# Coupled second order differential equations

I would like to solve this coupled second order differential equations

$$x^2E''(x)+xE'(x)-(p_2x^2+p_0\nu^2)E(x)+s\nu \,xG'(x)=0,$$

$$x^2G''(x)+xG'(x)-(q_2x^2-q_0\nu^2)G(x)+s\nu \,xE'(x)=0,$$

where $x$, $p_0$, $p_2$, $q_0$, $q_2$, $s$ are positive real numbers. $\nu$ is an integer. $x$ is from $x_0$ to $\infty$.

I am expecting to have two different type of solutions. The 1st type should be exponentially increasing, and the 2nd type should be exponentially decaying. for our application, we are only interested in the exponentially decaying solution, which has a physically meaning.

The equation can be discussed in two cases: $\nu=0$ and $\nu\neq 0$.

When $\nu=0$, the equation system becomes:

$$x^2E''(x)+xE'(x)-p_2x^2E(x)=0,$$

$$x^2G''(x)+xG'(x)-q_2x^2G(x)=0.$$

Therefore, $E(x)$ and $G(x)$ are uncoupled, which are both modified Bessel equations. The general solution of them are:

$$E(x)=C_1I_0(\sqrt{p_2}x)+C_2K_0(\sqrt{p_2}x)$$ $$G(x)=D_1I_0(\sqrt{q_2}x)+D_2K_0(\sqrt{q_2}x)$$

Since the $I_0(x)$ function will have the infinite value when $x\to \infty$, we have to remove it. Hence the solution will be $$E(x)=C_2K_0(\sqrt{p_2}x)$$ $$G(x)=D_2K_0(\sqrt{q_2}x)$$

However, when $\nu\neq 0$, things become complicated. We are expecting that the solution would be the combination of Bessel functions and the total behavior is similar to $K_{\nu}(x)$. But unfortunately, we don't know how to solve this problem.

Does anyone have an idea how to solve it? We will really thank you so much for the help!

P.S.

Some people mentioned that we can try the numerical method. However, the boundary condition $E(x_0)$, $E(x_0)'$, $G(x_0)$, $G(x_0)'$ contains another two unknowns variable $A$ and $B$. Hence I guess it is not possible to feed the boundary condition to the software?

• $s$ is also a real positive number. Could you tell me what math packages I can use for solving this problem? Also, I found out that some software requires boundary conditions, which I do not have at the moment. Also what is ICs? Sorry I am not a mathematician. – Ying Tang Mar 9 '17 at 18:12