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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?

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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
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Ah, OK. You might get better responses if you add this explanation to your question. –  Igor Rivin Oct 4 '11 at 15:15

1 Answer 1

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.

Thus, there is no integral curve of the kind that you would get from a solution to your problem.

I suspect that you have left out some hypotheses or not made a correct symmetry reduction of the system that you are trying to study.

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