M. B. KAPILEVICH

ON THE APPROXIMATION OF SINGULAR SOLUTIONS OF THE CHAPLYGIN EQUATION

(Presented by Academician I. N. Vekua on 24 VII 1963)

As follows from the results of the note \((^1)\), if for the Chaplygin equation

\[ z_{\theta\theta}-z_{\sigma\sigma}-b(\sigma)z_\sigma=0,\qquad b(\sigma)=\left[\ln \sqrt{K(\sigma)}\right]_\sigma =\sum_{n=0}^{\infty} b_n\sigma^{2/3n-1} \tag{1} \]

one considers, in the region \(\sigma\geq 0\), the singular Cauchy problem

\[ z(\theta,0)=\tau(\theta),\qquad z_\eta(\theta,0)=v(\theta),\qquad \eta=-\left({}^3/_2\sigma\right)^{2/3}, \tag{2} \]

and seeks its solution \(z(\theta,\sigma)\) in the integral form

\[ z(\theta,\sigma)= \int_{\theta-\sigma}^{\theta+\sigma} G(\theta-\alpha,\sigma)\tau(\alpha)\,d\alpha + \int_{\theta-\sigma}^{\theta+\sigma} \overline{G}(\theta-\alpha,\sigma)v(\alpha)\,d\alpha, \tag{3} \]

then \(G(\theta,\sigma)\) and \(\overline{G}(\theta,\sigma)\) can be approximated in a neighborhood of the line \(\sigma=0\) by the series

\[ G(\theta,\sigma)=\sum_{n=0}^{\infty}G_n(\theta,\sigma),\qquad \overline{G}(\theta,\sigma)=\sum_{n=0}^{\infty}\overline{G}_n(\theta,\sigma), \tag{4} \]

in which \(G_0=\overline{\gamma}_1\sigma^{2/3}r^{-5/3}\) and \(\overline{G}_0=-\gamma_2 r^{-1/3}\) are the values of the kernels \(G\) and \(\overline{G}\) for the case \(b(\sigma)=1/3\sigma\), \(G_1=-{}^3/_4 b_1(\sigma^{2/3}G_0-\overline{\gamma}_2r^{-1/3})\), \(G_2=c_0\sigma^{4/3}G_0+c_1\gamma_2\sigma^{2/3}r^{-1/3}+4c_2\overline{\gamma}_1 r^{1/3}\), \(G_3=2D_0\sigma^2G_0+2D_1\gamma_2\sigma^{4/3}r^{-1/3}+8D_2\gamma_1\sigma^{2/3}r^{1/3}+2D_3g_3(\theta,\sigma)\), and the functions \(\overline{G}_n\) \((n=1,2,3)\) have the form \(\overline{G}_1=-{}^3/_4 b_1\sigma^{2/3}\overline{G}_0\), \(\overline{G}_2=c_0\sigma^{4/3}\overline{G}_0+{}^8/_9 A_1c_2g_3\), \(\overline{G}_3=2D_0\sigma^2\overline{G}_0-{}^8/_5 t^2D_4r^{5/3}-4A_1D_3\sigma^{2/3}g_3\). Here \(\overline{\gamma}_1=2^{2/3}\gamma_1\), \(\gamma_0=({}^2/_3)^{2/3}\gamma_2\), \(A_1=-{}^1/_2({}^3/_2)^{1/3}\), \(r=\sqrt{\sigma^2-\theta^2}\); the constants \(c_n\) \((n=0,1,2)\) and \(D_n\) \((n=0,1,2,3,4)\) depend only on \(b_1,b_2,b_3\), while by \(g_3(\theta,\sigma)\) is denoted the difference

\[ g_3=\theta\left[I_{t^2}\left({}^7/_6,-{}^1/_2\right)-I_{t^2}\left({}^5/_6,-{}^1/_2\right)\right], \qquad t=\frac{r}{\sigma}. \]

