Laplacian inequality in Sobolev space

Question

Is the following assertion true?

For all $\alpha>0$ there exists some $\theta \in H^2(\Omega)\cap H_0^1(\Omega)$ such that $\|\frac{\Delta \theta}{\theta}\|_\infty \le \alpha.$

Thanks!

Answer 1

No, this is not true for all $\alpha>0$. The necessary and sufficient condition is $\alpha\ge \lambda_1$ where $\lambda_1>0$ is the first eigenvalue of the Laplacian with Dirichlet boundary condition. Indeed, if $|\Delta \theta| \le \alpha |\theta|$ in $\Omega$, then integration by parts yields $$\int_\Omega |\nabla \theta|^2 = - \int_\Omega \theta\,\Delta \theta \le \alpha \int_\Omega \theta^2 \le \frac{\alpha}{\lambda_1} \int_\Omega |\nabla \theta|^2 $$ hence $\alpha\ge \lambda_1$.

Conversely, if $\alpha\ge \lambda_1$ then the first eigenfunction of the Dirichlet Laplacian satisfies $|\Delta \theta|=\lambda_1|\theta|\le \alpha|\theta|$.

Answer 2

No. Suppose there is such a $\theta$. Then there is a $\varphi \in L^\infty$ so that $$ -\frac{\Delta \theta}{\theta} = \varphi \text{ a.e. in } \Omega,$$ which is equivalent to the PDE $$ \begin{cases} -\Delta \theta = \varphi \theta & \text{in }\Omega \\ \theta =0 &\text{on } \partial \Omega. \end{cases} $$ Multiply by $\theta$, integrate over $\Omega$, and IBP: $$ \int_\Omega |\nabla \theta|^2 = \int_\Omega \varphi \theta^2. $$ Recall the Poincare' inequality: there exists a constant $C_0>0$ so that $$ C_0 \int_\Omega |u|^2 \le \int_\Omega |\nabla u|^2 \text{ for all } u \in H^1_0(\Omega). $$ Then $$ C_0 \int_\Omega |\theta|^2 \le \int_\Omega |\nabla \theta|^2 = \int_\Omega \varphi \theta^2 \le \alpha \int_\Omega |\theta|^2, $$ and hence $\alpha \ge C_0.$