I am building a code to solve a diff. equation:
function dy = KIN1PARM(t,y,k)
%
% version : first order reaction
% A --> B
% dA/dt = -k*A
% integrated
So here comes the best which I can get right now. For my way I tread ordinatus values as time and the abscissa values as measured quantity which you try to model. Also, you seem to have set alot of options for the solver, which I all omitted. First comes your proposed solution using ode45()
, but with a non-zero y0 = 100
, which I just "guessed" from looking at the data (in a semilogarithmic plot).
function main
abscissa = [0;
240;
480;
720;
960;
1140;
1380;
1620;
1800;
2040;
2220;
2460;
2700;
2940];
ordinatus = [ 0;
19.6;
36.7;
49.0;
57.1;
64.5;
71.4;
75.2;
78.7;
81.3;
83.3;
85.5;
87.0;
87.7];
tspan = [min(ordinatus), max(ordinatus)]; % // assuming ordinatus is time
y0 = 100; % // <---- Probably the most important parameter to guess
k0 = -0.1; % // <--- second most important parameter to guess (negative for growth)
k_opt = fminsearch(@minimize, k0) % // optimization only over k
% nested minimization function
function e = minimize(k)
sol = ode45(@KIN1PARM, tspan, y0, [], k);
y_hat = deval(sol, ordinatus); % // evaluate solution at given times
e = sum((y_hat' - abscissa).^2); % // compute squarederror
end
% // plot with optimal parameter
[T,Y] = ode45(@KIN1PARM, tspan, y0, [], k_opt);
figure
plot(ordinatus, abscissa,'ko', 'markersize',10,'markerfacecolor','black')
hold on
plot(T,Y, 'r--', 'linewidth', 2)
% // Another attempt with fminsearch and the integral form
t = ordinatus;
t_fit = linspace(min(ordinatus), max(ordinatus))
y = abscissa;
% create model function with parameters A0 = p(1) and k = p(2)
model = @(p, t) p(1)*exp(-p(2)*t);
e = @(p) sum((y - model(p, t)).^2); % minimize squared errors
p0 = [100, -0.1]; % an initial guess (positive A0 and probably negative k for exp. growth)
p_fit = fminsearch(e, p0); % Optimize
% Add to plot
plot(t_fit, model(p_fit, t_fit), 'b-', 'linewidth', 2)
legend('location', 'best', 'data', 'ode45 with fixed y0', ...
sprintf ('integral form: %5.1f*exp(-%.4f)', p_fit))
end
function dy = KIN1PARM(t,y,k)
%
% version : first order reaction
% A --> B
% dA/dt = -k*A
% integrated form A = A0*exp(-k*t)
%
dy = -k.*y;
end
The result can be seen below. Quit surprisingly to me, the initial guess of y0 = 100
fits quite well with the optimal A0
found. The result can be seen below: