Second order differential equation given the form of the solution

Question

Consider the following second order differential equation: $\frac{d^2 u}{dr} + \frac{N-1}{r}\frac{du}{dr} + \frac{\lambda}{r^2}u = 0$ in $(0,1)$, where $\lambda = 1+\frac{1}{4}(N-2)^2$ and $N = 1,2,3$. Find all the solutions having the form $u(r) = r^{-\frac{N-2}{2}}v(log(r))$. Hint: make the change of variable $t = log(r)$ for $r \in (0,1)$ and find the ODE that $v(t) = v(log(r))$ solves.

I don't understand if I have to actually find the solutions by hand as if I didn't know their form or if I have to use the information of the form of the solution and substitute inside the ODE the generic solution $u(r)$ to verify the identity (I suppose the latter). And what does it mean to find the ODE that is solved by $v(t) = v(log(r))$, as if for that function there was an unique ODE having as a solution such function?

My first idea was to compute the first and second derivative of $u(r) = ...$ without doing the change of variable, and then substitute $u(r), \frac{du}{dr}(r), \frac{d^2(u)}{dr}(r)$ inside the ODE so I should get an identity but... does it make sense? Once I substituted them inside the ODE what should I do? Unfortunately I'm new to differential equations, One thing I read is that an ODE like the one above can be written as a system of differential equations of first order, but I can't figure out how to transform this one in particular, if only the term $\frac{\lambda}{r^2}u$ wasn't present I think it would have been easy. What would you suggest?

The exercise continues asking to find the exact solution with final values $u(1)=1$ and $\frac{du}{dr}(1) = 1$ but I think that once the general solutions are known, it is only a matter of substituting. For now if I try to substitute $r=1$ inside $u(r) = r^{-\frac{N-2}{2}}v(log(r))$ I can only conclude that $log(r=1)=0$ and that's it, because I don't know the function $v(t)=v(log(r))$.

Thanks for reading and for any type of help or suggestion.

Answer 1

The change of variable is a little tricky because you need to use the chain rule for second derivatives. Let $U(t) = u(r(t))$, with $r(t) = e^t$. Applying the chain rule twice, $$ U'(t) = u'(r(t))r'(t) = u'(r(t))e^t $$ and $$ U''(t) = u''(r(t)) (r'(t))^2 + u'(r(t)) r''(t) = u''(r(t))e^{2t} + u'(r(t))e^t. $$ The first equation is equivalent to $$ \frac{du}{dr} = e^{-t} \frac{du}{dt} $$ and the second equation is equivalent to $$ \frac{d^2 u}{dr^2} = e^{-2t}\left( \frac{d^2 u}{dt^2} - \frac{du}{dt}\right). $$ So the original ODE with independent variable $r$ is equivalent to the following linear second-order ODE with independent variable $t$: $$ \frac{d^2 u}{dt^2} + (N-2)\frac{du}{dt} + \lambda u. $$