[Math] Coupled second-order differential equations

ordinary differential equationssystems of equations

I am trying to solve the following system of coupled ODEs:
\begin{align}
-x^2 f'' – 3xf' + (1-2a)f – (a+1)x^2g'' + (2-4a)xg' + (4a-2)g &= 0,\\
(a-1)x^2 f'' + (4a+2)xf' + (12-6a)f + 12xg' + (12a-24)g &= 0,
\end{align}
where $f$ and $g$ are function of $x$ and $a$ is a constant.

What method do you suggest for solving this system? Any suggestion will be appreciated!

Thanks!

Best Answer

The idea:

  1. Following the @Michael Galuza suggestion you will get a system of two linear equations with constant coefficients wrt functions $F(t)=f(e^t)=f(x)$ and $G(t)=g(e^t)=g(x)$ $$ \left[\matrix{-1 & -(a+1)\\a-1 & 0}\right]\left[\matrix{F''\\G''}\right]+ \left[\matrix{-2 & 3(1-a)\\3(1+a) & 12}\right]\left[\matrix{F'\\G'}\right]+ \left[\matrix{1-2a & 4a-2\\12-6a & 12a-24}\right]\left[\matrix{F\\G}\right]= \left[\matrix{0\\0}\right] $$ that I write shortly as $$ AH''+BH'+CH=\left[\matrix{0\\0}\right]. $$
  2. The matrix $A=\left[\matrix{-1 & -a-1\\a-1 & 0}\right]$ is non-singular for $a\ne\pm 1$. So you can left-multiply with $A^{-1}$ to get $$ H''=DH'+EH, $$ which with the notation $Z=\left[\matrix{H\\H'}\right]$ is equivalent to the linear system $$ Z'=\left[\matrix{0 & I\\ E & D}\right]Z. $$
  3. If $A$ is singular then the equations are dependent, and the system can be reduced to a first order DE after eliminating the coefficients for the second derivatives. For example, if $a=1$ we get the second equations as $$ \left[\matrix{1 & 2}\right]H'+\left[\matrix{1 & -2}\right]H=0.\tag1 $$ Adding it to the first equation gives $$ \left[\matrix{-1 & -2}\right]H''+\left[\matrix{-1 & 2}\right]H'=0\tag2 $$ which is, in fact, dependent (minus the derivative of $(1))$. Solving the equation $(1)$ gives you the dependence between $F$ and $G$ (the whole subspace of solutions) $$ F'+2G'+F-2G=0\quad\Rightarrow\quad (F+2G)'+(F+2G)=4G\quad\Rightarrow\quad $$ $$ \Rightarrow\quad F(t)+2G(t)=4\int e^{s-t}G(s)\,ds. $$ Similar for $a=-1$.
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