A question in my PDE class is

Suppose $a(x) = a_{ij}(x)$ takes values in the class of symmetric, positive definite $n \times n$ matrices. Consider the PDE

$$u_t = \sum_{i,j=1}^n \frac{\partial}{\partial x_i}\left( a_{ij}(x) \frac{\partial u}{\partial x_j} \right)$$

in a bounded domain $\Omega$, with intial condition $u=u_0(x)$ at $t=0$ and a Dirichlet boundary condition $u=g$ at $\partial \Omega$. Use the "energy method" to show there can be at most one solution.

In the past, we have used the energy method to show uniqueness by multiplying both sides by $u$ and integrating:

$$\int_{\Omega}uu_t= \frac12 \frac{d}{dt}\int_{\Omega}u^2 = \int_{\Omega} u \sum_{i,j=1}^n \frac{\partial}{\partial x_i}\left( a_{ij}(x) \frac{\partial u}{\partial x_j} \right),$$

and attempting to show somehow that the right hand side must be negative, therefore if there are two solutions $u_1$ and $u_2$, their difference has zero initial state, and $\frac{d}{dt}\int_{\Omega}u^2 dx \leq 0$ shows that it must remain there, so their difference is the zero solution, and therefore the solution is unique.

Here I am trying to use the positive definiteness of $a(x)$ to show that the right hand side must be negative. If we can write it in the form $\int_{\Omega} -\langle a(x)\vec{v}, \vec{v}\rangle$ we will be done. But so far I have not been able to do so.

If we apply the chain rule we get

$$\frac12 \frac{d}{dt}\int_{\Omega}u^2 = \int_{\Omega} u \sum_{i,j=1}^n \left( \frac{\partial a_{ij}}{\partial x_i} \frac{\partial u}{\partial x_j} + a_{ij}(x)\frac{\partial^2 u}{\partial x_i \partial x_j}\right),$$

but I'm not getting anywhere with that. Any ideas?

  • 1
    $\begingroup$ Aren't you missing an integral in the second equation? $\endgroup$
    – dls
    Sep 15, 2013 at 4:52

1 Answer 1


As dls mentioned you're lacking an integral on the RHS. With this notice that

$$ \sum_{i,j=1}^n \partial_i(a_{ij} \partial_j u) =\text{div}(a \nabla u), $$

so that, integrating by parts and using the zero boundary conditions, we get that the RHS is equal to

$$ -\int_{\Omega} \langle \nabla u, a \nabla u \rangle dx \leq 0. $$

Edit: Notice that you want to prove that if $u(x,0)=0$ for every $x\in \Omega$ and $u(y,t)=0$ for every $y\in \partial \Omega$ and $t>0$ then $u\equiv 0$. So take such an $u$, then your argument gives

$$ \frac{1}{2}\frac{d}{dt} \int_{\Omega} |u|^2 dx = \int_{\Omega} u \text{div}(a\nabla u) dx. $$

Call $v=a\nabla u$. The divergence theorem, applied to the vector field $uv$, gives

$$ \int_{\Omega}u \text{div}(v)dx= -\int_{\Omega}\langle \nabla u, v \rangle dx + \int_{\partial \Omega}\langle uv,\nu \rangle d \sigma. $$

Since we have that $u(y,t)=0$ for every $y\in \partial \Omega$ the boundary integral is zero for every $t$. Therefore we have

$$ \frac{1}{2}\frac{d}{dt}\int_{\Omega} |u|^2 dx =- \int_{\Omega} \langle \nabla u,a\nabla u\rangle dx \leq 0. $$

  • $\begingroup$ I'm sorry, can you clarify the integration by parts? We have $$\int_{\Omega}\operatorname{div}{(a\nabla u)} = \int_{\partial \Omega}(a\nabla u \cdot \hat{n}dS),$$ and then what? Normally I would try to use $$\nabla \cdot (f \nabla g) = \langle \nabla f, \nabla g \rangle + f \Delta g,$$ but here $a$ is a matrix valued function... $\endgroup$
    – Eric Auld
    Sep 15, 2013 at 16:01
  • $\begingroup$ @EricAuld: I've edited, let me know if you still have doubts on the argument. $\endgroup$
    – Jose27
    Sep 15, 2013 at 18:44
  • $\begingroup$ Thank you very much! I forgot about the $u$. $\endgroup$
    – Eric Auld
    Sep 15, 2013 at 18:49

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