UDC 517.946+517.516

MATHEMATICS

M. B. KAPILEVICH

ON SINGULAR CAUCHY AND TRICOMI PROBLEMS

(Presented by Academician I. N. Vekua, March 1, 1967)

Consider in the half-plane \(s \geq 0\) the equation

\[ z_{xx}+\frac{c}{x}z_x=z_{ss}+\frac{a}{s}z_s+b^2z \qquad (a,b,c=\mathrm{const}) \tag{1} \]

and call \(z(x,s)\), \(\bar z(x,s)\) solutions of the Tricomi problems for (1), if

\[ z(x,0)=\tau(x),\qquad z(x,x)=\varphi(x),\qquad \tau(0)=\varphi(0); \tag{2a} \]

\[ \bar z_\eta(x,0)=\nu(x),\qquad \bar z(x,x)=\psi(x),\qquad \eta=-\left(\frac{s}{1-a}\right)^{1-a}. \tag{2b} \]

As shown in \((^1)\), \(z\) and \(\bar z\) can be written in quadratures if the Riemann function \(v(x,s;x_0,s_0)\) and the Green–Hadamard resolvents \(H(x,s;x_0,s_0)\), \(\bar H(x,s;x_0,s_0)\) of problems (1), (2) are known. Introducing, instead of \(v,H,\bar H\), the expressions
\(U=\Phi_0v,\ V=\Phi_0H,\ \bar V=\Phi_0\bar H,\ \Phi_0=x_0^{-c}s_0^{-a}\), we obtain:

1. Let \(a=2\beta,\ c=2\mu,\ \pi\delta=\sin\pi\beta,\ 4ss_0\omega=R^2,\ 4xx_0\lambda=-R^2,\)
   \(4\rho=b^2R^2,\ R^2=(x-x_0)^2-(s-s_0)^2\). Then for \(0<\beta<1\)

\[ U=\delta(xx_0)^{-\mu}(ss_0)^{-\beta} \int_0^1 \xi^{-\beta}(1-\xi)^{\beta-1}(1-\xi\omega)^{-\beta}Q_0(\xi)\,d\xi; \tag{3a} \]

\[ Q_0(\xi)=\Xi_2[\mu,1-\mu,\beta;\lambda(1-\xi),\rho(1-\xi)], \tag{3b} \]

where \(\Xi_2(\alpha,\beta,\gamma;x,y)\) is Humbert’s function. Further, for all \(\beta<1\)

\[ V=\delta(xx_0)^{-\mu}\omega^{a-1}\left(\frac{2}{R}\right)^a \int_0^1[\xi(1-\xi)]^{-\beta} \left(1-\frac{\xi}{\omega}\right)^{\beta-1}Q_1(\xi)\,d\xi; \tag{4a} \]

\[ Q_1(\xi)=\Xi_2[\mu,1-\mu,\beta;\lambda(1-\xi/\omega),\rho(1-\xi/\omega)], \tag{4b} \]

and, with the same notation and for \(\beta>0\),

\[ \bar V=\delta(xx_0)^{-\mu}\left(\frac{2}{R}\right)^a \int_0^1[\xi(1-\xi)]^{\beta-1} \left(1-\frac{\xi}{\omega}\right)^{-\beta}\bar Q_1(\xi)\,d\xi; \tag{5a} \]

\[ \bar Q_1(\xi)=\Xi_2[\mu,1-\mu,1-\beta;\lambda(1-\xi/\omega),\rho(1-\xi/\omega)]. \tag{5b} \]

From (3), (4), (5) it further follows that

\[ U=(xx_0)^{-\mu}(ss_0)^{-\beta} \sum_{n=0}^{\infty}A_n\rho^n F_3(\beta,\mu,1-\beta,1-\mu,1+n;\omega,\lambda), \tag{6a} \]

\[ V=\varkappa(xx_0)^{-\mu}\omega^{a-1} \left(\frac{2}{R}\right)^a \times \]

\[ \times \sum_{n=0}^{\infty}B_n\rho^n H_2(1-\beta-n,1-\beta,\mu,1-\mu,2-a;\omega^{-1},-\lambda), \tag{6b} \]

\[ \bar V=\bar\varkappa(xx_0)^{-\mu} \left(\frac{2}{R}\right)^a \sum_{n=0}^{\infty}\bar B_n\rho^n H_2(\beta-n,\beta,\mu,1-\mu,a;\omega^{-1},-\lambda). \tag{6c} \]

Here \((n!)^2A_n=1,\ n!(\beta)_nB_n=1,\ n!(1-\beta)_n\bar B_n=1\), and \(\varkappa,\bar\varkappa\) are indicated in \((^2)\).

