# Showing Solution to Some Random PDE Tends to Zero Uniformly

Another qual problem that is causing me some difficulty...

Consider the PDE $$\left\{\begin{array}{rl} u_{xxt}+u_{xx}-u^{3}=0&\text{in}\;[0,1]\times(0,\infty)\\ u(0,t)=u(1,t)&\text{on}\{x=0\}\times(0,\infty)\bigcup\{x=1\}\times(0,\infty)\\ u(x,0)=g(x).\end{array}\right.$$ Take $g(x)=x(x-1)$ and show that solutions tend to zero uniformly in $t$ as $t\infty$.

To solve the problem I used the usual energy argument, multiplying the PDE by $u$ and integrating both sides in space (applying to the periodic boundary conditions to eliminate boundary terms from the integration by parts) in order to easily obtain $$\frac{de}{dt}(t):=\frac{d}{dt}\int\limits_{0}^{1}|u_{x}(x,t)|^{2}\;dx=-2e(t)-2\int\limits_{0}^{1}|u(x,t)|^{4}\;dx\leq0$$ so that the energy decreases. Then, using the initial condition, I also have $$0\leq e(t)\leq e(0)=\frac{1}{3}$$ for all $t>0$. So I have a fairly sharp bound and the fact that $e(t)$ is decreasing in $t$. If I could show further that $e(t)$ actually tends to $0$ uniformly, then that would imply $u_{x}(x,t)\to0\;\text{as}\;t\to\infty$ uniformly and that would give me $u(x,t)\to\;\text{C}\;\text{as}\;t\to\infty$. Applying the periodic boundary conditions once more would then give me $C=0$ as required.

However, I do not know how to show $e(t)\to0$ as $t\to\infty$. I've tried computing additional derivatives of $e$ to see if I could glean any additional information on the decay of $e$, but the computations did not seem to lead to anything fruitful (even after playing around/substituting terms using the PDE).

Any hints are appreciated (a fully fleshed answer is not needed). Thanks!

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Thanks for pointing that out - it should have been $=u^{3}$ in the equation. – Taylor Martin Feb 24 '13 at 8:14

Hint: You can rewrite

$$u_{xxt} + u_{xx} = e^{-t} \partial_t (e^t u_{xx})$$

by integrating factors. So let $v = e^t u$ and $v_{xx} = e^{t} u_{xx}$ you can rewrite the equation in terms of $v$. If you can show that $v$ has bounded energy, then since the energy of $v$ is $e^{2t}$ times that of $u$, the energy of $u$ must decay exponentially.

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Thanks for the tip. It is easy to see the energy of $v$ is bounded since a similar computation shows $\frac{d}{dt}E_{v}(t)=-2e^{-2t}\int_{0}^{1}v^{4}dx\leq0$ (interesting that the middle term $u_{xx}=e^{-t}v_{xx}$ drops out in the PDE) and $0\leq E_{v}(t)\leq E_{v}(0)=e^{2t}\int_{0}^{1}|u_{x}(x,t)|^{2}dx\bigg|_{t=0}=E_{u}(0)=\frac{1}{3}$ as before. Is there any methodology you used in order to come up with this trick/substitution? I've had to use a similar substitution in a problem dealing with a modified diffusion equation, but sadly such ideas don't often come to me.... – Taylor Martin Feb 24 '13 at 8:38
"Is there any methodology..." It is called heuristics. It you spend a few more years looking at PDEs you'll also pick up these types of intuitions. – Willie Wong Feb 24 '13 at 22:39

You don't just have $\dfrac{de}{dt} \le 0$, you have $\dfrac{de}{dt} \le -2e$. That (with $e > 0$) implies $e(t)$ decreases at least as fast as the solutions to $\dfrac{dy}{dt} = - 2y$, which are $C \exp(-2t)$. In fact, $$\dfrac{d}{dt} (e(t) \, \exp(2t)) = \exp(2t) \left(\dfrac{de}{dt} + 2 e \right) \le 0$$ so $e(t) \le e(0) \exp(-2t)$.

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Good point. I actually noticed this, and in retrospect should have applied Gronwall's inequality. – Taylor Martin Feb 24 '13 at 9:38