# Half Line Wave Equation

• Oct 28th 2007, 05:07 PM
tbyou87
Half Line Wave Equation
Hi,

Solve $\ \ \ \frac{\partial{u}}{\partial{t}} = k(\frac{\partial^2{u}}{\partial{t}^2}) \ \ \ \ \ \ \ u(x,0) = e^{-x} \ \ \ \ \ \ \ \ u(0,t) = 0 \ \ \ \ \$ on the half line $0.

I know you are supposed to put this into the equation $\frac{1}{\sqrt{4\pi*kt}}\int_0^\infty[e^\frac{-(x-y)^2}{4kt} - e^\frac{-(x+y)^2}{4kt}]\Phi(y)dy$ but i'm not sure how to simplify it.

Thanks
• Oct 28th 2007, 08:01 PM
ThePerfectHacker
Quote:

Originally Posted by tbyou87
Hi,

Solve $\ \ \ \frac{\partial{u}}{\partial{t}} = k(\frac{\partial^2{u}}{\partial{t}^2}) \ \ \ \ \ \ \ u(x,0) = e^{-x} \ \ \ \ \ \ \ \ u(0,t) = 0 \ \ \ \ \$ on the half line $0.

I know you are supposed to put this into the equation $\frac{1}{\sqrt{4\pi*kt}}\int_0^\infty[e^\frac{-(x-y)^2}{4kt} - e^\frac{-(x+y)^2}{4kt}]\Phi(y)dy$ but i'm not sure how to simplify it.

Thanks

What is $\Phi (y)$?
• Oct 29th 2007, 10:27 AM
tbyou87
Basically $\Phi(y) = e^{-y}$
So you "just" need to simplify the integrand. I know you need to use the gaussian at some point and maybe competing the square for the exponents.
It thus becomes:
$\frac{1}{\sqrt{4\pi*kt}}\int_0^\infty[e^\frac{-(x-y)^2}{4kt} - e^\frac{-(x+y)^2}{4kt}]e^{-y}dy$
• Oct 29th 2007, 12:12 PM
ThePerfectHacker
Quote:

Originally Posted by tbyou87
Basically $\Phi(y) = e^{-y}$
So you "just" need to simplify the integrand. I know you need to use the gaussian at some point and maybe competing the square for the exponents.
It thus becomes:
$\frac{1}{\sqrt{4\pi*kt}}\int_0^\infty[e^\frac{-(x-y)^2}{4kt} - e^\frac{-(x+y)^2}{4kt}]e^{-y}dy$

Open up the paranthesis for $(x-y)^2$ and $(x+y)^2$ then distribute over by $e^{-y}$. That should do it.