I'm trying to read an "An Introduction to Stochastic PDEs" (see http://hairer.org/notes/SPDEs.pdf).
In the second chapter "Some Motivating Examples", they write down the Navier-Stokes equation for incompressible flow in two spatial dimensions as $$ \frac{du}{dt} = \nu\Delta u - (u\cdot\nabla)u - \nabla p + f, \tag{1}\label{eq:a} $$ where it is assumed that $f=0,\;\text{div}\;u=0$ (incompressibility condition) and that "for simplicity, we consider solutions that are periodic in space, so that we view $u$ as a function from $\mathbf{T}^2 × \mathbf{R}_+$ to $\mathbf{R}^2$".
The weak form of equation \ref{eq:a} is then given as $$ \frac{d}{dt}\int_{T^2}|u(x,t)|^2dx = -2\nu\int_{T^2}\text{tr}\;Du(x,t)^*Du(x,t)dx,\tag{2}\label{eq:b} $$ where I believe $f^*$ is the complex conjugate, tr the trace of a matrix and $Du$ the gradient of a vector, i.e. a matrix.
What I don't understand:
Where does the factor 2 in equation \ref{eq:b} come from? I've tried deriving it myself, but I end up with $\int_{T^2}(\Delta u)\cdot u^*dx=-\int_{T^2}\text{tr}\;Du^*\;Du\;dx$.
What I think I understand:
First of all, I believe that div $u =0 \implies(u\cdot\nabla)u=0$ and $\nabla p=0$, although I cannot proof it. Secondly, I can believe that you can distribute the second-order derivative (i.e. Laplacian) from equation \ref{eq:a} to get the two first-order derivatives in equation \ref{eq:b} and the minus using integration-by-parts (although I'm not completely sure how to do this for vector-valued functions). Using integration-by-parts, you get an integral over the boundary of $T^2$. I believe that cancels out due to periodicity of $u$ over $T^2$.
Not a particularly motivating example, if you ask me...
It comes from the left hand side: $$ \frac{1}{2}\frac{d}{dt}\int_{T^2} |u(x,t)|^2\, dx = \int_{T^2} \frac{du(x,t)}{dt}u(x,t)\, dx. $$