While working on a differential geometry problem I have come across a second order ODE that I am trying solve:\begin{equation}1+(h')^2-hh''=2ch(1+(h')^2)^{3/2}\end{equation} It turns out that is equation can't really be solved analytically but I am told that the following first order ODE gives all the solutions to the second order ODE:
\begin{equation}h^2\pm\frac{h}{c\sqrt{1+(h')^2}}=\pm b^2\end{equation} where b is some real constant. The problem is that this equation can't be solved analytically either.
I must show that this first order ODE does indeed give all the solutions to the second order equation. So far I haven't been successful in doing so but I will list what I have done (my attempts have been a little desperate and I haven't been very confident with any of them):
- I've isolated for h' in the first order ODE, calculated h'' and plugged this in to both sides of the second order ODE to see if both sides are equal. This was done using math software (maple) but it didn't seem like the sides were equal.
- I've assumed that h must satisfy some other condition such that the second order ODE is simplified in a way that gives us the first order equation. For instance, I've assumed h must also satisfy $h''=k^2h$ in which case this second order ODE becomes a first order one. I did this because I figured that as long as I get two linearly independent solutions to the second order ODE, I have spanned the entire solution space. If my first order equation above can get me these two solutions I have unlocked all the solutions to the second order equation.
Is there a general way that I can show that these two equations share solutions? What would you do from here? What can I try in order to show that the solutions to the first order equation give me the solutions to the second order equation?
PS: I am also quite confused about how a first order equation can have several linearly independent solutions. I don't see how it can. Am I expected to construct another linearly independent solution from the one that I get?
Best Answer
Compute $$ \frac{d}{dt}\frac{h}{\sqrt{1+h'^2}} =\frac{h'}{\sqrt{1+h'^2}}-\frac{hh'h''}{\sqrt{1+h'^2}^3} =\frac{h'}{\sqrt{1+h'^2}^3}(1+h'^2-hh'')=2chh'. $$ Now the left and right term are easy derivatives, so integrate to get $$ \frac{h}{\sqrt{1+h'^2}}=ch^2+d. $$ I can not see where the other sign variants should come from, so they are spurious solutions from another solution path.