This question is an exact duplicate of:

I have obtained the following coupled two ODEs as the subsidiary equations of two coupled PDEs in Laplace domain. I was trying to solve them by decoupling them and getting a higher order differential equation form for each ODE but the solution later became so complicated. I wonder if there is a method to solve them.

First ODE. $$ y_1''(x) + \frac{C_1}{x}y_1'(x)= C_2 y_1(x) + C_3 y_2(x)$$

Second ODE. $$ y_2''(x) + \frac{C_4}{x}y_2'(x)= C_5 y_2(x) + C_6 y_1(x)$$


marked as duplicate by Namaste, Rebellos, ArsenBerk, José Carlos Santos, GNUSupporter 8964民主女神 地下教會 Nov 7 '18 at 10:49

This question was marked as an exact duplicate of an existing question.


For: First ODE: $$ y_1''(x) + \frac{C_1}{x}y_1'(x)= C_2 y_1(x) + C_3 y_2(x)$$ Second ODE: $$ y_2''(x) + \frac{C_4}{x}y_2'(x)= C_5 y_2(x) + C_6 y_1(x)$$

proceed as follows: multiply the first by $C_{6}$ and the second by $C_{2}$ and subtract to obtain $$C_{6} \, \left(y_{1}^{''} + \frac{C_{1}}{x} \, y_{1}^{'} \right) = \lambda = C_{2} \left( y_{2}^{''} + \frac{C_{4}}{x} \, y_{2}^{'} \right) + (C_{6} C_{3} - C_{2} C_{5}) y_{2},$$ where $\lambda$ is a separation constant. Now, \begin{align} y_{1}^{''} + \frac{C_{1}}{x} \, y_{1}^{'} &= \frac{\lambda}{C_{6}} \\ y_{2}^{''} + \frac{C_{1}}{x} \, y_{2}^{'} + \frac{C_{6}C_{3} - C_{2}C_{5}}{C_{2}} \, y_{2} &= \frac{\lambda}{C_{2}}. \end{align}

The $y_{1}$ equation can be solved as follows: \begin{align} y_{1}^{''} + \frac{C_{1}}{x} \, y_{1}^{'} &= \frac{\lambda}{C_{6}} \\ \frac{1}{x^{C_{1}}} \, \frac{d}{dx} \left( x^{C_{1}} \, y_{1}^{'} \right) &= \frac{\lambda}{C_{6}} \\ x^{C_{1}} \, \frac{d y_{1}}{dx} &= \frac{\lambda \, x^{C_{1} + 1}}{C_{6} \, (C_{1} + 1)} + d_{0} \\ \frac{d y_{1}}{dx} &= \frac{\lambda \, x}{C_{6} \, (C_{1} + 1)} + \frac{d_{0}}{x^{C_{1}}}\\ y_{1} &= \frac{\lambda \, x^{2}}{2 \, C_{6} \, (C_{1} + 1)} - \frac{d_{0}}{(C_{1} -1) \, x^{C_{1}-1}} + d_{1}. \end{align}

The second equation is solvable in terms of Bessel functions as seen by $$f'' + \frac{a}{x} \, f' + b f = c$$ has the solution $$f(x) = x^{(1-a)/2} \, \left(A_{0} \, J_{\beta}(\sqrt{b} \, x) + B_{0} \, Y_{\beta}(\sqrt{b} \, x) \right) + \frac{c}{b},$$
where $\beta = (1-a)/2$.

Given the conditions for the equations and the solution then the forms of solutions for $y_{1}$ and $y_{2}$ could be reduced.

