# Evaluating 'Constant' Term

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

$$u(\pi,t)=u_x(0,t)=u_x(\pi,t)\equiv0.$$

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

$$u_{xxt}+uu_{xxx}+u_xu_{xx}=0.$$

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.

• 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 $uu_{xx}$? Jul 28 '15 at 6:19
• 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 $-\int{uu_{xx}}$. 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:
$$u_{xt}(\pi,t)+u(\pi,t)u_{xx}(\pi,t)=h(t)=0\;.$$
• 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. Jul 28 '15 at 7:04