3

Suppose we have a PDE $$\langle u', v \rangle + a(u,v) = 0$$ where $a:V\times V \to \mathbb{R}$ is a bounded symmetric bilinear form, then if $u_0 \in V$ then $u \in L^2(0,T; V)$ with $u' \in L^2(0,T;H)$. This works because we get a bound on $u_n'$ (Galerkin approximation) by writing $a(u_n, u_n')$ as $\frac{1}{2}\frac{d}{dt}a(u_n, u_n)$, integrating and using the boundedness condition.

If $a$ is not symmetric (eg. $a = \int_{\Omega}\nabla \cdot (\textbf{b}u)v + \int_{\Omega}\nabla u \cdot \nabla v$), then we can't write it as a derivative alone. What can we do to get a uniform bound on $u_m'$ in $L^2(0,T;H)$ in this case? I know we can get a bound in $L^2(0,T;V^*)$ but I want it in a stronger space. Thanks.

(posted to https://mathoverflow.net/questions/125736/galerkin-method-for-existence-for-pde-with-nonsymmetric-bilinear-form too)

Community
  • 1
maximumtag
  • 889

1 Answers1

3

Let me derive an a priori estimate for the concrete case you mentioned. Write your equation as $$ \langle u',v\rangle + a(u,v) = b(u,v), $$ where $a(u,v)=\langle \nabla u,\nabla v\rangle$ and $b(u,v)=-(\nabla u,v)$. Let us apply the same procedure you used in the symmetric case. With the notation $|u|_V^2=a(u,u)$, We have $$ a(u,u')=\frac12\frac{\mathrm d}{\mathrm d t} a(u,u), \qquad |b(u,u')|\leq|u|_{V}\|u'\|_{H}, $$ which implies $$ \int_0^T\|u'\|_H^2 + \frac12a(u(T),u(T)) \leq \frac12a(u(0),u(0)) + \int_0^T|u|_{V}\|u'\|_{H}. \qquad\qquad(1) $$ We estimate the last term by the Cauchy-Bunyakowsky-Schwarz inequality: $$ \int_0^T|u|_{V}\|u'\|_{H} \leq |u|_{L^2V} \|u'\|_{L^2H} \leq\frac12\|u'\|_{L^2H}^2+\frac12|u|_{L^2V}^2, $$ where $|u|_{L^2V}^2=\int_0^T|u|_V^2$. Plugging this into (1) and taking into account the positivity of $a$, we have $$ \|u'\|_{L^2H}^2 \leq |u(0)|_{V}^2 + |u|_{L^2V}^2. $$ Deriving a bound for $|u|_{L^2V}$ is standard but let me include here for completeness. We test the original equation with $v=u$, to get $$ \frac12\frac{\mathrm d}{\mathrm d t}\|u\|_H^2+|u|_V^2 = b(u,u) \leq |u|_V\|u\|_H\leq\frac12|u|_V^2+\frac12\|u\|_H^2, \qquad\qquad(2) $$ which implies $$ \frac{\mathrm d}{\mathrm d t}\|u\|_H^2 \leq \|u\|_H^2. $$ Then by Gronwall's inequality we have $$ \|u(t)\|_H^2\leq e^t \|u(0)\|_H^2. $$ In particular, $u\in L^\infty([0,T],H)$ for any finite $T>0$. Now we integrate (2) in time, yielding $$ \frac12\|u(T)\|_H^2 + |u|_{L^2V}^2 \leq \frac12\|u(0)\|_H^2 + \frac12|u|_{L^2V}^2+\frac12\|u\|_{L^2H}^2. $$ This implies $$ |u|_{L^2V}^2 \leq \|u(0)\|_H^2 + \|u\|_{L^2H}^2 \leq \|u(0)\|_H^2 + T\|u\|_{L^\infty H}^2. $$

timur
  • 16,599
  • Thank you. Can you tell me how to do it for a general case? What is the perturbation thing you mentioned? – maximumtag Apr 05 '13 at 16:59
  • 1
    @maximumtag: This answer is what I mean by perturbation. You put the antisymmetric part of you bilinear form $a$ in the right hand side, and try to estimate it with what is to be estimated. This argument will work whenever the "antisymmetric part" (that is $b$) is of lower order than the principal part. So it is ok for differential equations. I leave it to you to figure out the precise assumptions. – timur Apr 05 '13 at 18:56
  • @timur I have posted a question very similar (but different to the one posted here) https://math.stackexchange.com/q/3186878/374907 I would greatly appreciate any comments that you could make. – AzJ Apr 14 '19 at 21:51