# Basic question about a first-order linear equation

I am just learning PDE. My lecture notes say the following: Consider the IVP

$$\begin{cases} u_t + a u_x = 0 \\ u(x,0) = \phi(x) \end{cases}$$

where $$a \in \mathbb{R}$$. Our goal is to reduce this pde to an ode. To do so, find some curve $$x(t)$$ so that $$\frac{d}{dt} u(x(t),t) = a u_x + u_t$$. Obviously $$x(t) = x(0) + at$$ is such a curve which obviously along this curve $$u_t = 0$$.

### Questions:

I understand that they get $$x = x(0) + at$$ basically by the chain rule because

$$\frac{ d }{dt} u(x(t),t) = \frac{ \partial u }{\partial x } \frac{ d x }{d t} + \frac{ \partial u}{\partial t} \frac{d t }{d t} = \frac{ d x }{d t} u_x + u_t$$

and so $$dx/dt = a \implies x = x(0) + at$$. However, the part $$u_t =0$$. Does it follow because we have

$$\frac{ d }{dt} u(x(t),t) = u_t + a u_x = 0$$

?? But this is confusing because dont we have $$u_t = - a u_x$$. Is this what my prof mean?

• I'm not sure what you're asking in regards to $u_{t} = 0$. You're application of the chain rule is correct, and so we know that if $$x' = a$$ then the total derivative of $u$ with respect to $t$, $u'(x(t),t)$, equals $0$. – Mattos Jan 16 at 6:24

What you are showing is that along the lines $$x=x_0+at$$ any solution to your equation is constant. This is used to be able to find the solution out of your initial data. Given any point on the $$(x,t)$$ plane, all you have to do is consider where the line of the given form through this point intersects $$t=0$$; let it be the point $$(x_1,0)$$. You are given the value of your solution at $$(x_1,0)$$, namely $$\phi(x_1)$$, so the solution at your initial point as well as along all that straight line has this very value. That is why this is called transport equation, it just "shifts" your initial data forward (or backwards, depending on $$a$$) in time.