2. With the aid of (6b), (6c), for \(\varphi(x)=\psi(x)\equiv 0\) we obtain

\[ z=\varkappa(1-a)(2s)^{1-a}\int_0^{x-s}\tau(x_0)r^{a-2}\left(\frac{x_0}{x}\right)^\mu \Xi_2(\mu,1-\mu,\beta;\lambda_1,\rho_1)\,dx_0, \tag{7a} \]

\[ \bar z=\frac12\varkappa[2(1-a)]^a\int_0^{x-s}\nu(x_0)r^{-a}\left(\frac{x_0}{x}\right)^\mu \Xi_2(\mu,1-\mu,1-\beta;\lambda_1,\rho_1)\,dx_0, \tag{7b} \]

where \(4xx_0\lambda_1=-r^2,\ 4\rho_1=b^2r^2,\ r=\sqrt{(x-x_0)^2-s^2}\). If, however, \(\tau(x)=x^\alpha\) \((\alpha>0)\), then

\[ z^{(\alpha)}=Dx^{-(c+\alpha+1)}s^{1-a}r^{a+c+2\alpha} \Xi_2\left(\frac{\alpha}{2}+1,\frac{\alpha+1}{2}+\mu,\gamma;\frac{r^2}{x^2},\frac14 b^2r^2\right), \tag{8} \]

\[ r^2=x^2-s^2,\qquad \gamma=\alpha+\beta+\mu+1,\qquad -c-2\alpha<a<1, \]

\[ \Gamma(\gamma)\Gamma(-\nu)D=\Gamma\left(\frac{\alpha}{2}+1\right) \Gamma\left(\frac{\alpha+1}{2}+\mu\right). \]

Since \(z(x,s)\) and \(\bar z(x,s)\) are connected by the equality

\[ \bar z[x,s;a,b,c;\nu(x)]\equiv \eta z[x,s;2-a,b,c;\nu(x)], \tag{9} \]

(8) and (9) give a convergent solution \(\bar z^{(\alpha)}(x,s)\) of problem (2b) for \(\nu(x)=x^\alpha\). In the more general case when \(\tau(x)\) and \(\nu(x)\) are given by power series

\[ \tau(x)=\sum_{m=0}^{\infty}A_mx^{m+\alpha},\qquad \nu(x)=\sum_{m=0}^{\infty}\bar A_mx^{m+\alpha} \quad(\alpha=\text{const}>0), \tag{10} \]

we arrive at expansions in the functions \(z^{(\alpha)}\) and \(\bar z^{(\alpha)}\):

\[ z(x,s)=\sum_{m=0}^{\infty}A_mz^{(m+\alpha)}(x,s),\qquad \bar z(x,s)=\sum_{m=0}^{\infty}\bar A_m\bar z^{(m+\alpha)}(x,s). \tag{11} \]

Finally, denoting by \(z_0(x,s)\) and \(\bar z_0(x,s)\) the solutions of the Cauchy problems for (1):

\[ z_0(x,0)=\tau(x),\qquad z_{0n}(x,0)=0,\qquad \bar z_0(x,0)=0,\qquad \bar z_{0n}(x,0)=\nu(x), \tag{12} \]

we obtain, for \(\beta>0\), \(4(1+\xi t)\omega_0=t^2(\xi^2-1),\ t=s/x,\ 4\sigma=b^2s^2(\xi^2-1)\),

\[ z_0(x,s)=\gamma\int_{-1}^{1}\tau(x+\xi s)(1-\xi^2)^{\beta-1}(1+\xi t)^\mu Q(x,s;\xi)\,d\xi, \tag{13a} \]

