Solving PDEs: Deriving Wave Equation from u(x ± ct)

  • Thread starter Thread starter KenBakerMN
  • Start date Start date
  • Tags Tags
    Pde
KenBakerMN
Messages
11
Reaction score
2
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.
 
Physics news on Phys.org
KenBakerMN said:
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)
Typo #1? g?
to

∂^2u/∂t^2 = (C^2)(∂^2u/∂t^2)
Typo # 2? Only one independent variable?

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.

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.
 
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.
 
KenBakerMN said:
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]
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.
 
Okay, I get it now. I needed to carry out the second PDEs one more step and "chain rule" it. Thanks for your help.
 
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...
Back
Top