how to fit the coupled differential equations and get the coefficients?

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영훈 임 2023년 11월 6일

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https://kr.mathworks.com/matlabcentral/answers/2043387-how-to-fit-the-coupled-differential-equations-and-get-the-coefficients

답변: Alex Sha 2023년 11월 20일

Hi! I'm trying to fit the coupled differential equations to my data and get the coefficients. Below is the equations that I use

d(N) = -(2/3)*L1*N^2*K - (1/3)*L2*N*K^2;

d(K) = -(1/3)*L1*N^2*K - (2/3)*L2*N*K^2;

Here, N and K are the data from the experiments, and L1 and L2 are what I want to get from the fit.

I'm not sure what method I should use to get L1 and L2.. Any tips are very appreciated!

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Answer 1

Sam Chak 2023년 11월 6일

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https://kr.mathworks.com/matlabcentral/answers/2043387-how-to-fit-the-coupled-differential-equations-and-get-the-coefficients#answer_1347267

편집: Sam Chak 2023년 11월 6일

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Hi @영훈 임

You can try the Batch Least Squares method to estimate values of the parameters L1 and L2.

Update: The solution is updated at your request to plot the comparisons. The Batch Least Squares algorithm does not change.

%% For generating data purposes (real data can skip)

tspan = linspace(0, 10, 101);

[t, y1] = ode45(@odefcn1, tspan, [1 0.5]);

N = y1(:,1); % data N

K = y1(:,2); % data K

%% Batch Least Squares method to estimate L1 & L2

dN = gradient(N)./gradient(t);

dK = gradient(K)./gradient(t);

phi1 = [-(2/3)*(N.^2).*K -(1/3)*N.*(K.^2)];

phi2 = [-(1/3)*(N.^2).*K -(2/3)*(N.*K.^2)];

phi = [phi1; phi2];

Y = [dN; dK];

L = ((phi'*phi)\eye(length(phi'*phi)))*phi'*Y

L = 2×1

1.9583 2.6825

%% Testing the fitness

tspan = linspace(0, 10, 101);

[t, y2] = ode45(@(t, y) odefcn2(t, y, L), tspan, [N(1) K(1)]);

subplot(211)

plot(t, N, 'o'), hold on

plot(t, y2(:,1), 'linewidth', 1.5), title('Comparison between data N and numerical N')

grid on, xlabel('t'), ylabel('N'), legend('data N', 'numerical N')

subplot(212)

plot(t, K, 'o'), hold on

plot(t, y2(:,2), 'linewidth', 1.5), title('Comparison between data K and numerical K')

grid on, xlabel('t'), ylabel('K'), legend('data K', 'numerical K')

%% differential equations (for generating data purposes)

function dydt = odefcn1(t, y)

dydt = zeros(2, 1);

L1 = 2; % true value of L1

L2 = 3; % true value of L2

N = y(1);

K = y(2);

dydt(1) = - (2/3)*L1*N^2*K - (1/3)*L2*N*K^2;

dydt(2) = - (1/3)*L1*N^2*K - (2/3)*L2*N*K^2;

end

%% differential equations (for testing purposes)

function dydt = odefcn2(t, y, L)

dydt = zeros(2, 1);

L1 = L(1); % estimated value of L1

L2 = L(2); % estimated value of L2

N = y(1);

K = y(2);

dydt(1) = - (2/3)*L1*N^2*K - (1/3)*L2*N*K^2;

dydt(2) = - (1/3)*L1*N^2*K - (2/3)*L2*N*K^2;

end

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영훈 임 2023년 11월 6일

Thanks for your kind tips! Could I plot the fitting function based on estimated L1 and L2 to the data?

Sam Chak 2023년 11월 6일

Hi @영훈 임

I have updated my answer to show the comparison. I hope this helps. Also study @Star Strider's approach.

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Answer 2

Star Strider 2023년 11월 6일

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https://kr.mathworks.com/matlabcentral/answers/2043387-how-to-fit-the-coupled-differential-equations-and-get-the-coefficients#answer_1347367

See: Parameter Estimation for a System of Differential Equations (and similar posts) for an easy way to estimate the parameters.

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영훈 임 2023년 11월 6일

Tbody2.m

Thank you for the information! One question is that when I wrote the code based on your comment (Igor_Moura.m), I got an error message "Function value and YDATA sizes are not equal.". What is my mistake on the code? I attach my code! thanks.

Star Strider 2023년 11월 6일

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Tbody2.m

The original problem was that ode45 was crashing (saturating) and stopped. The result was that the returned matrix had fewer rows than the data matrix, throwing the error.

This works (in the sense that it runs completely), however it does not come close to fitting the data —

Tbody2

Warning: Failure at t=4.001628e-01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (8.881784e-16) at time t.

