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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!

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    $\begingroup$ What is the $V$ norm? Norm of the gradient? I take it you have no issue with the right side application of Poincare? Can you explicitly write out what the Poincare application is on the left side? $\endgroup$ Dec 26, 2023 at 9:46
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    $\begingroup$ @CalvinKhor Thanks a lot! I think your questions clarified everything. I will try to update the question as soon I finish all the calculations and write everything in Latex. $\endgroup$
    – alfa
    Dec 26, 2023 at 10:14
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    $\begingroup$ That was the intent :) $\endgroup$ Dec 26, 2023 at 10:15

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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).

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    $\begingroup$ Thanks a lot! I will meditate a bit on this and then, if everything is 100% clear as it seems to me now, accept it as an answer. There is a typo in the first derivative, r instead of t. Thanks again $\endgroup$
    – alfa
    Dec 26, 2023 at 10:39

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