\[ \sqrt{\pi}\Gamma(\beta)\gamma=\Gamma(\nu+1),\qquad \nu=\beta-\frac12,\qquad Q=\Xi_2(\mu,1-\mu,\beta;\omega_0,\sigma), \]

and when \(\beta<1\),

\[ \sqrt{\pi}\Gamma(1-\beta)\bar\gamma=\Gamma(1-\nu),\qquad \bar Q=\Xi_2(\mu,1-\mu,1-\beta;\omega_0,\sigma): \]

\[ \bar z_0(x,s)=\bar\gamma\eta\int_{-1}^{1}\nu(x+\xi s)(1-\xi^2)^{-\beta}(1+\xi t)^\mu \bar Q(x,s;\xi)\,d\xi. \tag{13b} \]

This time the holomorphic data (10) correspond to formulas (11), in which

\[ z_0^{(\alpha)}(x,s)=x^\alpha \Xi_2\left(-\frac{\alpha}{2},\frac{1-\alpha}{2}-\mu,1+\nu;\frac{s^2}{x^2},-\frac14 b^2s^2\right); \tag{14a} \]

\[ \bar z_0^{(\alpha)}(x,s)=x^\alpha\eta \Xi_2\left(-\frac{\alpha}{2},\frac{1-\alpha}{2}-\mu,1-\nu;\frac{s^2}{x^2},-\frac14 b^2s^2\right). \tag{14b} \]

3. In analogous operators one solves problems (2), (12) for the equation

\[ z_{xx}+\frac{c}{x}z_x=z_{ss}+\frac{a}{s}z_s+b(1+\bar r^2)^{-2}z \qquad(a,b,c=\text{const}), \tag{15} \]

where \(\bar r^2=x^2-s^2\). Namely, here (3)—(6) likewise determine \(U,V,\bar V\), but now

\[ Q_0=F_3[\mu,\gamma,1-\mu,1-\gamma,\beta;\lambda(1-\xi),\rho(1-\xi)],\qquad (n!)^2A_n=\Gamma_n; \tag{16a} \]

\[ Q_1=F_3[\mu,\gamma,1-\mu,1-\gamma,\beta;\lambda(1-\xi/\omega),\rho(1-\xi/\omega)],\qquad n!(\beta)_nB_n=\Gamma_n. \tag{16b} \]

and \(\overline Q_1,\ \overline B_n\) arise from \(Q_1,\ B_n\) after replacing \(\beta\) by \(1-\beta\), where in (16)

\[ (1+\overline r^{\,2})(1+\overline r_0^{\,2})\rho=R^2,\qquad \overline r_0^{\,2}=x_0^2-s_0^2,\qquad 4\gamma(1-\gamma)=b,\qquad \Gamma_n=(\gamma)_n(1-\gamma)_n . \]

The solutions \(z\) and \(\overline z\) of the Tricomi problems (2), (15) for \(\varphi=\psi=0\) have the form

\[ z=\chi(1-a)(2s)^{1-a}\int_0^{x-s} \tau(x_0)r^{a-2}\left(\frac{x_0}{x}\right)^\mu F_3(\mu,\gamma,1-\mu,1-\gamma,\beta;\lambda_1,\rho_1)\,dx_0; \tag{17a} \]

\[ \overline z=\frac12\,\overline\chi[2(1-a)]^a\int_0^{x-s} \nu(x_0)r^{-a}\left(\frac{x_0}{x}\right)^\mu F_3(\mu,\gamma,1-\mu,1-\gamma,1-\beta;\lambda_1,\rho_1)\,dx_0, \tag{17b} \]

where \((1+\overline r^{\,2})(1+x_0^2)\rho_1=r^2\), and \(r,\lambda_1,\chi\) and \(\overline\chi\) are the same as in (7). With the aid of (3a), (16a) we further find for the Cauchy problems (12), (15) \(\bigl(r_1=|s^2-(x-x_0)^2|\bigr)\)

\[ z_0=\gamma_0s^{1-a}\int_{x-s}^{x+s} \tau(x_0)r_1^{a-2}\left(\frac{x_0}{x}\right)^\mu F_3(\mu,\gamma,1-\mu,1-\gamma,\beta;\lambda_2,\rho_1)\,dx_0; \tag{18a} \]