  • $\begingroup$ I have tried to follow your suggested method @leucippus but I have a problem with the selection of the separation constant. I have two boundary conditions per ODE which have been used to get the arbitrary constants. Do you think the separation constant can be a function of x ? $\endgroup$ – Galal Sep 21 '18 at 13:45
  • $\begingroup$ @Galal In the general case there are 6 constants for the two solutions with only 4 boundary conditions. Suppose, as is often the case that, the solutions be finite near the origin. This would eliminate $d_{0$ and $B_{0}$ and leave 4 constants with 4 conditions. Separations constants cannot be functions, but that can be set to zero if the solutions fit all the conditions of the equations, both mathematically and physically. $\endgroup$ – Leucippus Sep 21 '18 at 15:54

What's your background? I'm not sure how much detail to include.


Method One Note you can isolate y2(x) in the first ODE. Take the corresponding derivatives and plug them into the second ODE and you get a new ODE completely in terms of y1(x).


Method 2 Use the Method of Frobenius where y1 is expressed in terms of one set of coefficients and y2 is expressed as a series with a different set of coefficients, then proceed as normal. http://mathworld.wolfram.com/FrobeniusMethod.html


Method 3, I'm a bit fuzzy on this one, but in p Solve the equation you get if you assume C3=0 and substitute y0 in place of y1: y0''+C1/x * y0' = C2y0. Again you should be able to get this into Bessel's Equation. Let y2=v*y1, swapping into both equations. Simplify. Then swap y1=u*y0. You should get simpler equations.

  • $\begingroup$ @ TurlocTheRed Can you clarify the details of your third method, as I can't understand it? Please. $\endgroup$ – Galal Oct 19 '18 at 19:42
  • $\begingroup$ First, Solve $$x^2\frac{d^2y_p}{dx^2}+C_1x\frac{dy_p}{dx}-C_2x^2y_p=0$$ This can probably be put in the form of Bessel's Equation. Then Let $y_1=v(x)y_p$ and $y_2=u(x)y_p$back into the first equation. $y_p$ is known by this point and will probably simplify the equation. $$x^2(\frac{d^2v}{dx^2}y_p+2\frac{dv}{dx}\frac{dy_p}{dx}+)+C_1x(\frac{dv}{dx}y_p)=C_2vx^2y_p+C_3ux^2y_p$$ $$x^2(\frac{d^2u}{dx^2}y_p+2\frac{du}{dx}\frac{dy_p}{dx}+)+C_4x(\frac{du}{dx}y_p)=C_5x^2vy_p+C_6ux^2y_p$$ $\endgroup$ – TurlocTheRed Oct 19 '18 at 20:35
  • $\begingroup$ @ TurlocTheRed So, you mean the Method of Variation of Parameters, I have used it before, but the procedures became so complicated. $\endgroup$ – Galal Oct 20 '18 at 3:34

2 potential ideas:

(1) If you're interested in the behavior away from $x=0$:

Using the usual reduction of order substitutions $y_1'=y_3$ and $y_2'=y_4$, we can rewrite as a first order linear system $Y'(x) = A(x)Y(x)$, where

$$ Y(x) = \begin{pmatrix}y_1(x)\\y_2(x)\\y_3(x)\\y_4(x)\end{pmatrix}, \ Y'(x) = \begin{pmatrix}y_1'(x)\\y_2'(x)\\y_3'(x)\\y_4'(x)\end{pmatrix}, \text{ and } A(x) = \begin{pmatrix}0&0&1&0 \\ 0&0&0&1 \\ C_2&C_3&-C_1x^{-1}&0 \\ C_6&C_5&0&-C_4x^{-1} \end{pmatrix} $$

Let $B(x) = \int_{x_0}^x A(s)\text ds$ for the initial point $x_0$. Then, if $AB = BA$ for all $x$, $Y(x) = \exp(B(X))$ is the solution.

(2) If you're interested in the behavior near $x=0$:

Let $M(x) = xA(x)$, and rewrite the equation as $xY'(x) = M(x)Y(x)$. Now, $M(x)$ is holomorphic at $x=0$ and we can proceed with the method laid out in Chapter II of Wasow's "Asymptotic expansions for ordinary differential equations" $-$ essentially a matrix version of Frobenius/power series method.


Not the answer you're looking for? Browse other questions tagged or ask your own question.