We have the problem: $$\begin{cases}div(\alpha\nabla u)+\gamma u=f & \text{in } \Omega\\u=0 & \text{in } \Gamma_1\subset\partial \Omega\\\alpha \nabla u\cdot \vec{n}=0 & \text{in } \Gamma_2 \subset \partial \Omega\end{cases}$$

So the variational formulation for the problem is: find $u \in V:=\{v\in H^1(\Omega): v\vert_{\Gamma_1}=0\}$ such that

$$\int_{\Omega}\alpha \nabla u \nabla v+\gamma uv \;dx=\int_{\Omega}fv \;dx$$ for all $v\in V$.

We suppose that $\alpha>a_0>0$ is a funtion such that $\alpha \in L^{\infty}(\Omega)$, $\gamma\in L^{\infty}(\Omega)$ is a positive function and $f \in L^2(\Omega)$. I want to prove that the bilinear form associated to the variational form is bicontinous and coercive, in order to apply Lax Milgram theorem. These things are going to be proved in $V$. Is that correct?

If the answer is yes, how can I prove these? I suppose that the bilinear form associated is $$a(u,v)=\int_{\Omega}\alpha \nabla u \nabla v+\gamma uv \;dx$$

and then I define a norm in $V$ as $\vert v \vert_{V}=\lVert \nabla v \rVert_{L^2(\Omega)}$. How can I prove bicontinuity from this new norm?

And finally, Is there a simplier way to demonstrate Lax Milgram theorem?

Thanks a lot.

  • $\begingroup$ I just want to point out that Lax-Milgram is only applicable on Hilbert space. $\endgroup$ – Chee Han Feb 9 '17 at 8:00
  • $\begingroup$ $V$ is a HS with the inner product associated with the norm defined $\endgroup$ – Diego Vargas Feb 9 '17 at 12:02
  • $\begingroup$ I suppose that is easier to work with the induced inner product induced by $H^1(\Omega)$ over $V$. $\endgroup$ – Diego Vargas Feb 9 '17 at 12:59
  • $\begingroup$ Usually it is, but in this particular case, it is better to work with the standard $H^1$ norm on the space $V$. $\endgroup$ – Chee Han Feb 9 '17 at 17:09
  • $\begingroup$ Typically, some uniform ellipticity condition of the differential operator is required if you want to approach this problem with Lax-Milgram. The boundedness is fairly straightforward, but coercivity is difficult here since you do not have any lower bound for the function $a$. You should be able to find what you're looking for in any elliptic PDEs textbook, but see for example this excellent note by Hunter math.ucdavis.edu/~hunter/pdes/ch4.pdf $\endgroup$ – Chee Han Feb 9 '17 at 17:15

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