# weak solution of biharmonic equation

Consider $U$ a open and bounded subset of $R^n$, with smooth boundary. A weak solution for the problem :

$$\Delta^2 u = f \ \in \Omega \ and \ u=\frac{\partial u}{\partial\nu} = 0 \text{ in } \partial U$$

where $f \in L^{2}(U)$ is a function(say u )with $u \in H^{2}_{0}(U)$ with

$$\int_{U} \Delta u \Delta v dx = \int_{U}fv dx, \forall \ v \ \in H^{2}_{0}(U)$$.

I believe this definition of weak solution is obtained taking a smooth function $u$ satisfing pointwise the two conditions of the problem and before integrate by parts the first equation and using the second condition of the problem I obtain the integral identity of the weak solution for test functions. Am I right ? I trying to do this calculus but i am getting anywhere...

Someone can help me ?

$$\int_\Omega u\Delta v=-\int_\Omega\nabla u\cdot\nabla v+\int_{\partial\Omega}u\frac{\partial v}{\partial \nu},\ u,v\in H^2(\Omega)$$
$$\int_\Omega v\Delta^2 u=-\int_\Omega\nabla(\Delta u)\cdot\nabla v+\int_{\partial\Omega} v\frac{\partial\Delta u}{\partial\nu},\ \forall\ u\in H^4(\Omega),\ v\in H^1(\Omega)$$
Now, suppose that $u\in H^4(\Omega)$ satisfies pointwise the equation. Multiply the equation by $v\in H^2_0(\Omega)$ and integrate it. Can you conclude from here?
Remark: If $u\in H_0^2(\Omega)$ is a weak solution of yoru problem, then, it can be showed that $u\in H^4(\Omega)$, which implies by du Bois-Raymond theorem that $u$ satisfies your equation pointwise. For a better understanding on the subject, take a look on this book. This book were avaliable on internet for free here, nut the link does not seems to work now. Try it later.