# Characterization of functions in the Sobolev space $H_0^2(U)$ as zero trace functions in $H^2(U)$

Firstly, I am wondering if there exists a trace operator $$T:H^2(U)\rightarrow L^2(\partial U)$$ such that it satisfies analogous properties to that of the usual trace operator for functions in $$W^{1,p}(U)$$.

Secondly, I would like to know if the functions $$u\in H_0^2(U)$$ are those characterized by the following two conditions $$Tu=0,\quad\frac{\partial u}{\partial\nu}=0\quad\text{on }\partial U$$

Motivation: In problem 3 of section 6 of Evans' PDE book, second edition, it is asserted that the weak formulation corresponding to the boundary problem of the biharmonic equation $$\begin{cases} \Delta^2u&=f\quad\text{in }U\\ u=\frac{\partial u}{\partial\nu}&=0\quad\text{on }\partial U\end{cases}$$ is $$\int_U\Delta u\Delta vdx=\int_U fv$$ for each $$v\in H_0^2(U)$$. However, if we want to derive the latter identity, after integrating the equation of the problem, we find out through the third Green identity, that $$\int_U\Delta^2uvdx=\int_U\Delta u\Delta vdx+\int_{\partial U}v\frac{\partial\Delta u}{\partial\nu}dS-\int_{\partial U}\Delta u\frac{\partial v}{\partial\nu}dS$$ In order to obtain the idenity proposed by Evans, we have to impose the boundary conditions $$\frac{\partial v}{\partial\nu}=0,\quad v=0\quad\text{on}\partial U$$ for every $$v$$ in our space of weak solutions. However, how can we conclude that this completely characterizes functions in $$H_0^2(U)$$? I understand this should be a characterization of such a space, and via aproximation by functions with compact support I can get an idea of why this must be the case, but there is no mention of this fact in Evans' book. I would like to have some references to learn more about this, and the notion of the trace operator extended to other Sobolev spaces, because I am only familiar with the aforementioned book.

• This is certainly not an answer, but it is useful to think of functions $u \in H^2_0(U)$ as functions $u \in H^2(\mathbb{R}^n)$ satisfying $u \equiv 0$ outside $U$. In one direction, the implication is easy (any $u \in H^2_0(U)$ extended by zero lies in $H^2(\mathbb{R}^n)$), in another it requires some regularity of $U$. In problems similar to the one cited by you, one can (somewhat artificially) avoid discussing traces at all, by considering such extensions. Jul 1, 2021 at 10:23

Let's not worry about the optimal geometric conditions on $$U$$ and just assume throughout that $$U$$ is smooth.

The trace operator, the usual one, maps $$H^1(U)\to L^2(U)$$. This means that if $$u\in H^2(U)$$, then we can apply it to both $$u$$ and its first derivatives. In particular if $$u\in H^2(U)$$ then we can make sense of $$u|_{\partial U}$$ and $$\nabla u|_{\partial U}$$ (although at first they have no derivative relation on $$\partial U$$, but see point 2 below). This suggests that we define $$\partial_{\nu} u(x):= \nabla u(x) \cdot \nu(x) , \qquad x\in \partial U,$$ as an element of $$L^2(\partial U)$$.

Now the intuition behind the equivalence you mention is that if we define $$\nabla_{\|}u:= \nabla u-\partial_{\nu}u \nu$$ as the projection of $$\nabla u$$ onto the orthogonal complement of $$\nu$$, then $$\nabla_{\|}$$ (at least in the case $$u$$ is smooth) consists of the "tangential" derivatives to $$\partial U$$; i.e. $$\nabla_{\|}$$ is only seeing the behavior of $$u$$ on $$\partial U$$. What this does for us is the following: If $$u=0$$ on $$\partial U$$, then $$\nabla_{\|}u=0$$ and so in this case, by definition, we have $$\nabla u= \partial_{\nu} u\nu$$; or put another way: with the assumption that $$u=0$$, the vanishing of the full gradient (on $$\partial U$$) is equivalent to vanishing of the normal derivative.

With the above intuition in place, the proof of this has two main points.

1. Prove that $$H^2_0(U)=\{ u\in H^2(U): u=0, \, \nabla u=0\, \text{ on } \partial U\}$$. This is done as in Evans's result for $$H^1_0$$, with the important fact being that the vanishing of $$u$$ and $$\nabla u$$ at the boundary guarantees that the extension of $$u$$ by 0 outside of $$\bar{U}$$ defines an element of $$H^2(\mathbb{R}^n)$$.

2. Make rigorous the previous discussion about normal and tangential derivatives. This is kind of tedious by the nature of traces of Sobolev functions, but the main statement is that the trace of $$u$$ and it's gradient are actually related in the natural way: $$\nabla_{\|} u$$ is actually the weak derivative of $$u|_{\partial U}$$ on the boundary. More precisely the trace map defines a bounded map $$H^2(U)\to H^1(\partial U)$$, with the above formula for the "boundary gradient" of the trace (of course we have to define $$H^1(\partial U)$$, this is done via coordinate patches and straightening the boundary).

Hope this helps.

• I'm interested in a formal proof of the facts you mentioned. Could you please tell me a good reference/book where these aspects are introduced? Aug 14, 2021 at 13:13