Each of the functions \(G_n\) and \(\overline{G}_n\) \((n=0,1,2,\ldots)\) contains terms that become infinite on the characteristics \(\theta\pm\sigma=0\), and therefore, in order to improve the convergence of the series (4) near the lines \(\theta\pm\sigma=0\), it is expedient to consider another iteration method. Substituting in (1) \(z=\sigma^{1/6}\chi^{-1}u\), \(\chi=\sqrt[4]{K}\), we obtain \((^2)\)

\[ T[u]=u_{\theta\theta}-u_{\sigma\sigma}-\frac{1}{3\sigma}u_\sigma+c(\sigma)u=0, \tag{5} \]

where

\[ c(\sigma)=\frac{5}{36\sigma^2}+\chi^{-1}\chi_{\sigma\sigma} =\sum_{n=-1}^{\infty} c_n\sigma^{2n/3}. \]

We shall use the procedure indicated in \((^{1,3})\), and instead of (4) we shall seek, in the region \(\Omega\) \((\sigma\geq \theta\geq 0)\), the Riemann integrals

\[ W(\theta,\sigma)=\int_0^\theta G(\alpha,\sigma)\,d\alpha,\qquad \overline{W}(\theta,\sigma)=\int_0^\theta \overline{G}(\alpha,\sigma)\,d\alpha, \]

and then, differentiating them with respect to \(\theta\), compute the kernels themselves: \(G=W_\theta\), \(\overline{G}=\overline{W}_\theta\). The functions \(W(\theta,\sigma)\) and \(\overline{W}(\theta,\sigma)\), as well as \(G,\overline{G}\), satisfy equation (5), and they are uniquely determined in \(\Omega\) by their boundary data

\[ W(0,\sigma)=\overline{W}(0,\sigma)=0,\qquad W(\sigma,\sigma)=S(\sigma),\qquad \overline{W}(\sigma,\sigma)=\overline{S}(\sigma). \tag{6} \]

Here \(S(\sigma)\) and \(\overline{S}(\sigma)\) are the solutions of the Cauchy problem (2), (5) for the values \(\tau(\theta)=1/2,\ v(\theta)=0\) and \(\tau(\theta)=0,\ v(\theta)=1/2\) \((-\infty<\theta<\infty)\), respectively.

or, what is the same thing, the solutions of two initial problems

\[ S_{\sigma\sigma}+\frac{1}{3\sigma}S_\sigma-c(\sigma)S=0,\quad S(0)=\overline{S}_\eta(0)=\frac{1}{2},\quad S_\eta(0)=\overline{S}(0)=0. \tag{7} \]

Put \(W=\sum_{n=0}^{\infty} W_n(\theta,\sigma)\) and replace (5) by the recurrence system:

\[ E[W_0]=0,\quad E[W_1]=-c_{-1}\sigma^{-2/3}W_0,\quad E[W_2]=-c_{-1}\sigma^{-2/3}W_1-c_0W_0,\ldots, \tag{8} \]

where \(E[W]=W_{\theta\theta}-W_{\sigma\sigma}-\frac{1}{3\sigma}W_\sigma\). Equations (7) lead, near \(\sigma=0\), to the corresponding expansions for the initial functions \(S(\sigma)=\sum_{n=0}^{\infty}S_n(\sigma)\),

\[ \overline{S}(\sigma)=\sum_{n=0}^{\infty}\overline{S}_n(\sigma), \]

where

\[ S_0=\frac{1}{2},\quad \overline{S}_0=A_1\sigma^{2/3},\quad S_n=\mu_n\sigma^{2/3(n+1)},\quad \overline{S}_n=\overline{\mu}_n\sigma^{2/3(n+2)} \]

\[ (n=1,2,\ldots), \]

with

\[ \mu_1=\frac{9}{16}c_{-1},\quad \overline{\mu}_1=\frac{3}{8}A_1c_{-1},\ldots \]

Consequently, (6) will be satisfied if we require that, for all \(n=0,1,2,\ldots\),

\[ W_n(0,\sigma)=\overline{W}_n(0,\sigma)=0,\quad W_n(\sigma,\sigma)=S_n(\sigma),\quad \overline{W}_n(\sigma,\sigma)=\overline{S}_n(\sigma). \tag{9} \]

