UDC 517.944/.947+517.516

MATHEMATICS

M. B. KAPILEVICH

ON AN INHOMOGENEOUS SINGULAR CAUCHY PROBLEM

(Presented by Academician I. N. Vekua on 20 IX 1966)

Consider, in the \((n+1)\)-dimensional domain \(\Omega[-\infty < x < \infty,\ 0 \leq s < \infty,\ x=(x_1,\ldots,x_n)]\), the solution \(u(x,s;a,b,c)\) of the inhomogeneous singular Cauchy problem

\[ Xu=u_{ss}+\frac{a}{s}u_s+\left(b^2+\frac{c}{s^2}\right)u-\frac{c}{s^2}\tau(x),\quad u(x,0)=\tau(x),\quad u_s(x,0)=0, \tag{1} \]

where \(a=2\beta\geq 0;\ b,\ c\leq \nu^2;\ \nu=\beta-\frac12\) are constants; \(\tau(x)\in C^\infty\) is given on the entire hyperplane \(\mathfrak{S}_n(-\infty<x_k<\infty;\ k=1,2,\ldots,n)\); \(X\) is a linear differential operator independent of \(s\), acting with respect to \(x\), and \(X[0]=0\). Comparing \(u(x,s)\) with \(z(x,s;a,b)=u(x,s;a,b,0)\), we arrive at the following results:

1. Let \(2^{2m}m!(\nu+1)_m g_m(\nu)=(-1)^m,\quad b_0=\sqrt{b_1^2-b_2^2},\quad 4\sigma=b_0^2s^2.\) Then

\[ u(x,\lambda s;a_2,b_2,c)=\sum_{m=0}^{\infty} A_m s^{2m}(X-b_1^2)^m z(x,s;a_1+2m,b_1), \tag{2a} \]

\[ A_m=g_m(\nu_1)\sum_{n=0}^{m} \frac{(-m)_n(\nu_1+1)_n\lambda^{2n}} {(p_2+1)_n(q_2+1)_n}\, {}_1F_2(n+1;\ p_2+n+1,\ q_2+n+1;\ \lambda^2\sigma). \tag{2b} \]

Here \(\lambda=\mathrm{const};\ p_k\) and \(q_k\) are the roots of the equation \(\rho^2-\nu_k\rho+c/4=0\).

\[ u(x,\lambda s;a_2,b_2,c)=\sum_{m=0}^{\infty} \bar A_m s^{2m}(X-b_1^2)^m z(x,s;a_1+4m,b_1), \tag{3a} \]

\[ \bar A_m=\bar g_m(\nu_1)\sum_{n=0}^{m} \frac{(-m)_n(\nu_1+m)_n\lambda^{2n}} {(p_2+1)_n(q_2+1)_n}\, {}_1F_2(n+1;\ p_2+n+1,\ q_2+n+1;\ \lambda^2\sigma), \tag{3b} \]

where \(2^{2m}m!(\nu+m)_m\bar g_m(\nu)=(-1)^m\). With the aid of the equality

\[ s^{2m}(X-b^2)^m z(x,s;a+4m,b)=\sum_{n=0}^{m}\gamma_n z(x,s;a+2n,b), \tag{4} \]

\[ n!(\nu+1)_n\gamma_n=2^{2m}(-m)_n(\nu+1)_{2m}(\nu+m)_n \quad (m,n=0,1,2,\ldots), \]

one may pass from (3a) to an expansion in the family \(\{z(a+2n)\}\).

