Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 .

share|improve this question
2  
When we work with a PDE, we may ask a condition at the boundary of an open set. Unfortunately, it have a measure $0$ and the function of $W^{m,p}(\Omega)$ are defined up sets of measure $0$. The trace operator avoid these complications. –  Davide Giraudo Jun 3 '12 at 10:30
    
@DavideGiraudo : Can you please help me to understand this with an example :) –  Theorem Jun 3 '12 at 10:37
1  
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). –  Davide Giraudo Jun 3 '12 at 10:40

1 Answer 1

up vote 1 down vote accepted

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.