- #1

Zaare

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[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?