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Consider the following example: $$ - \Delta u = f \mbox{ in } \Omega, $$ $$ u = 0 \mbox{ on } \Gamma, $$ Here $\Gamma$ is boundary of $\Omega$. To produce weak formulation we multiply by arbitrary $v$ from $H^1(\Omega)$, integrate over $\Omega$ and apply integration by parts: $$ \int_{\Omega} \nabla u \nabla v dx - \int_{\Gamma} \frac{\partial u}{\partial n}vds = \int_{\Omega} f v dx. $$ Because we don't have information about $\partial u /\partial n$ on $\Gamma$ we restrict $v$ to lie in $V = \{ v \in H^1(\Omega): v|_\Gamma = 0 \}$. This new constructed space may not be a complete, so we need to complete it with respect to Sobolev norm (Otherwise we won't be able to apply Lax–Milgram theorem and prove existence of weak solution). But after this procedure we may have functions in $V$ which violate boundary conditions $v|_\Gamma = 0$. Why after completion of $V$ with respect to Sobolev norm we won't run into functions $v|_\Gamma \neq 0$? All textbooks I've consulted skip this moment, probably because it is obvious.

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  • $\begingroup$ By completion of $V$, you mean adh$(V)$ right ? Then if $v_n \to v$ in $H^1$ with $v_n = 0$ on $\Gamma$, could we have $v=0$ on $\Gamma$ ? $\endgroup$ – anonymus Jul 16 '16 at 13:56
  • $\begingroup$ @anonymus I'm not sure. Consider "hat" functions which are nonzero only on [a,a+b], and we have a region of interest [0,1]. Then we build series of hats [1/n, 1/n+b] which are in $H^1$ and yet converges to hat function [0,b] which is nonzero in $0$. $\endgroup$ – Moonwalker Jul 16 '16 at 14:01
  • $\begingroup$ Well, i'm not expert of multivariate pde analysis. But in dimension $1$, if $v_n \to v$ in $H^1_0(a,b)$, then $v_n \to v$ uniformly on $(a,b)$. $\endgroup$ – anonymus Jul 16 '16 at 14:03
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So first of all, $H^1$ does not really have a restriction map, even to interior points, much less boundary points (except in one dimension, where there is a continuous embedding into $C^0$). The better way of thinking about this is to start out with smooth test functions, then look at the equation that you get and identify the solution space and the test function space appropriately. The spaces that you get in these two cases are denoted by $H^1_0$. This is essentially what we mean by $\{ f \in H^1 : \left. f \right |_\Gamma = 0 \}$. More formally it can be identified as either the completion of $C^\infty_c$ in $H^1$ or as the kernel of the map $T : H^1(\Omega) \to L^2(\Gamma)$ which is the continuous extension of the restriction map. This map is called the trace, and its existence requires a little bit of regularity about the boundary $\Gamma$. (As I recall Lipschitz is enough, but references like Evans tend to assume $C^1$ or at least piecewise $C^1$ to simplify the proof.)

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  • $\begingroup$ And by $C_c^\infty$ you mean $C_0^\infty$? $\endgroup$ – Moonwalker Jul 16 '16 at 15:37
  • $\begingroup$ @Moonwalker Depends on the notation you're using. I write $C^\infty_c$ for compact support inside the domain and $C^\infty_0$ for going to zero as you approach the boundary of the domain (or infinity, in an unbounded domain). $\endgroup$ – Ian Jul 16 '16 at 15:39
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It is usual to assume some regularity property for $\Gamma$. For example, in Evan's PDE, if $\Omega$ is bounded and has a $C^1$ boundary, then there is (bounded) trace operator

$$T : H^1(\Omega) \to L^2(\Gamma)$$

so that $T u = u|_\Gamma$ if $u \in H^1(\Omega) \cap C(\overline\Omega)$. It is also proved that

$$\{ H^1(\Omega) : Tu = 0\} = H^1_0(\Omega),$$

thus $\{ H^1(\Omega) : Tu = 0\}$ is already a complete space.

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  • $\begingroup$ I'm interested in details of such profs. Could you recommend some textbook? I'm interested in heat conduction equation with very good boundaries. $\endgroup$ – Moonwalker Jul 16 '16 at 15:13
  • $\begingroup$ If you have good boundary, Evans' book is good enough. Take a look at the chapter on Sobolev spaces. $\endgroup$ – user99914 Jul 16 '16 at 15:52

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