Eigenfunction/Fourier Series Relationship
I'm having difficulty formulating a proof detailing that every eigenfunction of $D^2$ is either a constant or of the form $a$cos$(nx)$ + $b$sin($nx$) for some value $n$.
I understand that the Fourier basis B = {1/√2, cos(nx),sin(nx)} consists of eigenfunctions of D$^2$ on the space of piecewise smooth 2π-periodic functions with corresponding eigenvalues of $-n^2$, but I am confused as to how that relates to the given form above. Can anyone help?
$\endgroup$2 Answers
$\begingroup$A function $y:\>{\mathbb R}\to{\mathbb C}$ is an eigenfunction of $D^2$ if there is a $\lambda\in{\mathbb C}$ such that $$y''(x)=\lambda y(x)\tag{1}$$ for all $x\in{\mathbb R}$. Set up the full space $E_\lambda$ of all solutions of $(1)$ for a given $\lambda$, and find out which values of $\lambda$ lead to solutions of period $2\pi$ (since you seem to impose this condition). Finally you can sieve out the real solutions of this kind.
$\endgroup$ $\begingroup$If $\lambda = 0$, then $D^2f=\lambda f$ iff $f=A+ Bx$. So the eigenfunctions with eigenvalue $0$ are $1$ and $x$. For non-zero $\lambda$, $$ D^2f = \lambda f $$ iff $$ (D-\lambda)(D+\lambda)f = 0 \\ D(e^{-\lambda t}(D+\lambda)f)=0 \\ e^{-\lambda t}(D+\lambda)f = A,\;\;\; A=\mbox{constant} \\ (D+\lambda)f = Ae^{\lambda t} \\ e^{\lambda t}(D+\lambda)f = Ae^{2\lambda t} \\ D(e^{\lambda t}f)=Ae^{2\lambda t} \\ e^{\lambda t}f = \frac{1}{2\lambda}Ae^{2\lambda t}+B,\;\;\; B=\mbox{constant} \\ f = \frac{A}{2\lambda}e^{\lambda t}+Be^{-\lambda t}. $$ Because this is the case where $\lambda\ne 0$, then $\frac{A}{2\lambda}$ can be replaced with a constant $A'$.
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