Wave equation and odd extension with constant IC

[xdrgnh]

Solve U_xx=U_tt with c=1.
Dirchlet boundary conditions
U(x,0)=1 for 5<x<7
U(x,0)=0 for everywhere else
U_t(x,0)=0

I know that by taken an odd extension I can get rid of the boundary condition and then solve the initial value problem using the d'alembert solution and only care for x>0 Graphically I know it will be square wave that break off in opposite directions and has amplitude of 1/2. At the origin the square wave will die while on the other side it will go on forever. What is tripping me up are the constant initial conditions. When I plug them in I get for x>t U(x,t)=1 and for t>x I get 0. This doesn't sound right to me. Any help will be appreciated.

 

[LCKurtz]

Solve U_xx=U_tt with c=1.
Dirchlet boundary conditions
U(x,0)=1 for 5<x<7
U(x,0)=0 for everywhere else
U_t(x,0)=0

I know that by taken an odd extension I can get rid of the boundary condition and then solve the initial value problem using the d'alembert solution and only care for x>0 Graphically I know it will be square wave that break off in opposite directions and has amplitude of 1/2. At the origin the square wave will die while on the other side it will go on forever.

Are you sure about it dying at the origin? Won't it reflect back?

What is tripping me up are the constant initial conditions. When I plug them in I get for x>t U(x,t)=1 and for t>x I get 0. This doesn't sound right to me. Any help will be appreciated.

I'm not sure what you are saying here. The initial displacement isn't a constant. Have you neglected to tell us about a boundary condition such as $u(0,t)=0$?

 

[xdrgnh]

The initial displacement for at time =0 is 1 between x=5 and x=7. well I'm doing an odd extension so their is wave of equal amplitude but opposite direction on the x=-5 and x=-7 side and when they meet at x=0 they cancel each other keeping the boundary condition U(0,t) zero.

 

[LCKurtz]

The initial displacement for at time =0 is 1 between x=5 and x=7. well I'm doing an odd extension so their is wave of equal amplitude but opposite direction on the x=-5 and x=-7 side and when they meet at x=0 they cancel each other keeping the boundary condition U(0,t) zero.

That's true as they pass at the origin. But you are imagining an infinite string and that wave from the left keeps going to the right after it passes through the origin.

 

[xdrgnh]

Okay But What About The Initial Condition. They Are Constants Not Function Of X. How Do I Represent Them At A Function Of Time,? Sorry For Caps My Phone Is Weird.

 

[LCKurtz]

Okay But What About The Initial Condition. They Are Constants Not Function Of X. How Do I Represent Them At A Function Of Time,? Sorry For Caps My Phone Is Weird.

$u(x,0) = 1,~5<x<7$ and $0$ elsewhere is not a constant function. For example, it is $1$ when $x=6$ and $0$ when $x=4$. Its value depends on $x$.

 

[xdrgnh]

So A Step Function?

 

[LCKurtz]

No, it isn't a step function. It is a simple function which is a linear combination of step functions. What difference does it make what you call it? Call it $f(x)$ if you want to. I don't understand what your real question is here.

 

[xdrgnh]

So for my solution I can just represent f(x) as a piecewise function and then using the d'alembert get .5f(x+t) in one direction and .5f(x-t) in the other direction?

 

[LCKurtz]

So for my solution I can just represent f(x) as a piecewise function and then using the d'alembert get .5f(x+t) in one direction and .5f(x-t) in the other direction?

Yes. Or if you have been using the standard unit step function $u(x)$ in your class, you could express your $f(x)$ in terms of it.