$-\phi'' +c(x) \phi(x) = f(x), \phi (0) = \alpha $ and $\phi (1) = \beta $ If we use central difference method to discretize $\phi ''$, we know when $c(x) \geq 0 $ the matrix is invertible. Thus, we can solve the linear algebra problem to find the numerical solution.
My question is: If $c(x)<0$ ?How can we solve this problem numerically if use the finite difference method?
The linear system to be solved reduces to $$ \dfrac{-1}{2+h^2 c_i} \phi_{i+1} + \phi_i + \dfrac{-1}{2+h^2 c_i} \phi_{i-1} = \dfrac{h^2 f_i}{2+h^2 c_i} $$
When $c_i > 0$, the matrix is diagonally dominant and thus invertible, as you already know. This can be extended easily for $c_i \ge 0$.
However, these conditions on $c(x)$ are only sufficient conditions, not necessary conditions, they do noy rule out the possibility of applying de FD method for more general $c(x)$.
I direct you to this paper (Figure 1. is a must see). Based on their results, the condition $$ (2+h^2c_i)(2+h^2 c_{i+1})\ge 4 $$ guarantees invertibility. But it is still a sufficient condition.