For example, when \(a_2=a_1=a,\ b_2=b_1=b\), substituting (4) into (3a), we find:

\[ u(x,s;a,b,c)=\sum_{n=0}^{\infty} \frac{pq\Gamma(p+n)\Gamma(q+n)} {n!\,\Gamma(\nu+n+1)} z(x,s;a+2n,b). \tag{5} \]

Under the conditions \(a_2>a_1\geq 0,\quad \beta_0=\beta_2-\beta_1,\quad \Gamma(\beta_0)\Gamma(\nu_1+1)\mu=\)

\[ =2\Gamma(p_2+1)\Gamma(q_2+1), \]
(2), (3), and (5) generates the connection formula
\[ u(x,s;a_2,b_2,c)=\int_0^1 \xi^{a_1}(1-\xi^2)^{\beta_0-1}Q(\xi,s)z(x,\xi s;a_1,b_1)\,d\xi, \tag{6a} \]
\[ Q(\xi,s)=\mu \Xi_2[p_2,q_2,\beta_0;1-\xi^2,\sigma(1-\xi^2)]. \tag{6b} \]
When \(b_2=b_1=b\), (2b) and (3b) give
\[ \begin{aligned} A_m&=g_m(\nu_1)\,{}_3F_2(-m,\nu_1+1,1;\,p_2+1,q_2+1;\lambda^2),\\ \overline{A}_m&=\overline{g}_m(\nu_1)\,{}_3F_2(-m,\nu_1+m,1;\,p_2+1,q_2+1;\lambda^2), \end{aligned} \tag{7} \]
and (6b) becomes \(Q_1=\mu F(p_2,q_2;\beta_0;1-\xi^2)\). Conversely, in the case \(c=0\) \((p=0,\ q=\nu)\), where \(\Gamma(\beta_0)\Gamma(\nu_1+1)\mu_2=2\Gamma(\nu_2+1)\),
\[ A_m=g_m(\nu_1)\Xi_2(-m,\nu_1+1;\nu_2+1;\lambda^2,-\lambda^2\sigma), \tag{8a} \]
\[ \overline{A}_m=\overline{g}_m(\nu_1)\Xi_2(-m,\nu_1+m;\nu_2+1;\lambda^2,-\lambda^2\sigma), \tag{8b} \]
the series (6b) reduces to \(Q_2=\mu_2 \overline{I}_{\beta_0-1}(b_0s\sqrt{1-\xi^2})\) (1). For \(a_2=a_1=a,\ c=0\), from (3b) and (8b) there arises the addition theorem
\[ z(x,s;a,b_2)=\sum_{m=0}^{\infty}\overline{A}_m s^{2m}(X-b_1^2)^m z(x,s;a+4m,b_1), \tag{9a} \]
\[ (\nu+1)_{2m}\overline{A}_m=\overline{g}_m(\nu)\sigma^m \overline{I}_{\nu+2m}(b_0,s), \tag{9b} \]
inverting which, we arrive at the equality
\[ \overline{I}_{\nu}(b_0s)z(x,s;a,b_1)=\sum_{n=0}^{\infty}B_n(b_0s^2)^{2n}(X-b_1^2)^n z(x,s;a+4n,b_2), \tag{10} \]
where \((\nu+1)_{2n}2^{2n}B_n=(-1)^n g_n(\nu)\). It is also worth noting the case \(p_2=\beta_0,\ q_2=\nu_1\ (a_1=2q_2+1)\), when \(A_m=g_m(q)F(-m,1,p+1;\lambda^2)\),
\[ u(x,\lambda s;a,b,c)=\sum_{m=0}^{\infty}A_m s^{2m}(X-b^2)^m z[x,s;2(q+m)+1,b], \tag{11a} \]
\[ u(x,s;a,b,c)=p\int_0^1(1-\eta)^{p-1}z(x,s\sqrt{\eta};2q+1,b)\,d\eta\quad (p>0). \tag{11b} \]

