Solve the first order PDE with the initial condition. Show that $v(x,t)=f(x+ct)$ Solve the first order PDE with the initial condition
$$\frac{\partial}{\partial t} v(x,t)-c\frac{\partial}{\partial x}v(x,t)=0\\
v(x,0)=f(x)\\.$$
Show that $v(x,t)=f(x+ct)$.
We have been working on solutions to wave equations in class.  I thought this might be a transport equation, but the transport equation has a plus sign not a minus between the first derivatives. Can anyone help me?
 A: The principle of solving this equation is to write it as follow  :
Taking the full derivative of $v$ over time
$$ \dfrac{dv}{dt}=\dfrac{\partial v}{\partial t}+\dfrac{\partial x}{\partial t}\dfrac{\partial v}{\partial t} $$
Then the methods used is called the methods of caracteristics
Which means we search for $X(\cdot t)$ such as $v$ is constant along the line $(X(t),t)$ i.e.
$$\dfrac{dv(X(t),t)}{dt}=0$$
Which leads to :
$$ \dfrac{dv}{dt}(X(t),t)=\dfrac{\partial v}{\partial t}(X(t),t)+\dfrac{d X}{d t}\dfrac{\partial v}{\partial t}(X(t),t)=0 $$
Then from your equation :
$$ \dfrac{d X}{d t}=-c$$
For any real $x_0$ initial condition :
$$ X_{x_0}(\cdot t) = x_0 -ct $$
Note that the orbit of the $X_{x0}$ functions make a partioning of $\mathbb{R}^2$ with $x_0$ variating in $\mathbb{R}$.
Note
$v(X(t),t)$ is constant along a line where $\exists x_0 \, X(\cdot t)=X_{x_0}(\cdot t)=x_0 -ct $
So
$$ v(X_{x_0}(t),t)=v(X_{x_0}(0),0)=f(x)$$
Part 2
Now take a point $(x,t)$ where you want $v(x,t)$
Because the $X_{x0}$ are defining a partition :
$$ \exists x_o \in \mathbb{R}\ , x=X_{x0}(t)=x_0-ct $$
Then $$ v(x,t)=v(X_{x0}(t),t)=_{\text{constant along}}v(x_0,0)=v(x+ct,0)=f(x+ct)$$
Understanding the transport
So here you see that you're transported backwards due to the $-c$. It is still a transport equation but the direction of transport is changed.
It is shown by the $$\dfrac{dX}{dt}=-c$$
which notes how will evolve the position in time, here going backwards.