Solving successively the boundary-value problems (8), (9), we obtain

\[ W_0=\frac{1}{2}I_\xi\left(\frac{1}{2},\frac{1}{6}\right),\quad \overline{W}_0=\frac{1}{2}\eta I_\xi\left(\frac{1}{2},\frac{5}{6}\right),\quad W_1=\frac{9}{16}c_{-1}\sigma^{4/3}I_\xi\left(\frac{1}{2},\frac{7}{6}\right),\ldots, \tag{10} \]

where \(\xi=\theta^2/\sigma^2\). Continuing such calculations, one can, step by step, find the subsequent iterations \(W_n,\overline{W}_n\) \((n=2,3,\ldots)\), which are uniquely determined by conditions (8), (9) and are likewise expressed in the form of a linear combination of incomplete Euler beta-functions

\[ I_\xi\left(\frac{1}{2},\frac{1}{6}+n\right),\quad I_\xi\left(\frac{1}{2},\frac{5}{6}+n\right) \]

\((n=0,1,2,\ldots)\). We now differentiate (10) with respect to the variable \(\theta\); then, for the terms \(G_n=(W_n)_\theta\) and \(\overline{G}_n=(\overline{W}_n)_\theta\) of the series (4), we obtain

\[ G_0=\gamma_1\sigma^{1/3}r^{-5/3},\quad G_1=\frac{9}{2}c_{-1}\overline{\gamma}_1 r^{1/3},\quad \overline{G}_0=-\gamma_2 r^{-1/3},\quad \overline{G}_1=A_1c_{-1}g_3(\theta,\sigma),\ldots \tag{11} \]

These expressions show that, in contrast to the case (1), for (5), with the same iteration method, already the second approximations \(G_1(\theta,\sigma)\) and \(\overline{G}_1(\theta,\sigma)\) are not only bounded, but even vanish on the characteristics \(\theta\pm\sigma=0\). Improvement of the convergence of the series (4) in a neighborhood of the lines \(\theta\pm\sigma=0\) can also be achieved by means of another iterative process, in which (5) is compared with the generalized wave equation \({}^{(3)}\)

\[ Q[u]=u_{\theta\theta}-u_{\sigma\sigma}-\frac{a}{\sigma}u_\sigma+c_0u=0\quad (c_0=\mathrm{const},\ a=1/3). \tag{12} \]

For (12), the initial problems (7) (with \(c(\sigma)=c_0,\ \eta=-(\sigma/(1-a))^{1-a}\)) are solved in Bessel functions

\[ S(\sigma)=\frac{1}{2}J_{\beta-1/2}\left(\sigma\sqrt{c_0}\right),\quad \overline{S}(\sigma)=\frac{1}{2}\eta J_{1/2-\beta}\left(\sigma\sqrt{c_0}\right) \quad(\beta=a/2), \tag{13} \]

therefore, in order to construct the integrals \(W(\theta,\sigma)\) and \(\overline{W}(\theta,\sigma)\) from their boundary values (6), (13), we introduce in (12), instead of \(\theta,\sigma\), the variables \(\xi=\theta^2/\sigma^2\) and \(\zeta=-c_0\sigma^2/4\), and, moreover, put \(u=\sqrt{\xi}\,v\). Then the transformed operator \(Q\) vanishes, in particular, in the case when \(v(\xi,\zeta)\) satisfies simultaneously two equations:

\[ \zeta v_{\zeta\zeta}-\xi v_{\xi\zeta}+\frac{1}{2}av_\zeta+v=0, \]

\[ \xi(1-\xi)v_{\xi\xi}+\xi\zeta v_{\xi\zeta} +\frac{1}{2}[3-(5-a)\xi]v_\xi+\frac{1}{2}\zeta v_\zeta-\frac{1}{4}(2-a)v=0. \]

One of the solutions of such a system is, as is known \({}^{(4)}\), the function

\[ H_3(A,B,B+1;\xi,\zeta) =\delta \xi^{-B}\sum_{n=0}^{\infty} \frac{(-1)^n \xi^{-n}}{(B-A+1)_n\,n!}\, I_\zeta(B,n+1-A) \tag{14} \]

for \(|\xi|<1,\ A=1-\beta,\ B=\frac{1}{2},\ \delta=\Gamma(1-A)\Gamma(1+B)/\Gamma(B-A+1)\).

