Why i getting complex double in fitting ode with experimental data

I need to optimize kinetic parameter in following equations:

dca/dw=rA
dcb/dw=rB
dcc/dw=rC
dcd/dw=rD
dce/dw=rE
dcg/dw=rG
dP/dw=(-alfa / 2) * (var(8) / T0) * (var(7) / (var(7) / P0)) * (Ft / Ft0);
dT/dW=(((U*a*(Ta-var(8))/roB*10)+((-dHr1)*(-rA1)+(-dHr2)*(-rA2)+(-dHr3)*(-rC3)+(-dHr4)*(-rA4))))/(var(1)*cpA+var(2)*cpB+var(3)*cpC+var(4)*cpD+var(5)*cpE+var(6)*cpG);
where:
rA = -(rA1 + rA2 + rA4);
rB1 = 3.5 * rA1;
rB2 = 6.5 * rA2;
rB3 = 3 * rC3;
rB4 = 15 / 6 * rA4;
rB = -(rB1 + rB2 + rB3 + rB4);
rC1 =rA1;
rC = rC1+rC3;
rD1 = 4 * rA1;
rD2 = 5 * rA2;
rD3 = rC3;
rD4 = 14 / 6 * rA4;
rD = rD1 + rD2 + rD3 + rD4;
rE2 =4 * rA2;
rE3 =4 * rC3;
rE = rE2+rE3;
rG =8 / 6 * rA4;

I used this:

https://www.mathworks.com/matlabcentral/answers/uploaded_files/64693/Igor_Moura.m

and this:

https://www.mathworks.com/matlabcentral/answers/314965-parameter-estimation-for-a-system-of-differential-equations?s_tid=prof_contriblnk

to make function:

