# Trace operator

In mathematics, the concept of trace operator plays an important role in studying the existence and uniqueness of solutions to boundary value problems, that is, to partial differential equations with prescribed boundary conditions. The trace operator makes it possible to extend the notion of restriction of a function to the boundary of its domain to "generalized" functions in a Sobolev space.

## Informal discussion

Let $\Omega$ be a bounded open set in the Euclidean space $\mathbb {R} ^{n}$ with C1 boundary $\partial \Omega .$ If $u$ is a function that is $C^{1}$ (or even just continuous) on the closure ${\bar {\Omega }}$ of $\Omega ,$ its function restriction is well-defined and continuous on $\partial \Omega$ . In other words, it is clear what the values of $u$ on the boundary $\partial \Omega$ should be because the function is continuous as we move from the interior to the boundary of the domain. We then call that function $g:\partial \Omega \rightarrow {\mathbb {R} }$ the "trace" of $u$ that at each point $\mathbf {x}$ of the boundary has the same values as the limit of $u:\Omega \rightarrow {\mathbb {R} }$ when we move from the interior of $\Omega$ toward $\mathbf {x}$ . (Remember that $\Omega$ is an open set, so boundary points are not in $\Omega$ and consequently $u$ itself is not defined at boundary points.)

If however, $u$ is the solution to some partial differential equation, it is in general a weak solution and only belongs to some Sobolev space. Such functions are in general not continuous, and the operation "limit of $u:\Omega \rightarrow {\mathbb {R} }$ when we move from the interior of $\Omega$ toward $\mathbf {x} \in \partial \Omega$ " that we used above may not be allowed for some $\mathbf {x}$ because the limit does not yield a unique value for all sequences of points that converge to $\mathbf {x}$ . It follows that simple function restriction cannot be used to meaningfully define the trace of weak functions.

The way out of this difficulty is the observation that while an element $u$ in a Sobolev space may be ill-defined as a function, $u$ can be nevertheless approximated by a sequence $(u_{n})$ of $C^{1}$ functions defined on the closure of $\Omega .$ Then, the restriction $u_{|\partial \Omega }$ of $u$ to $\partial \Omega$ is defined as the limit of the sequence of restrictions $(u_{n}|_{\partial \Omega })$ . An alternative approach uses the fact that functions in most Sobolev spaces may be discontinuous at some points, but "not too many". For example, in 2d, functions in the space $H^{1}$ (i.e., the space in which the solution lies for many partial differential equations) can be discontinuous at individual points, but not along lines. Consequently, the limit of a function $u\in H^{1}$ may be undefined at individual points on the boundary, but not along the entire boundary, and one can define the trace as that function (or one of those functions) that matches the limit of $u$ "almost everywhere" on the boundary.

## Construction of the trace operator

To rigorously define the notion of restriction to a function in a Sobolev space, let $p\geq 1$ be a real number. Consider the linear operator

$T:C^{1}({\bar {\Omega }})\to L^{p}(\partial \Omega )$ defined on the set of all $C^{1}$ functions on the closure of $\Omega$ with values in the Lp space $L^{p}(\partial \Omega ),$ given by the formula

$Tu=u_{|\partial \Omega }.\,$ The domain of $T$ is a subset of the Sobolev space $W^{1,p}(\Omega ).$ It can be proved that there exists a constant $C$ depending only on $\Omega$ and $p,$ such that

$\|Tu\|_{L^{p}(\partial \Omega )}\leq C\|u\|_{W^{1,p}(\Omega )}$ for all $u$ in $C^{1}({\bar {\Omega }}).$ Then, since the $C^{1}$ functions on ${\bar {\Omega }}$ are dense in $W^{1,p}(\Omega )$ , the operator $T$ admits a continuous extension

$T:W^{1,p}(\Omega )\to L^{p}(\partial \Omega )\,$ defined on the entire space $W^{1,p}(\Omega ).$ $T$ is called the trace operator. The restriction (or trace) $u_{|\partial \Omega }$ of a function $u$ in $W^{1,p}(\Omega )$ is then defined as $Tu.$ This argument can be made more concrete as follows. Given a function $u$ in $W^{1,p}(\Omega ),$ consider a sequence of functions $(u_{n})$ that are $C^{1}$ on ${\bar {\Omega }},$ with $u_{n}$ converging to $u$ in the norm of $W^{1,p}(\Omega ).$ Then, by the above inequality, the sequence $u_{n|\partial \Omega }$ will be convergent in $L^{p}(\partial \Omega ).$ Define

$u_{|\partial \Omega }=\lim _{n\to \infty }u_{n\,|\partial \Omega }.\,$ It can be shown that this definition is independent of the sequence $(u_{n})$ approximating $u.$ ## Application

Consider the problem of solving Poisson's equation with zero boundary conditions:

${\begin{cases}-\Delta u=f{\text{ in }}\Omega \\u_{|\partial \Omega }=0.\end{cases}}$ Here, $f$ is a given continuous function on ${\bar {\Omega }}.$ With the help of the concept of trace, define the subspace $H_{0}^{1}(\Omega )$ to be all functions in the Sobolev space $W^{1,2}(\Omega )$ (this space is also denoted $H^{1}(\Omega )$ ) whose trace is zero. Then, the problem above can be transformed into its weak formulation

Find $u$ in $H_{0}^{1}(\Omega )$ such that
$\int _{\Omega }\!\nabla u(x)\cdot \nabla v(x)\,dx=\int _{\Omega }\!f(x)v(x)\,dx$ for all $v$ in $H_{0}^{1}(\Omega ).$ Using the Lax–Milgram theorem one can then prove that this equation has precisely one solution, which implies that the original equation has precisely one weak solution.

One can employ similar ideas to prove the existence and uniqueness of solutions for more complicated partial differential equations and with other boundary conditions (such as Neumann and Robin), with the notion of trace playing an important role in all such problems.