Let $u$ be the weak solution to $$u_t - \Delta u + cu = f$$ $$u(0) = 0$$ with zero Neumann data, on a bounded domain and time interval $[0,T]$. Here $c > 0$.
If $f$ is in $L^\infty$ in time and space, what kind of dependence can I get on the $L^\infty$ norm of $u$ on the constant $c$? Can I obtain something like $$\lVert u(t) \rVert_\infty \leq Me^{-ct}\lVert f \rVert_\infty?$$ ($M$ independent of $c$) Or something else giving me decay.
I'm looking for a sharp estimate giving me decay wrt the constant $c$. Basically I would like the RHS of the above inequality to tend to zero as $c \to \infty$.
I doubt that you can obtain the decay estimate that you want. But you can use this way to obtain one estimate of $\|u\|$. Multiplying both sides by $u$ and then integrating, one has $$ \frac{d}{dt}\|u\|^2+\|\nabla u\|^2+c\|u\|^2=(f,u). $$ By the Poincaré's inequality, $\|\nabla u\|^2\ge \lambda_1\|u\|^2$, where $\lambda_1$ is the first positive eigenvalue of $\Delta$, and by the inequality $2ab\le\frac{1}{\varepsilon}a^2+\varepsilon b^2$, one has $$ \frac{d}{dt}\|u\|^2+\lambda_1\|u\|^2+c\|u\|^2\le\frac{1}{2\varepsilon}\|f\|^2+\frac{1}{2}\varepsilon \|u\|^2. $$ Letting $\frac12\varepsilon=\lambda_1$ in the above inequality gives $$ \frac{d}{dt}\|u\|^2+c\|u\|^2\le\frac{1}{4\lambda_1}\|f\|^2 $$ which is equivalent to $$ (e^{ct}\|u\|^2)'\le\frac{1}{4\lambda_1}e^{ct}\|f\|^2. $$ Integrating both sides from 0 to t gives $$ e^{ct}\|u(t)\|^2\le\frac{1}{4\lambda_1}\int_0^te^{cs}\|f(s)\|^2ds\le\frac{\|f\|^2_\infty}{4\lambda_1c}(e^{ct}-1). $$ Thus one has the following estimate $$ \|u(t)\|^2\le\frac{\|f\|^2_\infty}{4\lambda_1c}(1-e^{-ct}). $$ So $$ \|u(t)\|^2\le\frac{\|f\|^2_\infty}{4\lambda_1c} $$ which gives $$ \lim_{c\to\infty}\|u(t)\|=0. $$