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

## 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

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

## 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:

LCKurtz
Homework Helper
Gold Member

## 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

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

## 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...

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?

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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:
LCKurtz
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