\[ \overline z_0=\overline\gamma_0\int_{x-s}^{x+s} \nu(x_0)r_1^{-a}\left(\frac{x_0}{x}\right)^\mu F_3(\mu,\gamma,1-\mu,1-\gamma,1-\beta;\lambda_2,\rho_1)\,dx_0; \tag{18b} \]

\[ 4xx_0\lambda_2=-r_1^2,\qquad \sqrt\pi\,\Gamma(\beta)\gamma_0=\Gamma(1+\nu), \]

\[ \sqrt\pi\,\Gamma(1-\beta)(1-a)^{1-a}\overline\gamma_0=-\Gamma(1-\nu). \]

Other integral representations are obtained for \(U,\ V,\ \overline V\) by substituting in (13) and (18) the initial data \(\tau(x),\nu(x)\) of these functions on the line \(s=0\).

4. In terms of the series \(E_2\), the Cauchy problem (12a) is solved for the more general, than (1), nonhomogeneous equation with a singular perturbing function:

\[ u_{xx}+\frac{c}{x}u_x=u_{ss}+\frac{a}{s}u_s+ \left(b^2+\frac{k}{s^2}\right)u-\frac{k}{s^2}\tau(x) \qquad (k=\mathrm{const}). \tag{19} \]

Here, under the condition \(\beta>\mu>0\), we find

\[ u_0(x,s)=\delta_1\int_{-1}^{1} \tau(x+\xi s)(1-\xi^2)^{\beta-1}(1+\xi t)^\mu T(x,s;\xi)\,d\xi; \tag{20a} \]

\[ T=\delta_2\int_0^1 \eta^{\mu-1}(1-\eta)^{\beta-\mu-1}(1-\eta\omega_0)^{\mu-1}Q(\eta)\,d\eta, \tag{20b} \]

\[ Q(\eta)=E_2[p,q,\beta-\mu;(1-\xi^2)(1-\eta),\sigma(1-\eta)], \tag{20c} \]

where \(p\) and \(q\) are the roots of the equation \(\rho^2-\nu\rho+k/4=0\); \(\omega_0,\sigma,t,\beta,\mu,\nu\) are the same as in (13a);

\[ \sqrt\pi\,\Gamma(\beta)\delta_1=\Gamma(p+1)\Gamma(q+1),\qquad \Gamma(\mu)\Gamma(\beta-\mu)\delta_2=\Gamma(\beta). \]

From (20b) it follows that

\[ T=\sum_{n=0}^{\infty}(p)_n(q)_n[(\beta)_n n!]^{-1} (1-\xi^2)^n E_2(\mu,1-\mu,\beta+n;\omega_0,\sigma). \tag{21} \]

When \(\beta=0\) \((\beta=1)\), \(\mu=0\) \((\mu=1)\), \(b=0\), (6) and (21) give Appell, Humbert, and Horn series with two arguments; while in the general case (6), (21) determine confluent hypergeometric functions of three variables, so that, for example, (6) may be written in the form

\[ \overline V=\overline\chi(xx_0)^{-\mu}(2/R)^a H_2^{(3)}(\beta,\beta,\mu,1-\mu,a;\omega^{-1},-\lambda,\rho). \tag{22} \]

For the purposes of regular continuation of the series (6b), (6c), the following autotransformation is used:

\[ H_2(a,\beta,\gamma,\delta,\varepsilon;x,y) =(1-x)^{-a}H_2\left[a,\varepsilon-\beta,\gamma,\delta,\varepsilon; \frac{x}{x-1},\,y(1-x)\right], \tag{23a} \]

as well as the quadratic transformation

\[ \begin{gathered} H_2(\alpha,\beta,\gamma,\delta;2\beta;x,y)=\\ =\left(1-\frac{x}{2}\right)^{-\alpha} H_7\left[\alpha,\gamma,\delta,\beta+\frac12;\frac{x^2}{4(2-x)^2},\,y\left(1-\frac{x}{2}\right)\right], \end{gathered} \tag{23b} \]

by which \(H_2\) can be transformed in each term of the series (6b), (6c).

