Points on ellipsoid with maximum Gaussian curvature/mean curvature.

Question

Find the points on the ellipsoid $$x^2/a^2+y^2/b^2+z^2/c^2=1$$ with maximum Gaussian curvature and mean curvature respectively.

I parametrized it as $(a\sin u\cos v,b\sin u\sin v, c\cos v)$ and managed to compute its Guassian and mean curvature, which are both very messy:$$ K(u,v)=\frac{a^2b^2c^2}{[a^2b^2\cos^2v+c^2(b^2\cos^2u+a^2\sin^2u)\sin^2v]^2 } $$ and $$H(u,v) =\frac{abc[3(a^2+b^2)+2c^2+(a^2+b^2-2c^2)\cos(2v)-2(a^2-b^2)\cos(2u)\sin^2v]}{8[a^2b^2\cos^2v+c^2(b^2\cos^2u+a^2\sin^2u)\sin^2v]^{3/2}}$$

How can I cleverly find points which maximize the complicated $K$ or $H$? (Perhaps in less than 2 pages of computation, by some geometric reasoning?)

Answer 1

Using the implicit definition of this surface may be a better approach. Let $$F(x,y,z) = \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} - 1.$$ The ellipsoid is then defined by $F(x,y,z) = 0$. Then it is known that the Gaussian curvature at a point $(x,y,z)$ is given by $$\kappa = \frac{g H^{ad}g^T}{|g|^4}.$$ and the mean curvature is given by $$h = \frac{gHg^T - \operatorname{trace}(H)|g|^2}{2|g|^3}.$$ where $g = \nabla F$ is the gradient, $H = \nabla g$ is the Hessian matrix, and $H^{ad}$ is its adjoint. Here $$ g = (2x/a^2, 2y/b^2, 2z/c^2), \, H = \begin{pmatrix} 2/a^2 & 0 & 0 \\ 0 & 2/b^2 & 0 \\ 0 & 0 & 2/c^2 \end{pmatrix} , \quad H^{ad} = \begin{pmatrix}4/(b^2c^2) & 0 & 0 \\ 0 & 4/(a^2c^2) & 0 \\ 0 & 0 & 4/(a^2b^2) \end{pmatrix} $$ and therefore $$ |g|^4 = 16\left( (x^2/a^4 + y^2/b^4 + z^2/c^4) \right)^2, \quad gH^{ad} g^T = 16\frac{x^2/a^2 + y^2/b^2 + z^2/c^2}{a^2b^2c^2} = \frac{16}{a^2b^2c^2} \, . $$

The Guassian curvature can now be maximized as follows. We obtain $$ \kappa = \frac{1}{a^2b^2c^2\left( (x^2/a^4 + y^2/b^4 + z^2/c^4) \right)^2} \, . $$ Now it is enough to find the extrema of the denominator, subject to the constraint $F(x,y,z) = 0$ which leads to the Lagrange multiplier equations $$ 2x/a^4 = 2\lambda x/a^2, \quad 2y/b^4 = 2\lambda y/b^2, \quad 2z/c^4 = 2\lambda z/c^2\, . $$ In the generic case where $a,b,c$ are all different, this means $x(\lambda - a^2) = y(\lambda - b^2) = z(\lambda - c^2) = 0$. This is only possible if any two of the unknowns are zero and the third one isn't. Therefore the critical points of $\kappa$ on the ellipsoid are $(\pm a,0,0), (0,\pm b, 0,0), (0,0,\pm c)$ with Gaussian curvatures $ \frac{a^2}{b^2c^2}, \frac{b^2}{a^2c^2},\frac{c^2}{a^2b^2}$. Pick the largest one and that's the maximum.

Answer 2

It is known extreme curvatures occur at ends of major/minor axes for each of $ xy,yz,zx $ sections.This fact should be used to evaluate only at the 6 principal points.

