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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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    @martini It is from a handout I got, it is talking about 1st integrals of the [Korteweg–de_Vries_equation](http://en.wikipedia.org/wiki/Korteweg–de_Vries_equation)2012-11-10

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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).