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Given $a\in\mathbb{R}^{mn\times n}$, find a $C\in\mathbb{R}^{n}$, $x\in\mathbb{R}^{m\times n}$ such that $$ 0 = f_{k}(\boldsymbol{C}, \boldsymbol{x}):=\sum_{i=1}^{m} C_{i} \left(\prod_{j=1}^{n} a_{kj}^{x_{i,j}} - 1\right) $$ for all $k\in\{1,\dots,mn\}$.

(The rows of $a_{kj}$ are vectors of solutions to a system of differential equations, and the goal of finding the roots of the above system is to find invariant quantities of those diff. eq.)

To show that the system has a singular Jacobian, I have a test setup where $a_{k1} = (1-2t_{k})^{-0.5}$, $a_{k2} = (1-2t_{k})^{-1}$ and the rest zero (only $i=1$, but it is still singular if you include the sum). It is in this case easy to check that the Jacobian is singular irregardless of the values of $\boldsymbol{C}$ and $\boldsymbol{x}$. The solution is any $x$ with $2x_{11} - x_{12}=0$ and $C$ anything, and I'm guessing that the Jacobian is singular because this does imply infinite solutions.

I just need to be able to find a couple of them (that are non-trivial such as $C_{i} =0$, and I can go from there. Most root-finding methods I found that can deal with singular Jacobians only mention singularities at the solution, not everywhere. Are there any methods that can tackle this problem? I should also mention that the system should eventually get quite large (order of 1000 equations).

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  • $\begingroup$ possible duplicate of Finding zeroes of an infinitely differentiable function of ~100 to ~1000 variables $\endgroup$ – Nico Schlömer Oct 17 '13 at 4:07
  • $\begingroup$ You don't need any gradient-based methods for finding roots of this function; see the answer to your previous question. $\endgroup$ – Nico Schlömer Oct 17 '13 at 4:08
  • $\begingroup$ But this is a system of equations. I understand I can write down solutions to a single of these equations, but how would I then find a solution to all equations? $\endgroup$ – RobVerheyen Oct 17 '13 at 6:57
  • $\begingroup$ For every $k$, the equation above is the same as the one you presented before, and since the $a_{k,*}$ aren't coupled, you can consider $N$ separate problems (or whatever the maximum value of $k$ is). $\endgroup$ – Nico Schlömer Oct 17 '13 at 16:36
  • $\begingroup$ I understand that. I can write down a solution for every of these equations, but the challenge is then to find a solution that fits all the equations. I don't see an easy way of doing that. Apologies if I'm not seeing obvious things here... $\endgroup$ – RobVerheyen Oct 17 '13 at 18:10

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