# Solving f(u) = 0 by Time Stepping

To solve an equation $f(u)=(f_1(u),f_2(u),...,f_N(u))=0$ of $N$ equations $f_i(u)=0$, $i=1,...,N,$ in $N$ unknowns $u=(u_1,u_2,...,u_N)$, with thus $f:R^n\rightarrow R^n$, it is natural to connect to solution of the IVP: Find $u(t)$ such that

• $\dot u(t)+f(u(t))=0\quad\mbox{for }t>0,\quad u(0)=u^0$,

with some given initial value $u^0$. If it turns out that as $t$ increases, the function $u(t)$ tends to some value $\hat u$, then $\dot u(t)$ could be expected to become small, and if so, we would have

• $f(u(t))\approx 0$,

and we would be led to set $\hat u =u(t)$ for some large $t$ and consider $\hat u$ to be an approximate solution of $f(u)=0$ with small residual $f(\hat u)$.

If $f(u)$ has several different solutions, which is often the case, then we could expect to capture different solutions by chosing different initial values $u^0$.

Computing $u(t)$ by Forward Euler with time step $dt=1$, we would have

• $u^{n+1}=u^n-f(u^n),\quad{for }\,\, n=0,1,2,...,$    (1)

If $\vert u^{n+1}-u^n\vert$ would become small for increasing $n$, then $f(u^n)$ would become small and thus $u^n$ would be an approximate solution of $f(u)=0$ with small residual $f(u^n)$.

# Solving u=g(u) by Fixed Point Iteration

With $f(u)=u-g(u)$, we are thus led to study the convergence of the iteration, referred to as Fixed Point Iteration:

• $u^{n+1}=g(u^n),\quad n=0,1,2,....,$   (2)

where

• $g(u)=u-f(u)$.

To this end we take the difference of (2) for two consecutive steps to get

• $e^{n+1}\equiv u^{n+1}-u^n=g(u^n)-g(u^{n-1})$.

If $g:R^N\rightarrow R^N$ is Lipschitz continuous with Lipschitz constant $L$, then

• $\vert e^{n+1}\vert =\vert g(u^n)-g(u^{n-1})\vert\le L\vert e^n\vert\le L^2\vert e^{n-1}\vert\le L^{n}\vert u^1-u^0\vert$.

We see that if $L<1$, then $\vert e^n\vert$ becomes vanishingly small as $n$ increases, which by (1) means that $f(u^n)$ becomes vanishingly small and thus $u^n$ may be viewed as an approximate solution of $f(u)$ in the sense that the residual $f(u^n)$ is small. In the next chapter we will consider the error in the approximate root $u^n$ compared to an exact root $\bar u$ with vanishingly small residual $f(\bar u)$.

We see that if $L<<1$ then the convergence is fast, and if $L\approx 1$ then the convergence is slow. If $L=\frac{1}{2}$ then the residual $\vert g(u^{n})-u^n\vert =\vert u^{n+1}-u^n\vert$ is reduced with a factor 2 in each iteration step, that is with a binary digit per step.