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I have the following small finite element non linear boundary value problem:

$$ -u''(x) = 1 + u^2(x) \quad \text{for} \quad 0<x<1 \quad \text{with} \quad u(0) = u(1) = 0 $$

with a grid with points at $\vec{x}= (0.2, 0.4, 0.6, 0.8)$

The system of equation can be written as followed :

$$ \frac{1}{0.2^2}\begin{bmatrix} +2 & -1 & 0 & 0 \\ -1 & +2 & -1 & 0 \\ 0 & -1 & +2 & -1 \\ 0 & 0 & -1 & +2 \\ \end{bmatrix} \begin{pmatrix} u_1 \\ u_2 \\ u_3 \\ u_4 \\ \end{pmatrix} = \begin{pmatrix} 1 + u_1^2 \\ 1 + u_2^2 \\ 1 + u_3^2 \\ 1 + u_4^2 \\ \end{pmatrix} $$ which can also be written as $\textbf{A}\cdot\vec{u} = 1 + \vec{u}^2$

Using the method of partial substitution to solve the problem with an initial vector $\vec{u_0}$ e.g. $\vec{u_0}=\vec{0}$. We can then use the iteration

$$\textbf{A}\cdot\vec{u}_{k+1} = 1 + \vec{u}_k^2$$ $$\vec{u}_{k+1} = \textbf{A}^{-1}\cdot\ (1 + \vec{u}_k^2)$$

my question is : why is there an increment of 1 in the factor $k$ in the last 2 equations? why is it not

$$\textbf{A}\cdot\vec{u}_{k} = 1 + \vec{u}_k^2$$

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  • $\begingroup$ How would you do the iterations if you had $u_k$ on both sides? $\endgroup$ – Yuriy S Nov 24 '18 at 19:27
  • $\begingroup$ thank you for your comments @Yuriy S. Regarding your second comment you're write. I corrected the error $\endgroup$ – ecjb Nov 24 '18 at 19:31
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You're making a sequence of progressively better solutions. $\vec{u}_{k+1}$ is the $k+1^{\text{th}}$ approximation and you have a recipe to construct it from the $k^{\text{th}}$ approximation.

You are linearly approximating a system nonlinearity. There's no reason to think that your initial guess at a solution is close to a real solution, nor that it obeys your nonlinear equation. You construct a sequence of approximate solutions, using the linear system approximation to inform the changes you make to your approximate solution. Ideally, this means you make smaller and smaller changes that converge to a "best" approximate solution.

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