convergence of finite difference method for boundary value ODE

Suppose we have to solve $d^2y/dx^2= f(y,x)$ where $f$ is Lipschitz and $y(0) =a, y(1) =b$, using finite difference method, i.e., by discretizing the problem into $y_{i+1} - 2y_i + y_{i-1} = h^2f(y_i,x_i)$, with equal spacing. How do we show it converges, i.e., $\lim_n \max_{i \le n} |y(x_i) - y_i| = 0$? You can assume there is a unique twice differentiable solution, but I would like to know what happens when the solution is not unique as well. This seems quite a bit harder than the initial value problem situation.

I understand this is probably a very basic question in numerical analysis, but I just couldn't find a good reference that covers this general case. There has been a paper on this subject, http://www.jstor.org/stable/2004339?seq=3, but I found their proof for the boundary value problem difficult to follow (note that the proof does not mention a or b at all).

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Your discretization is incomplete. Say that the $x_i$'s are equally spaced, with $x_{i+1}-x_i=h$. Then the right-hand side should be $h^2f(y_i,x_i)$. – Denis Serre Dec 9 '10 at 7:56
Yes you are right. I was in a rush. – John Jiang Dec 9 '10 at 7:58

If the solution of the BVP is non-unique, it is likely that the solution of the discretization is non unique as well. I should bet on the following. If a solution of the BVP has the (generic) property that the linearized equation $$\frac{d^2z}{d^2x}=\frac{\partial f}{\partial y}(y,x)z,\qquad z(0)=z(1)=0$$ admits only the solution $z\equiv0$, then for small $h$ ($h=x_{i+1}-x_i$) the discretization admits a solution $y^h$ which converges to $y$ as $h\rightarrow0$. Now, if your BVP has several solutions, they are likely to share the property above since it is generic, and you will have several disctint discretized solutions for small $h$.
The nonlinear example considered in the text is $y''(x)=\sin(x)$, and it is demonstrated that the numerical solution is non-unique (as is the analytic solution).