Elliptic PDE – How to Define Weak Solution with Non-Zero Dirichlet Boundary Condition

partial differential equations

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.

Best Answer

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.

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