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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?

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

1 Answer 1

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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.

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