fsolve/fzero: No solution found, appears regular - matlab

Question

I am trying to do the following algorithm using fsolve or fzero:
K5=8.37e-2
P=1
Choose an A
S2=(4*K5/A)^(2/3)
S6=3*S2
S8=4*S2
SO2 = (5*P)/149 - (101*S2)/149 - (293*S6)/149 - (389*S8)/149
H2O = (40*P)/149 + (556*S2)/447 + (636*S6)/149 + (2584*S8)/447
H2S = 2*SO2
newA = (H2O)^2/(SO2)^3
Repeat until newA=oldA
The main thing to solve is K5=1/4 * A * S2^3/2. It is from this that S2 is calculated in the first place.
So here is what I did in Matlab:
function MultipleNLEexample
clear, clc, format short g, format compact
Aguess = 300000; % initial guess
options = optimoptions('fsolve','Display','iter','TolFun',[1e-9],'TolX',[1e-9]); % Option to display output
xsolv=fsolve(#MNLEfun,Aguess,options);
[~,ans]=MNLEfun(xsolv)
%- - - - - - - - - - - - - - - - - - - - - -
function varargout = MNLEfun(A);
K5 = 8.37e-2;
S2 = (4*K5/A)^(2/3);
S6 = 3*S2;
S8 = 4*S2;
P=1; %atm
SO2 = (5*P)/149 - (101*S2)/149 - (293*S6)/149 - (389*S8)/149;
H2O = (40*P)/149 + (556*S2)/447 + (636*S6)/149 + (2584*S8)/447;
newA=H2O^2/SO2^3;
fx=1/4*newA*S2^(3/2)-K5;
varargout{1} = fx;
if nargout>1
H2S = 2*SO2;
varargout{2} = ((2*S2+6*S6+8*S8)/(2*S2+6*S6+8*S8+H2S+SO2)*100);
end
I cannot get my code to run, I get the following error:
No solution found.
fsolve stopped because the problem appears regular as measured by the gradient,
but the vector of function values is not near zero as measured by the
selected value of the function tolerance.
I have tried setting the tolerances as low as 1e-20 but that did not change anything.

Answer 1

The way your system is set up, it is actually convenient to plot it and observe its behavior. I vectorized your function and plottedf(x) = MLNEfun(x)-x, where the output of MLNE(x) is newA. Effectively, you are interested in a fixed point of your system.
What I observe is this:
There is a singularity, and a root crossing, at A ~ 3800. We can use fzero since it is a bracketed root solver, and give it very tight bounds on the solution fzero(#(x)MLNEfun(x)-x, [3824,3825]) which produces 3.8243e+03. This is a couple order of magnitudes from your starting guess. There is no solution to your system near ~3e5.
Update
In my haste, I failed to zoom in on the plot, which shows another (well-behaved) root at 1.3294e+04. It is up to you to decide which is the physically meaningful one(s). Everything I say below still applies. Just start your guess near the solution you're interested in.
In response to comment
Since you want to perform this for varying values of K, then your best bet is to stick with fzero so long as you are solving for one variable, instead of fsolve. The reason behind this is that fsolve uses variants of Newton's method, which are not bracketed and will struggle to find solutions at singular points like this. fzero on the other hand, uses Brent's method which is guaranteed to find a root (if it exists) within a bracket. It is also much more well behaved near singular points.
MATLAB's implementation of fzero also searches for a bracketing interval before carrying out Brent's method. So, if you provide a single starting guess sufficiently close to the root, it should find it for you. Sample output from fzero below:
fzero(#(x)MLNEfun(x)-x, 3000, optimset('display', 'iter'))
Search for an interval around 3000 containing a sign change:
Func-count a f(a) b f(b) Procedure
1 3000 -616789 3000 -616789 initial interval
3 2915.15 -433170 3084.85 -905801 search
5 2880 -377057 3120 -1.07362e+06 search
7 2830.29 -311972 3169.71 -1.38274e+06 search
9 2760 -241524 3240 -2.03722e+06 search
11 2660.59 -171701 3339.41 -3.80346e+06 search
13 2520 -109658 3480 -1.16164e+07 search
15 2321.18 -61340.4 3678.82 -1.7387e+08 search
17 2040 -29142.6 3960 2.52373e+08 search
Search for a zero in the interval [2040, 3960]:
Func-count x f(x) Procedure
17 2040 -29142.6 initial
18 2040.22 -29158.9 interpolation
19 3000.11 -617085 bisection
20 3480.06 -1.16224e+07 bisection
21 3960 2.52373e+08 bisection
22 3720.03 -4.83826e+08 interpolation
....
87 3824.32 -5.46204e+48 bisection
88 3824.32 1.03576e+50 bisection
89 3824.32 1.03576e+50 interpolation
Current point x may be near a singular point. The interval [2040, 3960]
reduced to the requested tolerance and the function changes sign in the interval,
but f(x) increased in magnitude as the interval reduced.
ans =
3.8243e+03