TITLE: Approximation of Hessian=$J^TJ$ for general non-linear optimization problems
QUESTION [7 upvotes]: My question is: when is the aprroximation of Hessian matrix $H=J^TJ$ reasonable?
One truth is that it is reasonable to approximate Hessian with first order derivatives (jacobian), i.e.,  $H=J^TJ$ when we are solving a non-linear least square problem (which is called Gauss-Newton method). In other words, that is the case when the cost function (energy function) is in quadratic form. This can be derived from Newton's method. 
See wiki: https://en.wikipedia.org/wiki/Gauss–Newton_algorithm
But is there any other cases when the aprroximation of Hessian matrix $H=J^TJ$ reasonable? Such as, for some of the general non-linear optimization problems?
I have noticed that some papers (in the field of Computer Vision) used Gauss-Newton or Levenberg–Marquardt (L-M) algorithm to solve non-linear non-least-square (i.e. general non-linear optimization) problems. (Which, in fact, is using the approximation of $H=J^TJ$) But none of them have actually explained why it is reasonable.
I have used this strategy in my own research too, and the experiments proved it to be efficient. But still, I don't know how to justify the hessian approximation mathemetically. (And I was asked by a reviewer in my recent journal paper submission.)
So again, is there any hints about how to justify the aprroximation of Hessian matrix $H=J^TJ$ for some of the general non-linear optimization problems?
Thank you very much for your kind help!

REPLY [3 votes]: It has been noted by many researchers that the Gauss-Newton Hessian often performs much better than theory predicts. For example, one can show that the Gauss-Newton Hessian will perform well when the residual is small, but it tends to perform well when the residual is large, too. Since the Gauss-Newton Hessian is positive definite, one may expect the Gauss-Newton method to outperform Newton's method when moving through regions of space where the true Hessian is indefinite, but often the Gauss-Newton method outperforms Newton's method even regions where the Hessian is positive definite. No one knows why this is (to my knowledge), but the following paper proposes an explanation and has some more general discussion and literature review of the phenomenon:
Chen, Pei. "Hessian matrix vs. Gauss–Newton hessian matrix." SIAM Journal on Numerical Analysis 49.4 (2011): 1417-1435. https://epubs.siam.org/doi/abs/10.1137/100799988
I've been trying to understand this myself. I suspect a complete understanding will come from analysis of the continuous flows associated with the Newton and Gauss-Newton methods. That is, the continuous limit of the (Gauss-)Newton method as the step size goes to zero and the number of steps goes to infinity. Here are some pictures I made comparing the continuous flows associated with the Gauss-Newton method, Newton method, and some others, on a version of the Rosenbrock function that is regularized so that there is a residual at the optimal point.