Burgers' equation / Perturbation theory

Question

Let the Burgers' equation: $$u_t+u u_x=\epsilon u_{xx} ,\qquad x\in\mathbb{R}, \; t>0.$$ With the initial condition: $u(x,0)=1_{\{x<0\}}(x).$

Studying it in the theory of perturbation:

I did not know how to find the shock wave $s(t)$

Answer 1

Let us seek traveling wave solutions of the form $u(x,t) = v(\xi)$ with $\xi = x-ct$. According to the chain rule, we have $u_x = v'$ and $u_t = -cv'$. Thus, the PDE leads to the ODE $$ -cv' + vv' = \epsilon v'' \, , $$ with the boundary conditions $v(-\infty) = 1$ and $v(+\infty) = 0$. Integrating once w.r.t. $\xi$ gives Riccati's equation $$ \tfrac12 v^2 - c v + c - \tfrac12 = \epsilon v' \, , $$ which bounded solutions are of the form $$ v(\xi) = \frac{1}{1 + A e^{\xi/(2\epsilon)}} \, , $$ where $A>0$ is a constant, and where we have set $c=1/2$ to satisfy the boundary conditions. These traveling wave solutions are displayed below for several values of $\epsilon$:

While the solution converges towards a discontinuity with infinite slope as $\epsilon \to 0$, the speed of the wave $c=1/2$ remains unchanged. Therefore, the shock trajectory in the $x$-$t$ plane is $s(t) = t/2$ for $\epsilon \to 0$. However, note that the traveling wave isn't a Riemann solution as it doesn't satisfy the discontinuous boundary condition. Nevertheless, it does satisfy the original problem in the limit of vanishing viscosity $\epsilon\to 0$.

With classical perturbation theory, we make the power series Ansatz $$ u = u^0+\epsilon u^1 +\dots $$ and inject it in the PDE. By separating powers of $\epsilon$, the following system is obtained: $$ \begin{aligned} u^0_t + u^0 u^0_x &= 0\\ u^1_t + u^0 u^1_x &= u^0_{xx} - u^0_x u_1 \\ &\dots \end{aligned} $$ However, solving at 0th order amounts to solving the inviscid Burgers' equation (i.e, the case $\epsilon = 0$), which doesn't make much sense in the present strong form... So the traveling wave approach is more appropriate here.