MHB Solving Coupled ODEs: Analytical Solutions?

[Pascal1]

Hello everyone,
I have the following coupled ODEs ($$r\geq 0$$)

$$g^2v^2f(r)h^2(r)+\frac{6}{r}f'(r)-3f''(r)=0$$,
$$r^2h''(r)-4f^2(r)h(r)=0$$,

with boundary conditions

$$f(\infty)=1$$,
$$h(\infty)=1$$.

The other 2 boundary conditions are arbitrary. Also v and g are constants, that could be set to a fixed value, if this helps to find a special case analytical solution.

I was wondering if anyone has an idea on how to tackle them. Or maybe someone has an argument that shows the non-existence of an analytical solution.

 

[Ackbach]

Let's examine the first equation, and take the limit as $r\to\infty$. Note that, since
\begin{align*}
\lim_{r\to\infty}f(r)&=1 \quad \text{and} \\
\lim_{r\to\infty}h(r)&=1,
\end{align*}
it follows (under sufficient smoothness conditions) that
$$\lim_{r\to\infty}f'(r)=\lim_{r\to\infty}f''(r)=\lim_{r\to\infty}h'(r)=\lim_{r\to\infty}h''(r)=0.$$
So, we have that
$$\lim_{r\to\infty}\left[g^2v^2 f(r)h^2(r)+\frac6r f'(r)-3f''(r)\right]=g^2v^2=0.$$
Hence, it must be that $gv=0$. This reduces the first equation to
$$\frac6r f'(r)-3f''(r)=0.$$
This is uncoupled, and a Cauchy-Euler differential equation, and its solution is
$$f(r)=C_1+\frac{C_2 r^{3}}{3}.$$
The only way to satisfy the boundary condition is if $C_2=0$ and $C_1=1$. Hence, $f(r)=1$. Plugging this into the other equation yields
$$r^2 h''(r)-4h(r)=0,$$
with solution (this is also Cauchy-Euler)
$$h(r)=A_1 r^{\frac12 \left(1+\sqrt{17}\right)}+A_2 r^{\frac12 \left(1-\sqrt{17}\right)}.$$
And here we have a problem. The boundary condition forces $A_1=0$, otherwise $h(r)$ would blow up at infinity. That leaves
$$h(r)=A_2 r^{\frac12 \left(1-\sqrt{17}\right)}.$$
But this function goes to zero as $r\to\infty$.

Therefore, I conclude, there is no solution to this boundary value problem.