function C=kinetics(par,W)
var0=[0.001808;0.02258704;0;0.00057911;0.00003447;0;134;431.15];
[W,Cv]=ode45(@DifEq,W,var0);
function dC=DifEq(W,var)
dcdt=zeros(8,1);
T0=431.15;
a=190.48;
Ta=673.15;
du=0.021;
U=1.22*((var(8)-Ta)/du)^(1/4);
P0=134;
pi = 3.14;
dv = 25 / 1000;
delta = 2 / 1000;
Hkat = 3.25;
roB = 0.965517241/ ((pi / 4) * ((dv - 2 * delta) ^ 2 * Hkat));
Ft=var(1)+var(2)+var(3)+var(4)+var(5)+var(6);
Ct0=P0/(8.314*T0);
Ca= Ct0 * (var(1) / Ft) * (T0 / var(8)) * (var(7) / P0);
Cb= Ct0 * (var(2) / Ft) * (T0 / var(8)) * (var(7) / P0);
Cc= Ct0 * (var(3) / Ft) * (T0 / var(8)) * (var(7) / P0);
Cd= Ct0 * (var(4) / Ft) * (T0 / var(8)) * (var(7) / P0);
Ce= Ct0 * (var(5) / Ft) * (T0 / var(8)) * (var(7) / P0);
Cg= Ct0 * (var(6) / Ft) * (T0 / var(8)) * (var(7) / P0);
FA0= 0.001808;
FB0= 0.02258704;
FC0= 0;
FD0= 0.00057911;
FE0= 0.00003447;
FG0 = 0;
Ft0=FA0+FB0+FC0+FD0+FE0+FG0;
G = 0.0007189;
epsilon = 0.45;
roO = 1.160147;
Dp = 0.003461538;
Ac = 0.00035;
viskA = ((261.5278 + 12.55191 * (var(7) / 100) - 0.55997 * (var(8) - 273.15)) / 1000000) * 3600;
yA = 0.016405;
Ma = 58;
viskB = ((17.87683461 + 4.238289936 * (var(7) / 100) + 0.016904644 * (var(8) - 273.15)) / 1000000) * 3600;
yB = 0.773086;
Mb = 28;
viskC = ((28.30049 + 5.541702 * (var(7) / 100) - 0.00358 * (var(8) - 273.15)) / 1000000) * 3600;
yC = 0.000313;
Mc = 44;
viskE = ((730.3278 + 11.11223 * (var(7) / 100) - 1.11653 * (var(8) - 273.15)) / 1000000) * 3600;
yE = 0.005255;
Me = 18;
viskD = ((-5.38734 + 2.683369 * (var(7) / 100) + 0.877204 * (var(8) - 273.15)) / 1000000) * 3600;
yD = 0.204942;
Md = 32;
viskosmj = (((viskA * yA * (Ma ^ 0.5)) + (viskB * yB * (Mb ^ 0.5)) + (viskC * yC * (Mc ^ 0.5)) + (viskD * yD * (Md ^ 0.5)) + (viskE * yE * (Me ^ 0.5))) / ((yA * (Ma ^ 0.5)) + (yB * (Mb ^ 0.5)) + (yC * (Mc ^ 0.5)) + (yD * (Md ^ 0.5)) + (yE * (Me ^ 0.5)))) / 1000;
alfa = ((2 * G * (1 - epsilon)) / (roO * Dp * epsilon ^ 3 * roB * Ac * P0)) * ((150 * (1 - epsilon) * viskosmj) / Dp + 1.75 * G) * (760 / (60 * 10));
cpA1 = 9.487;
cpA2 = 3.313e-1;
cpA3 = -0.0001108;
cpA4 = -0.000000002821;
cpB1 = 28.106;
cpB2 = -0.00000368;
cpB3 = 1.475e-5;
cpB4 = -0.00000001065;
cpC1 = -13.075;
cpC2 = 0.3484;
cpC3 = -0.0002184;
cpC4 = 0.00000004839;
cpD1 = 32.243;
cpD2 = 0.001923;
cpD3 = 0.00001055;
cpD4 = -0.000000003596;
cpE1 = 19.975;
cpE2 = 0.07343;
cpE3 = -0.00005601;
cpE4 = 0.00000001715;
cpG1 = 1.742;
cpG2 = 3.19e-1;
cpG3 = -0.0002352;
cpG4 = 6.975e-8;
cpA = cpA1 + cpA2 * var(8) + cpA3 * var(8) ^ 2 + cpA4 * (var(8) ^ 3);
cpB = cpB1 + cpB2 * var(8) + cpB3 * var(8) ^ 2 + cpB4 * (var(8) ^ 3);
cpC = cpC1 + cpC2 * var(8) + cpC3 * var(8) ^ 2 + cpC4 * (var(8) ^ 3);
cpD = cpD1 + cpD2 * var(8) + cpD3 * var(8) ^ 2 + cpD4 * (var(8) ^ 3);
cpE = cpE1 + cpE2 * var(8) + cpE3 * var(8) ^ 2 + cpE4 * (var(8) ^ 3);
cpG = cpG1 + cpG2 * var(8) + cpG3 * var(8) ^ 2 + cpG4 * (var(8) ^ 3);
rA1=((par(1) * par(2) * Ca * (Cb^par(3)))) / (1 + par(2) * Ca);
rA2=(par(4) * (Cb ^ par(3)));
rC3=(par(5) * Cc * ((Cb ^ par(6)))/ (Ca ^ par(7)));
rA4 =((par(8) * par(9)*Ca* (Cb ^ par(10)))) / (1 + par(9)*Ca);
dHr1 = -1240070 + (8.039 * (var(8) - 298.15) + (0.024805 / 2) * (var(8) ^ 2 - 298.15 ^ 2) - (0.00012 / 3) * (var(8) ^ 3 - 298.15 ^ 3) + (7.4102e-8 / 4) * (var(8) ^ 4 - 298.15 ^ 4));
dHr2 = -2646628.37 + (48.939 * (var(8) - 298.15) - (0.0279411 / 2) * (var(8) ^ 2 - 298.15 ^ 2) - (0.0001551 / 3) * (var(8) ^ 3 - 298.15 ^ 3) + (1.22801e-7 / 4) * (var(8) ^ 4 - 298.15 ^ 4));
dHr3 = -1428408.55 + (40.9 * (var(8) - 298.15) - (0.026385 / 2) * (var(8) ^ 2 - 298.15 ^ 2) - (1.19333 / 3) * (var(8) ^ 3 - 298.15 ^ 3) + (1.2141E-8 / 4) * (var(8) ^ 4 - 298.15 ^ 4));
dHr4 = -5322.22+(1005.905*(var(8)-298.15)+4.595567/2*(var(8)^2-298.15^2)-0.00224/3*(var(8)^3-298.15^3)+4.37E-7/4*(var(8)^4-298.15^4));
rA = -(rA1 + rA2 + rA4);
rB1 = 3.5 * rA1;
rB2 = 6.5 * rA2;
rB3 = 3 * rC3;
rB4 = 15 / 6 * rA4;
rB = -(rB1 + rB2 + rB3 + rB4);
rC1 =rA1;
rC = rC1-rC3;
rD1 = 4 * rA1;
rD2 = 5 * rA2;
rD3 = rC3;
rD4 = 14 / 6 * rA4;
rD = rD1 + rD2 + rD3 + rD4;
rE2 =4 * rA2;
rE3 =4 * rC3;
rE = rE2+rE3;
rG =8 / 6 * rA4;
dcdt(1)=rA;
dcdt(2)=rB;
dcdt(3)=rC;
dcdt(4)=rD;
dcdt(5)=rE;
dcdt(6)=rG;
dcdt(7)=(-alfa / 2) * (var(8) / T0) * (var(7) / (var(7) / P0)) * (Ft / Ft0);
dcdt(8)=(((U*a*(Ta-var(8))/roB*10)+((-dHr1)*(-rA1)+(-dHr2)*(-rA2)+(-dHr3)*(-rC3)+(-dHr4)*(-rA4))))/(var(1)*cpA+var(2)*cpB+var(3)*cpC+var(4)*cpD+var(5)*cpE+var(6)*cpG);
dC=dcdt;
    end
C=Cv;
end

