In dynamical systems, we are often interested in finding stable points, or equilibria. Some systems have multiple equilibria. As an example, take the lake problem, which is modeled by the equation below where X t is the lake P concentration, a t are the anthropogenic P inputs, Y t ~LN(μ,σ2) are random natural P inputs, b is the P loss rate, and q is a shape parameter controlling the rate of P recycling from the sediment. The first three terms on the right hand side make up the “Inputs” in the figure, while the last term represents the “Outputs.” A lake is in equilibrium when the inputs are equal to the outputs and the lake P concentration therefore is not changing over time.

For irreversible lakes this occurs at three locations, even in the absence of anthropogenic and natural inputs: an oligotrophic equilibrium, an unstable equilibrium (called the critical P threshold) and a eutrophic equilibrium (see figure below).

The unstable equilibrium in this case is called the critical P threshold because once it is crossed, it is impossible to return to an oligotrophic equilibrium by reducing anthropogenic and natural P inputs alone. In irreversible lakes like this, we would therefore like to keep the lake P concentration below the critical P threshold. How do we find the critical P threshold? With a root finding algorithm!

As stated earlier, the system above will be in equilibrium when the inputs are equal to the outputs and the P concentration is not changing over time, i.e. when

Therefore we simply need to find the zero, or “root” of the above equation. Most of the methods for this require either an initial estimate or upper and lower bounds on the location of the root. These are important, since an irreversible lake will have three roots. If we are only interested in the critical P threshold, we have to make sure that we provide an estimate which leads to the unstable equilibrium, not either of the stable equilibria. If possible, you should plot the function whose root you are finding to make sure you are giving a good initial estimate or bounds, and check afterward to ensure the root that was found is the one you want! Here are several examples of root-finding methods in different programming languages.

In MATLAB, roots can be found with the function fzero(fun,x0) where ‘fun’ is the function whose root you want to find, and x0 is an initial estimate. This function uses Brent’s method, which combines several root-finding methods: bisection, secant, and inverse quadratic interpolation. Below is an example using the lake problem.

myfun = @(x,b,q) x^q/(1+x^q)-b*x; b = 0.42; q = 2.0; fun = @(x) myfun(x,b,q); pcrit = fzero(fun,0.75);

This returns pcrit = 0.5445, which is correct. If we had provided an initial estimate of 0.25 instead of 0.75, we would get pcrit = 2.6617E-19, basically 0, which is the oligotrophic equilibrium in the absence of anthropogenic and natural P inputs. If we had used 1.5 as an initial estimate, we would get pcrit = 1.8364, the eutrophic equilibrium.

In R, roots can be found with the function uniroot, which also uses Brent’s method. Dave uses this on line 10 of the function lake.eval in his OpenMORDM example. Instead of taking in an initial estimate of the root, this function takes in a lower and upper bound. This is safer, as you at least know that the root estimate will lie within these bounds. Providing an initial estimate that is close to the true value should do well, but is less predictable; the root finding algorithm may head in the opposite direction from what is desired.

b <- 0.42 q <- 2.0 pcrit <- uniroot(function(x) x^q/(1+x^q) - b*x, c(0.01, 1.5))$root

This returns pcrit = 0.5445145. Good, we got the same answer as we did with MATLAB! If we had used bounds of c(0.75, 2.0) we would have gotten 1.836426, the eutrophic equilibrium.

What if we had given bounds that included both of these equilibria, say c(0.5, 2.0)? In that case, R returns an error: ‘f() values at end points not of opposite sign’. That is, if the value returned by f(x) is greater than 0 for the lower bound, it must be less than 0 for the upper bound and vice versa. In this case both f(0.5) and f(2.0) are greater than 0, so the algorithm fails. What if we gave bounds for which one is greater than 0 and another less, but within which there are multiple roots, say c(-0.5,2.0)? Then R just reports the first one it finds, in this case pcrit = 0.836437, the eutrophic equilibrium. So it’s important to make sure you pick narrow enough bounds that include the root you want, but not roots you don’t!

In Python, you can use either scipy.optimize.root or scipy.optimize.brentq, which is what Jon uses on line 14 here. scipy.optimize.root can be used with several different algorithms, but the default is Powell’s hybrid method, also called Powell’s dogleg method. This function only requires an initial estimate of the root.

from scipy.optimize import root b = 0.42 q = 2.0 pcrit = root(lambda x: x**(1+x**q) - b*x, 0.75)

scipy.optimize.root returns an object with several attributes. The attribute of interest to us is the root, represented by x, so we want pcrit.x. In this case, we get the correct value of 0.54454. You can play around with initial estimates to see how pcrit.x changes.

Not surprisingly, scipy.optimize.brentq uses Brent’s method and requires bounds as an input.

from scipy.optimize import brentq as root b = 0.42 q = 2.0 pcrit = root(lambda x: x**(1+x**q) - b*x, 0.01, 1.5)

This just returns the root itself, pcrit = 0.5445. Again, you can play around with the bounds to see how this estimate changes.

In C++, Dave again shows how this can be done in the function ‘main-lake.cpp’ provided in the Supplementary Material to OpenMORDM linked from this page under the “Publications” section. On lines 165-168 he uses the bisect tool to find the root of the function given on lines 112-114. I’ve copied the relevant sections of his code into the function ‘find_Pcrit.cpp’ below.

#include <stdio.h> #include <stdlib.h> #include <math.h> #include <iostream> #include <boost/math/tools/roots.hpp> namespace tools = boost::math::tools; using namespace std; double b, q, pcrit; double root_function(double x) { return pow(x,q)/(1+pow(x,q)) - b*x; } bool root_termination(double min, double max) { return abs(max - min) <= 0.000001; } int main(int argc, char* argv[]) { b = 0.42; q = 2.0; std::pair<double, double> result = tools::bisect(root_function, 0.01, 1.0, root_termination); pcrit = (result.first + result.second)/2; cout << pcrit << endl; }

This yields the desired root of pcrit = 0.54454, but of course, changing the bounds may result in different estimates. In case you missed it, the take home message is to be careful about your initial estimate and bounds ;).