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
    Commented Sep 10, 2013 at 11:23

1 Answer 1


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
    Commented Sep 10, 2013 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
    Commented Sep 10, 2013 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
    Commented Sep 10, 2013 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. $\endgroup$ Commented Sep 10, 2013 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
    Commented Sep 10, 2013 at 12:42

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