2.6.8 So­lu­tion herm-h

Ques­tion:

A com­plete set of or­tho­nor­mal eigen­func­tions of $\vphantom{0}\raisebox{1.5pt}{$-$}$${\rm d}^2$$\raisebox{.5pt}{$/$}$​${\rm d}{x}^2$ on the in­ter­val 0 $\raisebox{-.3pt}{$\leqslant$}$ $x$ $\raisebox{-.3pt}{$\leqslant$}$ $\pi$ that are zero at the end points is the in­fi­nite set of func­tions

$$\frac{\sin(x)}{\sqrt{\pi /2}}, \frac{\sin(2x)}{\sqrt{\pi /2}... ...in(3x)}{\sqrt{\pi /2}}, \frac{\sin(4x)}{\sqrt{\pi /2}}, \ldots$$

Check that these func­tions are in­deed zero at $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 and $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\pi$, that they are in­deed or­tho­nor­mal, and that they are eigen­func­tions of $\vphantom{0}\raisebox{1.5pt}{$-$}$${\rm d}^2$$\raisebox{.5pt}{$/$}$​${\rm d}{x}^2$ with the pos­i­tive real eigen­val­ues

$$1, 4, 9, 16, \ldots$$

Com­plete­ness is a much more dif­fi­cult thing to prove, but they are. The com­plete­ness proof in the notes cov­ers this case.

An­swer:

Any eigen­func­tion of the above list can be writ­ten in the generic form $\sin(kx)$$\raisebox{.5pt}{$/$}$​$\sqrt{\pi /2}$ where $k$ is a pos­i­tive whole num­ber, in other words where $k$ is a nat­ural num­ber (one of 1, 2, 3, 4, ....) If you show that the stated prop­er­ties are true for this generic form, it means that they are true for every eigen­func­tion.

First check the end points. The graph of the sine func­tion, [1, item 12.22], shows that a sine is zero when­ever its ar­gu­ment is a whole mul­ti­ple of $\pi$. That makes both $\sin(k0)$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\sin(0)$ and $\sin(k\pi)$ zero. So $\sin(kx)$$\raisebox{.5pt}{$/$}$​$\sqrt{\pi /2}$ must be zero at $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 and $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\pi$ too.

Now check that the norm of the eigen­func­tions is one. First find the norm of $\sin(kx)$ by it­self:

$$\vert\vert\sin(kx)\vert\vert = \sqrt{\langle\sin(kx)\vert\sin(kx)\rangle} = \sqrt{\int_0^\pi\sin(kx)^* \sin(kx) { \rm d}x}.$$

Since the sine is real, the com­plex con­ju­gate does not do any­thing, and you get

$$\vert\vert\sin(kx)\vert\vert = \sqrt{\int_0^\pi\sin^2(kx) { \rm d}x} = \sqrt{\pi /2}$$

us­ing [1, item 18.26]. Di­vid­ing this by $\sqrt{\pi /2}$, the norm $\left\vert\left\vert\sin(kx)/\sqrt{\pi /2}\right\vert\right\vert$ be­comes one; every eigen­func­tion is nor­mal­ized.

To ver­ify that $\sin(kx)$$\raisebox{.5pt}{$/$}$​$\sqrt{\pi /2}$ is or­thog­o­nal to every other eigen­func­tion, take the generic other eigen­func­tion to be $\sin(lx)$$\raisebox{.5pt}{$/$}$​$\sqrt{\pi /2}$ with $l$ a nat­ural num­ber dif­fer­ent from $k$. You must then show that the in­ner prod­uct of these two eigen­func­tions is zero. Since the nor­mal­iza­tion con­stants do not make any dif­fer­ence here, you can just show that $\langle\sin(kx)\vert\sin(lx)\rangle$ is zero. You get

$$\langle\sin(kx)\vert\sin(lx)\rangle = \int_0^\pi\sin(kx) \sin(lx) { \rm d}x = 0$$

us­ing again [1, item 18.26].

Ahem. Com­plete­ness. Well, just don't worry about it. There are a heck of a lot of func­tions here. In­fi­nitely many of them, to be pre­cise. Surely, with in­fi­nitely many func­tions, you should be able to ap­prox­i­mate any given func­tion to good ac­cu­racy?

(This state­ment is, of course, de­lib­er­ately lu­di­crous. In fact, if you leave out a sin­gle eigen­func­tion, say the $\sin(x)$ func­tion, the re­main­ing in­fi­nitely many func­tions $\sin(2x)$, $\sin(3x)$, ...can sim­ply not re­pro­duce it by them­selves. The best they can do is be­ing zero and not try to ap­prox­i­mate $\sin(x)$ at all. Still, if you do in­clude $\sin(x)$ in the se­quence, any (rea­son­able) func­tion can be de­scribed ac­cu­rately by a com­bi­na­tion of the sines. It was hard to prove ini­tially; in fact, Fourier in his the­sis did not. The first proof is due to Dirich­let.)