Background: In our PDE class we explored the heat equation with Dirichlet boundary condition

$$u_t - \Delta u = 0 \;\text{ in } \Omega \subset \mathbb{R}^n \;\text{bounded}\\ u = u_0(x) \;\,\text{at} \; \,t=0\\ u = 0 \;\, \text{at} \;\, \partial \Omega$$

Multiplying by $u$ on both sides and using the divergence theorem gives us

$$\frac12 \frac{d}{dt}\int_{\Omega}u^2 + \int_{\Omega}|\nabla u|^2 = 0 \tag{1}$$

Now since the function $u$ is zero on the boundary, we can employ Poincare's inequality and say that

$$-\int_{\Omega}|\nabla u|^2 \leq \frac{-1}{C_{\Omega}}\int_{\Omega}|u|^2,$$

which together with (1) gives that

$$\frac12 \frac{d}{dt}\int_{\Omega}u^2 \leq \frac{-1}{C_{\Omega}}\int_{\Omega}|u|^2,$$

so $\int_{\Omega}u^2$ decays exponentially to $0$ as $t \to \infty$.

Question: Having concluded that $\int_{\Omega}u^2$ decays exponentially, we are asked to prove an analogous statement for a homogeneous Neumann boundary condition which says that

$$\frac{\partial u}{\partial \vec{\nu}}=0 \;\, \text{at} \;\, \partial\Omega, $$

for $\vec{\nu}$ the normal vector to the boundary. We are given the hint that there is a constant $M_{\Omega}$ with the property that if $u: \Omega \to \mathbb{R}$ has mean value $0$ then $\int_{\Omega} u^2 \, dx \leq M_{\Omega} \int_{\Omega} |\nabla u|^2 \, dx$.

Now, I understand $u$ having mean value zero to mean that $\int_{\Omega} u \, dx=0$. One way to achieve that might be to subtract (the function of $t$) $\int_{\Omega}u\,dx$ from $u$, so that $v : = u - \int_{\Omega}u\,dx$ has mean value zero for all $t$. But this seems to screw things up when we try to square $v$.

How can we use the Neumann boundary condition to find out something regarding the mean value of $u$? For instance, the divergence theorem would tell us

$$\int_{\partial \Omega} u\cdot \vec{\nu} \, ds = \int_{\Omega} \nabla \cdot u \, dx.$$

The left side here is zero, I believe, but I'm not sure that tells us anything about the mean value.

Could someone help me see the next step? I'm stuck.

  • 1
    $\begingroup$ I think your last equation should be $$ \int_{\partial\Omega} \nabla u \cdot \vec\nu ds = \int_{\Omega} \Delta u dx. $$ $\endgroup$ – Tunococ Sep 10 '13 at 11:23

The divergence theorem is a good idea, but the way you wrote it down doesn't make sense - you're treating $u$ as a vector field when it is actually a scalar function. The vector field you want to study is $\nabla u$ - the divergence theorem is $$\int_{\partial \Omega} \nabla u \cdot \nu = \int_\Omega \Delta u.$$ The LHS is $0$ by the boundary conditions and the RHS is $$\int_\Omega \frac{\partial u}{\partial t} = \frac d{dt} \int_\Omega u$$ by the PDE (and some regularity assumptions + convergence theorem to exchange the derivative and integral). Thus the mean value of $u$ is constant in time.

  • $\begingroup$ OK, great. So to apply the hint, we can let $\alpha$ be the (constant) mean value of $u$, and write that $$\int_{\Omega} (u-\alpha)^2 \, dx = \int_{\Omega} u^2\, dx - 2\alpha \int_{\Omega} u + \alpha A_{\Omega} \leq M_{\Omega}\int_{\Omega} |\nabla u|^2,$$ where $A_{\Omega}$ is the volume of the set $\Omega$. Now how can we show some analogous condition about the behavior as $t\to \infty$? We might want to derive with respect to time, but that doesn't seem to help...(continued) $\endgroup$ – Eric Auld Sep 10 '13 at 12:23
  • $\begingroup$ We could use the divergence theorem (with the boundary conditions) again and get $$\int_{\Omega} (u-\alpha)^2 \, dx = \int_{\Omega} u^2\, dx - 2\alpha \int_{\Omega} u + \alpha A_{\Omega} \leq M_{\Omega}\int_{\Omega} |\nabla u|^2 = -M_{\Omega} \int_{\Omega} u \Delta u,$$ but I'm not seeing anything there... $\endgroup$ – Eric Auld Sep 10 '13 at 12:26
  • $\begingroup$ Here's something: if we substitute the original equation in, we seem to get $$-M_{\Omega} \int_{\Omega} u \Delta u = -M_{\Omega} \frac{\partial}{\partial t} \int_{\Omega} u^2,$$ so $$\frac{\partial}{\partial t} \int_{\Omega} u^2 \leq \frac{-1}{M_{\Omega}} \int_{\Omega} (u-\alpha)^2 = \frac{-1}{M_{\Omega}} \left( \int_{\Omega}u^2 - 2\alpha \int_{\Omega} u + \alpha A_{\Omega} \right).$$ But I'm still not seeing how that can tell about the growth of $\int_{\Omega} u^2$ as $t\to \infty$, unless we can say something about $-2\alpha \int_{\Omega} u + \alpha A_{\Omega}$ somehow... $\endgroup$ – Eric Auld Sep 10 '13 at 12:40
  • $\begingroup$ @EricAuld: Intuitively we expect the heat equation with insulated boundary conditions (i.e. no loss of $\int u$) to smooth out to a constant; so what you should be trying to show is that $\int (u-\alpha)^2$ decays exponentially. You're basically there - just show that $\partial_t \int u^2 = \partial_t \int (u-\alpha)^2$ and apply the exact argument you used in the Dirichlet case. $\endgroup$ – Anthony Carapetis Sep 10 '13 at 12:41
  • $\begingroup$ I know, how can I remove it?! There's no "x" showing up for me! It's broken even its own ability to be deleted... $\endgroup$ – Eric Auld Sep 10 '13 at 12:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.