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Consider the initial value problem associated to the parabolic equation $u^\epsilon_t + u^\epsilon_x = \epsilon u^\epsilon_{xx}$ and the corresponding hyperbolic problem $u_t + u_x = 0$.

What is a simple way to prove the convergence estimate $$\|u^\epsilon - u\|_{L^2(\mathbb R)} \le C \sqrt{\epsilon t} $$ (for a suitable constant $C$)?

Is this error estimate the optimal one?

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  • $\begingroup$ The convergence depends on the norm you measure the difference in and how many derivatives you are willing to lose on the data. As Scott points out below, the equation is equivalent to $u_t = \varepsilon u_{xx}$ after a change of variables. If you ask for initial data in $H^1$, you should get your estimate with the difference measured in $L^2$. If you ask for initial data in $H^2$, you should get the better estimate $O(\varepsilon t)$ with the difference measured in $L^2$. $\endgroup$
    – sharpend
    Commented Jan 20, 2022 at 23:20

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To remove the transport term, consider the change of variables $v^\epsilon(t,x) := u^\epsilon(t,t+x)$. Notice that $v^\epsilon$ satisfies the heat equation $\partial_t v^\epsilon = \epsilon \partial^2_x v^\epsilon$. Thus $v^\epsilon(t,x)$ is just the convolution of $v$ with the standard heat kernel $\Phi(s,\cdot)$ at time $s=\epsilon t$, which is a function living at length scale $\sqrt{\epsilon t}$. I think you can then get an easy estimate for $v^\epsilon-v$ and translate it back into one for $u^\epsilon-u$. You will presumably need some info involving the smoothness of the initial data.

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