- #1
Zaare
- 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?
[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?