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.

[Edit]: 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

$$|\Delta u|_{0,\Omega}^2=\int_\Omega (\Delta u)^2 \text{d}x =\int_\Omega \Delta u \Delta u\text{ d}x= \int_\Omega \sum_{i} \partial_{ii} u \sum_j \partial_{jj} u \text{ d}x,$$

which is nothing else than Ciarlet wrote in his book.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.