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Why i getting complex double in fitting ode with experimental data

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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:
and this:
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.

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Walter Roberson
Walter Roberson on 9 May 2019
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
Walter Roberson on 9 May 2019
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.

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Accepted Answer

Stephane Dauvillier
Stephane Dauvillier on 9 May 2019
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.

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More Answers (4)

Stephane Dauvillier
Stephane Dauvillier on 9 May 2019
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

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Stephane Dauvillier
Stephane Dauvillier on 9 May 2019
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);

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Stephane Dauvillier
Stephane Dauvillier on 9 May 2019
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

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Stephane Dauvillier
Stephane Dauvillier on 9 May 2019
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.

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