Quick recall of the transport equation: I have the Cauchy problem \begin{cases} u_t(x,t)+ \langle c, \nabla_xu(x,t)\rangle=0, \text{ in }\mathbb{R}^n\times (0,a) \\ u(x,t)=g(x), \text{ on } \mathbb{R}^n. \end{cases} The solution is then $u(x,t)=g(x-ct)$, but I'm trying to verify that by simply putting it in the main equation: $u_t(x,t)=-g_t(x-ct)\cdot c$, which seems to be a vector to me, and since $u_k(x,t)=g_k(x,t)$ we have that $\nabla_x u(x,t)=\nabla_x g(x-ct)$ which is also a vector.
My problem is that not only can't I prove that $c \cdot g_t(x,t)+ \langle c, \nabla_xg(x,t)\rangle=0$, but also I'm adding a vector to a sclar... Please help!
You are confused because there is no such thing as $g_t$: $g$ is a function on $\mathbb R^n$, so
\begin{align} u_t (x, t) &= \frac{d}{dt} g (x-ct) \\ &= \nabla_x g (x-ct) \cdot (-c)\\ &= - \nabla_x g (x-ct)\cdot c \\ &= -\nabla_x u (x, t) \cdot c. \end{align}