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I am trying to solve for $y(x)$ in terms of $f(x)$ in a convenient space (eg. $\dot{H}^2(\mathbb{T})$-zero mean). Here is the ode: $y(x)+y(x)y'(x)=f(x)$. I think a contraction mapping argument will work, but this is not clear to me. In trying to compute the $\dot{H}^2(\mathbb{T})$ norm, I get $\int_{\mathbb{T}} (1+ \frac{5}{2}y')(y'')2\, dx = \int_{\mathbb{T}}y'' f''\, dx$. Here I am just trying to show that if $\vert h'' \vert _{\dot{H}^2(\mathbb{T})}\leq C$, where $C$ is a constant then the operator $T(y):=y(x)+y(x)y'(x)$ maps the ball of radius $C$ into a ball of radius $C$. But $T$ as defined above is not a contraction.... May I have some suggestions, please?

Thank you! Rosa

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  • $\begingroup$ Also note that $u:=1/y$ solves the ODE in normal form $u'=u^2(u-f)$, which suggest to translate your problem in terms of $u$. $\endgroup$
    – Pietro Majer
    Commented May 30, 2011 at 17:57
  • $\begingroup$ I get $u'=u^2(1-fu)$. $\endgroup$
    – Michael Renardy
    Commented May 30, 2011 at 21:16
  • $\begingroup$ yes sorry, a typo. $\endgroup$
    – Pietro Majer
    Commented May 30, 2011 at 23:08

2 Answers 2

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In general your problem does not have a solution. Suppose f has zero average and one simple zero $x_0$ in the interval. If $y$ is to be smooth with zero average, then it must also have a simple zero at $x_0$, and we easily check that $y'(x_0)+(y'(x_0))^2=f'(x_0)$. If $f'(x_0)<-1/4$, there are therefore no solutions. If $f'(x_0)>-1/4$, there are two solutions, at least locally near $x_0$, but there is no reason why either of these should have zero average.

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But if you take fourier ttransform of your initial ecuation you can construct a contraction on $l^2$. I think I am not mistaken.

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