A drumhead is a disk. A disk is round. Solving a PDE on it in x,y coordinates fights the domain's own shape, at every step. Separation of variables works cleanly on a disk or a ball only once the Laplacian itself is rewritten. It needs the coordinate system that matches the domain. That's exactly the machinery this section is reviewing, not re-deriving.
Ordinary (x,y) coordinates happen to have scale factors both equal to 1. A unit step in x is always a unit of real distance. So the plain Laplacian f_{xx}+f_{yy} never has to correct for that. Polar coordinates don't have it so easy. They use the angle \theta, the Greek letter theta, to measure how far around a circle a point sits. A unit step in that angle covers more real distance far from the start point than close to it. Near the center, a small turn barely moves you at all. Far out, that same small turn sweeps a much bigger arc. The extra \tfrac1r and \tfrac1{r^2} terms in the polar Laplacian are exactly that correction, built right in.
Take an orthogonal curvilinear coordinate system (u,v,w). In it, the coordinate curves cross at right angles everywhere. Each coordinate also carries its own local scale factor, h_u,h_v,h_w. A scale factor converts a small coordinate step into actual arc length. The Laplacian in such a system is \nabla^2f = \frac{1}{h_uh_vh_w}\left[\frac{\partial}{\partial u}\!\left(\frac{h_vh_w}{h_u}\frac{\partial f}{\partial u}\right)+\cdots\right], one term per coordinate. This reduces to the familiar f_{rr}+\tfrac1rf_r+\tfrac1{r^2}f_{\theta\theta} in polar coordinates. That happens once h_r=1 and h_\theta=r are substituted in.
Polar coordinates: right angles everywhere
Drag the highlighted cell. Its two edge lengths ARE the scale factors h_r and h_theta at that point -- watch h_theta grow as the point moves outward, while h_r stays fixed at 1.
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Worked example
Scale factors for polar coordinates
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1
Find the scale factors h_r,h_\theta for x=r\cos\theta, y=r\sin\theta.
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2
h_r = \left|\left(\dfrac{\partial x}{\partial r},\dfrac{\partial y}{\partial r}\right)\right| = |(\cos\theta,\sin\theta)| = 1. h_\theta = \left|\left(\dfrac{\partial x}{\partial\theta},\dfrac{\partial y}{\partial\theta}\right)\right| = |(-r\sin\theta,r\cos\theta)| = r.
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3\boxed{h_r=1,\qquad h_\theta = r}
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Worked example
A harmonic function's Laplacian, checked in polar form
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1
Confirm f=r^2\cos2\theta satisfies \nabla^2f=0 using the polar Laplacian.
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2
f_r=2r\cos2\theta, f_{rr}=2\cos2\theta; f_{\theta\theta}=-4r^2\cos2\theta. \nabla^2f = 2\cos2\theta+\frac1r(2r\cos2\theta)+\frac1{r^2}(-4r^2\cos2\theta)
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3= 2\cos2\theta+2\cos2\theta-4\cos2\theta = \boxed{0}
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Where this goes next
Bessel and Legendre functions
Splitting any of this course's three main PDEs on a disk or a ball starts the same way. You swap in the curvilinear Laplacian reviewed here, in place of \nabla^2. That swap matters a lot. It's what turns the part of the equation that depends on r into Bessel's or Legendre's equation, in ch.09, "Special functions in curvilinear domains." Without it, that part would turn into something with sines and cosines instead.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Advanced Multivariable Calculus
What defines an orthogonal curvilinear coordinate system?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Scale factors can vary freely (like h_theta=r in polar) in an orthogonal system -- equaling 1 everywhere is specific to Cartesian, not required for orthogonality.
Countless other orthogonal systems exist (parabolic, elliptic, and more) -- cylindrical and spherical are just the two most commonly named.
Polar coordinate curves are circles and rays, not straight lines in general, yet the system is still orthogonal.
Why
Orthogonality is specifically about the ANGLE between the two coordinate curve families at their crossings being 90 degrees -- it says nothing about the scale factors' values or the curves' shapes.
Focus on the ANGLE at which the two coordinate curve families cross each other -- not on the scale factors' sizes or the curves' shapes.
From Advanced Multivariable Calculus
For the map x=u,\ y=uv, compute \partial\mathbf r/\partial u \cdot \partial\mathbf r/\partial v at (u,v)=(2,1) to check orthogonality.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
dr/du = (1,v), dr/dv=(0,u). Dot product = (1)(0) + (v)(u) = uv. At (2,1): uv=2, nonzero -- this system is NOT orthogonal (a nonzero dot product means the basis vectors aren't perpendicular).
Correct answer
Also accepted:
Why
Differentiate r(u,v)=(x,y)=(u,uv) componentwise with respect to u and, separately, with respect to v -- that gives the two tangent (basis) vectors to dot together.
dr/du = (1, v); dr/dv = (0, u). Dot them, then substitute u=2, v=1.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Advanced Multivariable Calculus
How is the curvilinear Laplacian formula actually derived, according to this section?
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No new argument is needed -- it's pure composition of two formulas already derived earlier in the chapter.
The curl of ANY gradient is always the zero vector (a fact from chapter 6/7) -- that's unrelated to the Laplacian, which is div of grad, not curl of grad.
The formula is derived directly in general curvilinear coordinates, with Cartesian as just the trivial h_i=1 special case.
Why
The Laplacian is defined as div(grad f) -- substituting the gradient formula's curvilinear components directly into the divergence formula (in place of F_i) produces the Laplacian formula with no additional argument needed.
Start from the Laplacian's own definition, nabla^2 f = nabla . (nabla f) -- which two formulas, both already derived earlier in this chapter, does that combine?
From Advanced Multivariable Calculus
Find \nabla^2 f for f = r^2 in polar coordinates (using h_r=1, h_\theta=r).
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
df/dr = 2r. r * df/dr = 2r^2. d/dr(2r^2) = 4r. Divided by r: 4r/r = 4, constant everywhere -- matching the Cartesian check, since the Laplacian of x^2+y^2 is 2+2=4.
Correct answer
Also accepted:
Why
Differentiate f=r^2 with respect to r first, to get df/dr -- that's the piece the Laplacian formula's r * df/dr term needs.
Laplacian = (1/r) d/dr(r * df/dr), since f has no theta dependence.