The results obtained generate various basis representations for \(z,\bar z,z_0,\bar z_0,u_0\). Thus, starting from (11a), (14a) and the addition theorem for \(\Xi_2(\alpha,\beta,\gamma;x,y)\) with respect to \(\gamma\) and \(y\), we find, for \(\beta_0=\beta_2-\beta_1>-n,\ b_0=\sqrt{b_1^2-b_2^2},\ \nu_k=\beta_k-\frac12\),

\[ z(x,s;a_2,b_2,c)= \sum_{m=0}^{n-1}(-1)^m g_m(\beta_2)(b_0s)^{2m} z(x,s;a_2+2m,b_1,c)+R_n; \tag{24a} \]

\[ R_n=\delta\int_0^1 \xi^{a_1}(1-\xi^2)^{\beta_0+n-1} U_n(s,\xi)\,z(x,\xi s;a_1,b_1,c)\,d\xi . \tag{24b} \]

Here \(U_n\) is the normalized Lommel function:

\[ (\beta_0)_n n! U_n=(b_0s/2)^{2n}\, {}_1F_2\left[1,\beta_0+n,n+1;\frac14 b_0^2s^2(1-\xi^2)\right], \tag{25} \]

\[ m!(\nu+1)_m2^{2m}g_m(\beta)=(-1)^m,\qquad \delta\Gamma(\beta_0)\Gamma(\nu_1+1)=2\Gamma(\nu_2+1). \]

Comparing (12), (19) with the homogeneous problem (1), (12), we arrive at the equality

\[ u(x,s;a_2,b,c,k)= \sum_{m=0}^{n-1}D_m z(x,s;a_2+2m,b,c)+\overline{R}_n; \tag{26a} \]

\[ \overline{R}_n=\int_0^1 \xi^{a_1}(1-\xi^2)^{\beta_0+n-1} Q_n(\xi)\,z(x,\xi s;a_1,b,c)\,d\xi, \tag{26b} \]

\[ Q_n(\xi)=\varkappa_n\,{}_3F_2(1,p_2+n,q_2+n;n+1,\beta_0+n;1-\xi^2), \tag{26c} \]

\[ \varkappa_n\Gamma(\beta_0+n)\Gamma(\nu_1+1)n! =2\Gamma(\nu_2+n+1)n!D_n =2\Gamma(p_2+1)\Gamma(q_2+1)(p_2)_n(q_2)_n, \]

where \(\beta_0=\beta_2-\beta_1>-n,\ a_1>0\). Other basis representations arise when substituting into (13a), (18a), (20a) the expansions \((2\alpha=a+1-n)\)

\[ \tau(x+\xi s)= \sum_{n=0}^{\infty}\frac{1}{(\nu)_n}\left(\frac{s}{2}\right)^n C_n^\nu(\xi)D_x^n z_0(x,s;a+2n,0,0); \tag{27a} \]

\[ \tau(x+\xi s)= \sum_{n=0}^{\infty}\frac{1}{(a)_n}\left(\frac{s}{2}\right)^n C_n^a(\xi)D_x^n z_0(x,s;a+n,0,0), \tag{27b} \]

as well as, for the confluent case, the series (27a)

\[ \tau(x+\xi s)=\sum_{n=0}^{\infty}\frac{s^n}{n!} H_n\left(\frac{\xi}{2}\right)D_x^n w(x,s^2), \]

where \(w(x,s)\) is the solution of the Cauchy problem \(w_{xx}=w_s,\ w(x,0)=\tau(x)\). Finally, note that the functions \(V,\bar V\), which possess logarithmic singularities on the characteristics \(R=0\), may be used as fundamental solutions in boundary-value problems for the corresponding equations (1) and (15) of elliptic type. In particular, they make it possible to generalize the mean-value theorems investigated earlier in \({}^3\).

Moscow Evening
Metallurgical Institute

Received
2 IX 1966

REFERENCES

1. S. Gellerstedt, Ark. math., astr. och fys., 25A, No. 29, 1 (1937).
2. M. B. Kapilevich, DAN, 170, No. 6, 1251 (1966).
3. M. B. Kapilevich, DAN, 125, No. 2, 251 (1959).