I am working through a a priori estimates related to parabolic equations, and I've encountered a conceptual challenge regarding the use of the Poincaré inequality. The context and steps I'm referring to are takenfrom Alfio Quarteroni's book, "Numerical Models for Differential Problems," specifically from page 126.
We start from this equation: $$ \int_{\Omega} \frac{\partial u(t)}{\partial t} u(t) d \Omega+a(u(t), u(t))=\int_{\Omega} f(t) u(t) d \Omega \quad \forall t>0 . $$
Breaking down the individual terms, we find:
$$ \int_{\Omega} \frac{\partial u(t)}{\partial t} u(t) d \Omega=\frac{1}{2} \frac{\partial}{\partial t} \int_{\Omega}|u(t)|^2 d \Omega=\frac{1}{2} \frac{\partial}{\partial t}\|u(t)\|_{L^2(\Omega)}^2 . $$
Assuming the bilinear form is coercive (with coercivity constant $( \alpha )$, we have:
$$ a(u(t), u(t)) \geq \alpha\|u(t)\|_V^2, $$
and by the Cauchy-Schwarz inequality:
$$ (f(t), u(t)) \leq\|f(t)\|_{\mathrm{L}^2(\Omega)}\|u(t)\|_{\mathrm{L}^2(\Omega)} . $$
Additionally, we often use Young's inequality:
$$ \forall a, b \in \mathbb{R}, \quad a b \leq \varepsilon a^2+\frac{1}{4 \varepsilon} b^2 \quad \forall \varepsilon>0, $$
which comes from:
$$ \left(\sqrt{\varepsilon} a-\frac{1}{2 \sqrt{\varepsilon}} b\right)^2 \geq 0 . $$
Using the first Poincaré inequality and Young's inequality, we arrive at:
$$ \begin{aligned} \frac{1}{2} \frac{d}{d r}\|u(t)\|_{L^2(\Omega)}^2+\alpha\|\nabla u(t)\|_{L^2(\Omega)}^2 & \leq\|f(t)\|_{L^2(\Omega)}\|u(t)\|_{L^2(\Omega)} \\ & \leq \frac{C_{\Omega}^2}{2 a}\|f(t)\|_{L^2(\Omega)}^2+\frac{\alpha}{2}\|\nabla u(t)\|_{L^2(\Omega)}^2 . \end{aligned} $$
My question is regarding the application of the Poincaré inequality in this context. It appears to be used on both the right-hand side and left-hand side of the final inequality. How is it possible to ensure the validity of this inequality in the last step, given its simultaneous application on both sides? Is there an underlying assumption or step that I'm missing which justifies this application?
Any insights or clarifications on this matter would be greatly appreciated. Thank you!
I believe the computation with some added steps should be \begin{aligned} \frac{1}{2} \frac{d}{d t}\|u(t)\|_{L^2(\Omega)}^2+\alpha\|\nabla u(t)\|_{L^2(\Omega)}^2 & \overset{\color{red}{\fbox{A}} }=\frac{1}{2} \frac{d}{d t}\|u(t)\|_{L^2(\Omega)}^2+\alpha\|u(t)\|_{V}^2\\ & \le \frac{1}{2} \frac{d}{d t}\|u(t)\|_{L^2(\Omega)}^2+a(u(t),u(t))\\ & = \int_{\Omega} f(t) u(t) d\Omega \\ &\leq\|f(t)\|_{L^2(\Omega)}\|u(t)\|_{L^2(\Omega)} \\ &\leq \frac{C_\Omega^2}{2\alpha}\|f(t)\|_{L^2(\Omega)}^2 + \frac1{4C_\Omega^2/(2\alpha)}\|u(t)\|_{L^2(\Omega)}^2 \\ & \overset{\color{blue}{\fbox{B}} }\leq \frac{C_{\Omega}^2}{2 \alpha}\|f(t)\|_{L^2(\Omega)}^2+\frac{\alpha}{2}\|\nabla u(t)\|_{L^2(\Omega)}^2 . \end{aligned}
$\color{blue}{\fbox{B}} $ is Poincaré, but $V=H^1_0$, so for $\color{red}{\fbox{A}}$ we take $\|v\|_V := \|\nabla v\|_{L^2(\Omega)}$ which is equivalent to the $H^1$ norm on that space (that is, both upper and lower bounds).