I am trying to remember the equation for the laplacian in polar coordinates: $$ \Delta u=u_{rr}+\frac{u_r}{r}+\frac{u_{\theta \theta}}{r^2} $$ without having to derive it everytime and hopefully gain some intuition in the process.

I am also hoping to get some understanding for the formula in $\mathbb{R}^n$ in hyperspherical coordinates: $$ \Delta u=u_{rr}+\frac{(n-1)u_r}{r}+\frac{\Delta_s u}{r^2} $$ where the $\Delta_su$ represents the laplace beltrami operator, which only depends on angular coordinates and which I definitely do not want to derive, ever.

When this was introduced in my PDE class, the professor drew a circle on the board with a tangential vector and talked about a correction term on the order of $r^2$ from traveling along the circle vs this tangential vector. No one really got what he meant, and maybe it is unclear. Anyway, I a mostly looking for some intuition of this kind for why this expression makes sense.

I would be happy to learn about the polar version but definitely would be happy if someone can shed light on the n dimensional one as well. Thanks!

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    $\begingroup$ The "physical" intuition is that the terms must have same dimension (in physics sense , $\dfrac{1}{length^2}$, and the angle is dimensionless. $\endgroup$ – user58697 Apr 23 '17 at 20:28
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    $\begingroup$ @user58697 that's quite helpful for remembering! thanks for the comment $\endgroup$ – qbert Apr 23 '17 at 20:30

This answer explains what I'm guessing the "correction term" was about.

Consider the point $(r,0)$ at the rightmost extent of the circle - this is no loss of generality since you can just rotate your coordinate system. We want a formula for $\Delta u = u_{xx} + u_{yy}$ in terms of radial and angular coordinates. At this point the radial direction is horizontal, so we have $u_{rr} = u_{xx}$, and thus the hard part is relating $u_{yy}$ (the second derivative along the tangent line $x = r$ at this point) to $u_{\theta \theta}$ (the second derivative along the circle at this point).

Since the equation of the circle is $\gamma(\theta) = r (\cos \theta, \sin \theta)$, we can get the desired formula by differentiating the composition $(u\circ \gamma)(\theta) = u(r \cos \theta, r \sin \theta)$.

The first derivative is (using the chain rule)

$$ u_\theta =-u_xr\sin\theta+u_yr\cos\theta.$$ Now, differentiate this again (using the chain and product rules) and substitute $\theta = 0$ (since we are interested in the point $(r,0)$) to get

$$ u_{\theta \theta}|_{\theta =0} = -ru_x + r^2 u_{yy} = -ru_r + r^2u_{yy}.$$

Thus we have $$u_{yy} = \frac 1{r^2} u_{\theta \theta} + \frac 1 r u_r$$ as desired.

  • $\begingroup$ this is excellent and certainly what was meant. Thanks! $\endgroup$ – qbert Apr 24 '17 at 1:42

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