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What is the importance of a Trace operator in PDE . Although I have read the Wiki page on this but I am not able to connect it to the aspect of solving PDE's.
Particularly why do we define Trace operator as $T\colon W^{1,p}(u) \to L^p(\partial U)$. Also why do we deal with only exponent $1$ . What is great about operator from $W^{1,p}$ to $L^p$?

Please help me to understand this concept of trace operator. Thank you very much .

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    For example $\Omega$ is the unit disk in $\Bbb R^2$. Then we can have a PDE of the form $-\operatorname{div}(u)=f$ on $\Omega$ and $u=g$ on the boundary. In fact, we have $T(u)=g$ and it's well-defined (but it's not necessarily the restriction of $u$ at the boundary, except when $u$ is regular).2012-06-03

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We take $U$ a smooth open set, regular enough in order to have density of $C^{\infty}_0(\overline U)$ in $W^{1,p}(U)$.

We define $T$ on $C^{\infty}_0(\overline U)$ by taking the restriction of these functions to the boundary (it's well-defined, these one are not equivalence classes of functions). We check that this operator is continuous on this space of functions, then we extend it by continuity to $W^{1,p}(U)$.

We get functions in $L^p(\partial U)$, which is a space of which can be defined using charts. Thanks to that, we can define an element of the Sobolev space on the boundary, even if this one has measure $0$, and it extends the concept of trace for function.