and i made script:

W=[0
0.118832891
0.148541114
0.178249337
0.20795756
0.237665782
0.267374005
0.297082228
0.326790451
0.356498674
0.356498675
0.386206896
0.386206897
0.415915119
0.415915120
0.445623342
0.475331565
0.505039788
0.53474801
0.564456233
0.594164456
0.623872679
0.653580902
0.683289124
0.712997347
0.74270557
0.772413793
0.831830238
0.89124668
0.95066313
0.965517241];
var=[0.001808	0.022587	0	        0.000579	0.00003447	0	        134	    431.15
0.0017568	0.0225492	0.000048	0.00162	    0.0003141	0.000979	131.75	685.15
0.0017056	0.0225114	0.000096	0.002661	0.00059373	0.001958	129.5	685.15
0.0016544	0.0224736	0.000144	0.003702	0.00087336	0.002937	127.25	686.15
0.0016032	0.0224358	0.000192	0.004743	0.00115299	0.003916	125	    683.15
0.001552	0.022398	0.00024	    0.005784	0.00143262	0.004895	122.75	685.15
0.0015008	0.0223602	0.000288	0.006825	0.00171225	0.005874	120.5	685.15
0.0014496	0.0223224	0.000336	0.007866	0.00199188	0.006853	118.25	686.15
0.0013984	0.0222846	0.000384	0.008907	0.00227151	0.007832	116	    685.15
0.0013472	0.0222468	0.000432	0.009948	0.00255114	0.008811	113.75	685.15
0.001296	0.022209	0.00048	    0.010989	0.00283077	0.00979	    111.5	684.15
0.0012448	0.0221712	0.000528	0.01203	    0.0031104	0.010769	109.25	686.15
0.0011936	0.0221334	0.000576	0.013071	0.00339003	0.011748	107	    684.15
0.0011424	0.0220956	0.000624	0.014112	0.00366966	0.012727	104.75	684.15
0.0010912	0.0220578	0.000672	0.015153	0.00394929	0.013706	102.5	685.15
0.00104	    0.02202	    0.00072	    0.016194	0.00422892	0.014685	100.25	686.15
0.0009888	0.0219822	0.000768	0.017235	0.00450855	0.015664	98	    685.15
0.0009376	0.0219444	0.000816	0.018276	0.00478818	0.016643	95.75	684.15
0.0008864	0.0219066	0.000864	0.019317	0.00506781	0.017622	93.5	685.15
0.0008352	0.0218688	0.000912	0.020358	0.00534744	0.018601	91.25	683.15
0.000784	0.021831	0.00096	    0.021399	0.00562707	0.01958	    89	    686.15
0.0007328	0.0217932	0.001008	0.02244	    0.0059067	0.020559	86.75	685.15
0.0006816	0.0217554	0.001056	0.023481	0.00618633	0.021538	84.5	685.15
0.0006304	0.0217176	0.001104	0.024522	0.00646596	0.022517	82.25	685.15
0.0005792	0.0216798	0.001152	0.025563	0.00674559	0.023496	80	    686.15
0.000528	0.021642	0.0012	    0.026604	0.00702522	0.024475	77.75	686.15
0.0004768	0.0216042	0.001248	0.027645	0.00730485	0.025454	75.5	683.15
0.0004256	0.0215664	0.001296	0.028686	0.00758448	0.026433	73.25	686.15
0.0003744	0.0215286	0.001344	0.029727	0.00786411	0.027412	71	    686.15
0.0003232	0.0214908	0.001392	0.030768	0.00814374	0.028391	68.75	683.15
0.000272	0.021453	0.00144	    0.031809	0.00842337	0.02937	    66.5	685.15
]; 
par0=[0.0001;2250;0.2;0.00001;0.00001;0.1;1;0.001;1;1];
[par,Rsdnrm,Rsd,ExFlg,OptmInfo,Lmda,Jmat]=lsqcurvefit(@kinetics,par0,W,var);
fprintf(1,'\tRate Constants:\n')
for k1 = 1:length(par)
    fprintf(1, '\t\tpar(%d) = %8.5f\n', k1, par(k1))
end
Wv = linspace(min(W), max(W));
Cfit = kinetics(par, Wv);
figure(1)
plot(W, var, 'p')
hold on
plot(Wv,Cfit);
hold off
grid
xlabel('Mass, kg')
ylabel('All, K')

