(x)+\frac{\mu}{x+iq}, \tag{25} \]

\[ F^+(x)\in \overline{M^+(1)},\qquad F^-(x)\in \overline{M^-(0)}, \tag{26} \]

where \(\lambda\), \(K(x)\), and \(G^+(x)\) are the same as in (5), and \(\mu\) is an arbitrary constant. Taking into account the example from § 3, we arrive at the conclusion that the solution \(F^+(x)\) of problem (5), (6) belongs to the space \(A_1\), i.e., the solution of equation (1) is representable in the form (12). Suppose that the corresponding to this

the constant \(D\) needed for the solution is known*). Let us solve problem (22), (23) and choose the constant \(\mathrm{const}\) so that \((T,\widetilde F^{+}(x))=D\). This can be done, since \((T,\widetilde F^{+}_{0})\ne 0\).

Now the following assertions of the theorem come into force: \(f-\widetilde f\in A_{0}\),

\[ K\widetilde f-g\in B_{0} \quad\text{and}\quad \|f-\widetilde f\|_{A_{0}} \le \frac{\|\widetilde K^{-1}\|\cdot \|\widetilde K\widetilde f-g\|_{B_{0}}} {1-\|\widetilde K^{-1}\|\cdot\|K-\widetilde K\|}. \]

This means, first, that \(F^{+}(x)-\widetilde F^{+}(x)\in \overline{M^{+}(0)}\), where \(F^{+}(x)\) is the solution of problem (5), (6), and, second,

\[ \left( \int_{-\infty}^{\infty} |F^{+}(x)-\widetilde F^{+}(x)|^{2}\,dx \right)^{1/2} \le \]

\[ \le \frac{\|\widetilde K^{-1}\|}{1-\sigma} \left( \int_{-\infty}^{\infty} \left| \frac{x^{2}+q^{2}}{x^{2}} \left\{[(\lambda+K(x))\widetilde F^{+}(x)]^{+}-G^{+}(x)\right\} - \right. \]

\[ \left. -\frac{c_{0}}{\sqrt{2\pi}\,x^{2}} +\frac{i c_{1}}{\sqrt{2\pi}\,x} \right|^{2} dx \right)^{1/2}. \tag{27} \]

Here the constants \(c_{0}\) and \(c_{1}\) are chosen so that the integral on the right-hand side converges (the choice is possible because \(K\widetilde f-g\in B_{0}\)). Applying Parseval’s equality, we obtain**

\[ \left( \int_{0}^{\infty} |f_{+}(x)-\widetilde f_{+}(x)|^{2}\,dx \right)^{1/2} \le \]

\[ \le \frac{\|\widetilde K^{-1}\|}{1-\sigma} \left( \int_{0}^{\infty} \left| \lambda \widetilde f_{+}(x) + \frac{1}{\sqrt{2\pi}} \int_{0}^{\infty} k(x-t)\widetilde f_{+}(t)\,dt - g_{+}(x) - \right. \right. \]

\[ \left. \left. - q^{2}\int_{0}^{x} (x-t) \left\{ \lambda\widetilde f_{+}(t) + \frac{1}{\sqrt{2\pi}} \int_{0}^{\infty} k(t-\tau)\widetilde f_{+}(\tau)\,d\tau - g_{+}(t) \right\} dt + c_{0}x+c_{1} \right|^{2} dx \right)^{1/2}. \tag{28} \]

Let us emphasize once more that here \(f_{+}(x)\) is the exact solution of equation (1), while \(\widetilde f_{+}(x)\) is its approximate solution, defined by the equality

*) In a number of practically important cases the number \(D\) is found without great difficulty. Thus, in the example of § 3,

\[ D= \sqrt{\frac{q}{\pi}}\, \frac{|\sqrt{-ia}-iq|} {a\,|\sqrt{-ia}|} \exp\left\{ -\frac{a}{2\pi} \int_{-\infty}^{\infty} \ln\Omega(t)\, \frac{dt}{t(t+ia)} \right\}. \]

**) In inequality (28) the expression
\[ \lambda\widetilde f_{+}(x) + \frac{1}{\sqrt{2\pi}} \int_{0}^{\infty} k(x-t)\widetilde f_{+}(t)\,dt \]
has the same meaning as in (1).

\[ \tilde f_+(x)=V^{-1}\tilde F^+=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\tilde F^+(t)e^{-ixt}\,dt, \tag{29} \]

where \(\tilde F^+(x)\) is the solution of problem (22), (23), in which the constant \(\mathrm{const}\) should be disposed of as indicated above.

6. On the choice of \(\tilde\lambda+\tilde K(x)\), \(\tilde G^+(x)\), and \(q\). First of all, these quantities must satisfy all the requirements set forth in 1), 3), 4), and 8) of Sec. 4. In addition, in order to reduce the mean-square error when choosing the indicated quantities, one should strive to make small the left-hand side of inequality (17) and the following integral:

\[ \int_{-\infty}^{\infty} \left| \frac{x^2+q^2}{x^2}\,(G^+-\tilde G^+) \right|^2 dx = \int_{0}^{\infty} \left| g_+(x)-\tilde g_+(x)-q^2\int_{0}^{x} (x-t)\,[g_+(t)-\tilde g_+(t)]\,dt \right|^2 dx. \]

One can arrive at this conclusion by considering inequality (27).

It is also desirable that the functions \(\tilde\lambda+\tilde K(x)\) and \(\tilde G^+(x)\) have a structure for which the solution \(\tilde F^+(x)\) of the Riemann problem (22), (23) can be constructed elementarily, without quadratures. This latter requirement is met, for example, by functions of the form

\[ \tilde\lambda+\tilde K(x)= \frac{|x|P_1}{\sqrt{x^2+c^2}\,P_2}, \qquad \tilde G^+(x)= \left[ \frac{|x|P_3}{\sqrt{x^2+c^2}\,P_4} +\frac{P_5}{P_6} \right]^+, \tag{30} \]

where \(c\) is a constant, and \(P_1,\ldots,P_6\) are polynomials.

In the form (30), however, the function \(\tilde\lambda+\tilde K(x)\) cannot always be chosen. If this form proves unsuitable, it is natural to change the point of view on the role of the functions \(\dot X^+(x)\), \(\tilde X^-(x)\), and \(\tilde\lambda+\tilde K(x)\), and to suppose that first the functions \(\dot X^+(x)\) and \(\tilde X^-(x)\) have been found by approximately computing the Cauchy-type integral in formula (9)

\[ \tilde X^{\pm}(x)\simeq \exp\left[ \mp\frac{1}{2}\ln\Omega(x) -\frac{x}{2\pi i} \int_{-\infty}^{\infty} \ln\Omega(t)\,\frac{dt}{t(t-x)} \right], \tag{31} \]

and then the function \(\tilde\lambda+\tilde K(x)\) is constructed by the formula

\[ \tilde\lambda+\tilde K(x)\simeq \frac{|x|\tilde X^-(x)} {\sqrt{x^2+q^2}\,\dot X^+(x)}. \]

With such an approach, the requirements in 1), 3), 4), and 8) of Sec. 4 turn into conditions that should be taken into account in the process of approximation (31).

References

1. Cherskii Yu. I., DAN SSSR, 150, No. 2, 271–274, 1963.

Received by the editors
January 13, 1965.

Odessa State University
named after I. I. Mechnikov