6
$\begingroup$

I thought that the definition of a harmonic function f such that

$\nabla^2f=0$

In one dimension, doesn't this mean a function who's second derivative is zero? IE

$\frac{d^2f}{dy^2}=0$

However, the sine nor cosine function's second derivative does not equal zero, but in many textbooks they are refereed to as harmonic functions.

What's going on here?

$\endgroup$
  • 1
    $\begingroup$ The might be referring to the complex plane. ​ ​ $\endgroup$ – user57159 Dec 9 '16 at 3:58
  • 1
    $\begingroup$ in many textbooks they are refereed to as harmonic functions Is it the trig functions themselves which are called harmonic, or maybe some related context like harmonic motion? $\endgroup$ – dxiv Dec 9 '16 at 4:03
  • $\begingroup$ What textbook specifically? In what context? $\endgroup$ – Robert Israel Dec 9 '16 at 4:55
5
$\begingroup$

Laplace's equation in polar coordinates takes the form $$\frac{\partial^2 f}{\partial r^2} + \frac{1}{r} \frac{\partial f}{\partial r} + \frac{1}{r^2} \frac{\partial^2 f}{\partial \theta^2} = 0.$$ A first attempt at solving this is to assume that $f$ has the form $f(r,\theta) = u(r)v(\theta)$ for some functions $u$ and $v$; by adding together functions of this form, you hope to be able to deal with enough possible boundary conditions.

This procedure (separation of variables) leads to $$u''(r)v(\theta) + \frac{1}{r}u'(r) v(\theta) + \frac{1}{r^2}u(r)v''(\theta) = 0.$$ When $f$ is sufficiently nonzero, you can rearrange this to get $$\frac{r^2 u''(r) + ru'(r)}{u(r)} = -\frac{v''(\theta)}{v(\theta)} = \lambda,$$ a constant independent of $r$ and $\theta$; in particular, depending on the sign of $\lambda$, $v(\theta)$ will be some combination of $\sin$ and $\cos$.

$\sin$ and $\cos$ were already known as "harmonics" due to their appearance in the harmonic osciallator; the more general term "spherical harmonics" came later, and because of this connection all solutions of Laplace's equation eventually became known as "harmonic" functions.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.