Thus,

\[ W(\beta,\theta,\sigma)=\bar{\gamma}_{1}\frac{\theta}{\sigma} H_{3}\left(1-\beta,\;{}^{1}/_{2},\;{}^{3}/_{2};\frac{\theta^{2}}{\sigma^{2}},-\frac{c_{0}\sigma^{2}}{4}\right), \tag{15} \]

where \(\bar{\gamma}_{1}=2^{1-a}\gamma_{1}=\Gamma(\beta+{}^{1}/_{2})[\sqrt{\pi}\Gamma(\beta)]^{-1}\). In an analogous way we find

\[ \bar W(\beta,\theta,\sigma)=\bar{\gamma}_{2}\eta\frac{\theta}{\sigma} H_{3}\left(\beta,\;{}^{1}/_{2},\;{}^{3}/_{2};\frac{\theta^{2}}{\sigma^{2}},-\frac{c_{0}\sigma^{2}}{4}\right), \tag{16} \]

where \(\bar{\gamma}_{2}=(1-a)^{1-a}\gamma_{2}=\Gamma({}^{3}/_{2}-\beta)[\sqrt{\pi}\Gamma(1-\beta)]^{-1}\).

Finally, differentiating (15) and (16) with respect to \(\theta\), we arrive at the expressions

\[ G=\bar{\gamma}_{1}\sigma^{1-a}r^{-a}\bar I_{\beta-1}(r\sqrt{c_{0}}),\qquad \bar G=-\bar{\gamma}_{2}r^{-a}\bar I_{-\beta}(r\sqrt{c_{0}}). \]

We shall further seek the solution of equation (5) in the form of the series \(u=\sum_{n=0}^{\infty}u_n\), whose terms are determined from the recurrence sequence

\[ Q[u_0]=0,\qquad Q[u_1]=-c_{-1}\sigma^{-2/3}u_0,\qquad Q[u_2]=-c_{-1}\sigma^{-2/3}u_1-c_1\sigma^{2/3}u_0,\ldots, \tag{17} \]

where \(Q[u]\) is operator (12) for \(a={}^{1}/_{3}\). Let us first expand in an analogous series the boundary value \(S(\sigma)=W(\sigma,\sigma)\) (and, correspondingly, \(\bar S(\sigma)=\bar W(\sigma,\sigma)\)). For this we put in (17) \(u=S(\sigma)\), \(u_n=S_n(\sigma)\), which gives

\[ \nabla[S_0]=0,\qquad \nabla[S_1]=c_{-1}\sigma^{-2/3}S_0,\qquad \nabla[S_2]=c_{-1}\sigma^{-2/3}S_1+c_1\sigma^{2/3}S_0,\ldots, \tag{18} \]

where \(\nabla[S]=S''+\frac{1}{3\sigma}S'-c_0S\). Adjoining to (18), by virtue of requirements (7), the equalities

\[ S_0(0)={}^{1}/_{2},\qquad S_n(0)=\frac{d}{d\eta}S_{n-1}(0)=0\quad (n=1,2,\ldots), \]

we obtain

\[ S_0={}^{1}/_{2}\bar I_{-1/3}(\sigma\sqrt{c_0}),\qquad S_1={}^{3}/_{2}\frac{c_{-1}}{c_0}\sigma^{1/3}S_0' ={}^{9}/_{16}c_{-1}\sigma^{4/3}\bar I_{2/3}(\sigma\sqrt{c_0}). \tag{19} \]

