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

[barefeet]

Homework Statement

Solve

\frac{\partial{w}}{\partial{t}} + c \frac{\partial{w}}{\partial{x}} =0 \hspace{3 mm} (c>0)
for x>0 and t>0 if

w(x,0) = f(x)
w(0,t) = h(t)

Homework Equations

The Attempt at a Solution

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

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

 

[LCKurtz]

Homework Statement

Solve

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



for x>0 and t>0 if

w(x,0) = f(x)
w(0,t) = h(t)

Homework Equations

The Attempt at a Solution

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

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...

 

[barefeet]

Now I am confused, if I start out with g(x-ct) then applying the first initial condition w(x,0) = f(x) =g(x-0)=g(x). So g(x-ct) must be f(x-ct), right?

 

[barefeet]

I might have overlooked the fact that I have to give a solution for : x>0 and t>0
So g(x-ct)=f(x-ct) is only valid for x-ct>0 and g(t-\frac{1}{c}x)=h(t-\frac{1}{c}x ) is only valid for t-\frac{1}{c}x>0

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

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

So the solution would be:
w(x,t) = f(x-ct) \hspace{3 mm} (x>ct)
w(x,t) = h(t-\frac{1}{c}x) \hspace{3 mm} (x<ct)
This would be ok right?

 

[LCKurtz]

Yes, I think that does it.

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