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Can anybody see why $${d\over dt}\int_{-\infty}^\infty -f_{xx}+f^2\,\,\, dx=0$$ where $f=f(x,t)$, follows from $$f_t+f_{xxx}+6ff_x=0$$?

I tried differentiating under the integral sign, but things got ugly.

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Where is this problem from? –  martini Nov 10 '12 at 3:12
    
I don't see how this could follow independently of the trajectory $x(t)$. Are you sure you intended that to be a total derivative with respect to time and not a partial derivative? –  joriki Nov 10 '12 at 4:57
    
@joriki unfortunately so... –  Simple Nov 10 '12 at 8:31
    
@martini It is from a handout I got, it is talking about 1st integrals of the Korteweg–de_Vries_equation –  Simple Nov 10 '12 at 8:33

1 Answer 1

This is false. A counterexample is afforded by $f(x,0)=\mathrm e^{-x^2/2}$. At $t=0$, we have $f_x(x,0)=-x\mathrm e^{-x^2/2}$, $f_{xx}(x,0)=(x^2-1)\mathrm e^{-x^2/2}$, and $f_{xxx}(x,0)=(3x-x^3)\mathrm e^{-x^2/2}$. Thus $f_t(x,0)=(x^3-3x)\mathrm e^{-x^2/2}+6(1-x^2)\mathrm e^{-x^2}$. Then at $t=0$

$$ \begin{align} \frac{\mathrm d}{\mathrm dt}\int_{-\infty}^\infty\left(-f_{xx}+f^2\right)\mathrm dx &= \frac{\mathrm d}{\mathrm dt}\int_{-\infty}^\infty f^2\mathrm dx \\ &= \int_{-\infty}^\infty \frac{\partial}{\partial t}\left(f^2\right)\mathrm dx \\ &= \int_{-\infty}^\infty ff_t\,\mathrm dx \\ &= \int_{-\infty}^\infty\left((x^3-3x)\mathrm e^{-x^2}+6(1-x^2)\mathrm e^{-3x^2/2}\right)\mathrm dx \\ &=\sqrt{\frac{32\pi}3} \end{align} $$

(computation).

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