Partial Differential Equation of Spheres in 3-Space

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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:

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

$\blacksquare$


Sources