# Level curves on ellipsoid

Let $a,b,c>0$ with $a\leq b\leq c$. Let $E$ be the ellipsoid determined by $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}=1$$

Is there a function $f:E\rightarrow \mathbb{R}$ such that $f\in C^{\infty}(E)$ and all level curves of $f$ are circles?

Thanks

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My guess is that you could find a plane that intersects the elipsoid in a circle, then any translation of that plane should also intersect in a circle, and the constant term of the normalized equation for this collection of planes would give you a fitting $C^\infty$-function. – Arthur Oct 1 '12 at 13:17

Elaborating my comment on joriki's answer: The intersection of a plane with an ellipsoid is an ellipse. Consider the plane passing through the origin and spanned by the two orthonormal vectors \begin{align} \hat u &= \hat x \cos\theta + \hat z \sin\theta, \\ \hat v &= \hat y. \end{align} By symmetry, its intersection with the ellipsoid is an ellipse whose principal axes lie along $\hat u$ and $\hat v$. The lengths of its semi-axis along $\hat v$ is clearly $b$, while that along $\hat u$ varies continuously from $a$, when $\theta=0$, to $c$, when $\theta=\pi/2$. As $a<b<c$, there is some $\theta$ between $0$ and $\pi/2$ for which both semi-axes equal $b$, yielding a circle. As Arthur's comment states, translations of this plane also yield circles upon intersection with the ellipsoid. The corresponding function $f$ is simply the dot product with the plane's normal, $f(\vec p) = \vec p \cdot (\hat u \times \hat v)$.

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So easy. Thanks you. – Tomás Oct 6 '12 at 12:50

Generalizing @Rahul's solution (but, as we'll see, not-generalizing that solution), consider the plane, through the origin, with normal $(p,q,r)$ such that $p^2+q^2+r^2=1$. Of course, obtaining circular level curves in the case $p^2+q^2 = 0$ requires $a=b$ and is easily dispatched; similarly, $q^2+r^2 = 0$ and $r^2 + p^2 = 0$; so, from here on, we assume that at most one of $p$, $q$, $r$ is zero.

The plane is spanned by orthogonal unit vectors $$\mathbf{u} := \frac{(q,-p,0)}{\sqrt{p^2+q^2}} \qquad \mathbf{v} := \frac{(-pr,-qr,p^2+q^2)}{\sqrt{p^2+q^2}}$$

The point $u \mathbf{u} + v \mathbf{v}$ lies on the ellipse iff

\begin{align} 1 &= \frac{(uq-vpr)^2}{a^2} + \frac{(-up-vqr)^2}{b^2} + \frac{v^2(p^2+q^2)^2}{c^2} \\ &= u^2 \left( \frac{q^2}{a^2}+\frac{p^2}{b^2} \right) + 2uvpqr\frac{a^2-b^2}{a^2b^2} + v^2 \left(\frac{p^2r^2}{a^2}+\frac{q^2r^2}{b^2}+\frac{(p^2+q^2)^2}{c^2}\right) \\ &=: A u^2 + 2 B u v + C v^2 \end{align}

For the parameterized curve to be a circle, we must have $B = 0$ and $A=C$.

For $B=0$, we have either that $a=b$, or that (at most) one of $p$, $q$, $r$ is zero. Together, $a=b$ and $A=C$ (and $p^2+q^2+r^2=1$) imply $a=c$, so that the ellipsoid is in fact a sphere. On the other hand, $pqr=0$ and $A=C$ allow us to either solve for $p$, $q$, $r$ or derive other constraints on $a$, $b$, $c$:

Case $p=0$: We have, from $A=C$, that $\frac{1}{a^2} = \frac{r^2}{b^2}+\frac{q^2}{c^2}$, so that $$q^2 a^2(c^2-b^2) = -c^2(b^2-a^2) \qquad r^2 a^2(c^2-b^2)= b^2(c^2-a^2)$$

Since $c^2-b^2$, $b^2-a^2$, and $c^2-a^2$ are all non-negative, the $q$ equation requires $a=b=c$ and we have another sphere.

Case $q=0$: Here, $\frac{1}{b^2} = \frac{r^2}{a^2}+\frac{p^2}{c^2}$, which is the same as the above case, with $p \to q \to r \to p$ and $a\to b\to c \to a$. Thus,

$$r^2 b^2(c^2-a^2) = a^2(c^2-b^2) \qquad p^2 b^2(c^2-a^2)= c^2(b^2-a^2)$$

which allows solutions for $p$ and $r$ even for non-spheres. (This is @Rahul's case.)

Case $r=0$: Here, $\frac{q^2}{a^2}+\frac{p^2}{b^2}=\frac{1}{c^2}$, which is the same as the above case under the mentioned cycling of variables, so that $$p^2 c^2(b^2-a^2) = b^2(c^2-a^2) \qquad q^2 c^2(b^2-a^2)= -a^2(c^2-b^2)$$ Again, the $q$ equation implies that the ellipse is a sphere.

Consequently, @Rahul's circles are the only ones for non-spherical $E$.

Note. The fact that the three cases are cyclically related suggests that there should be an insightful, cyclically-symmetric way of writing the equation $A=C$, but I'm not seeing it.

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As Rahul rightly pointed out in a comment, the conclusion in the last paragraph, and therefore the answer, is wrong. I'll fix the answer later if I get around to it before someone posts a correct one.

This is impossible because there are no circles on such an ellipsoid.

A circle can be described parametrically by

$$r(\phi)=r_0+u\cos\phi+v\sin\phi\;,$$

where $r_0$ is the centre and $u$ and $v$ are orthonormal vectors. Denote vectors transformed using $\operatorname{diag}(1/a,1/b,1/c)$ by a prime. Substituting $r(\phi)$ into the equation of the ellipsoid yields an equation with five independent Fourier components, leading to the five equations

\begin{align} r_0'^2+\frac12u'^2+\frac12v'^2&=1\;,\\ r_0'\cdot u'&=0\;,\\ r_0'\cdot v'&=0\;,\\ u'\cdot v'&=0\;,\\ u'^2&=v'^2\;. \end{align}

The last two equations, which arise from the two Fourier components with argument $2\phi$, say that the transformed vectors $u'$ and $v'$ are also orthonormal. That means that angles in the $u$-$v$-plane are preserved by the transformation, so the transformation is just a scaling in that plane, which can only be the case if two of the semi-axes of the ellipsoid coincide.

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The intersection of a plane with an ellipsoid is an ellipse, right? Fix the plane to contain the $y$-axis and rotate it about the same. The intersection is an ellipse one of whose principal axes has half-width $b$, while the other varies continuously between $a$ and $c$. It seems to me that the case with both half-widths equal to $b$ has to be attained in between the two extremes. – Rahul Oct 1 '12 at 15:29
Aha, that is precisely the case where the transformation is a scaling in the plane despite all semi-axes being distinct. – Rahul Oct 1 '12 at 15:31
@Rahul: I see, thanks. My mistake was that I didn't realize that the transformation could be a scaling and a rotation of the plane. It's true that it can't be just a scaling in the same plane when the semi-axes are distinct. – joriki Oct 1 '12 at 15:41