For homogeneous Dirichlet boundary condition, for example $$ \left\{\!\! \begin{aligned} &-\Delta u+c(x)u=f(x),x\in\Omega\\ &u|_{\partial\Omega}=0 \end{aligned} \right. $$ The weak solution is defined as a function $u\in H_0^1(\Omega)$ satisfying $$ \int_\Omega\left(\sum_{i=1}^n\frac{\partial u}{\partial x_i} \frac{\partial v}{\partial x_i}+c(x)uv\right)\,dx=\int_\Omega fv\,dx $$ for every $v\in H_0^1(\Omega)$.

I wonder how to define weak solution for an elliptic PDE with non-zero Dirichlet boundary condition.

For example, $$ \left\{\!\! \begin{aligned} &-\Delta u+c(x)u=f(x),x\in\Omega\\ &u|_{\partial\Omega}=g \end{aligned} \right. $$

Evans's Partial Differential Equations (1st edition, Section 6.1.2) says:

... is is necessary for $g$ to be the trace of some $H^1$ function, say $w$. But then $\tilde u:=u-w$ belongs to $H_0^1(\Omega)$, and is a weak solution of the boundary-value problem $$ \left\{\!\! \begin{aligned} &-\Delta \tilde u+c(x)\tilde u=\tilde f(x),x\in\Omega\\ &\tilde u|_{\partial\Omega}=0 \end{aligned} \right. $$ where $\tilde f:=f-(-\Delta w+c(x)w)$

The problem is: how to find the function $w$, in a constructive way?

Can you please help? Thank you.

  • $\begingroup$ More succinctly, the problem boils down to finding or recovering an $H^1$ function given its values on $\partial\Omega$. Remind me again what $H^1$ denotes? $\endgroup$
    – anon
    Jul 19, 2011 at 11:57
  • $\begingroup$ @anon: $H^1(\Omega)$ (Sobolev Space) is defined as: $H^1(\Omega)=\{u\in L_2(\Omega)\mid \frac{\partial u}{\partial x_i}\in L_2(\Omega),i=1,2,\ldots,n\}.$ You've reminded me that my question seems not well asked. Need I rewrite it more concisely? $\endgroup$
    – Roun
    Jul 19, 2011 at 12:27
  • $\begingroup$ I don't see providing background behind a problem as a negative - just that the PDE stuff isn't necessary to understand the second bolded question. Also, I have a habit of never remembering the labels of various continuity/differentiability classes. In any case, if $\Omega$ happens to be a ball, then you can find a harmonic function $w$ using the Poisson kernel - otherwise I'm not sure what explicit constructions exist. $\endgroup$
    – anon
    Jul 19, 2011 at 13:04

2 Answers 2


You don't find $w$. $w$ is a "given", in the following sense. For the weak formulation of the problem to make sense, the statement

$$ u|_{\partial\Omega} = g $$

is in fact the following statement: $\exists$ a fixed $w \in H^1(\Omega)$ such that the trace of $w$ to $\partial \Omega$ is equal to the trace of $u$.

The relevant section in Evans is trying to explain this. Basically what he is trying to say is that the intuition for the Dirichlet problem with "strong" solutions, where you prescribe boundary value as some continuous function $g$ on the boundary, must be replaced by an appropriate weak version defined relative to the trace operator to hypersurfaces, when you consider the weak formulation of the problem. This is because a solution $u$, as an object in the space $W^{1,2} = H^1$, is only an equivalent class of functions defined up to sets of measure zero. If $\Omega$ is a sufficiently regular open set, $\partial\Omega$ has measure zero, so it is meaningless to state that $u$ coincides with $g$ on $\partial\Omega$, since $u$ can always be modified on just $\partial\Omega$ to give any value you want there.

You should compare this to, for example, Theorem 8.3 in Gilbarg and Trudinger, Elliptical partial differential equations of second order, which states

Let [$L$ be an elliptic operator]. Then for $\psi \in W^{1,2}(\Omega)$ and $g,f^i\in L^2(\Omega)$, $i = 1 , \ldots, n$, the generalized Dirichlet problem, $Lu = g + D_i f^i$ in $\Omega$, $u = \psi$ on $\partial\Omega$ is uniquely solvable.


I ran across this old question and I would like to say something about it. While the point of view provided by Willie Wong is satisfactory, there is a theory of "inverse traces" which answers OP's question in bold more or less directly. For an example, have a look at the following theorem taken from Kufner, John, Kufcik, "Function Spaces".

6.9.2 Theorem Let $p>1$ and $\Omega\in C^{0, 1}$ [meaning that its boundary is Lipschitz and bounded]. Then there exists a continuous linear mapping $T$ from $W^{1-1/p, 1}(\partial \Omega)$ into $W^{1,p}(\Omega)$ such that for $v=Tu$ we have $v=u$ on $\Omega$ [in trace sense].

The idea of the proof is simpler than one might expect (at least, it is simpler than I expect). Suppose that $$\Omega=\mathbb{R}^{n+1}_+=\{(x, t)\in \mathbb{R}^n\times \mathbb{R}\ :\ t>0\}.$$ If you have a function $u\in C^{\infty}_0(\mathbb{R}^n)$ you can construct a function $v\in C^\infty(\mathbb{R}^{n+1}_+)$ by convolution against the heat kernel: $$v(x,t)=\frac{1}{(4\pi t)^{n/2}}\int_{\mathbb{R}^n} e^{-\frac{\lvert x-y\rvert^2}{4t}}u(y)\, dy.$$ Then the trace of $v$ on the boundary $\{t=0\}$ is $u$, as we already know.

If $\Omega$ is a domain with a nice enough boundary, we can use charts to perform this construction locally and then patch everything up with a partition of unity. This is the idea. Unfortunately, there are some technical difficulties which require the introduction of the cumbersome fractionary order Sobolev spaces $W^{1-1/p, p}$.

  • $\begingroup$ What is the reason for this construction involving the heat kernel? Why prefer this over a simpler construction such as $v(x,t)=u(x)\chi(t)$ where $\chi:[0,+\infty)\to[0,1]$ is a smooth bump function with say $\chi\equiv 1$ on $[0,1]$ and $\chi\equiv0$ on $[2,+\infty[$? $\endgroup$ Nov 25, 2020 at 23:06
  • $\begingroup$ @OlivierBégassat: your question is interesting, but it's been a long time, and I am afraid I cannot answer right now. The construction I hinted at in this answer is Lemma 6.9.1 of the aforementioned book. $\endgroup$ Nov 26, 2020 at 0:04
  • $\begingroup$ @OlivierBégassat: I thought a little bit at your question and I have a first, rough answer. If we define $v(x, y)$ via heat kernel, like the book does, then for each $y>0$ the function $v(\cdot, y)$ is $C^\infty$. This helps when it is time to choose a partition of unity and patch together all these functions $v$. $\endgroup$ Nov 28, 2020 at 1:56

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .