I am not getting anywhere with this task:
Consider the following PDE: $$-\frac{d^2u}{dx^2}(x)+cu(x)\frac{du}{dx}(x) = f(x),\,\, 0 < x < 1,\,\, u(0)=0,\,\, u(1)=1$$ Divide the interval $[0,1]$ into $n$ subintervals of width $h=1/n$. Use central difference approximation $$\frac{d^2u}{dx^2}(x_i) \approx \frac{u_{i+1}-2u_i +u_{i-1}}{h^2}$$ and $$\frac{du}{dx}(x_i) \approx \frac{u_{i+1}-u_{i-1}}{2h}$$ to find a nonlinear system for $u_2,...,u_n$.
Then state the nonlinear Gauss-Seidel for the obtained system.
Maybe you could lead me through the task or give some hints what to do. I don't know what Finite Differences have got to do with finding a nonlinear system, and I only know linear Gauss-Seidel so far. Google research wasn't helpful to me. I would be so grateful for at least a rough solution to this problem.
You are being asked to present a modification of the fixed point method that uses the latest values of the already computed components. For instance, lets say that the nonlinear equation $$ -\frac{1}{h^2}(u_{i+1}-2u_i+u_{i-1})+ \frac{c}{2h} u_i (u_{i+1}-u_{i-1}) = f_i $$
is recasted as
$$ u_i = \dfrac{2h^2 f_i+2(u_{i+1} + u_{i-1})}{4+c h (u_{i+1}-u_{i-1})}. $$
The usual fixed point would read $$ u_i^{(k+1)} = \dfrac{2h^2 f_i+2\left(u_{i+1}^{(k)} + u_{i-1}^{(k)}\right)}{4+c h \left(u_{i+1}^{(k)}-u_{i-1}^{(k)}\right)} $$
However, assuming that when you are computing $u_i^{(k+1)}$ you have already computed $u_{i-1}^{(k+1)}$, you can consider the iteration
$$ u_i^{(k+1)} = \dfrac{2h^2 f_i+2\left(u_{i+1}^{(k)} + \color{red}{u_{i-1}^{(k+1)}}\right)}{4+c h \left(u_{i+1}^{(k)}-\color{red}{u_{i-1}^{(k+1)}}\right)}. $$