Differential Equations for Shortest Path on 3d Sphere/Cartesian Coordinates
Theorem
Let $M$ be a $3$-dimensional Euclidean space.
Let $S$ be a sphere embedded in $M$.
Let $\gamma$ be a curve on $S$.
Let the chosen coordinate system be Cartesian.
Let $\gamma$ begin at $\paren {x_0, y_0, z_0}$ and terminate at $\paren {x_1, y_1, z_1}$.
Let $\map y x$, $\map z x$ be real functions.
Let $\gamma$ connecting both endpoints be of minimum length.
Then $\gamma$ satisfies the following equations of motion:
- $2 y \map \lambda x - \dfrac \d {\d x} \dfrac {y'} {\sqrt {1 + y'^2 + z'^2} } = 0$
- $2 z \map \lambda x - \dfrac \d {\d x} \dfrac {z'} {\sqrt {1 + y'^2 + z'^2} } = 0$
Proof
In $3$-dimensional Euclidean space the length of the curve is:
- $\ds \int_{x_0}^{x_1} \sqrt {1 + y'^2 + z'^2} \rd x$
The sphere satisfies the following equation:
\(\ds \map g {x, y, z}\) | \(=\) | \(\ds x^2 + y^2 + z^2 - a^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 0\) |
Consider its partial derivatives with respect to $y$ and $z$.
- $\dfrac {\partial g} {\partial y} = 2y$
- $\dfrac {\partial g} {\partial z} = 2z$
$g_y$ and $g_z$ vanish for $y = 0$ and $z = 0$ respectively.
Substitution of this into the sphere equation tells us that $x^2 = a^2$.
Therefore, the following analysis should exclude points with $x = \pm a$.
By Simplest Variational Problem with Subsidiary Conditions for Curve on Surface, the length functional is replaced by the following auxiliary functional:
- $\ds \int_{x_0}^{x_1} \sqbrk {\sqrt {1 + y'^2 + z'^2} + \map \lambda x \paren {x^2 + y^2 + z^2} } \rd x$
It follows that:
- $\map {\dfrac {\partial} {\partial y'} } {\sqrt {1 + y'^2 + z'^2} + \map \lambda x \paren {x^2 + y^2 + z^2} } = \dfrac {y'} {\sqrt {1 + y'^2 + z'^2} }$
- $\map {\dfrac {\partial} {\partial y} } {\sqrt {1 + y'^2 + z'^2} + \map \lambda x \paren {x^2 + y^2 + z^2} } = 2 y \lambda$
Analogous relations hold for $\map z x$.
Then by Euler's Equations the following equations of motion hold:
- $2 y \map \lambda x - \dfrac \d {\d x} \dfrac {y'} {\sqrt{1 + y'^2 + z'^2} } = 0$
- $2 z \map \lambda x - \dfrac \d {\d x} \dfrac {z'} {\sqrt{1 + y'^2 + z'^2} } = 0$
$\blacksquare$
Sources
- 1963: I.M. Gelfand and S.V. Fomin: Calculus of Variations ... (previous) ... (next): $\S 2.12$: Variational Problems with Subsidiary Conditions