Galerkin method for existence for PDE with nonsymmetric bilinear form 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)
 A: 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
$$
|u|_{L^2V}^2 \leq \|u(0)\|_H^2 + \|u\|_{L^2H}^2 \leq \|u(0)\|_H^2 + T\|u\|_{L^\infty H}^2.
$$
