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For example, if we consider the Dirichlet energy $\int\frac 12 |\nabla u|^2$ and the solution space as follows:

$$X=\{u\in W^{1,2}(\Omega) \text{ | } u = 0 \text{ on } \partial\Omega \}$$

, then the test function is $C_0^\infty$. My question is that:

Is the test function related to the boundary condition $u = 0 \text{ on } \partial\Omega$?

If $u = g \text{ on } \partial\Omega$, should the test function be changed?

jakeoung
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1 Answers1

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My understanding of your question is that you want to prove existence of a minimizer of the problem $\min_X I(u),$ where $I(u) = \int_{\Omega}\frac 12 |\nabla u|^2$. One way to approach the problem is by considering variations of the form $I(u + \epsilon\varphi)$, for $\varphi \in C^{\infty}_0(\Omega)$. Now what you should be careful about is that $u + \epsilon \varphi$ needs to belong to $X$ in order to get something that is meaningfully related to the problem we want to solve.

If you are now interested in minimizing over the space $Y = \{u \in W^{1,2} : \operatorname{Tr}(u) = g\}$, you should still consider variations of the form $u + \epsilon\varphi$, with $\varphi \in C^{\infty}_0$. Indeed $\operatorname{Tr(u + \epsilon\varphi)} = g + 0$ and hence $u + \epsilon\varphi \in Y$ if and only if $\operatorname{Tr}(\varphi) = 0.$

Giovanni
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  • Thanks, I didn't notice that $u+\epsilon \phi \in Y$, even if $\phi \notin Y$. Then, the test function doesn't need to be in the solution space, does it? – jakeoung Jan 22 '16 at 03:59
  • correct: all you care about is that $u + \epsilon\varphi$ belongs to the right space, and for this to happen you need the test functions to be $0$ on the boundary. This, together with a density argument, is the reason why $C^{\infty}_0$ is the right space to look at. I hope it is clear now :) – Giovanni Jan 22 '16 at 04:07