I am faced with the task of applying Newton's method to the following problem:
$$ \text{min} ~~~~~ 8x_1x_2+\frac{1}{4}(x_1-x_2)^4 $$
where $x \in \mathbb{R}^2$. For clarification, the Newton method to which I am referring is:
$$ x^{k+1} = x^{k} - \left(\nabla^2f(x^k)\right)^{-1}\nabla f(x^k) $$
where $f$ represents the objective function to be minimized, $\nabla f$ represents the gradient of $f$ as usual, and $\nabla^2f$ represents the Hessian of $f$.
I have already shown that for a choice of an initial $x^0 := (x_1^0,x_2^0)$ such that $3(x_1^0-x_2^0)^2 \neq4$ all the resulting iterates $x^k := (x_1^k,x_2^k)$ lie on the line $x_1^k+x_2^k=0$.
Where I am running into a roadblock is attempting to show two things:
Firstly: if one chooses a step length of one, then it converges to $(0,0)$ (a saddle point) cubically if $|x_1^0-x_2^0| < \frac{2}{\sqrt{5}}$.
Secondly: again using a unit step length it converges to a global minimum (which is not unique for this problem - it is attained at both $(1,-1) ~ \text{and} ~ (-1,1)$) if we take $3(x_1^0-x_2^0)^2>4$.
I fought with some unruly algebra to achieve the one result I've mentioned and I can't seem to make anything meaningful come out in the wash to show the remaining results.