Abstract Generated abstract
This note solves the third order linear ordinary differential equation with variable coefficients x squared y triple prime plus 3 x y double prime plus y prime plus x squared y equals 0 by an operational method. After a change of variables, the equation is related to an operator of the form d/dt, t d/dt, t d/dt, and its Laplace transform is reduced to a third order equation in the transform variable. By embedding this transformed equation in a parameter dependent family involving Bessel and Poisson functions, the paper handles the degenerate case at parameter zero and derives a third independent solution through limiting differentiation. The result is an explicit fundamental system of three linearly independent solutions, expressed as convergent series with logarithmic terms and gamma function derivatives.
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MATHEMATICS
A. P. PRUDNIKOV
ON FUNCTIONS SATISFYING THE DIFFERENTIAL EQUATION \(x^2y''' + 3xy'' + y' + x^2y = 0\)
(Presented by Academician A. A. Dorodnitsyn on 19 XII 1961)
In the present note, by an operational method we solve the linear ordinary differential equation with variable coefficients
\[ x^2y''' + 3xy'' + y' + x^2y = 0. \tag{1} \]
After the change of variable by the formula \(x = 3\sqrt[3]{t}\), equation (1) is reduced to the form
\[ \frac{d}{dt}\,t\,\frac{d}{dt}\,t\,\frac{d}{dt}\,y + y = 0. \tag{2} \]
This equation is of importance in constructing an operational calculus for the operator
\[
T=\frac{d}{dt}\,t\,\frac{d}{dt}\,t\,\frac{d}{dt}
\]
\((^1)\).
To obtain three linearly independent solutions of the latter equation by the operational method, it is expedient to consider the equation
\[ t^3y''' + 3t^2y'' + ty' + ty = 0, \tag{3} \]
which is obtained from (2) by multiplying by \(t\).
Let
\[ \varphi(p)=\int_0^\infty e^{-pt}y(t)\,dt. \]
Then equation (3) is reduced to the operational equation
\[ p^3\varphi''' + 6p^2\varphi'' + (7p+1)\varphi' + \varphi = 0. \tag{4} \]
After the change of variable by the formula \(z=\frac{1}{p}\), equation (4) becomes
\[ z^3\varphi''' + z(z+1)\varphi' - \varphi = 0. \tag{5} \]
Equation (5) is a special case of an equation depending on a parameter \(\nu\):
\[ z^3\varphi''' + \left[z(z+1)-\left(\frac{\nu}{2}\right)^2 z\right]\varphi' -\left[1-\left(\frac{\nu}{2}\right)\right]\varphi = 0, \tag{6} \]
when this parameter is equal to zero. If the parameter \(\nu\) is not an integer, then the functions
\[ \widetilde{\varphi}_1(z,\nu)=zJ_\nu(2\sqrt{z}); \tag{7} \]
\[ \widetilde{\varphi}_2(z,\nu)=zY_\nu(2\sqrt{z}); \tag{8} \]
\[ \widetilde{\varphi}_3(z,\nu)=z\Pi_\nu(2\sqrt{z}), \tag{9} \]
where
\[ J_\nu(z)=\sum_{m=0}^{\infty}\frac{(-1)^m(z/2)^{\nu+2m}}{m!\Gamma(\nu+m+1)},\qquad Y_\nu(z)=\frac{\cos(\pi\nu)J_\nu(z)-J_{-\nu}(z)}{\sin(\pi\nu)} \]
are Bessel functions,
\[ \Pi_\nu(z)=\cos\left(\frac{\pi\nu}{2}\right) \sum_{m=0}^{\infty} \frac{(-1)^m(z/2)^{2m}} {\Gamma(m+1+\nu/2)\Gamma(m+1-\nu/2)} \]
is a Poisson function, form a fundamental system of solutions of equation (6). For \(\nu=0\) this no longer holds, since as \(\nu\to0\) the Poisson function \(\Pi_\nu(z)\) degenerates into the Bessel function \(J_\nu(z)\), and the triple of solutions (7), (8), (9) of equation (6) degenerates into a pair of linearly independent solutions of equation (5)
\[ \varphi_1(z)=\lim_{\nu\to0}\widetilde{\varphi}_1(z,\nu) =\lim_{\nu\to0}\widetilde{\varphi}_3(z,\nu) =zJ_0(2\sqrt z); \tag{10} \]
\[ \varphi_2(z)=\lim_{\nu\to0}\widetilde{\varphi}_2(z,\nu) =zY_0(2\sqrt z). \tag{11} \]
To find the third linearly independent solution of equation (5), let us consider
\[ \lim_{\nu\to0} \frac{\widetilde{\varphi}_1(z,\nu)-\widetilde{\varphi}_3(z,\nu)}{\nu} = \frac12\varphi_2(z). \]
We find
\[ \lim_{\nu\to0} \frac{ \dfrac{d}{d\nu}\{\widetilde{\varphi}_1(z,\nu)-\widetilde{\varphi}_3(z,\nu)\} -\dfrac12\varphi_2(z)} {\nu} = \frac{\pi^2}{4}\varphi_1(z)+\varphi_3(z), \]
where
\[ \varphi_3(z)=\frac14\sum_{m=0}^{\infty} \frac{(-1)^m z^{m+1}}{(m!)^2} \{\ln^2 z-4\psi(m+1)\ln z+4\psi^2(m+1)-2\psi'(m+1)\}, \]
\[ \psi(m+1)=\frac{d}{dm}\ln\Gamma(m+1). \]
The function \(\varphi_3(z)\) satisfies equation (5) and is linearly independent of \(\varphi_1(z)\) and \(\varphi_2(z)\). After passing from the images
\(\varphi_1\left(\frac1p\right)\), \(\varphi_2\left(\frac1p\right)\), \(\varphi_3\left(\frac1p\right)\) to the originals, we obtain, respectively, three linearly independent solutions of equation (3):
\[ y_1(t)=\sum_{m=0}^{\infty}\frac{(-1)^m}{(m!)^3}t^m =J_{0,0}^{(2)}(3\sqrt[3]{t}), \]
\[ y_2(t)=J_{0,0}^{(2)}(3\sqrt[3]{t})\ln t -3\sum_{m=0}^{\infty}\frac{(-1)^m}{(m!)^3}t^m\psi(m+1), \]
\[ y_3(t)=\frac14\sum_{m=0}^{\infty}\frac{(-1)^m}{(m!)^3}t^m \{\ln^2 t-6\psi(m+1)\ln t+9\psi^2(m+1)-3\psi'(m+1)\}. \]
Computing Center
Academy of Sciences of the USSR
Received
18 XII 1961
CITED LITERATURE
- A. P. Prudnikov, DAN, 142, No. 4 (1962).