I would like to calculate the weak formulation of the following equation:
$$ \alpha v(x)+\beta\Delta^2 v(x)=\gamma $$
Which brings me to this formula:
$$ \int_\Omega \left(\alpha v(x) + \beta \Delta^2 v(x)\right)w(x)dx=\int_\Omega \gamma w(x)dx $$
The solution should only contain the first derivative of $v(x)$, but i always have problems to change the second summand:
$$ ...+\beta\int_\Omega\Delta\Delta v(x)w(x)dx=.... $$
Maybe someone knows how to do this:)?
Thank you very much.
Best regards
Kevin
To summarize for 2nd order problems: You first multiply your strong formulation with a test function $v\in C^\infty_0(\Omega)$, where $\Omega\subset\mathbb{R}^d$ is your domain (usually it should have a Lipschitz boundary). Then you apply Green's theorem to this result to obtain the weak formulation.
Green's theorem states (take care of the assumptions that have to be made)
$$\int_\Omega \Delta u v\,dx = -\int_\Omega \nabla u\nabla v + \int_{\partial\Omega} v\nabla u\cdot\nu$$ where $\nu$ is the outer normal of $\Omega$
However, this formula does not apply directly to your problem in this form, as you have the 4th derivative (bilaplacian, $\Delta\Delta$) inside
Green's formula can be modified to apply also to fourth order problems (including $\Delta^2 u$):
$$\int_\Omega\Delta^2 uv\,dx = \int_\Omega\Delta u\Delta v\,dx + \int_{\partial\Omega} \partial_\nu\Delta u v\,dS - \int_{\partial\Omega} \Delta u\partial_\nu v\,dS$$
This is the formula you have to use for the $\Delta^2 u$ part (it holds for all $u\in H^4(\Omega), v\in H^2(\Omega)$.
At this point you just need to use your boundary conditions to get the weak formulation