# existence and unicity of solution of an Variational Formulation

Let $\Omega=\mathbb{R}_+$ and let $\Gamma=\{(x,0), x \in \mathbb{R}\}$ We consider the problem $$\begin{cases} & - \mathrm{div} (A \nabla u) + \lambda u = f, \quad (x,y) \in \Omega\\ & u(x,0)=0 \end{cases}$$ with $f \in L^2(\Omega), \lambda \in \mathbb{R}_+^*$ and $A=(a_ij)_ij$ such as there exists $\alpha >0$ then $$\sum_{1 \leq i,j \leq 2} a_{ij} \xi_i \xi_j \geq \alpha ||\xi||^2, \forall \xi = (\xi_1,\xi_2) \in \mathbb{R}^2$$

How we can prouve thant the variational problem associate to the problem, admits a unique solution on $V = H^1_0(\Omega)$? My problem is to prouve coercivity and continuity. The variational formulation associate is $$\displaystyle\int\displaystyle\int_{\Omega} A \nabla u \nabla v + \lambda \displaystyle\int\displaystyle\int_{\Omega} A \nabla u \nabla v + \lambda \displaystyle\int\displaystyle\int_{\Omega} uv = \displaystyle\int\displaystyle\int_{\Omega} fv, \quad \forall v \in H^1_0$$ to prouve the continuité of $a$, we prouve that there exist $M >0$ then $|a(u,v)|\leq M ||u||_V ||v||_V$ and tou prouve the coercivity, we have to prouve the existance of $\alpha >0$ the $a(v,v)\geq \alpha ||v||^2_V$

But my problem is to prouve the coercivity and continuity of $a$

For coercivity: $a(v,v)=\displaystyle\int\displaystyle\int A (\nabla v)^2 + \lambda \displaystyle\int\displaystyle\int_{\Omega} v^2$ We know that $\lambda >0$, so $\lambda \displaystyle\int\displaystyle\int_{\Omega} v^2\geq \displaystyle\int\displaystyle\int_{\Omega}v$ but how about $\displaystyle\int\displaystyle\int_{\Omega} A(\nabla v)^2$? how we can use the hypothesis of $A$?

• Write down the weak formulation, then prove coercivity and continuity of the resulting bilinear form. – daw Mar 26 '14 at 12:17
• i edit my post. – ubun Mar 26 '14 at 12:24
• to prove coercivity: set u=v and the properties of $A$ and $\lambda$ – daw Mar 26 '14 at 12:27
• $a(v,v)=\displaystyle\int\displaystyle\int A (\nabla v)^2 + \lambda \displaystyle\int\displaystyle\int_{\Omega} v^2$ We know that $\lambda >0$, so $\lambda \displaystyle\int\displaystyle\int_{\Omega} \geq \displaystyle\int\displaystyle\int_{\Omega}$ but how about $\displaystyle\int\displaystyle\int_{\Omega} A(\nabla u)^2$? how we can use the hypothesis of $A$? – ubun Mar 26 '14 at 12:48
• It is not $A(\nabla u)^2$, it is $\nabla u^TA\nabla u$. Compare that with the assumption containing $\alpha$. – daw Mar 26 '14 at 13:09