# Simple PDE Question

1. Feb 11, 2013

### KenBakerMN

It's been a little too long since I've has to do this. Can someone please remind me, how do you get from:

∂u/∂t = C(∂u/∂g)

to

∂^2u/∂t^2 = (C^2)(∂^2u/∂t^2)

The notation here is a little clumsy, but I'm just taking the second PDE of each side. How does the C^2 get there? Seems like it ought to be C but I can't put my finger on a proof either way.

By the way, this comes up in a derivation of the wave equation:

∂^2u/∂x^2 = (1/c^2)(∂^2u/∂t^2)

starting from

u(x,t) = u(x ± ct)

I'm sure someone out there knows this. Thanks for your help.

2. Feb 11, 2013

### LCKurtz

Typo #1? g???
Typo # 2? Only one independent variable?

So you are trying to show u(x,t) satisfies the wave equation? If so, that should be straightforward. Show us what you have done so far, starting at the beginning.

3. Feb 11, 2013

### KenBakerMN

LCKurtz, thanks for the response. Alright, here goes.

Starting from a general function u(x - ct), define g=x - ct. [1]

So we have ∂u/∂x = (∂u/∂g)(∂g/∂x) and ∂u/∂t = (∂u/∂g)(∂g/∂t) . [2]

The PDEs from [1] are: ∂g/∂x = 1, and ∂g/∂t = - c . [3]

So from [2] and [3], ∂u/∂x = ∂u/∂g . [4]

The second PDE from [4] is ∂2u/∂x2 = ∂2u/∂g2, is that correct? [5]

Also from [2] and [3], ∂u/∂t = -c(∂u/∂g) . [6]

Now, to get from [5] and [6] to the wave equation ∂2u/∂x2 = (1/c2)(∂2u/∂t2)
seems to require, from [6], ∂2u/∂t2 = (c2)(∂2u/∂g2)

It's that last step I don't quite get, unless - which is by no means unlikely - I'm making an error someplace else. Seems like the c2 should just be c .

The context here is I'm an electrical engineer trying to understand the physics or ultrasound transmission through a waveguide. This derivation comes from "Basics of Biomedical Ultrasound for Engineers", Axhari, 2010.

4. Feb 11, 2013

### LCKurtz

I think the notation is giving you problems. You are trying to show that for any differentiable function $g$, the function $u(x,t)=g(x\pm ct)$ satisfies the wave equation $u_{tt}=c^2u_{xx}$. Note that $g$ has a single argument. So when you differentiate both sides of with respect to $x$ you would get $u_x(x,t) = g'(x\pm ct)\cdot 1$, and when you differentiate both sides with respect to $t$ you get $u_t(x,t) = g'(x\pm ct)\cdot (\pm c)$ with the sign on the $c$ agreeing. What you want to do next is differentiate these two equations again to get $u_{xx}(x,t)$ and $u_{tt}(x,t)$. Then just look at those two expressions and see if they work. I think you will see where the $c^2$ comes from.

5. Feb 13, 2013

### KenBakerMN

Okay, I get it now. I needed to carry out the second PDEs one more step and "chain rule" it. Thanks for your help.