I'm studying for a qualifying exam and I can't figure this problem out:

Suppose $B \subset \mathbb R^n$ is the unit ball centered at the origin and that $u$ is a smooth solution of \begin{align*} u_{tt} + a(x) u_t - \Delta u &= 0 \,\,\,\,\,\,\,\,\,\,\,\,\, \text{ in } B \times (0,\infty) \\ u(x,t) &= 0 \,\,\,\,\,\,\,\,\,\,\,\, \text{ on } \partial B\times (0,\infty) \\ u(x,0) = g(x), \,\,\, u_t(x,0) &=h(x) \,\,\,\,\, \text{ in } B\end{align*} where $a(x),g(x),h(x)$ are smooth with $g(x) = h(x) = 0$ on $\partial B$ and $0 < \lambda < a(x)$ in $B$. Show that $\int_B u(x,t)^2 dx \le C e^{-\lambda t}$ for all $t > 0$ where $C$ is a constant depending on $g,h$ and the dimension $n.$

This problem shouts 'energy methods' to me, but I can't make it work. First, I defined $$E(t) = \int_B u(x,t)^2 dx$$ since this is what we are trying to bound but this energy doesn't lend itself naturally to the wave equation and indeed differentiating gives nothing too useful (as far as I can see). Next, I tried the natural energy for the wave equation: $$E(t) = \int_B u_t^2 + \lvert \nabla u \rvert^2 dx$$ but this doesn't seem to help since it doesn't include $u^2$. Finally, I tried the energy $$E(t) =\int_B u_t^2 + \lvert \nabla u \rvert^2 + a(x) u^2 dx$$ but again, this didn't seem to lead to anything useful.

Am I simply not seeing the correct energy to use or is there some other method that I should be trying? Any help would be appreciated.


I was too hasty before. Here's one solution (which likely could be simplified):

You need in fact two different energy estimates.

  1. Multiply by $u_t$ and integrating by parts gives you that for (as you noted) $$ E(t) = \int_B u_t^2 + |\nabla u|^2 ~\mathrm{d}x $$ you have $$ \dot{E}(t) = - 2 \int_B a u_t^2 ~\mathrm{d}x $$

  2. Multiply by $u$ and integrating by parts you get that for $$ K(t) = \int_B a u^2 + 2 u_t u $$ that $$ \dot{K}(t) = 2 \int_B u_t^2 - |\nabla u|^2 ~\mathrm{d}x $$ Now, $K$ is not positive definite. But observing that $$ \lambda^2 u^2 + 2 \lambda u_t u + u_t^2 = (\lambda u + u_t)^2 \geq 0 $$ we have that $$ \lambda K(t) + E(t) \geq \int_B |\nabla u|^2 ~\mathrm{d}x $$ On the other hand, we also have $$ K(t) \leq \int_B (a + 1) u^2 + u_t^2 \leq C E(t) \tag{KB}$$ for some universal constant $C$ depending on $n$ using Poincare's inequality.

  3. So combining everything we get, for $S(T) = \lambda K(T) + 2 E(T)$, that $$ \dot{S}(T) = -2 \int_B (2a - \lambda) u_t^2 + \lambda |\nabla u|^2 ~\mathrm{d}x \leq -2\lambda E(T) \leq - \frac{2}{2 + \lambda C} \lambda S(T) $$ which implies that $$ S(T) \leq \tilde{C}(g,h) \exp( - \frac{2\lambda T}{2 + \lambda C} ) $$ Using again that $$ S(T) \geq \int_B |\nabla u|^2 $$ and Poincare inequality we have that $$ \int_B u^2~\mathrm{d}x \leq \hat{C}(g,h,n) \exp(- \frac{2\lambda T}{2 + \lambda C}) \tag{FB}$$ where $C$ depends on the constant in the Poincare inequality as well as the $\sup_B a$.

(In fact, the constant $C$ can be taken to be $$ \frac{\sup_B a + \sqrt{ (\sup_B a)^2 + 4 \kappa_1}}{2 \kappa_1} $$ where $\kappa_1$ is the first Dirichlet eigenvalue of $-\triangle$ on $B$.)

Now, the conclusion above get you exponential decay, but at a worse rate than claimed by your problem. I claim that your practice problem is in fact incorrect: the sharp rate cannot be attained for all $\lambda$.

Consider the following scenario: let $a(x) = \lambda > 0$. Take the separation-of-variables ansatz and assume that for any fixed $t$ that $u(t,x)$ is a Dirichlet eigenfunction of the Laplacian on $B$, with eigenvalue $-\kappa < 0$. Then a particular solution of the equation is

$$ u(t,x) = e^{\omega t} u(0,x) $$

with $\omega \in \mathbb{C}$ satisfying $$ \omega^2 + \lambda \omega + \kappa = 0 $$ This implies that $$ \omega = \frac{-\lambda \pm \sqrt{\lambda^2 - 4\kappa}}{2} $$

Now, when $\lambda$ is small compared to the eigenvalue $\kappa$, we have that the real part of $\omega$ is $-\lambda / 2$ and so $\int |u|^2$ decays like $e^{-\lambda t}$ as claimed. But when $\lambda \geq 2 \sqrt{\kappa}$, you are in the overdamped case and in the worst case scenario you have decay rate asymptotic to $\omega \approx \kappa / \lambda$.

(You should compare this to the answer given above: the constant $C$ that appears in the final estimate arises from (KB) and, when $\lambda$ is large, is given by $$ \frac{a+1}{\kappa_1} \approx \frac{\lambda}{\kappa_1} $$ where $\kappa_1$ is the first Dirichlet eigenvalue of your domain. Inserting it into our final bound (FB) we have that asymptotically the decay rate for $|u|^2$ found to be, as $\lambda \nearrow +\infty$, $$ \frac{2\lambda}{2 + \lambda C} \approx \frac{2}{C} = 2 \frac{\kappa}{\lambda} \approx 2 \omega $$ as predicted. On the other hand, when $\lambda\searrow 0$, we have that $C $ approaches $\max(1, \frac{1}{\kappa_1})$, and thus the decay rate $$ \frac{2\lambda}{2 + \lambda C} \approx \lambda \approx 2\omega $$ again as predicted by the separation-of-variables ansatz.)

  • $\begingroup$ Thanks for the notes! If I define $$E(t) = \int_B u_t^2 + \lvert \nabla u \rvert^2 dx$$ then I find that $$\dot E(t) = -2 \int_B a(x) u_t^2 dx$$ whence I know that $\dot E(t) \le 0$ and so $E(t) \le E(0)$ for all $t$. This is all I have been able to conclude thus far (that the desired quantity is bounded). How do I get the exponential decay? $\endgroup$ – User8128 Jul 20 '16 at 14:15
  • $\begingroup$ @User8128: I was mistaken. The question turns out to be quite a bit more delicate. In particular the stated decay rate $\exp -\lambda t$ is in fact false. I edited the answer to give the full analysis and justification showing that asymptotically the decay rate that I got is sharp. $\endgroup$ – Willie Wong Jul 20 '16 at 18:11

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