1. Let us replace in (1) \(s,a,c\) by \(2\sqrt{\varepsilon s},\,2\varepsilon-1,\,4c\varepsilon\) and pass to the limit as \(\varepsilon\to\infty\). Then (1) is reduced to the Cauchy problem
   \[ Xv=v_s+\left(b^2+\frac{c}{s}\right)v-\frac{c}{s}\tau(x),\qquad v(x,0)=\tau(x), \tag{12a} \]
   \[ v(x,s;b,c)=\lim_{\varepsilon\to\infty}u(x,2\sqrt{\varepsilon s};2\varepsilon-1,b,4c\varepsilon). \tag{12b} \]
   Conversely, conflating \(u\) and \(z\) according to the rule
   \[ \lim_{\varepsilon\to\infty}u(x,2\sqrt{\varepsilon s};2\varepsilon-1,b,c) = \lim_{\varepsilon\to\infty}z(x,2\sqrt{\varepsilon s};2\varepsilon-1,b) =w(x,s;b), \tag{13} \]
   we arrive at the problem, regular with respect to \(s\), with initial condition
   \[ Xw=w_s+b^2w,\qquad w(x,0)=\tau(x),\qquad (x,s)\in\Omega. \tag{14} \]
   For \(a_1=2\varepsilon-1,\ a_2=a,\ s=2\sqrt{\varepsilon s_1},\ \lambda=\sqrt{\lambda_1/\varepsilon},\ \varepsilon\to\infty\), (7) and (8a) give
   \[ u(x,2\sqrt{\lambda_1s_1};a,b,c)=\sum_{m=0}^{\infty}A_m^{(1)}s_1^m(X-b^2)^m w(x,s_1;b), \tag{15a} \]
   \[ z(x,2\sqrt{\lambda_1s_1};a,b_2)=\sum_{m=0}^{\infty}\overline{A}_m^{(1)}s_1^m(X-b^2)^m w(x,s_1;b_1), \tag{15b} \]
   where
   \[ m!A_m^{(1)}=(-1)^m{}_2F_2(-m,1;\,p+1,q+1;\lambda_1), \]
   \(\overline{A}_m^{(1)}\) is a Humbert function
   \[ \overline{A}_m^{(1)}=\frac{(-1)^m}{m!}\Phi_3(-m,\nu+1;\lambda_1,b_0^2\lambda_1s_1). \tag{15c} \]

Putting in (2) and (3) \(a_1=a,\ a_2=2\varepsilon-1,\ b_1=b_2=b,\ c=4\varepsilon c_1,\ \lambda=2\lambda_1\sqrt{\varepsilon},\ \varepsilon\to\infty\), we obtain

\[ v(x,\lambda_1^2s^2;b,c_1) =\sum_{m=0}^{\infty} A_m^{(2)}s^{2m}(X-b^2)^m z(x,s;\,a+2m,b), \tag{16a} \]

\[ v(x,\lambda_1^2s^2;b,c_1) =\sum_{m=0}^{\infty} \overline{A}_m^{(2)}s^{2m}(X-b^2)^m z(x,s;\,a+4m,b), \tag{16b} \]

\[ A_m^{(2)}=\overline{g}_{m3}F_1(-m,\nu+1,1;\,c_1+1;\,4\lambda_1^2), \]

\[ \overline{A}_m^{(2)}=\overline{g}_{m3}F_1(-m,\nu+m,1;\,c_1+1;\,4\lambda_1^2). \]

Let us also note a relation, similar to (15b), with the Humbert function \(\Phi_1\):

\[ v(x,\lambda s;\,b_2,c)=\sum_{m=0}^{\infty} A_m s^m (X-b_1^2)^m w(x,s;\,b_1), \tag{17a} \]

\[ m!A_m=(-1)^m\Phi_1(1,-m;\,c+1;\,\lambda,\ b_0^2\lambda s). \tag{17a} \]

3. When \(b_2=b_1=b\), (8b) reduces to Jacobi polynomials
\(\overline{A}_m=\overline{g}_m(\nu_1)G_m(\nu_1,\nu_2+1;\lambda^2)\), and therefore, if \(z(x,\lambda s;\,a_2,b)=f(\lambda)\), then

\[ s^{2m}(X-b^2)^m z(x,s;\,a_1+4m,b) =\delta_m\int_{0}^{1}\lambda^{\alpha_2}(1-\lambda^2)^{-\beta_0-1}\overline{A}_m(\lambda)f(\lambda)\,d\lambda, \]

\[ \Gamma(m-\beta_0)\Gamma(\nu_2+1)\overline{g}_m(\nu_1)\delta_m =(-1)^m2^{2m+1}(\nu_2+1)_m\Gamma(\nu_1+2m+1). \tag{18} \]

