Solving Highly Irregular Non-Linear Equations

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Alex on 13 Feb 2012
Commented: Walter Roberson on 18 Dec 2021
Hey all, I'm trying to solve a system of 3 highly irregular nonlinear equations using Matlab, but the solutions I get are not the expected values. Here are the equations in the most basic form, and the variables I am currently using:
The answers are never close to the expected values, which are n=1.5, k=0, and d=300. I'm using the fsolve command to perform the calculations. Any tips or suggestions?

Answers (2)

Walter Roberson
Walter Roberson on 13 Feb 2012
I used the equivalent expressions in Maple,
fcns[1] := ((n0-n+I*k)/(n0+n-I*k)+(n-I*k-n21+I*k21)*exp(-2*(2*I)*Pi*(n-I*k)*d/y1)/(n-I*k+n21-I*k21))/(1+(n0-n+I*k)*(n-I*k-n21+I*k21)*exp(-2*(2*I)*Pi*(n-I*k)*d/y1)/((n0+n-I*k)*(n-I*k+n21-I*k21)))-R1;
fcns[2] := ((n0-n+I*k)/(n0+n-I*k)+(n-I*k-n22+I*k22)*exp(-2*(2*I)*Pi*(n-I*k)*d/y2)/(n-I*k+n22-I*k22))/(1+(n0-n+I*k)*(n-I*k-n22+I*k22)*exp(-2*(2*I)*Pi*(n-I*k)*d/y2)/((n0+n-I*k)*(n-I*k+n22-I*k22)))-R2;
fcns[3] := ((n0-n+I*k)/(n0+n-I*k)+(n-I*k-n23+I*k23)*exp(-2*(2*I)*Pi*(n-I*k)*d/y3)/(n-I*k+n23-I*k23))/(1+(n0-n+I*k)*(n-I*k-n23+I*k23)*exp(-2*(2*I)*Pi*(n-I*k)*d/y3)/((n0+n-I*k)*(n-I*k+n23-I*k23)))-R3;
When I substitute your proposed n, k, d values in to that, the expressions come out non zero and not equal:
[0.5428451377e-1+.2713792967*I, 0.4862923664e-1+.2764299574*I, 0.4291778332e-1+.2813827716*I]
If I did not make an editing error in doing the conversion (and that is a possibility) then your proposed values are not solutions ?
Walter Roberson
Walter Roberson on 16 Feb 2012
With those values, your equations cannot produce anywhere near the desired result.
You have given 3 lines of values, with a header line that does not directly correspond to many of the variables in the equations. For example there is no "y" in the equation, so what are we to do with the y value from the table? The seemingly obvious thing to do would be to guess that the first line corresponds to y1, the second line to y2, the third to y3, and so on for the other values. But then we get to the variables "n", "k" and "d", with there being no k1, k2, k3, d1, d2, d3. But the three lines have identical k and d values so we can speculate that we should use that constant value for k, and the constant value for d. But now what do we do with "n" ? The values of "n" are different for the three lines and there is no series n1/n2/n3, and your search is looking for "n" and "k" and "d". So we hold all three n, k, and d as being variables to be solved for and hope we get the values from the table out... a plausibility for k and d, but only one single "n" value can be calculated at best, and there are three different "n" values in the table.
If you do the variable splitting as I describe and use n=1.548756 from the first line of the table, and use k = 0, and then individually solve each of the equations for d to determine whether you can get something "close to" the expected 300, the answers for k that come out are values that are not so far apart -- but they are all complex values, and all have magnitude on the order of 80.5 which is a long way from 300.
If you substitute in k=0 and d=300 and try to solve the individual equations for n, then you get complex values. In the _approximate_ area of the solution you are looking for, at roughly n = 1.25, there is a point at which the imaginary part is 0... but then the real part of the expression is non-zero which should not be the case if n was a solution to the equation. At least it is positive there, though.
Roughly speaking, the imaginary portion for n follows the pattern of a sine wave of increasing frequency, but soon after n=1.3, the real portion goes negative and becomes a decaying sine-like wave approximately centered around -0.7, probably never crossing 0 again.
Your equations appear to be, to use the technical term, "borked".

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Siamak Lotfi
Siamak Lotfi on 18 Dec 2021
if its does work , you may consider give a try with blocks in Simulink
  1 Comment
Walter Roberson
Walter Roberson on 18 Dec 2021
Simulink cannot calculate things that have no solution. Simulink could at most be used to numerically find a point of local minima of error relative to the expected values, which is something that could be done directly in MATLAB if that were the task.

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