Consider the initial-value problem associated to the PDE $u^\epsilon\_t + u^\epsilon\_x - \epsilon u^\epsilon\_{xx} + \epsilon u^\epsilon\_{xxx} = 0$.


To prove that, as $\epsilon \to 0$, the weak solution $u^{\epsilon}$ converges in $L^2$ to the weak solution $u$ of the IVP for $u\_t + u\_x=0$ (with the same initial data), does it suffice to observe that $$\sup\_{t\in (0,T)}\Vert u^\epsilon(t,\cdot)\Vert\_{L^2(\mathbb R)} \le \Vert u(0,\cdot) \Vert\_{L^2}$$
and use weak convergence? Or do we need something more?



Limit of $u^\epsilon_t + u^\epsilon_x - \epsilon u^\epsilon_{xx} + \epsilon u^\epsilon_{xxx} = 0$ as $\epsilon \to 0$
Since your equation is linear, this is sufficient. The convergence holds in the sense of distributions.


Notice the following: the limit, as a solution of the hyperbolic equation $u\_t+u\_x=0$ is unique, being actually $u(t,x)=u(0,x-t)$. In particular
$$\|u(t)\|\_{L^2({\mathbb R})}\equiv\|u(0)\|\_{L^2({\mathbb R})}.$$
Since on the other hand
$$\|u\_\epsilon(t)\|\_{L^2({\mathbb R})}\le\|u(0)\|\_{L^2({\mathbb R})}$$
and $u\_\epsilon\rightharpoonup u$, we conclude that the weak convergence is actually a strong one. More precisely $\|u\_\epsilon(t)-u(t)\|\_{L^2({\mathbb R})}\rightarrow0$ for almost every $t$.


