# The linear backward Euler's Method is given by y <mrow class="MJX-TeXAtom-ORD">

The linear backward Euler's Method is given by
${y}^{j+1}={y}^{j}+\tau {k}_{1}$
$\left(I-\tau J\right){k}_{1}=f\left({y}^{j}\right),J:={f}^{\prime }\left({y}^{j}\right)$
To show, is that ${y}^{j+1}$ is equal to $\stackrel{~}{{y}^{j+1}}$ the Output of the Backward Euler's Method, if you proccess one Newton-Step with
$g-\left({y}^{j}+\tau f\left(g\right)\right)=0$
for $\stackrel{~}{{y}^{j+1}}=g$
${g}^{0}={y}^{j}$
This is what I started with :
${g}^{1}={g}^{0}+\tau \frac{1}{I-\tau {f}^{\prime }\left({g}^{0}\right)}f\left({g}^{0}\right)$
$\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}{y}^{j+1}={y}^{j}+\tau \frac{1}{I-\tau {f}^{\prime }\left({y}^{j}\right)}f\left({y}^{j}\right)$
$\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}{y}^{j+1}-{y}^{j}=\tau \frac{1}{I-\tau {f}^{\prime }\left({y}^{j}\right)}f\left({y}^{j}\right)$
$\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}\tau {k}_{1}=\tau \frac{1}{I-\tau {f}^{\prime }\left({y}^{j}\right)}f\left({y}^{j}\right)$
$\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}{k}_{1}\left(I-\tau {f}^{\prime }\left({y}^{j}\right)\right)=f\left({y}^{j}\right)$
Did I miss something or is there anything wrong? Any help is very appreciated
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Gornil2
You are mixing the two iterations, one is the time step of the Euler method and the other the update of the Newton iteration. In the Euler step you want
${y}_{j+1}={y}_{j}+hf\left({y}_{j+1}\right)$
or
$\begin{array}{rl}{k}_{1}& =f\left({y}_{j}+h{k}_{1}\right)\\ {y}_{j+1}& ={y}_{j}+h{k}_{1}\end{array}$
and for the method description it is immaterial how you solve the implicit equation, you just assume that a solution close to yj is somehow obtained.

One way to solve the implicit equation for ${k}_{1}$ is via the (modified) Newton method where you can simplify to take the Jacobian for $0=F\left({k}_{1}\right)={k}_{1}-f\left({y}_{j}+h{k}_{1}\right)$ only once at ${y}_{j}$ resp. ${k}_{1}=0$. With $J={f}^{\prime }\left({y}_{j}\right)$ you then get the iteration
${k}_{1}^{m+1}={k}_{1}+\left(I-hJ{\right)}^{-1}\left(f\left({y}_{j}+h{k}_{1}^{m}\right)-{k}_{1}^{m}\right).$
Usually you start with ${k}_{1}^{0}=f\left({y}_{j}\right)$.
If you only perform one Newton step, you get what is sometimes called the most simple Rosenbrock method.
###### Did you like this example?
Michelle Mendoza
Newton's method for $F\left(g\right):=g-\left({y}^{\left(j\right)}+\tau f\left(g\right)\right)$ leads to following first step:
${F}^{\prime }\left({g}^{\left(0\right)}\right)d=-F\left({g}^{\left(0\right)}\right),$
${\stackrel{~}{y}}^{\left(j+1\right)}={g}^{\left(0\right)}+d,$
with the initial value given by ${g}^{\left(0\right)}:={y}^{\left(j\right)}$.

Hence:
$\left(I-\tau {f}^{\prime }\left({y}^{\left(j\right)}\right)\right)d=\tau f\left({y}^{\left(j\right)}\right),$
${\stackrel{~}{y}}^{\left(j+1\right)}={y}^{\left(j\right)}+d.$
Replacing ${f}^{\prime }\left({y}^{\left(j\right)}\right)$ by $J$ on the one hand, $d$ by $\tau {k}_{1}$ on the other, you will end up with the linear Backward Euler.