# Partial Differential Equation of Spheres in 3-Space

## Theorem

The set of spheres in real Cartesian $3$-dimensional space can be described by the system of partial differential equations:

$\dfrac {1 + z_x^2} {z_{xx} } = \dfrac {z_x z_x} {z_{xy} } = \dfrac {1 + z_y^2} {z_{yy} }$

and if the spheres are expected to be real:

$z_{xx} z_{yy} > z_{xy}$

## Proof

From Equation of Sphere, we have that the equation defining a general sphere $S$ is:

$\paren {x - a}^2 + \paren {y - b}^2 + \paren {z - c}^2 = R^2$

where $a$, $b$ and $c$ are arbitrary constants.

We use the technique of Elimination of Constants by Partial Differentiation.

Taking the partial first derivatives with respect to $x$ and $y$ and simplifying, we get:

 $\ds \paren {x - a} + \paren {z - c} \dfrac {\partial z} {\partial x}$ $=$ $\ds 0$ $\ds \paren {y - b} + \paren {z - c} \dfrac {\partial z} {\partial y}$ $=$ $\ds b$

$2$ equations are insufficient to dispose of $3$ constants, so the process continues by taking the partial second derivatives with respect to $x$ and $y$:

 $\ds 1 + \paren {\dfrac {\partial z} {\partial x} }^2 + \paren {z - c} \dfrac {\partial^2 z} {\partial x^2}$ $=$ $\ds 0$ $\ds \dfrac {\partial z} {\partial x} \dfrac {\partial z} {\partial y} + \paren {z - c} \dfrac {\partial^2 z} {\partial x \partial y}$ $=$ $\ds 0$ $\ds 1 + \paren {\dfrac {\partial z} {\partial y} }^2 + \paren {z - c} \dfrac {\partial^2 z} {\partial y^2}$ $=$ $\ds 0$

Eliminating $z - c$:

$\dfrac {1 + z_x^2} {z_{xx} } = \dfrac {z_x z_x} {z_{xy} } = \dfrac {1 + z_y^2} {z_{yy} }$

Let $\lambda = \dfrac {1 + z_x^2} {z_{xx} } = \dfrac {z_x z_y} {z_{xy} } = \dfrac {1 + z_y^2} {z_{yy} }$.

Then:

 $\ds \lambda^2$ $=$ $\ds \dfrac {1 + z_x^2} {z_{xx} } \dfrac {1 + z_y^2} {z_{yy} }$ $\ds \leadsto \ \$ $\ds \lambda^2 \paren {z_{xx} z_{yy} }$ $=$ $\ds 1 + z_x^2 + z_y^2 + z_x^2 z_y^2$ $\ds \leadsto \ \$ $\ds \lambda^2 \paren {z_{xx} z_{yy} } - z_{xy} \dfrac {z_x z_y} {z_{xy} }$ $=$ $\ds 1 + z_x^2 + z_y^2$ $\ds \leadsto \ \$ $\ds \lambda^2 \paren {z_{xx} z_{yy} - z_{xy} }$ $=$ $\ds 1 + z_x^2 + z_y^2$ $\ds$ $>$ $\ds 0$ $\ds \leadsto \ \$ $\ds z_{xx} z_{yy} - z_{xy}$ $>$ $\ds 0$

and so:

$z_{xx} z_{yy} > z_{xy}$

$\blacksquare$