Given the finite difference approximation for black scholes with zero interest rate,
$$ \frac{V_n^{m+1}-V_n^m}{\Delta t} + \frac{1}{2}\sigma^2S^2 \frac{V_{n+1}^{m}-2V_n^m+V_{n-1}^{m}}{\Delta S^2}=0\\ $$
We can write it as,
$$ \begin{align} V_n^{m+1} &= -\frac{1}{2}\sigma^2 n^2 \Delta t V_{n-1}^m + \left( 1+\sigma^2 n^2 \Delta t \right)V_n^m -\frac{1}{2}\sigma^2 n^2 \Delta t V_{n+1}^m\\ &=a_nV_{n-1}^m + b_nV_{n}^m+c_nV_{n+1}^m \end{align} $$
Given that $V_n^{m+1} \ge 0 $ for all $n$, what are sufficient conditions for $a_n$, $b_n$, $c_n$ which ensures $V_n^{m} \ge 0 $ for all $n$?
Can someone kindly point me to any available materials on this?
Observe that $a_n + b_n + c_n = 1,\; a_n, c_n < 0$. Suppose that $V_n^m$ is negative and is a local minimum. $$ V_n^m < 0\\ V_{n+1}^m \geq V_n^m\\ V_{n-1}^m \geq V_n^m. $$ Then $$ V_n^{m+1} = a_n V_{n-1}^m + b_n V_n^m + c_n V_{n+1}^{m} = -a_n (-V_{n-1}^m) + b_n V_n^m - c_n (-V_{n+1}^m) \leq -a_n (-V_n^m) + b_n V_n^m - c_n (-V_n^m) = V_n^m. $$ Since $V_n^{m+1} > 0$ then $V_n^m \geq V_n^{m+1} > 0$. So $V_n^m$ can't be negative, contradiction.