Analogously to this, from the subsequent equations (18) the solutions \(S_2(\sigma), S_3(\sigma),\ldots\) are also uniquely determined. Now, in order to compute \(W(\theta,\sigma)\) from its data (5), (6), we require that the terms of the series \(W=\sum_{n=0}^{\infty}W_n\) satisfy equations (17) in the domain \(\Omega\), and on the boundaries of this domain assume the values (9), (19). Then, first of all, from (15) for \(\beta={}^{1}/_{6}\) we find \(W_0=W({}^{1}/_{6},\theta,\sigma)\). Substituting this expression into the right-hand side of the second equation of the sequence (17), we obtain

\[ W_1(\theta,\sigma)={}^{9}/_{2}c_{-1}\bar{\gamma}_{1}\sigma^{1/3}\frac{\theta}{\sigma} H_{3}\left(-{}^{1}/_{6},\;{}^{1}/_{2},\;{}^{3}/_{2};\frac{\theta^{2}}{\sigma^{2}},-\frac{c_{0}\sigma^{2}}{4}\right). \tag{20} \]

Continuing such computations, one can also reduce the solutions \(W_2,W_3,\ldots\) of the subsequent boundary-value problems (9), (17) to combinations of known particular integrals of the Horn hypergeometric system of equations \((^{4})\). In the same way, first from (16) for \(\beta={}^{1}/_{6}\) we obtain \(\bar W_0(\theta,\sigma)=\bar W({}^{1}/_{6},\theta,\sigma)\), and then (9) and (17) will give the further iterations \(\bar W_1,\bar W_2,\ldots\). Differentiating the functions \(W_n\) and \(\bar W_n\) obtained with respect to \(\theta\), we find

\[ \begin{gathered} G_0=\sigma^{2/3}\bar{\gamma}_{1}r^{-5/3}\bar I_{-5/6}(r\sqrt{c_0}),\qquad G_1={}^{9}/_{2}c_{-1}\bar{\gamma}_{1}r^{-1/3}\bar I_{1/6}(r\sqrt{c_0}),\\ \bar G_0=-\bar{\gamma}_{2}r^{-1/3}\bar I_{-1/6}(r\sqrt{c_0}),\ldots \end{gathered} \tag{21} \]

We now replace \(\sigma\) in (1) by \(is\) \((i=\sqrt{-1})\); then we obtain

\[ z_{\theta\theta}+z_{ss}+\bar b(s)z_s=0,\qquad \bar b(s)=\frac{1}{s}\left(b_0-\sum_{n=1}^{\infty}b_ns^{2n/3}\right). \tag{22} \]

Let us find a solution of this equation, bounded in the half-plane \(s\geq 0\) and tending for \(s=0\) to the prescribed piecewise-continuous function

\(\tau(\theta)\), finite on the entire line \(s=0\):

\[ z(\theta,0)=\tau(\theta), \qquad -\infty<\theta<\infty . \tag{23} \]

It can be shown that the solution of this singular Dirichlet problem for the half-plane has the form

\[ z(\theta,s)=\int_{-\infty}^{\infty} H(\theta-\alpha,s)\tau(\alpha)\,d\alpha, \tag{24} \]

where \(H(\theta,s)=H_0(\theta,s)h(\theta,s)\), with
\(H_0=\gamma s^{2/3}r^{-5/3}\), \(\gamma=\Gamma(5/6)/\sqrt{\pi}\Gamma(1/3)\),
\(r=\sqrt{\theta^2+s^2}\), corresponds to the case \(\bar b=1/3s\) (5), while \(h(\theta,s)\) is a function bounded in the whole half-plane \(s\geq 0\) with a finite limit \(\lim_{\theta\to\infty} h(\theta,s)\) not identically equal to zero. We shall call the kernel of the Duamel integral here \(w(\theta,s)\), the solution of the particular discontinuous Dirichlet problem (23):

\[ w(\theta,0)=\frac12\operatorname{sign}\theta \quad (-\infty<\theta<\infty). \]

