suppose I have a pde $$u_{xt}(x,t)+u(x,t)u_{xx}(x,t)=h(t),\,\,\,\, x\in[0,\pi],\,\, t>0$$ for some unspecified function $h(t)$. This question is about finding what $h(t)$ is. Please, you may assume all computations are correct and that there is enough regularity for them to be justified. I just want to check if the idea behind the procedure is correct.

Suppose I have boundary condition


Note $u(0,t)$ not specified. Integrating the equation between $0$ and $\pi$ and using integration by parts gives

$$h(t)=-u_t(0,t)-\int_0^{\pi}{u_x^2 dx}.$$

Now, differentiating the pde in $x$ gives


Multiplying this last equation through by $x$, integrating between $0$ and $\pi$, and using integration by parts, yields

$$u_t(0,t)=\int_0^{\pi}{uu_{xx} dx}=uu_x\big|_{x=0}^{x=\pi}-\int_0^{\pi}{u_x^2}=-\int_0^{\pi}{u_x^2}.$$

Substituting this in the equation for $h(t)$ gives

$$h(t)\equiv 0.$$

That's it. Seems pretty straightforward to me.

  • $\begingroup$ You missed a factor of $\pi$ in integrating $h(t)$ over $[0,\pi]$. I don't understand the last step; could you explicate how you end up with that integral over $uu_{xx}$? $\endgroup$
    – joriki
    Jul 28 '15 at 6:19
  • $\begingroup$ Yeah I forgot that $\pi$, thanks. It comes from integrating $\int{xuu_{xxx}}+\int{xu_xu_{xx}}$ by parts, actually just the first term. The boundary terms are zero and we get $-\int{u_{xx}\partial_x(xu)}+\int{xu_xu_{xx}}$, which is just $-\int{uu_{xx}}$. $\endgroup$
    – user257319
    Jul 28 '15 at 6:33

I think what you've done is a terrible detour that consists mostly of undoing things you did before. The one substantial step involved is substituting $x=\pi$ into the differential equation, which you could have done directly:

$$ u_{xt}(\pi,t)+u(\pi,t)u_{xx}(\pi,t)=h(t)=0\;. $$

If I may take the liberty to quote you: That's it. Seems pretty straightforward to me. ;-)

  • $\begingroup$ Yes of course, that also works. I actually made up this pde to try and ask my question, which is about the procedure to determine $h(t)$. What I wanted to check was that all the things I did were justified and that not one single approach, say for instance the one where you integrate the original equation, should be used for some mathematical reason. Thanks anyway. $\endgroup$
    – user257319
    Jul 28 '15 at 7:04
  • $\begingroup$ @user257319: Well, it would have helped if you'd written all that in the question. What a wild goose chase. $\endgroup$
    – joriki
    Jul 28 '15 at 7:11
  • $\begingroup$ Sorry, I thought I did with ''I just want to check if the idea behind the procedure is correct.'' $\endgroup$
    – user257319
    Jul 28 '15 at 8:15

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