Proposition 1.1, Section 1.2 of Majda and Bertozzi reads as follows:
Let $v,p$ be a solution to the Euler or Navier Stokes Euquations. Then the following tranformations also yield solutions: For any $\textbf{c} \in \mathbb{R}^n$, $v^c = v(x-\mathbf{c}t,t)+\mathbf{c}$ and $p^c = p(x-\mathbf{c}t,t)$
I'm having trouble showing this. As an example I thought to consider one space dimension (instead of 3) the burgers equations: $\frac{\partial v}{\partial t} + v\frac{\partial v}{\partial x} = 0$.
For starters, I'm having trouble understanding whether I need to show. Let x' = x - ct and t' = t then either I want to show:
- $$\frac{\partial v(x,t)}{\partial t} + v(x,t)\frac{\partial v(x,t)}{\partial x} = 0 \implies \frac{\partial v^c(x',t')}{\partial t} + v^c(x',t')\frac{\partial v^c(x',t')}{\partial x} = 0$$
or
- $$\frac{\partial v(x,t)}{\partial t} + v(x,t)\frac{\partial v(x,t)}{\partial x} = 0 \implies \frac{\partial v^c(x',t')}{\partial t'} + v^c(x',t')\frac{\partial v^c(x',t')}{\partial x'} = 0$$
Either way, I'm getting stuck with the computation. This is related to the question posed here.
Let me use a slightly different notation.
Given $$u_t+uu_\xi=0 $$ for $u(\xi,t)$ show that $U(x,t)=u(x-c t,t)+c$ satisfies $$ U_t+UU_x=0. $$ Now it is the chain rule: $$ U_t=u_\xi (-c)+u_t,\quad U_x=u_\xi, $$ whence $$ U_t+UU_x=u_\xi(-c)+u_t+(u+c)u_\xi=u_t+uu_\xi=0 $$ as required.