Solving a particular 2d recurrence

Question

I'd be interested in a solution to the recurrence \begin{align*} \frac{u\left[m,n-1\right]-u\left[m,n\right]}{h}+\frac{u\left[m-1,n\right]-2u\left[m,n\right]+u\left[m+1,n\right]}{h^{2}} & =\left(mh\right)^{2} & & \text{for }n>0\\ u\left[m,n\right] & =0 & & \text{for }n=0 \end{align*} for $h>0$ and integers $m$ and $n>0$. The bilateral Z-transform of $m^{2}$ is divergent. Any suggestions?

Answer 1

This is a discretized heat equation. The source term is quadratic, so its second derivative is constant, so the solution should be a linear combination of a constant and a quadratic term:

$$ u [m,n]=a_n+b_nm^2\;. $$

Substituting this into your equation yields

$$ \frac{a_{n-1}+b_{n-1}m^2-a_n-b_nm^2}h+\frac{2b_n}{h^2}=(mh)^2\;. $$

This equation must be satisfied separately for the terms constant and quadratic in $m$:

$$ h(a_{n-1}-a_n)+2b_n=0\;,\\ b_{n-1}-b_n=h^3\;. $$

Together with $a_0=b_0=0$, this gives $b_n=-nh^3$ and $a_n=-(n^2+n)h^2$, so the solution is

$$ u[m,n]=-nh^2\left(n+1+hm^2\right)\;. $$

Edit:

You rightly pointed out that I said nothing about the uniqueness of this solution. For negative $n$ the solution is uniquely determined by the prescribed values at $n=0$, since solving for $u [m,n-1]$ expresses it in terms of values on the next time slice. The same isn't true for positive $n$; we can add any solution of the recurrence

$$u[m+1,n]=(2+h)u [m,n]-u[m-1,n]$$

to any time slice with $n\gt0$. The two positive real eigenvalues of this recurrence are reciprocal to each other, so they correspond to geometric sequences that grow exponentially in one or the other direction (of $m$) and are thus, as you suggested, not functions of polynomial growth.

Wikipedia has a link to some considerations on the uniqueness of solutions of the continuous heat equation up to exponentially growing solutions. It's interesting that your particular discretization has eliminated this non-uniqueness in the negative time direction but not in the positive time direction – in a sense it is "anti-causal", and you could make it causal by replacing $u[m,n-1]-u[m,n]$ by $u[m,n]-u[m,n+1]$.