Sam Auyeung

Sam Auyeung

An Application of Differential Equations to Plane Geometry

2 minute read

Published:

This is a somewhat different post from my previous ones about applying topology. However, the spirit of those posts was to illustrate unexpected applications and this post will remain in this spirit, if only because we’ll be doing things in a backwards and needlessly complicated way which perhaps breaks expectations. Here is a question: given the ellipse defined by $(ax+by)^2+(cd+dy)^2 = r^2$ in $\mathbb{R}^2$, what are the lines which contain its major and minor axes? I actually never took a formal geometry class in high school nor college so I don’t know what the usual pedagogical approach is to learning this.

Here is a Desmos link which graphs ellipses with sliders for the parameters $a,b,c,d$. If we write $H(x,y) = \frac{1}{2}(ax+by)^2 + \frac{1}{2}(cx+dy)^2$, then we can form a Hamiltonian system

\[\begin{cases} \frac{dx}{dt} = \partial_y H = b(ax+by) + d(cx+dy)\\ \frac{dy}{dt} = -\partial_x H = -a(ax+by) - c(cx+dy) \end{cases}\]

Being a Hamiltonian system, its solution curves live in the level sets of $H$ which are ellipses since they are preserved by $X_H$, the Hamiltonian vector field. For example, $H^{-1}(r^2)$ contains solution curves to this linear system of ODEs and exactly trace out the ellipses.

If we instead use $\nabla H = \langle \partial_x H, \partial_y H\rangle$, at every point, this vector field is orthogonal to $X_H$ and gives the directions of steepest ascent. In particular, the origin is a source which means the solutions all eminate out of the origin. Moreover, there are two straight-line solutions determined by the eigenvectors of the matrix and since they must eminate from the origin and also intersect the ellipses orthogonally, they must contain the major and minor axes of the ellipses. The computations aren’t hard but get to be a bit unwieldy. However, it’s not problem for WolframAlpha. The eigenvectors are:

\[v_1=((-a^{2}+b^{2}-c^{2}+d^{2}-\sqrt{(a^{4}+2a^{2}b^{2}+b^{4}+2a^{2}c^{2}-2b^{2}c^{2}+c^{4}+8abcd-2a^{2}d^{2}+2b^{2}d^{2}+2c^{2}d^{2}+d^{4})})/(2(ab+cd)),-1)\] \[v_2=((-a^{2}+b^{2}-c^{2}+d^{2}+\sqrt{(a^{4}+2a^{2}b^{2}+b^{4}+2a^{2}c^{2}-2b^{2}c^{2}+c^{4}+8abcd-2a^{2}d^{2}+2b^{2}d^{2}+2c^{2}d^{2}+d^{4})})/(2(ab+cd)),-1)\]

Let’s do an easy example with $a=b=d=1,c=0$. So $2H(x,y) = (x+y)^2+y^2$ and $\nabla H = \langle x+y,x+2y \rangle$. The eigenvalues of the associated matrix are $(3\pm \sqrt{5})/2$ and the eigenvectors are $(-1\pm \sqrt{5},2)$. So the lines containing the axes are $y= 2x/(-1\pm \sqrt{5})$. With some algebra, it’s not too hard to determine which is the major and which is the minor axis. Of course, when $a=b=c=d$, then we have a circle which has no major or minor axis. But also in this case, every ray eminating out from the origin is a solution to the ODE involving $\nabla H$.

Of course, if we want to ask a similar question for ellipsoids in $n\geq 3$ dimension, this technique easily generalizes; we’d only need to find more eigenvectors. For generic coefficients, we could just find $n-1$ of them and use Gram-Schmidt for the last since the eigenvectors will form an orthogonal basis. If there are instances which give rise to certain slices being circles, we can also use Gram-Schmidt and still get a nice set of bases.

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