For \(b_2=b_1=b\), it follows from (15c) that
\((\nu+1)_m\overline{A}_m^{(1)}=(-1)^mL_m^{(\nu)}(\lambda_1)\), and the Fourier coefficients of the function \(z(x,2\sqrt{\lambda_1}s;\,a,b)=f(\lambda_1)\) are

\[ s_1^m(X-b^2)^m w(x,s_1;\,b) =\frac{(-1)^m m!}{\Gamma(\nu+1)} \int_{0}^{\infty}\lambda_1^\nu e^{-\lambda_1}L_m^{(\nu)}(\lambda_1)f(\lambda_1)\,d\lambda_1. \tag{19} \]

Replacing the bases of the expansions (15), (16b), (17a) by their integral representations (18) and (19), we arrive at analogous relations for \(u\) and \(v\). For example, (17a) and (19) give

\[ v(x,s;\,b_2,c)=\int_{0}^{\infty}\lambda^{\nu/2}R(\lambda,s)\, z(x,2\sqrt{\lambda}s;\,a,b_1)\,d\lambda, \tag{20a} \]

\[ \Gamma(\nu+1)R=\int_{0}^{\infty}\xi^{\nu/2}e^{-\xi} J_\nu(2\sqrt{\lambda\xi})\Phi_2(1,c,c+1;\,b_0^2s,\xi)\,d\xi. \tag{20b} \]

Let us also note the relation, similar to (19) and (20a),

\[ v(x,s;\,b,p)=\frac{1}{\Gamma(q+1)} \int_{0}^{\infty}\lambda^q e^{-\lambda} u(x,2\sqrt{\lambda}s;\,a,b,c)\,d\lambda \quad(q>-1), \tag{21} \]

and, in the case \(c_2>c_1>-1\), the addition theorem:

\[ v(x,s;\,b,c_2) =\frac{\Gamma(c_2+1)}{\Gamma(c_1+1)\Gamma(c_2-c_1)} \int_{0}^{1}\xi^{c_1}(1-\xi)^{c_2-c_1-1} v(x,\xi s;\,b,c_1)\,d\xi. \tag{22} \]

If \(X=\partial/\partial x_1+\cdots+\partial/\partial x_n\), then
\(w=e^{-b^2s}\tau(x_1+s,\ldots,x_n+s)\), and here (15), (17), and (22) (for \(c_1=0\)) give the solutions \(u,z,v\), while (16) (for \(c_1=0\)), (19), and (20) (for \(c=0\)) invert the corresponding resolving operators with respect to \(\tau(x_1+s,\ldots,x_n+s)\). Suppose, further, that \(a=n-1,\ b=0,\ X\) is the Laplacian

\[ X[z]=\Delta_x[z]=\sum_{i=1}^{n} z_{x_i x_i}. \]

Then \(z(x,s;\)

\(n-1,0)=M[x,s;\tau(x)]\). This means that (6a) in the case \(a_1=n-1\), \(b_1=0\), \(a_2>n-1\), \(X=\Delta_x\), and also (20a) for \(a=n-1\), \(b_1=0\), give solutions \(u\) and \(v\) of problems (1), (12a). For example, here (6), when \(n=1\), take the form

\[ u(x,s;a,b,c)=\int_{-1}^{1}(1-\xi^2)^{\beta-1}Q_0(\xi,s)\tau(x+\xi s)\,d\xi, \tag{23a} \]

\[ Q_0(\xi,s)=\mu_0\overline{\Xi}_2[p,q,\beta;1-\xi^2,-\sigma_0(1-\xi^2)], \tag{23b} \]

where \(\sqrt{\pi}\Gamma(\beta)\mu_0=\Gamma(p+1)\Gamma(q+1)\), \(4\sigma_0=b^2s^2\), and from (20a) it follows that

\[ v(x,s;0,c)=\frac{\Gamma(c+1)}{\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-\xi}\Psi(c,1/2;\xi^2)\tau(x+2\xi\sqrt{s})\,d\xi . \tag{24} \]

Analogous assumptions reduce (2a), (3a), (8), and (15b) to expansions of the means \(M(x,s;\tau)\) in the basic series over the systems \(\{z(a+2m)\}\), \(\{z(a+4m)\}\), \(\{(X-b^2)^m w\}\).

