Hi @영훈 임
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
%% 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
