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I encountered the following ode in the attempt to solve the cauchy problem of Liouville equation. I have tried for a long time to give it a solution but failed.

$(K e^f h + f'h'-2h'')^2=g^2((h')^2-\frac{K}{2}h^2 e^f)$ where $K$ is a given constant, $f,g$ are known smooth functions. The problem is to solve $h$ from this equation.

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    $\begingroup$ What are you hoping for? A miracle? $\endgroup$
    – Michael Renardy
    Commented Dec 9, 2014 at 10:16

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Solving the linear first-order $$ h'=h K^{1/2}e^{f/2}/\sqrt 2 \tag{$\ast$}$$ makes the rhs to vanish. We get then \begin{align} 2h''-f'h'-Ke^f h&=h' K^{1/2}e^{f/2}\sqrt 2+h K^{1/2}e^{f/2}f'(\sqrt 2)^{-1}-f'h K^{1/2}e^{f/2}(\sqrt 2)^{-1}-Ke^f h \\ &= h' K^{1/2}e^{f/2}\sqrt 2-Ke^f h=h K^{1/2}e^{f/2}(\sqrt 2)^{-1} K^{1/2}e^{f/2}\sqrt 2-hKe^f=0. \end{align} As a result, if $F$ is an antiderivative of $K^{1/2}e^{f/2}/\sqrt 2$, $h=e^F$ solves your equation.

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  • $\begingroup$ If we interpret $K^{1/2}$ as (any) square root of $K$ the same computation holds true, therefore $e^{-F}$ is a solution too. $\endgroup$
    – Pietro Majer
    Commented Dec 10, 2014 at 17:44
  • $\begingroup$ I can't possibly express my gratitude for your help. Thx u! $\endgroup$
    – larry
    Commented Dec 11, 2014 at 7:18

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