# Non-homogeneous boundary value problem - weak solution

Consider the following boundary value problem on a bounded subset $U\subset\mathbb{R}^n$: $$-\Delta u = f,$$ for $u: U \rightarrow \mathbb{R}$ where $u = g \in C^2$ on the boundary of $U$. I have proven that the homogeneous problem has a weak solution. Now I would like to show existence of a weak solution for this non-homogeneous problem. In the book I am using (Evans) this problem is stated as well, but there the author assumes that $g$ is in the image of the Trace operator. Using this, the problem reduces to a homogeneous problem, which I already solved. The problem in this case is that the Trace operator has as its range $L^2(\partial U)$. Now I am wondering how to reconcile this with the given function $g$. Is $g\in L^2$ if it is in $C^2$? Or is there another way to tackle this more specific non-homogeneous boundary value problem?

• If a function defined on a closed curve ($\partial U$) is continuous, can you integrate its square? Mar 25 '13 at 19:30
• If you ask the question like that, the answer is probably yes, but I don't really see why. Mar 25 '13 at 19:40
• If you are talking about $g\in L^2(\partial\Omega)$, then that is obvious, when $\partial\Omega$ has finite measure and compact. if you are talking about $g$ extended as a function on $\Omega$ to be in $L^2(\Omega)$, then you need to check with trace theorem. Mar 25 '13 at 19:51
• @Erik, think about what makes a function not integrable. For the record, I think this is a great way to build some intuition about the space $L^p$. Mar 25 '13 at 19:58
• A continuous function on a closed set cannot diverge at any point, hence is integrable, is that correct? Another question: if g is in $C^2$ and we have $Tw = g$, with $T$ the trace operator, can we conclude that also $w \in C^2$? Mar 25 '13 at 20:08

I got an answer. In general $C(U) \subset L^2(U)$, for a bounded, closed domain U. (For the 1d case there is a lemma stating that any continuous functions on a closed interval has an existing Lebesgue integral.