lsqcurvefit for two different equations and variables

Hello,

i have a data which consists of one input value (i) and two output values (Y) values, say Y1 and Y2. I want to fit my experimental data to a model using lsqcurvefit. what i know is that i can fit my data using only one X variable and one Y variable. is there a way that i can optimize my functions to fit both variables at the same time?

i = [1.17451E+00
1.87115E+00
2.98100E+00
4.74915E+00
7.56606E+00
1.20538E+01
1.92034E+01
3.05936E+01
4.87399E+01
7.76494E+01
1.23706E+02
1.97081E+02
3.13978E+02
5.00210E+02
7.96904E+02
1.26958E+03
2.02262E+03
3.22231E+03
5.13358E+03
];

y1 = [242.5088007
444.8689004
802.8335979
1425.305568
2489.312239
4277.004598
7229.179527
12405.99648
20430.99265
33235.46262
53403.34191
87814.94488
138006.2801
213871.8819
326838.3103
492534.4889
731921.9324
1072549.39
1549866.402
];

y2 = [3377.017069
5268.932921
8164.853585
12566.38759
19209.17965
29163.75918
43975.87673
64027.8721
95789.07289
142240.1411
209646.5318
298137.6129
433807.5582
626647.3615
898659.2993
1279418.677
1808323.209
2537378.074
3534598.31
];

% i have 3 parameters 

% define the equations
y1_theo = @(x,xdata)x(1).*(1+(x(2)./xdata).^(log10(2)/x(3))).^(-x(3)./log10 (2));
% define d
y2_theo = @(x,xdata)90./(1+(xdata./x(2)).^(log10(2)./x(3)));
% initial values
x0 = [1E+9 1000 1];
% specify the bounds
lb=[0 0 0];
ub = [inf inf inf];

% optimize
[x,resnorm,~,exitflag,output] = lsqcurvefit([y1_theo y2_theo],x0,i,[y1 y2],lb,ub)

I have tried to write the code as above, but it is not working, can someone please help me with this?

Answer 1

[Star Strider]

This works, however it only fits one of the curves:

y_theo = @(x,xdata) [x(1).*(1+(x(2)./xdata).^(log10(2)/x(3))).^(-x(3)./log10 (2)),  90./(1+(xdata./x(2)).^(log10(2)./x(3)))];

x0 = [1E+9 1000 1];
% specify the bounds
lb=[0 0 0];
ub = [inf inf inf];
% optimize
[x,resnorm,~,exitflag,output] = lsqcurvefit(y_theo,x0,i,[y1 y2],lb,ub)

figure
plot(i, [y1 y2], 'p')
hold on
plot(i, y_theo(x,i), '-r')
hold off
grid

It may be necessary to review both parts of ‘y_theo’ to be certain they are coded correctly, and that ‘x0’ is appropriate. (I have used this sort of construction in other problems, and it worked as it should.)

[Mohammad Aljarrah]

Thank you very much for your quick answer, try using these values. i saw 2 lines. does this mean it is working now? is it optimizing for both equations now?

i = [4.15029E-06
6.71874E-06
1.08767E-05
1.76078E-05
2.85046E-05
4.61449E-05
7.47022E-05
1.20932E-04
1.95772E-04
3.16928E-04
5.13061E-04
8.30574E-04
1.34458E-03
2.17669E-03
3.52375E-03
5.70446E-03
9.23471E-03
1.49497E-02
2.42015E-02
3.91787E-02
6.34249E-02
1.02676E-01
1.66218E-01
2.69083E-01
4.35608E-01
7.05188E-01
1.14160E+00
1.84809E+00
2.99179E+00
4.84329E+00
7.84061E+00
1.26928E+01
2.05479E+01
3.32642E+01
5.38500E+01
8.71756E+01
1.41125E+02
2.28461E+02
3.69847E+02
5.98730E+02
9.69259E+02
1.56909E+03
2.54014E+03
4.11213E+03
6.65696E+03
1.07767E+04
1.74459E+04
2.82425E+04
4.57206E+04
7.40153E+04
1.19820E+05
1.93972E+05
3.14014E+05
5.08344E+05
8.22937E+05
1.33222E+06
2.15667E+06
3.49135E+06
5.65201E+06
9.14980E+06
1.48122E+07
2.39789E+07
3.88185E+07
6.28417E+07
1.01732E+08
1.64689E+08
2.66609E+08
4.31602E+08
];

y1 = [0.04419469
0.071367491
0.11521246
0.185930771
0.299942579
0.483659931
0.779533878
1.255734734
2.021627159
3.252470998
5.22878017
8.398917017
13.4784238
21.60727031
34.59810921
55.32701539
88.34638555
140.8426286
224.1261891
355.9386529
564.0057471
891.4760084
1405.189562
2113.056409
3395.65126
5403.130988
8513.425194
13283.82459
20526.72838
30705.14291
46647.46912
70559.47955
106265.2087
158446.5577
234778.5043
343613.8495
496663.7065
695167.8705
985034.4395
1383562.26
1926031.889
2656887.5
3631225.728
4916202.083
6592218.989
8336330.414
10690016.43
13518977.34
16861214.85
20742955.71
25175872.77
30155158.84
35658692.04
41647403.6
48066815.5
54849581.03
61918758.77
69191493.49
76582769.24
84008934.93
91390767.86
98655922.32
105740691.8
112591086.8
119163285.5
125423551.2
131347731.3
136920453.1
];

y2 = [89.56
89.51
89.45
89.38
89.31
89.22
89.13
89.02
88.90
88.77
88.62
88.46
88.27
88.07
87.84
87.58
87.29
86.97
86.61
86.22
85.77
85.28
84.74
86.28
85.54
84.72
83.81
82.82
81.77
80.37
79.17
77.93
76.65
74.75
73.43
72.12
70.84
68.88
67.66
66.49
65.38
64.35
63.41
62.56
61.81
47.76
45.16
42.57
39.99
37.44
34.94
32.51
30.15
27.88
25.70
23.64
21.68
19.84
18.11
16.49
14.99
13.61
12.33
11.15
10.07
9.08
8.18
7.36
];

[Star Strider]

My pleasure!

With the same code (except for using a loglog plot):

loglog(i, [y1 y2], 'p')

: and those data, it appears to be estimating both columns correctly, producing:

x =
          195548569.332446          17879.4178046969          1.25650432614129

and:

The only way to show both of them completely is with a loglog plot, since the magnitudes span a significant range.

[Mohammad Aljarrah]

That look amazing. Thank you very much for your help and time.

[Star Strider]

As always, my pleasure!