# weak solution of PDE and apply Lax-Milgram

Can someone help me for this problem?

Write the weak formulation of: \left\{\begin{align} -\frac{\partial^2u}{\partial x^2}-5\frac{\partial^2u}{\partial y^2}=f\quad&\text{in}\quad\Omega\subset\mathbb R^2\\ u=0 \quad&\text{in}\quad\partial\Omega \end{align}\right. Then apply Lax-Milgram to show existence of weak solution.

Using Green identity I found the following weak formulation:

Find $$u\in H_0^1(\Omega)$$ solution of $$\int_\Omega u_x v_x+5\int_\Omega u_y v_y =\int_\Omega fv,\quad\forall\ v\in H_0^1(\Omega).$$

Is this correct? How can I now apply Lax-Milgram?

Can I transform this equation using gradient or Laplace operator? Thanks

• Please, check if my edits in your post are correct. – Pedro Nov 26 '18 at 0:03
• Also, see this can help. – Pedro Nov 26 '18 at 0:09

How can I now apply Lax-Milgram?

You have to identify what are the bilinear form and the linear functional which correspond to your weak formulation.

In this case, the bilinear form is $$B:H_0^1\times H_0^1\to\mathbb R$$ defined by $$B[u,v]=\int_\Omega u_x v_x+5\int_\Omega u_y v_y.$$

And the linear functional is $$\Lambda:H_0^1\to\mathbb R$$ defined by $$\Lambda[v]=\int_\Omega fv.$$

Now, in order to prove existence of weak solution, you have to show that $$B$$ is continuous and coercive.

• i know this B[u,v]=$∫u_x v_x=(u_x ,v_x )≤||u_x ||_(L^2 ) |v_x ||_(L^2 )$ can I use that $||u_x ||_(L^2) |v_x ||_(L^2 )≤||∇u||_(L^2) ||∇v||_(L^2)$ – Mary Nov 26 '18 at 0:20
• i know this B[u,v]=$∫u_x v_x=(u_x ,v_x )≤||〖u_x ||〗_(L^2 ) |〖v_x ||〗_(L^2 )$ can I use that $||u_x ||_(L^2) |v_x ||_(L^2 )≤||∇u||_(L^2) ||∇v||_(L^2)$ – Mary Nov 26 '18 at 0:26
• Your first equality is not correct because there are other term. – Pedro Nov 26 '18 at 0:27
• Can I say that the norm of a component of the vector is smaller than the norm of the whole vector? (in this case the gradient) ||vx||<=||gradv|| – Mary Nov 26 '18 at 0:34
• @Mary According to the usual notation, $$\|\nabla v\|_{L^2}^2:=\int_\Omega |\nabla v|^2=\int_\Omega (v_x^2+v_y^2)=\|v_x\|_{L^2}^2+\|v_y\|^2_{L^2}\geq \|v_x\|^2_{L^2}.$$ Therefore, $\|v_x\|_{L^2}\leq \|\nabla v\|_{L^2}$. Analogously, $\|v_y\|_{L^2}\leq \|\nabla v\|_{L^2}$. – Pedro Nov 26 '18 at 0:44