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I have a system of equations that cannot be solved for in closed form:

$F_1(x_1,x_2,\beta)=0 ~\&~ F_2(x_1,x_2,\beta)=0 $

I want to solve for functions $x_1=x_1(\beta) ~\&~ x_2=x_2(\beta)$

My approach so far:

  1. Fix $\beta=\beta_1$
  2. Solve system numerically for $x^1_1,x^1_2$.
  3. Repeat for a different $\beta$ on some pre-defined finite support
  4. Generate a vector of $\mathbf{x_1}=(x^1_1...x^n_1)$ with corresponding $\mathbf{b}=(\beta_1...\beta_n)$
  5. Approximate the relationship between $x_1$ and $\beta$ using Sieve to get $\hat{x}_1(\beta)$ [repeat for $x_2$]

The problem with this method is that it gets computationally out of hand once I have many equations and parameters due to step 3.

I do know that the system has a solution for any $\beta$, which motivated this method. I also can easily get the Jacobian of $F$ analytically.

Is there a preferable method that is not as computationally burdensome?

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    $\begingroup$ Are you plotting bifurcation diagrams? Bifurcation diagrams are a very very difficult problem to get correct in the most general case. You can look at arclength continuation algorithms, but you should probably use a dedicated software for this because every bifurcation needs special handling. MATLAB = matcont, Python = PyDSTool, Julia = DifferentialEquations.jl, anything else you can use AUTO/XPPAUT $\endgroup$ – Chris Rackauckas May 22 '17 at 21:29
  • $\begingroup$ @ChrisRackauckas I am not. The context is a system of first order conditions that solve for optimal policy functions. $\endgroup$ – VCG May 22 '17 at 21:31
  • $\begingroup$ Well you can probably re-phrase this as a bifurcation problem. You're just finding the steady states of F given each beta. $\endgroup$ – Chris Rackauckas May 22 '17 at 21:33
  • $\begingroup$ Or better, if there are no bifurcations at all, you can just implement path following (continuation) methods that should make solving the problem for one $\beta_k$ much cheaper if you already have a solution for a nearby $\beta_{k-1}$. $\endgroup$ – Wolfgang Bangerth May 22 '17 at 22:59
  • $\begingroup$ @WolfgangBangerth You mean something like a Newton's Method approach? The problem with that is I don't know how to generate the $x(\beta)$ function with such an approach. $\endgroup$ – VCG May 22 '17 at 23:03

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