## The Dimension of the Set of Potential Functions in the Plane

A funtion
$\phi$
in the
$xy$
plane is a potential function if it satisfies
$\frac{\partial^2 \phi}{\partial x^2} + \frac{\partial^2 \phi}{\partial y^2} =0$
.
The set of potential functions forma a vector space, since if
$\phi, \psi$
are poteial functions
1.
$\frac{\partial^2 0}{\partial x^2} + \frac{\partial^2 0}{\partial y^2} =0$
, where 0=0(xy) is the function which returns zero for aal
$x,y$

2.
$a \frac{\partial^2 \phi}{\partial x^2} + \frac{\partial^2 \phi}{\partial y^2} + b \frac{\partial^2 \psi}{\partial x^2} + \frac{\partial^2 \phi}{\partial y^2} = \frac{\partial^2 (a \phi + b \psi )}{\partial x^2} + \frac{\partial^2 \phi}{\partial y^2} =0$
.
We can write
$\phi =\sum_{n=0}^{\infty} \sum_{i=0}^n a_{i, n-i} x^i y^{n-i}$
then substitute into Poisson's equation to give
$\sum_{n=0}^{\infty} \sum_{i=2}^n a_{i, n-i} i(i-1)x^{i-2} y^{n-i}+ \sum_{n=0}^{\infty} \sum_{i=0}^{n-2} a_{i, n-i} (n-i)(n-i-1) x^i y^{n-i-2} =0$
.
Reindexing gives
\begin{aligned} & \sum_{n=0}^{\infty} \sum_{i=0}^{n-2} a_{i+2, n-i-2} (i+2)(i+1)x^{i} y^{n-i-2}+ \sum_{n=0}^{\infty} \sum_{i=0}^{n-2} a_{i, n-i} (n-i)(n-i-1) x^i y^{n-i-2} \\ &= \sum_{n=0}^{\infty} \sum_{i=0}^{n-2} ( a_{i+2, n-i-2} (i+2)(i+1) +a_{i, n-i} (n-i)(n-i-1)) x^i y^{n-i-2} =0 \end{aligned}
.
Hence
$a_{i+2, n-i-2} (i+2)(i+1) +a_{i, n-i} (n-i)(n-i-1)=0$
.
Put
$i=1, i=2$
to get
$a_{3,n-3} =- \frac{a_{1,n-1} (n-1)(n-2)}{3 \times 2}$
and
$a_{4,n-4} =- \frac{a_{2,n-2} (n-2)(n-3)}{4 \times 3}$
.
These are independent recurrence relations which determine
$\phi$
so the dimension of the vector space is 2.