Hello

I'm trying to answer this question, but am completely stuck.

Argue that in analyzing the error in a stationery linear relaxation scheme applied to $Au=f$, it is sufficient to consider $Au=0$ with arbitrary initial guess, (say $v_0$).

Any ideas?

I'm not even sure what the author is trying to say, is he saying when studying the error produced by the relaxation method (eg. the jacobi method) it is sufficient to study the error of the associated homogenous system $Au=0$ with arbitrary initial guesses (say $v_0$). But in this case the error will simply be $e = -v_0$. (since the error is defined as $e=u-v_0$)

Maybe the residual equation plays a part, $Ae=r$ ... ???

Any help, would be much appreciated.

Incidently stationary linear relaxation schemes (eg. gauss seidel and jacobi) have the form

$v^{(k+1)}=(v^{k}+Br^{(0)})$. Not sure if this is usefull. here $r=f-Av$ is the residual vector, and $v^{(k)}$ is the kth iterate of the scheme

1

# Relaxation Scheme for $Au=f$ error analysis

Hello

I'm trying to answer this question, but am completely stuck.

Argue that in analyzing the error in a stationery linear relaxation scheme applied to $Au=f$, it is sufficient to consider $Au=0$ with arbitrary initial guess, (say $v_0$).

Any ideas?

I'm not even sure what the author is trying to say, is he saying when studying the error produced by the relaxation method (eg. the jacobi method) it is sufficient to study the error of the associated homogenous system $Au=0$ with arbitrary initial guesses (say $v_0$). But in this case the error will simply be $e = -v_0$. (since the error is defined as $e=u-v_0$)

Maybe the residual equation plays a part, $Ae=r$ ... ???

Any help, would be much appreciated