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Periodic solution to DE

  • Thread starter Zaare
  • Start date
  • #1
54
0
I don't know how to use the extra condition given in the this problem:


[tex]f^{\prime\prime}+\lambda f = 0[/tex], [tex]f=f\left(r\right)[/tex]
[tex]f\left(r\right) = f\left(r+\pi\right)[/tex]

For [tex]\lambda = 0[/tex], the solution is some constant.
For other [tex]\lambda[/tex], I put [tex]\lambda=k^2[/tex], and get


[tex]f^{\prime\prime}+k^2 f = 0[/tex],

which has the solution


[tex]f\left(r\right)=A\cos\left(kr\right)+B\sin\left(kr\right).[/tex]

The condition now gives


[tex]A\cos\left(kr\right)+B\sin\left(kr\right)=
A\cos\left(kr+k\pi\right)+B\sin\left(kr+k\pi\right).[/tex]

How can I use this to further specify the solution?
 

Answers and Replies

  • #2
LeonhardEuler
Gold Member
858
1
Well, you know that [tex]\cos{x} + \sin{x} = \cos{(x+2n\pi)} + \sin{(x+2n\pi)}[/tex] for all x only when n is an integer. What does that tell you about k, and therefore [itex]\lambda[/itex]?
 
  • #3
54
0
Oh, right, so k must be an even integer.
Thank you. :smile:
 

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