Then, for an odd function \(w(\theta,s)\) of the variable \(\theta\), we have \(w(0,s)=0\), and therefore it will be constructed in the domain \(D\) \((0\leq\theta<\infty,\ 0\leq s<\infty)\), if for (22) we solve the discontinuous Dirichlet problem for the quadrant \(D\):

\[ w(\theta,0)=\frac12,\qquad w(0,s)=0 \quad (0\leq\theta<\infty,\ 0\leq s<\infty). \tag{25} \]

Furthermore, since \(H(\theta,s)=H(-\theta,s)\), we have
\(w(\theta,s)=\int_0^\theta H(\xi,s)\,d\xi\), whence \(w_\theta=H\), i.e., knowing the solution \(w(\theta,s)\) of problem (25), one can then compute the kernel \(H(\theta,s)=w_\theta\) in the domain \(D\), and consequently, in view of the evenness of the function \(H(\theta,s)\) with respect to the variable \(\theta\), also in the entire half-plane \(s\geq 0\). We shall seek the solution of problem (22), (25) in the form of the series \(w=\sum_{n=0}^{\infty} w_n\), whose terms satisfy the equations
\[ E[w_0]=(w_0)_{\theta\theta}+(w_0)_{ss}+\frac{1}{3s}(w_0)_s=0; \]
\[ E[w_1]=-b_1s^{-1/3}(w_0)_s;\qquad E[w_2]=-b_1s^{-1/3}(w_1)_s-b_2s^{1/3}(w_0)_s,\ldots \]
and the boundary conditions
\[ w_0(\theta,0)=\frac12,\qquad w_{n+1}(\theta,0)=w_n(0,s)=0\quad (n=0,1,2,\ldots). \]

Then first of all we obtain
\[ 2w_0=1-I_{s^2/r^2}\!\left(\frac13,\frac12\right) =I_{\theta^2/r^2}\!\left(\frac12,\frac13\right), \]
or, equivalently,
\[ 2w_0=\gamma \sqrt[4]{27}\,F(\varphi,k), \]
where
\[ \cos\varphi=\bigl[\sqrt3-1+(s/r)^{2/3}\bigr]\bigl[\sqrt3+1-(s/r)^{2/3}\bigr]^{-1}, \]
and \(F(\varphi,k)\) is the elliptic integral of the first kind with modulus
\[ k=\frac12\sqrt{2+\sqrt3}=\sin 75^\circ . \]
The subsequent iterations here have the form:
\[ w_1=\frac34 b_1s^{2/3}w_0, \]
\[ w_2=c_1s^{4/3}w_0+c_2\left[2\theta(2r)^{1/3}+(3\theta-r)(r+\theta)^{1/3}-(3\theta+r)(r-\theta)^{1/3}\right], \]
where
\[ c_1=\frac{9}{16}b_1^2,\qquad 16\sqrt[3]{2}\,c_2=9\gamma\left(\frac34 b_1^2-b_2\right). \]

In the elliptic domain \(s\geq 0\) in the case (5) (with \(\sigma=is\)), along with the first iteration method, one can also apply a second iterative process, in which the first approximations \(H_0(\theta,s)\), \(\bar H_0(\theta,s)\) to the kernels \(H(\theta,s)\) and \(\bar H(\theta,s)\) of the singular Dirichlet and Neumann problems are the integrals of equation (12) \((\sigma=is,\ b^2=-c_0)\) (3):

\[ H_0=\gamma s^{2/3}r^{-5/3}\bar K_{5/6}(br),\qquad \bar H_0=\gamma r^{-1/3}\bar K_{1/6}(br). \]

Moscow Evening
Metallurgical Institute

Received
16 VII 1963

REFERENCES

1. M. B. Kapilevich, DAN, 146, No. 3, 527 (1962).
2. P. Germain, R. Bader, C. R., 231, No. 4, 268 (1950).
3. M. B. Kapilevich, Matem. sborn., 30, No. 1, 11 (1952).
4. A. Erdélyi, Higher Transcendental Functions, 1, 1953.
5. I. N. Vekua, DAN, 56, No. 3, 229 (1947).