Good afternoon all.
Using the greens first identity I am finding weak formulation. Can someone check I did this correct with below mentioned greens first identity
$$\int_U (\nabla \cdot \nabla \mathbf{u}) \cdot \, \vec{\omega} \, dU = \int_{\partial U} \frac{\partial {\mathbf{u}}}{\partial n} \cdot \vec{\omega}\, dS – \int_U \nabla \mathbf{u} : \nabla \vec{\omega} \, dU $$
My answer is :
Best Answer
You have the equation: $$-9u_{xx}-5u_{yy}=f$$ Now multiplying by a test function $v\in H_0^1(\Omega)$ both sides and integrating over $\Omega$ you get. $$-9\int\limits_{\Omega}{\frac{\partial}{\partial x}\left(\frac{\partial u}{\partial x}\right)vdxdy}-5\int\limits_{\Omega}{\frac{\partial}{\partial y}\left(\frac{\partial u}{\partial y}\right)vdxdy}=\int\limits_{\Omega}{fvdxdy}$$ $$9\int\limits_{\Omega}{\frac{\partial u}{\partial x}\frac{\partial v}{\partial x}dxdy}+5\int\limits_{\Omega}{\frac{\partial u}{\partial y}\frac{\partial v}{\partial y}dxdy}=\int\limits_{\Omega}{fvdxdy}\quad\quad (*)$$ which is the weak formulation that you should get. We used $$-\int\limits_{\Omega}{\frac{\partial}{\partial x}\left(\frac{\partial u}{\partial x}\right)vdxdy}=-\int\limits_{\Omega}{\frac{\partial}{\partial x}\left(\frac{\partial u}{\partial x}v\right)dxdy}+\int\limits_{\Omega}{\frac{\partial u}{\partial x}\frac{\partial v}{\partial x}dxdy}=-\underbrace{\int\limits_{\partial\Omega}{\frac{\partial u}{\partial x}vn_1ds}}_{=0}+\int\limits_{\Omega}{\frac{\partial u}{\partial x}\frac{\partial v}{\partial x}dxdy}=\int\limits_{\Omega}{\frac{\partial u}{\partial x}\frac{\partial v}{\partial x}dxdy}$$ Here, the tool that we used is the divergence theorem (with which is actually derived the Green's first identity). Note that the surface integral is $0$ because $v$ is zero on $\partial \Omega$ (to be more speciffic, it is zero in the trace sense). Or if you want to avoid the use of the divergence formula in the weak sense, then you can start by multiplying by a smooth test function $v\in C_0^\infty(\Omega)$, derive the equation $(*)$ above, and then use density argument to show that $(*)$ is actually satisfied also for all $v\in H_0^1(\Omega)$. The integral with the $y$-derivative is done in similar way, just having in mind that in the surface integral there will be the second component $n_2$ of the normal vector $n$ to $\partial\Omega$ .