We want to find $u \in H^{1}_{0}(\Omega)$, such that $$\int_\Omega \lambda \nabla u\nabla v dx =\int_\Omega fvdx \text{ } \forall v \in H^{1}_{0}(\Omega) $$ We then define the bilinear form $a(u,v) :=\int_\Omega \lambda \nabla u\nabla v dx $, which is bounded: $$ |a(u,v)| = |\lambda| |\int_\Omega \nabla u\nabla v dx| \leq ||u||_{H^{1}_{0}(\Omega)} ||v||_{H^{1}_{0}(\Omega)}$$

We can now say, that the functional $v \rightarrow a(u,v)$ lives in the dualspace $H^{1}_{0}(\Omega)^*$. By our first equation, we get $$\int_\Omega fvdx = a(u,v)\text{ }\forall v \in H^{1}_{0}(\Omega)$$ Since $H^{1}_{0}(\Omega)$ is a Hilbert Space, we can identify it with its dual, so we can now say: $$f(v) = a(u,v)$$ Is that correct? I One thing that confuses me, is that my book does the derivation of the weak formulation of the stationary heat conduction in almost the same way, but uses a testfunction $v \in C^{\infty}_0$ instead of $H^{1}_{0}(\Omega)$. What makes the difference in that? Bonusquestion: I always assumed that the 0 in the subscript of $H^{1}_{0}(\Omega)$ claims compact support. But does it really??



The $0$ subscript does not mean compact support. The definition is

$$H^1_0(\Omega) = \text{Closure}(C^\infty_0(\Omega)),$$

where closure is under the $H^1$ norm. Thus, $C^\infty_0(\Omega)$ is dense in $H^1_0(\Omega)$, which means any $u\in H^1_0(\Omega)$ can be approximated arbitrarily well (in the $H^1$ norm) by a smooth function with compact support, even though $u$ may not itself have compact support.

Since $C^\infty_0(\Omega)$ is dense in $H^1_0(\Omega)$, it does not matter whether your test functions belong to $C^\infty_0(\Omega)$ or $H^1_0(\Omega)$, the notion of weak solution is the same. Indeed, by density you can show that if the weak formulation held for just $C^\infty_0$ test functions, then it would hold for $H^1_0$ functions as well (by approximation).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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