# Existence/Uniqueness of solution of singular ODE

Consider the following singular ODE $$(t^{N-1}g(y'))'=t^{N-1}f(y)$$ with initial condition $$y(0)=y_0$$ $$y'(0)=0$$. where $g$ is and increasing function. How one can prove that this problem has a solution?

• Why do you believe it has one? – Igor Rivin Oct 4 '11 at 14:26
• It is difficult to understand both variables and functions, please, define them. In any case, it looks like homework. – mikitov Oct 4 '11 at 14:28
• $g$ and $f$ are increasing function. $t$ is the variable and $y$ the unknown function. – Lugica Oct 4 '11 at 14:32
• @mikitov's theory that this is homework is still not disproved. Voting to close until it is... – Igor Rivin Oct 4 '11 at 14:35
• Ah, OK. You might get better responses if you add this explanation to your question. – Igor Rivin Oct 4 '11 at 15:15

One cannot prove existence in the generality that you have stated it. For example, if $g(p) = e^p$, there is no solution when $N>1$ and $f$ is differentiable.
To see this, observe that, if there were a solution $y(t)$ to your problem in a neighborhood of $t=0$, then the curve $(t,y,p) = (t,y(t),y'(t))$ would be an integral curve of the system $$dy - p\ dt = d(t^{N-1}g(p)) - t^{N-1}f(y)\ dt = 0$$ that passes through the point $(t,y,p) = (0,y_0,0)$.
Now, off the hypersurface $t=0$ in $typ$-space, this is equivalent to the system $$dy - p\ dt = \bigl((N{-}1)g(p)-tf(y)\bigr)\ dt + tg'(p)\ dp = 0.$$ As long as $g(0)\not=0$ and $N>1$, this system has rank $2$ at the point $(t,y,p) = (0,y_0,0)$, so that there is a unique integral curve of the system passing through this point. However, inspection shows that the curve is $(t,y,p) = (0,y_0,p)$ is an integral curve that passes through the point in question. By uniqueness, it is the only one.