Wave equation

2
The next four problems concern the wave equation u = c u on the
tt XX
finite space interval 0 X L with boundry conditions u ( 0 , t) = 0 = u ( L , t).
To get ready, read section 4.4 of Haberman, followed by your lecture
notes on the wave equation in 1 space dimension. In your solution,
you may quote any material from pages 142-144 of Haberman.

Problem - 1
Suppose that the initial conditions are u ( X , 0) = f ( X) and u ( X , 0) = 0, where f
t
is defined on [0, L]. Let fodd denote the function on [-L, L] obtained by

extending f oddly to [-L, 0). Also, let fodd be the periodic extension of fodd

to all X. At any point where f has
odd
a jump discontinuity, define it's y
value there to be the midpoint of the jump. Show that the solution u ( X , t)
can be manipulated into the following form:

1
u ( X , t) = f ( X + c t) + f ( X - c t)
2 odd odd

You may look at Ex. 4.4.7 for hints. You MUST pinpoint when the
oddness of the extension, and the Fourier convergence theorem, get used.

___________________________________________________________

Ex. 4.4.7) If a vibrating string satisfying 4.4.1 - 4.4.3 is initially at rest,
g ( X) = 0, show that:
1
u ( X , t) = ( F ( X - c t) + F ( X + c t) )
2


Where F ( X) is the odd periodic extension of f ( X). Hints:

N X
1) For all X, F ( X) =
A sin
N L
N = 1
1
2) sin ( a) cos ( b ) = ( sin ( a + b) + sin ( a - b ) )
2


4.4.1 = u = c2 u 4.4.2 = ( u ( 0 , t) = 0 = u ( L , t) ) 4.4.3 = u ( X , 0) = f ( X)
tt XX
u ( X , 0) = g ( X)
t
___________________________________________________________


You MUST pinpoint when the oddness of the extension, and the
Fourier convergence theorem, get used.
Test IV Page - 3 of 18
Equation 4.4.11:


A sin N X cos N c t + B sin N X sin N c t
u ( X , t) =
N


L

L N
L

L
N = 1

We also have the identity:
1
sin ( A) cos ( B) = ( sin ( A + B) sin ( A - B) )
2


1
My problem is not showing that u ( X , t) = f ( X + c t) + f ( X - c t ) .
2 odd odd
All I need to do that, is just plugin the identity above and realize that
the the odd periodic extension of any function f(x) is:


N X
f ( X) =
A sin
N L
N = 1
this is by the forier convergence theorem....

My problem however, is that I don't know where or why to invoke the
"oddness" described above in the description of the problem. I belive I
invoke it around the time that the B sub N term drops off... but I'm not
exactly sure how?..

If there's not enough info to do the problem PLEASE DON'T GUESS!!!
Just let me know and I'll provide more info.




Awnser here

Test IV Page - 4 of 18