Limit of $u^\epsilon_t + u^\epsilon_x - \epsilon u^\epsilon_{xx} + \epsilon u^\epsilon_{xxx} = 0$ as $\epsilon \to 0$

Question

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?

Answer 1

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 $t$.