# Existence and uniqueness of weak solutions to the homogeneous biharmonic equation.

I am currently trying to prove boundedness and coercivity of the bilinear operator $a \colon H^2_0(\Omega) \times H^2_0(\Omega) \to \mathbb{R}$ defined by

$$a(u, v) = \int_\Omega \Delta u \Delta v \, \mathrm{d}x$$ for all $u, v \in H^2_0(\Omega)$. This operator props up in the derivation of the weak formulation of the biharmonic equation. By showing boundedness and coercivity, the Lax-Milgram theorem establishes existence and uniqueness of the weak solution. The $H^2$-norm and seminorm is defined as

$$\|v\|_{2, \Omega} = \int_\Omega \sum_{|\alpha|\leq 2} |D^\alpha v|^2 \, \mathrm{d}x, \\ |v|_{2, \Omega} = \int_\Omega \sum_{|\alpha| = 2} |D^\alpha v|^2 \, \mathrm{d}x.$$

I've had some ideas that have not been fruitful for the boundedness:

1. Show that $a$ defines an inner product on $H^2_0(\Omega)$ and then using the Cauchy-Schwartz yields boundedness. However, the positive definiteness of $a$ is probably not satisfied.

2. Using the Hölder-inequality with $p = q = 2$ gives me an inequality, but with the wrong power.

: Denote by $\langle u, v \rangle_{L^2}$ the $L^2$ inner product. Then we see that $a(u, v) = \langle \Delta u, \Delta v \rangle_{L^2}$. Consequently, we can apply Cauchy-Schwartz yielding $$a(u, v) \leq |\Delta u|_{0, \Omega} |\Delta v|_{0, \Omega} \leq \|u\|_{2, \Omega} \|v\|_{2, \Omega}$$ and consequently, the boundedness is proven.

For the coercivity, I have a hunch that I need to use the Poincare-Friedrichs inequality, however I am not sure how this inequality, which holds in $H^1_0(\Omega)$ relates to $H^2_0(\Omega)$.

I've looked at the proof of both boundedness and coercivity in The finite element method for elliptic problems by Ciarlet (1978), and his proof relies on the fact that $$|\Delta v|_{0, \Omega} = |v|_{2, \Omega}$$ and he uses this to deduce that the seminorm $v \mapsto |\Delta v|_{0, \Omega}$ is a norm equivalent to the full norm $\|\cdot\|_{2, \Omega}$. The reason I do not understand his reasoning is due to how he defines his norms, which he does as follows:

$$|\Delta v|_{0, \Omega}^2 = \int_{\Omega} \sum_{i=1}^n (\frac{\partial^2 v}{\partial x_i^2})^2 + \sum_{i \neq j} \frac{\partial^2 v}{\partial x_i \partial x_i} \frac{\partial^2 v}{\partial x_j \partial x_j} \mathrm{d}x.$$

Based on the definition used above however, I would have assumed that this norm would be equal to:

$$|\Delta v|_{0, \Omega}^2 = \int_{\Omega} \sum_{i=1}^n \frac{\partial^2 v}{\partial x_i^2} \, \mathrm{d}x,$$ i.e., only the diagonal terms.

If someone could provide me with some help, or some nudges in the general direction, that would be greatly appreciated. Thanks in advance!

We have $\Delta u = \sum_i \partial_{ii} u$ but that does not imply $|\Delta u|_0^2=\int \sum_i \partial_{ii} u \text{ d}x$ as you assumed. You should try to write out the integral. Indeed, we have
$$|\Delta u|_{0,\Omega}^2=\int_\Omega (\Delta u)^2 \text{d}x =\int_\Omega \Delta u \Delta u\text{ d}x= \int_\Omega \sum_{i} \partial_{ii} u \sum_j \partial_{jj} u \text{ d}x,$$