Showing Solution to Some Random PDE Tends to Zero Uniformly

Question

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!

Answer 1

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.

Answer 2

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 $e(t) \le e(0) \exp(-2t)$.