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First order partial wave eqaution, one boundary and one initial condit

  • Thread starter barefeet
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Homework Statement




Solve

[tex] \frac{\partial{w}}{\partial{t}} + c \frac{\partial{w}}{\partial{x}} =0 \hspace{3 mm} (c>0) [/tex]



for x>0 and t>0 if

[tex] w(x,0) = f(x) [/tex]
[tex] w(0,t) = h(t) [/tex]


Homework Equations



The Attempt at a Solution


I know how to solve for the conditions separately and that would give
[tex] w(x,t) = f(x-ct) [/tex] and
[tex] w(x,t) = h(t-\frac{1}{c}x) [/tex]

but how do you solve it for both? And when does x>0 or t>0 matter?
 
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Answers and Replies

  • #2
LCKurtz
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Homework Statement




Solve

[tex] \frac{\partial{w}}{\partial{t}} + c \frac{\partial{w}}{\partial{x}} =0 \hspace{3 mm} (c>0) [/tex]



for x>0 and t>0 if

[tex] w(x,0) = f(x) [/tex]
[tex] w(0,t) = h(t) [/tex]


Homework Equations



The Attempt at a Solution


I know how to solve for the conditions separately and that would give
[tex] w(x,t) = f(x-ct) [/tex] and
[tex] w(x,t) = h(t-\frac{1}{c}x) [/tex]

but how do you solve it for both? And when does x>0 or t>0 matter?
Let's not call your solution ##w(x,t) = f(x-ct)## because you are using ##f## in the initial condition. So begin by stating your general solution is ##w(x,t) = g(x-ct)## for arbitrary, but as yet unknown, ##g##. Now what happens when you apply your first initial condition to that. Does it tell you what ##g## must be? Then continue...
 
  • #3
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Now I am confused, if I start out with [tex] g(x-ct) [/tex] then applying the first initial condition [tex] w(x,0) = f(x) =g(x-0)=g(x) [/tex]. So [tex]g(x-ct)[/tex] must be [tex]f(x-ct)[/tex], right?
 
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  • #4
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I might have overlooked the fact that I have to give a solution for : [itex] x>0[/itex] and [itex] t>0[/itex]
So [itex] g(x-ct)=f(x-ct) [/itex] is only valid for [itex]x-ct>0[/itex] and [itex]g(t-\frac{1}{c}x)=h(t-\frac{1}{c}x )[/itex] is only valid for [itex]t-\frac{1}{c}x>0[/itex]

I think I have the right visual picture in my head now. So, in the [itex]x,t[/itex] plane my solution only considers the first quadrant of the plane.

And the triangle made by the lines [itex] x=ct [/itex], the [itex]x[/itex]-axis and the line [itex]x=\infty[/itex] is defined by the initial condition at the positive [itex]x[/itex]-axis.
And the triangle made by the lines [itex] x=ct [/itex], the [itex]t[/itex]-axis and the line [itex]t=\infty[/itex] is given by the boundary condition at the positive [itex]t[/itex]-axis.

So the solution would be:
[tex]w(x,t) = f(x-ct) \hspace{3 mm} (x>ct)[/tex]
[tex]w(x,t) = h(t-\frac{1}{c}x) \hspace{3 mm} (x<ct)[/tex]
This would be ok right?
 
Last edited:
  • #5
LCKurtz
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Yes, I think that does it.

Disclaimer: I'm not a PDE expert, so if any such experts disagree, speak up now....:uhh:
 
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