# $\int_0^T\int_Ω\langle(u_n\cdot\nabla)w,u_n\rangle dxdt\to\int_0^T\int_Ω\langle(u\cdot\nabla)w,u\rangle dxdt$ if $u_n\to u$ weakly and strongly

Good day,

Currently I am reading the book "Navier-Stokes equations: Theory and Numerical analysis" by R. Temam. I am trying to make this question here self-contained so if something unclear then please ask. I am in section 3 where the existence of weak solutions of the homogeneous incompressible Navier-Stokes equations is proved.

The Galerkin approximation is used and it is proved that the subsequence of the solution of the Galerkin equations is converging weakly to the weak solution of the Navier-Stokes equations. In this limit process we pass the limit in some linear terms which is no problem. Further there is a nonlinear term for which Temam states the following Lemma:

Lemma 3.2: If $u_n$ converges to $u$ in $L^2(0,T;V)$ weakly and in $L^2(0,T;H)$ strongly then for any continuous differentiable vector function $w$ $$\int_0^T b(u_n(t),u_n(t),w(t)) dt \longrightarrow \int_0^T b(u(t),u(t),w(t)) dt$$

But first let me introduce some definitions:

• $V=\{ u \in H_0^1 (\Omega): \text{div} u=0\}$
• $H=\{ u \in L^2(\Omega) : \text{div} u = 0, u|_{\partial \Omega}=0 \}$
• $b(u,v,w)=\int_\Omega \langle(u\bullet\nabla)v,w \rangle~dx=\sum_{i,j=1}^d \int_\Omega u_i (D_i v_j) w_j dx$ (where $D_i := \partial / \partial x_i$)
• $V \subset H \equiv H' \subset V'$

And now..

Proof: \begin{align*}\int_0^T b(u_n, u_n, w) ~d t = - \int_0^T b(u_n,w,u_n)~d t &= - \sum_{i,j=1}^d \int_0^T \int_{\Omega} u_{n,i} (D_i w_j) u_{n,j} ~d x d t \\ &\color{red}{\to - \sum_{i,j=1}^d \int_0^T \int_{\Omega} u_i (D_i w_j) u_j ~d x d t} \\ &= -\int_0^T b(u,w,u) ~d t = \int_0^T b(u,u,w) ~d t\end{align*}

My problem is this limit process. If it is possible I would prefer an answer where I could see this step detailed using the definitions of weak and strong convergence.

I know the following:

$u_n \to u$ in $L^2(0,T;V)$ weakly $\Longleftrightarrow f(u_n) \to f(u)$ for all $f \in L^2(0,T;V')$

$u_n \to u$ in $L^2(0,T;H)$ strongly $\Longleftrightarrow$ $$||u_n-u||^2_{L^2(0,T;H)}=\int_0^T ||u_n(t)-u(t)||_{L^2(\Omega)}^2 ~dt=\int_0^T \int_\Omega |u_n(x,t)-u(x,t)|^2 ~dxdt \to 0$$

But I don't know how to connect these two concepts to get the wished convergence. Maybe there is a similar concept in these Bochner spaces to the multiplication of weak and strong converging sequences. For example I know that if $v_k \to v$ in $L^2$ and $w_k \overset{w}{\to} w$ in $L^2$ then $\int v_k w_k \to \int v w$.

It would be great if someone could help me with this.

Marvin

• What is the regularity of $w$? Which convergences do you have for $u_n$? – gerw Oct 25 '16 at 6:39
• @gerw This should be contained in the stated lemma i.e. $u_n \to u$ in $L^2(0,T;V)$ weakly and in $L^2(0,T;H)$ strongly. Further $w$ is a continuous differentiable vector function. Or did you mean something else? – Fritz Oct 25 '16 at 8:04
• My fault. I overlooked the assumptions in the lemma... – gerw Oct 25 '16 at 9:27

You have $u_{n,i} \to u_i$ in $L^2((0,T) \times \Omega)$. Hence, $u_{n,i} \, u_{n,j} \to u_i \, u_j$ in $L^1((0,T)\times\Omega)$. Since $D_i w_j \in L^\infty((0,T) \times \Omega)$, you obtain the convergence of this integral.
I do not see, however, where the weak convergence of $u$ comes into play...
• Thanks a lot for your answer. Just be clear, the convergence follows since (let's denote $f_n := u_{n,i} u_{n,j} \to u_i u_j =: f$) $$||D_i w_j f_n - D_j w_j f||_1 \leq ||D_i w_j||_\infty ||f_n -f ||_1 \to 0$$ and $D_i w_j$ is bounded since it is continuous on a bounded set. Then I understood it. This weak convergence of $u$ was used in the according theorem, maybe Temam copied by accident in this lemma as an assumption. – Fritz Oct 25 '16 at 12:37