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I have to solve the following linear system using an iterative method : $\begin{cases} &4x_1+x_2 = b_1 \\ &x_{i-1}+4x_i + x_{i+1}=b_i,\ i=2 \cdots n-1\\ &x_{n-1} + 4x_n=b_n \end{cases}$

The iterative method is

$\begin{cases} x_1^{k+1} = \alpha x_1^{(k)} + \frac{\alpha-1}{4}(x_2^{(k)}-b_1)\\ x_i^{k+1} = \alpha x_i^{(k)} + \frac{\alpha-1}{4}(x_{i-1}^{(k)}+x_{i+1}^{(k)}-b_i),\ i=2 \cdots n-1\\ x_n^{k+1} = \alpha x_n^{(k)} + \frac{\alpha-1}{4}(x_{n-1}^{(k)}-b_n)\\ \end{cases}$

where $\alpha$ is a real and $x^{(0)}=0$.

I have to prove first that $\parallel x^{(k+1)}-x \parallel_\infty \leq \left(|\alpha| + \left|\frac{\alpha-1}{2}\right|\right) \parallel x^{(k)}-x \parallel_\infty $

I tried lot of things but without success. I am sure that I have to use a proof by induction, but I cannot prove that the initialization is true for $k=0$. Can you help me?

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  • $\begingroup$ Begin your work by expressing your iteration using matrices and vectors. Exploit that your matrix is strictly diagonally dominant. $\endgroup$
    – Carl Christian
    Commented Feb 20, 2022 at 1:36

1 Answer 1

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The $\|\cdot\|_{\infty}$ norm of you iteration matrix is precisely $|\alpha| +\frac{|\alpha-1|}{2}$, so the result comes directly from the fixed point theorem.

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