I write t instead of w.

I got complex double, and i got only good agreement for 8th equation.

Please, try my MATLAB code and if you know what is problem, i will appreciate help.

Thanks in advance.

[Walter Roberson]

If var(8) becomes less than Ta then after you subtract that would be negative and raising to 1/4 would be complex.

ode45 does not have any bounds for variables except what you enforce with event functions.

[Walter Roberson]

Inside DifEq, var(8) is your 8th boundary condition -- the current value of the 8th parameter, what-ever role that might happen to have in your equations.

You give ode45 initial values for the boundary conditions, but after that they evolve according to the differential equations, and ode45 makes no attempt to control them. If the differential equations happen to have var(8) become 1E10 in the 17th time step, and -1E20 in the 29th time step, then ode45 might insert extra time steps to ensure that it meets integration tolerances, but it will make no attempt to say "Hold on, var(8) has to be strictly between 0 and 634!"

The only way to get ode45 to enforce "var(8) has to be strictly between 0 and 634" is to use the ode options to add an "event function" to test that the values are inside expected bounds, and to signal termination of the ode if not. See the ballode example, which uses this to bounce a ball off of the floor (velocity and accerlation say that the position of the ball should go negative, but the event function catches negative position and the controlling routine responds by changing changing the initial conditions.

[Walter Roberson]

Your fitting parameters, par, are all complex valued. Your program misleads you about that because it uses fprintf(), and fprintf() only displays the real portion of a value.

The problem starts with the very first call to kinectics, invoked by lsqcurvefit with the initial parameters. You invoke ode45 passing in boundary conditions var0, which starts var(8) as 431.15 . When you go to calculate U, you calculate ((var(8)-Ta)/du)^(1/4) . du is a postive constant from a few lines above, so the base value is negative of var(8)-Ta is negative. Ta is 673.15 which is less than 431.15 so indeed ((var(8)-Ta)/du) is negative, giving you a complex value when raised to 1/4, leading to complex U, which polutes everything else.