Finally, let \(b=0\), \(Xz=\mathfrak{B}_x[z]=z_{xx}+\dfrac{2a}{x}z_x\). Then

\[ z=\gamma\int_{-1}^{1}\tau(x+\xi s)(1-\xi^2)^{\beta-1}(1+\xi t)^\alpha F(\alpha,1-\alpha,\beta;\omega)\,d\xi, \tag{25} \]

\[ \sqrt{\pi}\Gamma(\beta)\gamma=\Gamma(\nu+1),\qquad t=-s/x,\qquad 4(1+\xi t)\omega=t^2(\xi^2-1). \]

Substituting (25) into (6a) (\(b_1=0\)), (11b), (19), and (20a) (\(b=0\)), we obtain \(u,v,w\) for \(X=\mathfrak{B}_x\).

1. To power data \(\tau(x)=x^\alpha\) (\(\alpha=\mathrm{const}\)) for \(X=\partial/\partial x\) there correspond the solutions

\[ u_1=x^\alpha\,{}_2F_2(-\alpha,1;p+1,q+1;-t_1),\qquad w_1=e^{-\sigma_1}(x+s)^\alpha, \]

\[ z_1=x^\alpha\Phi_3(-\alpha,\nu+1;-t_1,-\sigma_0),\qquad v_1=x^\alpha\Phi_1(-\alpha,1,c+1;-t,-\sigma_1), \]

where \(t_1=s^2/4x\), \(\sigma_1=b^2s\). If, however, \(X=\partial^2/\partial x^2\), then

\[ z_2=x^\alpha\Xi_2\left(-\frac{\alpha}{2},\frac{1-\alpha}{2},\nu+1;t^2,-\sigma_0\right), \]

\[ u_2=x^\alpha\,{}_3F_2\left(-\frac{\alpha}{2},\frac{1-\alpha}{2},1,p+1,q+1;t^2\right). \]

Finally, confluencing \(z_2\) and \(u_2\) according to the rules (12b), (13), we obtain

\[ w_2=x^\alpha e^{-\sigma_1}\times {}_2F_0\left(-\frac{\alpha}{2},\frac{1-\alpha}{2};t_2\right), \]

\[ v_2=x^\alpha\,{}_3F_1\left(-\frac{\alpha}{2},\frac{1-\alpha}{2},1,c+1;t_2\right), \]

where \(t_2=4s/x^2\).

In the case \(\mathfrak{B}_x z=z_{ss}+\dfrac{a}{s}z_s\), the solutions \(z\) and \(\overline{z}\) of Tricomi’s problems from \((2)\) are written in quadratures with the aid of the Green–Hadamard functions \(H(x,s;x_0,s_0)\), \(\overline{H}(x,s;x_0,s_0)\), which here have the form:

\[ H=\chi t^{1-a}\left(\frac{2s_0}{R}\right)^a \left(\frac{x}{x_0}\right)^\alpha H_2(1-\beta,1-\beta,\alpha,1-\alpha,2-a;t,\rho), \]

\[ \overline{H}=\overline{\chi}\left(\frac{2s_0}{R}\right)^a \left(\frac{x}{x_0}\right)^\alpha H_2(\beta,\beta,\alpha,1-\alpha,a;t,\rho), \]

where \(tR^2=4ss_0\), \(4xx_0\rho=R^2\), \(R=\sqrt{(x-x_0)^2-(s-s_0)^2}\); \(\chi\) and \(\overline{\chi}\) are indicated in \((2)\).

Moscow Evening
Metallurgical Institute

Received
6 IX 1966

REFERENCES

1. M. N. Olevskii, Dokl. Akad. Nauk SSSR, 101, No. 1, 21 (1955).
2. M. B. Kapilevich, Dokl. Akad. Nauk SSSR, 170, No. 6 (1966).