Differentiating $x^2/a^2+y^2/b^2=1$ twice, curvature at end of minor axis = $ b/a^2 $. In the perpendicular direction $ b/c^2$ by symmetry.

Product: Gaussian curvatures $ (b/ac)^2 $ and two other cyclic symmetry curvatures.

Similarly,

Half-sum: Mean curvatures $ \dfrac {b(a^2 +c^2)}{2 a^2 c^2 } $ and two cyclic symmetry curvatures.

Depending upon relative magnitudes of $a,b,c$ maximum curvatures can be selected.

EDIT1:

Although it is known that maximum curvatures occur at ends of major/minor axes due to two-fold symmetry of coordinate axes, it can be shown by differential calculus method for natural equations of conics (1) , (2) below established by me earlier as properties of conics. Looks tedious for what is obvious.

Notation:

Slope of tangent to x-axis $= \phi$, angle between radius vector and ellipse in Newton eccentric focus polar coordinate model $\epsilon= $ eccentricity, $p$ semi-latus rectum. Now,

$$ \frac{\cos \psi}{ \cos \phi}= \epsilon \tag{1}$$ $$ p k_g = \sin ^{3}\psi \tag{2}$$ Differentiating (2)with respect to elliptic arc $s$ $$p k_g^{'} = 3 \sin ^{2}\psi \cos \psi \psi^{'} \tag{3}$$

Liouville's formula for curvature polar coordinates

$$ k_g = \psi^{'} + \frac{\sin \psi}{r} \tag{4} $$

From (3) and (4) $$ p k_g^{'} = 3 \sin \psi ^{3} \cos \psi \left( \frac{\sin ^{2}\psi }{p} -\frac{1}{r} \right) \tag {5} $$

Multiply by $p$ both sides

$$ p^2 k_g^{'} = 3 \sin \psi ^{3} \cos \psi \left( \sin ^2\psi - p/r\right ) \tag{6} $$

$ \psi = \pi/2$ at ends of major axes,$ \, \because $ coefficient and derivative vanish curvature is a maximum.

Newton's canonical ellipse equation

$$ \frac{p}{r} = 1 - \epsilon \cos \theta \tag {7} $$

The quantity in brackets of (6) becomes

$$ -( \epsilon \cos \theta - \cos^2\psi )\tag{8} $$

The situation at ends of minor axis can now be tackled:

$$ \phi = 0, \theta = \psi \tag{9} $$

Quantity in brackets

$$ -\cos \psi ( \epsilon -\cos \psi) \tag {11}$$

From (1) at $\phi =0 , \epsilon = \cos \psi \tag{12}$

From (11) and (12) the derivative vanishes at $y=b$ also, curvature here also is an extremum.

The logic holds for other two $ yz, zx $ plane sections as well.

Sign of next derivative for max/min of curvature can be verified by reader :).

EDIT2:

By virtue of pairwise application Euler's theorem for normal curvature

$$ \kappa_{n1,2}= \kappa_1 \cos^2 \psi_{12} + \kappa_2 \sin^2 \psi_{12} ;\ \kappa_{n1,3}= \kappa_1 \cos^2 \psi_{13} + \kappa_3 \sin^2 \psi_{13}$$

the lines of principal curvature ( simply called curvature in differential geometry) have extrema combinations at these six " cardinal" points.

As far as I am concerned it is demonstrated.

EDIT3

Having said that there a much simpler way to show it. The necessary and sufficient requirement for maximization is $ \dfrac{d \,\kappa_n}{d \psi} = 0 $ and for $ \kappa_{n}= \kappa_1 \cos^2 \psi + \kappa_2 \sin^2 \psi $ the derivative is:

$$\frac{d\, \kappa_n}{d \psi} = -2 (\kappa_1 -\kappa_2 )\cos \psi \sin \psi $$

This is (twice negative) geodesic torsion $\tau_g $ that should vanish at all the 6 points ( $\psi =\pi/2 $ ) at ends of major and minor axes. As usual Mohr's circle of curvature depicts these nicely.