Has a unique solution for the elliptic equation over special case $\mu>-\lambda_1$

Question

My question is :

Let $\mu=\mathbb{R}\setminus\left\{ 0\right\} $. Consider $$\left(*\right)\begin{cases} -\triangle u+\mu u=f & U\\ u=0 & \partial U. \end{cases} $$ Define $$ \lambda_{1}=\inf_{u\in H_{0}^{1}}\frac{\int_{U}|Du|^{2}\ dx}{\int_{U}u^{2}\ dx} $$ If $\mu>-\lambda_{1}$, want to prove that $\left(*\right)$ has a unique weak solution $u\in H_{0}^{1}\left(U\right)$ for each $f\in > L^{2}\left(U\right)$.

I know the Lax-Milgram theorem and some existence theorem in Chapter 6, Evans.

If I let $$ B[u,v]=\int_{U}-\triangle u\cdot v+\mu uv\ dx, $$ then $$ B[u,v]=\int_{U}Du\cdot Dv\ dx+\mu\int_{U}uv\ dx. $$ To show the assumption on Lax-Milgram theorem, clearly it satisfies symmetry and uniform ellipticity. And readily, I proved that $$ |B[u,v]\leq\alpha\|u\|_{H_{0}^{1}}\|v\|_{H_{0}^{1}}. $$ So, $B[u,u]\geq\beta\|u\|_{H}^{2}$ is left to me.

For this, $$ B[u,u]=\int_{U}|Du|^{2}+\mu\int_{U}u^{2}>\int_{U}|Du|^{2}-\lambda_{1}\int_{U}u^{2}\geq0. $$

This one never gives me anyting.

How can I go further? Or, do I have to take other way?

Answer 1

@TrialAndError's answer is almost correct but it gives an estimate with the wrong norm.

Note that there's nothing to do if $\mu \geq 0$. In the case $-\lambda_1 < \mu < 0$ the last line of your calculations can be slightly improved: decompose $1$ as $\frac{-\mu}{\lambda_1} + \frac{\lambda_1+\mu}{\lambda_1}$ to obtain $$ B[u,u]=\frac{\lambda_1+\mu}{\lambda_1} \int_{U}|Du|^{2} - \mu \left( \frac{1}{\lambda_1} \int_{U}|Du|^{2} - \int_{U}u^{2} \right) \geq \frac{\lambda_1+\mu}{\lambda_1} \int_{U}|Du|^{2}. $$

Answer 2

Using your definition of $\lambda_1$, and assuming $\mu > -\lambda_1$, \begin{align} \langle -\Delta f+\mu f,f\rangle & = \|Df\|^2+\mu\|f\|^2 \\ & \ge \lambda_1\|f\|^2+\mu\|f\|^2 \\ & = (\lambda_1+\mu)\|f\|^2 = \epsilon\|f\|^2, \end{align} where $\epsilon = \mu+\lambda_1 > 0$.