1

I am trying to apply the Lax Milgram theorem to a System of PDE. Therfore I would like to apply it on the cartesian Product $H_{0}^{1}\times H_{0}^{1}$. This is a Hilbert space with the Norm $\sqrt{\|\cdot\|^{2}_{H_{0}^{1}}+\|\cdot\|^{2}_{H_{0}^{1}}}$. I could solve my Problem if I could use the Norm $\|\cdot\|_{H_{0}^{1}}+\|\cdot\|_{H_{0}^{1}}$. I came across this post uniqueness of system of PDE using Lax Milgram but sadly have some problems in understanding the given answer. Can someone explain how they got this estimation $$ |B((\phi,\phi'),(u,v))| = \left|\int_{\Omega} -\nabla \phi \nabla u - \int_{\Omega} \phi v + \int_{\Omega} \nabla \phi' \nabla v + \int_{\Omega} \phi' u \right| \\ \leq \int_{\Omega} |\nabla \phi| |\nabla u| + \int_{\Omega} |\phi| |v| + \int_{\Omega} |\nabla \phi'| |\nabla v| + \int_{\Omega} |\phi'||u| \\ \leq \|\phi\|\|u\| + \|\phi\|\|v\| + \|\phi'\|\|u\| + \|\phi'\|\|v\| \\ \leq (\|(\phi,\phi')\|_{H_0^1 \times H_0^1}) (\|(u,v)\|_{H_0^1 \times H_0^1}) $$ There it seems to me like they are just using the second Norm, but then Lax Milgrams Theorem, at least the version for Hilbert spaces, can't be applied.

joanna
  • 11
  • The first inequality is simply Cauchy-Schwarz, the last inequality comes from the definition of the norm in $H_0^1\times H_0^1$, say, for example, $|(\phi, \phi')|{H_0^1\times H_0^1} = |\phi|{H_0^1} + |\phi'|_{H_0^1}$. – Chee Han May 13 '22 at 15:37
  • Thank you, I understand that. But we can't apply the theorem of Lax Milgram to that norm or can we? Because $H_{0}^{1}\times H_{0}^{1}$ is not a Hilbert space with that norm. – joanna May 13 '22 at 15:47
  • It actually is: https://math.stackexchange.com/questions/2606158/is-the-cartesian-product-of-two-hilbert-spaces-a-hilbert-space – Chee Han May 13 '22 at 16:01

0 Answers0