Change of Variables and Chain Rule

[Stalker_VT]

Homework Statement

I am trying to solve the transport PDE using a change of variables and the chain rule, and my problem seems to be with the chain rule. The PDE is:

$\frac{\partial u}{\partial t}$+c$\frac{\partial u}{\partial x}$ = 0 ......(1)

The change of variables (change of reference frame) is:

$\xi$ = x - ct......(2)

From this we know that

u(t,x) = v(t, x - ct) = v(t, $\xi$)......(3)

The Attempt at a Solution

Taking the total derivative of both sides of (3)

$\frac{d}{dt}$ [u(t,x) = v(t,$\xi$)]

using chain rule yields

$\frac{\partial u}{\partial t}$ = $\frac{\partial v}{\partial t}$ - c$\frac{\partial u}{\partial x}$......(4)

I think the next step is to PROVE that

$\frac{\partial u}{\partial x}$ = $\frac{\partial v}{\partial \xi}$.....(5)

and then substitute (4) into (5) to get

$\frac{\partial v}{\partial t}$ = 0

but i am not sure how to do this...Any help Greatly Appreciated

Thanks!

 

[LCKurtz]

and then substitute (4) into (5) to get

$\frac{\partial v}{\partial t}$ = 0

but i am not sure how to do this...Any help Greatly Appreciated

Thanks!

If I understand correctly, I think you are basically done. You have $$
\frac{\partial v(t,\xi)}{\partial t}=0$$Take the anti-partial derivative of this with respect to t. Here your constant of integration is an arbitrary function of the other variable:$$
v(t,\xi) = f(\xi) = f(x-ct)$$so you have$$
u(x,t)=v(t,\xi)=f(\xi)=f(x-ct)$$Unless, of course, you already understood that and I misunderstand what your question is.