# 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?

• Think about the gradient of $v$. It is orthogonal to the level curves of $v$. Feb 11, 2023 at 0:39
• Redefine $a=-c$, which would mean that $v_t+av_x=0$, which you seem to know how to solve. Feb 11, 2023 at 0:39

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.

$$\dfrac{dv(X(t),t)}{dt}=0$$

$$\dfrac{dv}{dt}(X(t),t)=\dfrac{\partial v}{\partial t}(X(t),t)+\dfrac{d X}{d t}\dfrac{\partial v}{\partial t}(X(t),t)=0$$

$$\dfrac{d X}{d t}=-c$$

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.