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The linear backward Euler's Method is given by

$$ y^{j+1}=y^j+ \tau k_1 $$ $$ (I- \tau J) k_1= f(y^j), J:= f'(y^j)$$

To show, is that $y^{j+1}$ is equal to $\tilde{y^{j+1}} $ the Output of the Backward Euler's Method, if you proccess one Newton-Step with $$ g-(y^j+ \tau f(g))=0 $$ for $\tilde{y^{j+1}} =g $
( $ g^0= y^j$)

This is what I started with : $$ g^1= g^0 + \tau \frac{1}{I-\tau f'(g^0)}f(g^0) $$ $$ \iff y^{j+1} = y^j + \tau \frac{1}{I-\tau f'(y^j)}f(y^j) $$ $$ \iff y^{j+1}-y^j =\tau \frac{1}{I-\tau f'(y^j)}f(y^j) $$ $$ \iff \tau k_1 = \tau \frac{1}{I-\tau f'(y^j)}f(y^j) $$ $$ \iff k_1(I-\tau f'(y^j))= f(y^j) $$

Did I miss something or is there anything wrong? Any help is very appreciated

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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 $$ y_{j+1}=y_j+hf(y_{j+1}) $$ or \begin{align} k_1&=f(y_j+hk_1)\\ y_{j+1}&=y_j+hk_1 \end{align} and for the method description it is immaterial how you solve the implicit equation, you just assume that a solution close to $y_j$ 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 $0=F(k_1)=k_1-f(y_j+hk_1)$ only once at $y_j$ resp. $k_1=0$. With $J=f'(y_j)$ you then get the iteration $$ k_1^{m+1}=k_1+(I-hJ)^{-1}\bigl(f(y_j+hk_1^m)-k_1^m\bigr). $$ Usually you start with $k_1^0=f(y_j)$.

If you only perform one Newton step, you get what is sometimes called the most simple Rosenbrock method.

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Newton's method for $ F(g) := g - (y^{(j)} + \tau f(g)) $ leads to following first step: $$ F'(g^{(0)}) d = - F(g^{(0)}), $$ $$ \tilde{y}^{(j+1)} = g^{(0)} + d, $$ with the initial value given by $ g^{(0)} := y^{(j)} $.

Hence: $$ (I - \tau f'(y^{(j)})) d = \tau f(y^{(j)}), $$ $$ \tilde{y}^{(j+1)} = y^{(j)} + d. $$

Replacing $ f'(y^{(j)}) $ by $ J $ on the one hand, $ d $ by $ \tau k_1 $ on the other, you will end up with the linear Backward Euler.

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