Raising Ta does not help, as that makes (var(8)-Ta) more negative.

In my tests, no matter which Ta value I choose, at some point var(8) become less than it. Even if I made Ta 0, then eventually var(8) would become negative. If I made du negative against the off-chance that you had reversed the slope, then var(8) would increase until there was a problem.

[Walter Roberson]

You can add options to lsqcurvefit to permit to try longer. That might help. But so far 15000 function evaluations hasn't been enough.

[Walter Roberson]

Modified version attached. But one million function evaluations was not enough to get the curve fitting to converge.

[Walter Roberson]

Please put a breakpoint in at line 72 and when it is encountered, ask

size(XDATA)
size(YDATA)

I get 31 x 1 and 31 x 8 and it is running for me.

Which MATLAB release are you using?

[Walter Roberson]

If the curve fitting ran out of iterations then it might return the initial parameters or it might return the best it had found.

The + 0I is good and is telling you that the imaginary components of the parameters are exactly 0. I put in the code to display the imaginary components of the parameters because earlier we had not realized that we were getting complex valued parameters. It is too much bother to create code that displays the imaginary component only if it is nonzero, easier to always display it.

[Walter Roberson]

Earlier I created the symbolic form of the equations with those var0 boundary conditions. I had hoped that it might be practical to dsolve to find theoretical solutions that could then be optimized. Unfortunately most of the variables involved a ratio of sums of some of the terms, raised to 1/5 or 3/5. dsolve for maple was taking too long so I stopped that. I do not know for sure that closed form solutions are not available but it seems likely. Which leaves simulation.

I have a suspicion that this all can be rephrased as a boundary value problem, but I do not know if doing that would help.

I could perhaps rip it apart and feed it to my pet optimization program, but with 10 parameters the search would not be fast.

You mentioned that the parameters must be positive. Are there any other known bounds? Even order if magnitude would help: sending some of the parameters large enough would drive a number of terms to effectively zero, resulting in a fitting around the remaining terms, which is not necessarily invalid but you can waste a lot of time searching towards infinity looking for a combination that gets you a one bit improvement in results.

[Rena Berman]

(Answers Dev) Restored edit

Answer 1

[Stephane Dauvillier]

Sorry, I may have been confusing. The LB and UB will be the lower and upper bound of your parameter. Youtr parameter is a 10 elements vector so LB and UB should have 1à elements (that's why you should have a warning message saying

Warning: Length of lower bounds is < length(x); filling in missing lower bounds with -Inf.

I was confused by the 2 variables named both var.

Knowing that, can you set boundaries or do you need to specify constraint (see the docuement page here) ?

For your problem, the column is the "Least Squares" but depending on your constraint you lay have to change the solver.

Answer 2

[Stephane Dauvillier]

Hi it's seems that the variable U is the first to become complex (just put a breakpoint on the following line):

U=1.22*((var(8)-Ta)/du)^(1/4);

In fact on first iteration (var(8)-Ta)/du is negative, which explains why you have a complex number for the variable U

You may want to specify boundaries on var to avoid this

Answer 3

[Stephane Dauvillier]

Hi, You can specify lower and upper bounds for your var.

By the way is it normal that your function DifEq compute the following variable but don't use it :

        Cd= Ct0 * (var(4) / Ft) * (T0 / var(8)) * (var(7) / P0);
        Ce= Ct0 * (var(5) / Ft) * (T0 / var(8)) * (var(7) / P0);
        Cg= Ct0 * (var(6) / Ft) * (T0 / var(8)) * (var(7) / P0);

Answer 4

[Stephane Dauvillier]

Yes, except you have to modify LB and UB to specify these boundaries. These should be vectors (for instance LB(1) and UB(1) will be the lower and upper boundaries for var(1) and so on).

If one of the element has no lower or upper boundaries, just set it to -inf or +inf

Answer 5

[Stephane Dauvillier]

Have follow the hyperlink:

Knowing that, can you set boundaries or do you need to specify constraint (see the docuement page here) ?

Your objective function is least square since your want to minimize the square root norm.