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

1. Oct 13, 2013

### barefeet

1. The problem statement, all variables and given/known data

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

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

2. Relevant equations

3. 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?

Last edited: Oct 13, 2013
2. Oct 13, 2013

### LCKurtz

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. Oct 13, 2013

### 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?

Last edited: Oct 13, 2013
4. Oct 13, 2013

### 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?

Last edited: Oct 13, 2013
5. Oct 13, 2013

### LCKurtz

Yes, I think that does it.

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