- #1

Eclair_de_XII

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## Homework Statement

"The solution ##u(x,t)=f(x+ct)+g(x-ct)## solves the PDE, ##u_{tt}=c^2u_{xx}##. By graphing the solution ##u(x,t)=f(x+ct)## on the ##ux##-plane, please show that as ##t## increases, the graph shifts to the left at a velocity ##c##. Conversely, show that for ##u(x,t)=g(x-ct)##, the graph shifts to the right as ##t## increases."

## Homework Equations

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## The Attempt at a Solution

The first thing I was going to do was to take the partial derivative of ##u## with respect to ##t##. Then I can observe that ##u_t(x,t)=c⋅f'(x+ct)## and that ##u_t=c⋅u_x##. I honestly don't understand how I would apply this derivative to the concept of shifting the graph of ##u(x,t)## to the left or to the right. I have a factor of ##c## there. So I guess the instantaneous rate at which ##u(x,t)## changes with respect to ##t## is ##c## times greater than the instantaneous rate at which ##u(x,t)## changes with respect to ##x##? I don't know how to start this problem, honestly.

I mean I could try setting ##u(x,t)=f(x+ct)=α## and graphing it. From there I could then set ##u(x,0)=α##. Then I guess I could set ##u(x,1)=α## and by changing the boundary conditions each time, I could perhaps show that as ##t## increases, this arbitrary point of ##u(x,t)## moves to the left for each increment of ##t##. And I guess I could use the partial derivative of ##u(x,t)##, ##c⋅f'(x+ct)## to show that ##f(x+ct)## changes at an instantaneous rate of ##c##. The problem is, I don't exactly understand how ##f'(x+ct)## would correspond to a shift in the graph of ##u## in the ##ux##-plane...

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