Transport equation $u_t+c(x)u_x=0$

Question

Let c $\in$ $C^1$$\mathbb(R)$. Consider the transport equation.

$u_t+c(x)u_x=0$.

Prove that if $x=f(x)$ is a characteristic curve, then so are all horizontally translated curves $ x=f(t+a)$ for any $a \in R$ and

Why a solution wave $u(t,x)$ cannot change its direction?

Can someone help me with these questions i dont now how to do it. please

Answer 1

What does it mean for $(t,f(t))$ to be a characteristic curve? It means that $$ \frac{d}{dt}u(t,f(t))=u_t+f'(t)u_x=0 $$ Which in your equation imposes that $f$ satisfies $f'(t)=c(f(t))$ for any $t$ in the domain.

Now take $f(t+a)$ and differentiate it to find by the chain rule $$ f'(t+a)\frac{d}{dt}(t+a)=f'(t+a) $$ but we assumed $$ f'(t+a)=c(f(t+a)) $$ so $(t,f(t+a))$ is also a characteristic curve.