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
\[\begin{equation} \begin{aligned} & \sum_{n=0}^{\infty} \sum_{i=0}^{n-2} a_{i+2, n-i-2} (i+2)(i+1)x^{i} y^{n-i-2}+ \sum_{n=0}^{\infty} \sum_{i=0}^{n-2} a_{i, n-i} (n-i)(n-i-1) x^i y^{n-i-2} \\ &= \sum_{n=0}^{\infty} \sum_{i=0}^{n-2} ( a_{i+2, n-i-2} (i+2)(i+1) +a_{i, n-i} (n-i)(n-i-1)) x^i y^{n-i-2} =0 \end{aligned} \end{equation}\]
.
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.

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