Solving system of equations with boundary conditions

Question

I am taking a graph theory class this quarter, and while this specific question below is not a homework question, proving it is required for a homework question that I am working on. I assume it may be a commonly asked question, if so apologies for the repeat:

Show that the system of equations for $0 < k < n:$

$$f_k = \frac{1}{2} (1 + f_{k-1}) + \frac{1}{2} (1 + f_{k+1})$$

together with the boundary conditions $f_0 = f_n = 0$ has a unique solution $f_k = k(n − k).$

I've written the equation out for $f_1, f_2, f_3$, for $f_1$ plugging in $f_0 = 0$ and solving in terms of $f_2$, repeating etc., and although I see some sort of pattern, I am having trouble connecting it to the solution $k(n-k)$. Any thoughts on how I can connect the dots, as well as a better approach than simply extending the problem out one calculation at a time.

Thanks!

Answer 1

As an engineer, I typically use $z$-transform for such problems, which looks like a straightforward approach. But that doesn't work with such boundary conditions. Your equation $$f_k = \frac{1}{2} (1 + f_{k-1}) + \frac{1}{2} (1 + f_{k+1})$$ implies that $$f_{k+2}-2f_{k+1}+f_{k}=-2$$ Now take a look at the notation used in here which looks interesting: $$\Delta^2(f_k)=-2$$ This problem is somehow equivalent to finding a continuous $f(t)$ such that $$\ddot{f}=-2$$ which means $f(t)=-t^2+c_1 t+c_2$, and considering the equivalent boundary conditions, i.e. $$f(0)=f(a)=0$$ it boils down to $f(t)=t(a-t)$.

Of course this approach is careless and has some caveats. But I think it gives you the tools to start with. For a more detailed and precise explanation, take a look at here.