I need to use finite difference method without introducing fiction points to solve the following problem:
$−\mu u′′(x)+\eta u′(x)+\sigma u(x)=f(x)$, $a<x<b,$ subject to the boundary conditions $u′(a) = \alpha$ and $u′(b) = \beta $.
I discretize the derivatives with second order central differences formulas at interior points ($a < x < b$), but use one-sided forward and backward differences to approximate the boundary conditions (at $x = a$ and at $x = b$).
$u_1 - u_0 = \alpha h$;
$u_N - u_{N-1} = \beta h$
At $i=0$ I have to introduce a fiction point of $x_{-1}$:
$(-\mu + \eta h/2)u_{1} + (2 \mu + \sigma h^2)u_0 + (-\mu - \eta h/2)u_{-1}=f_0 h^2$
What am I doing wrong? Thank you
Those one-sided differences aren't what you want because the central difference approximations to the derivatives have error $O(h^2)$ and the one-sided differences have error $O(h)$. So you can expand in Taylor series about $x=a$: $$\begin{align}x_0&=x_0\\ x_1&=x_0+hx_0^{\prime}+\frac12h^2x_0^{\prime\prime}+O(h^3)\\ x_2&=x_0+2hx_0^{\prime}+2h^2x_0^{\prime\prime}+O(h^3)\end{align}$$ You want $c_0x_0+c_1x_1+c_2x_2=x_0^{\prime}+O(h^2)$. Comparing coefficients of like derivatives of $x_0$, we get $$\begin{align}c_0+c_1+c_2&=0\\ c_1+2c_2&=\frac1h\\ \frac12c_1+2c_2&=0\end{align}$$ With solution $c_0=-\frac3{2h}$, $c_1=\frac2h$, and $c_2=-\frac1{2h}$, so now your boundary condition reads $$\frac{-3u_0+4u_1-u_2}{2h}=\alpha$$ The other boundary condition reads something like $$\frac{u_{N-2}-4u_{N-1}+3u_N}{2h}=\beta$$ You don't need the equations with $h^2f_0$ and $h^2f_N$ because you are using the boundary conditions at $x=a$ and $x=b$.