Equality involving characteristic curve I want to prove some equality involving the characteristic curve associated with the transport equation, namely that the solution given by $u_0(\varphi(-t,x))$ is a solution to the PDE.
Let's begin with some notation :
The characteristic curve $\varphi(t,x)$ associated with the transport equation
\begin{equation}
\left\{ \begin{aligned} 
  u_t(t,x) + c(t,x)u_x(t,x) &=0\\
  u(0,x)&=u_0(x)
\end{aligned} \right.
\end{equation}
is the solution of
\begin{equation}
\left\{ \begin{aligned} 
  \varphi_t(t,x) &= c(t,\varphi(t,x))\\
  \varphi(0,x) &= x
\end{aligned} \right.
\end{equation}
Then we show that $x \mapsto \varphi(t,x)$ is a $C^1$-diffeomorphism with inverse $x \mapsto \varphi(-t,x)$. So far, so good.
The next theorem states that $u_0(\varphi(-t,x))$ is a solution to the PDE. Let's call it $u(t,x)$.
What I have done :
\begin{equation}
\begin{aligned} 
  u_t(t,x) &= u_0'(\varphi(-t,x))\cdot (-\varphi_t(-t,x))) \\
  u_x(t,x) &= u_0'(\varphi(-t,x))\cdot \varphi_x(-t,x))
\end{aligned}
\end{equation}
Then $u_t(t,x)+c(t,x)u_x(t,x) = u_0'(\varphi(-t,x))[-\varphi_t(-t,x)+c(t,x)\varphi_x(-t,x)]$
We know that $\varphi_t(-t,x) = c(-t,\varphi(-t,x))$ but from now on, I cannot prove that $u(t,x)$ is indeed a solution of the PDE.
Have I missed something obvious ?
 A: I always find these computations confusing, but I think your mistake is that $\varphi(-t,x)$ is not the inverse of $\varphi$ (as shown below it solves a different equation). I would instead approach the problem by working with $\varphi^{-1}$ directly as follows.
First, start with $x=\varphi^{-1}(t,\varphi(t,x))$. Then apply $\partial_t$ to this equation to get
$$
0=\left(\frac{\partial (\varphi^{-1})}{\partial t}+\frac{\partial (\varphi^{-1})}{\partial x}\cdot c\right)\mid_{(t,\varphi(x))}.
$$
If we then replace $x$ by $\varphi^{-1}(t,x)$, we have that for all $t,x$
$$
0=\left(\frac{\partial (\varphi^{-1})}{\partial t}+\frac{\partial (\varphi^{-1})}{\partial x}\cdot c\right)\mid_{(t,x)}.
$$
Then, we will define $u(t,x):=u_0(\varphi^{-1}(t,x))$. If we apply the equation, we get
$$
\partial_t u(t,x)+c(t,x)\partial_x u(t,x)=u_0'(\varphi^{-1}(t,x))\cdot \left(\left(\frac{\partial (\varphi^{-1})}{\partial t}+\frac{\partial (\varphi^{-1})}{\partial x}\cdot c\right)\mid_{(t,x)} \right).
$$
By the equation above, the right hand side is zero as desired.