Warning: Failure at t=4.001927e-01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (8.881784e-16) at time t.

Warning: Failure at t=4.000002e-01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (8.881784e-16) at time t.

Local minimum possible. lsqcurvefit stopped because the final change in the sum of squares relative to its initial value is less than the value of the function tolerance. Rate Constants: Theta(1) = 2.78455E-05 Theta(2) = -2.72145E-05 Theta(3) = 6.32045E+05 Theta(4) = 4.79119E+05

Warning: Failure at t=4.000002e-01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (8.881784e-16) at time t.

Cfit = 1×2

1.0e+05 * 5.6884 4.3121

Error using plot
Vectors must be the same length.

Error in solution>Tbody2 (line 65)
hlp = plot(tv, Cfit);

function Tbody2

t = [0.4

0.6

0.8

1

1.2

1.5

1.8

2

2.5

3

3.5

4

4.5

5

7

10

15

20

30

50];

% c(:,1) = [770211; 715721.400000000;779650.800000000;807933.200000000;718145.600000000;746019.400000000;680262.600000000;643406.400000000;654130.600000000;597111.400000000;662146;578453;531229.400000000;574800.200000000;513836.800000000;400240.200000000;287214.400000000;308158.400000000;216808;191333.200000000];

% c(:,2) = [556098.903900000;597496.722600000;564025.111900000;506661.652400000;517884.943100000;500070.435600000;500966.444800000;517265.509000000;448318.326500000;509527.419300000;432720.902200000;434608.807900000;468438.727900000;401627.602100000;339789.050100000;351852.917000000;337195.730400000;241086.271400000;236918.419200000;161586.537200000];

c = [770211 556098.9039

715721.4 597496.7

779650.8 564025.1

807933.2 506661.6

718145.6 517884.9

746019.4 500070.4

680262.6 500966.4

643406.4 517265.5

654130.6 448318.3

597111.4 509527.4

662146 432720.9

578453 434608.8

531229.4 468438.7

574800.2 401627.6

513836.8 339789.0

400240.2 351852.9

287214.4 337195.7

308158.4 241086.2

216808 236918.40

191333.2 161586.5];

% theta0 = [1e-12; 1e-12];

theta0 = [rand(2,1)*1E-14; c(1,:).'];

[theta,Rsdnrm,Rsd,ExFlg,OptmInfo,Lmda,Jmat]=lsqcurvefit(@Loss,theta0,t,c);

fprintf(1,'\tRate Constants:\n')

for k1 = 1:length(theta)

fprintf(1, '\t\tTheta(%d) = %13.5E\n', k1, theta(k1))

end

tv = linspace(min(t), max(t));

Cfit = Loss(theta, tv)

figure(1)

plot(t, c, 'p')

hold on

hlp = plot(tv, Cfit);

hold off

grid

xlabel('Time')

ylabel('Concentration')

legend(hlp, 'C_1(t)', 'C_2(t)', 'C_3(t)', 'C_4(t)', 'Location','N')

function C=Loss(theta,t)

% c0=[1e6; 5e5];

c0 = theta(3:4)*0.9;

[T,Cv]=ode45(@DifEq,t,c0);

%

function dC=DifEq(t,c)

dcdt=zeros(2,1);

dcdt(1) = -theta(1).*(-2/3)*c(1).^2.*c(2) - theta(2).*(-1/3)*c(1).*c(2).^2;

dcdt(2) = -theta(1).*(-1/3)*c(1).^2.*c(2) - theta(2).*(-2/3)*c(1).*c(2).^2;

dC=dcdt;

end

C=Cv;

end

The changes I made specifically adding ther initial conditions for the differential equation as the last two elements of‘theta’ since that usually results in a successful fit, however that is only marginally successful here.

I am not certain what the problem is, since I do not understand the system you are modeling and estimating. One option may be to re-scale the data so that it is possible to estimate the parameters (that are currently approaching the eps value and so are likely not actually doing anything) to fit the data and then scale the parameters appropriately after estimating them. The model and the data are currently stretching the limits of floating-point computations.

You might also consider using ode15s even though your system is not specifically ‘stiff’. (I got good results with it in MATLAB Online, and better than I got here, however I was still not pleased with the results.)

.

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Answer 3

Alex Sha 2023년 11월 20일

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https://kr.mathworks.com/matlabcentral/answers/2043387-how-to-fit-the-coupled-differential-equations-and-get-the-coefficients#answer_1356107

MATLAB Online에서 열기

Refer to the results below:

Sum Squared Error (SSE): 46715596564.0427
Root of Mean Square Error (RMSE): 35062.1990798768
Correlation Coef. (R): 0.977599960330532
R-Square: 0.955701682438257
Parameter	Best Estimate       
---------	-------------       
L1       	3.57062147972612E-13
L2       	7.93804069320317E-14