Uniformly elliptic implies existence of unique weak solution

Solution 1:

You have the correct bilinear form. You should then say $u \in H^1_0(\Omega)$ is a weak solution if $B[u,v]=(u,v)_{L^2(\Omega)}$ for all $v\in H^1_0(\Omega)$ which I expect is what you had in mind.

Your calculation shows that if $c \geqslant -\mu$ for your chosen $\mu>0$ then there is a unique weak solution by Lax-Milgram. But you have $c\geqslant 0 >-\mu$ regardless of what $\mu$ is which proves your claim. (Maybe your issue is that you have a small typo - after you set $\mu = \frac \theta {2C}$ you go on to state 'if $c \geqslant \frac \theta {2C}$ ...' when you mean 'if $c \geqslant -\frac \theta {2C}$ ...')

However, you don't need to introduce the $\mu$ at all since you are assuming $c\geqslant 0$. Indeed,

$$ \theta \int_\Omega \vert Du \vert^2 \, dx \leqslant \sum_{i,j}a^{ij}D_{i}u D_ju = B[u,u]-\int_\Omega c u^2 \, dx \leqslant B[u,u]. $$ Therefore, $$\| u \|_{H^1(\Omega)}^2 \leqslant C \| Du \|_{L^2(\Omega)}^2\leqslant CB[u,u]. $$ ($C>0$ changes each line).