ode45求解差分方程，并进一步拟合结果 [英] ode45 solving of diff.equation with further fitting to exp.results
问题描述
我正在构建用于解决差异的代码.等式:
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 form A = A0*exp(-k*t)
%
dy = -k.*y;
end
我希望对该方程进行数值求解，并希望将结果(y作为t和k的函数)用于实验值的最小化，以获得参数k的最佳值.
I want this equation to be solved numerically and the results (y as a function of t, and k) to be used for minimization with respect to the experimental values to get the optimal value of parameter k.
function SSE = SSE_minimization_1parm(tspan_inp,val_exp,k_inp,y0_inp)
f = @(Tt,Ty) KIN1PARM(Tt,Ty,k_inp); %function to call ode45
size_limit = length(y0_inp);
options = odeset('NonNegative',1:size_limit,'RelTol',1e-4,'AbsTol', 1e-4);
[ts,val_theo] = ode45(f, tspan_inp, y0_inp,options); %Cexp is the state variable predicted by the model
err = val_exp - val_theo;
SSE = sum(err.^2); %sum squared-error
绘制实验数据和计算数据的主要代码是:
The main code to plot the experimental and calculated data is:
% Analyzing first order kinetics
clear all; clc;
figure_title = 'Experimental Data';
label_abscissa = 'Time [s]';
label_ordinatus = 'Concentration [mol/L]';
%
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];
%
title_string = [' Time [s]', ' | ', ' Complex [mol/L] ', ' '];
disp(title_string);
for i=1:length(abscissa)
report_raw_data{i} = sprintf('%1.3E\t',abscissa(i),ordinatus(i));
disp([report_raw_data{i}]);
end;
%---------------------/plotting dot data/-------------------------------------
%
f = figure('Position', [100 100 700 500]);
title(figure_title,'FontName','arial','FontWeight','bold', 'FontSize', 12);
xlabel(label_abscissa, 'FontSize', 12);
ylabel(label_ordinatus, 'FontSize', 12);
%
grid on; hold on;
%
marker_style = { 's'};
%
plot(abscissa,ordinatus, marker_style{1},...
'MarkerFaceColor', 'black',...
'MarkerEdgeColor', 'black',...
'MarkerSize',4);
%---------------------/Analyzing/----------------------------------------
%
options = optimset('Display','iter','TolFun',1e-4,'TolX',1e-4);
%
CPUtime0 = cputime;
Time_M = abscissa;
Concentration_M = ordinatus;
tspan = Time_M;
y0 = 0;
k0 = rand(1);
[k, fval, exitflag, output] = fminsearch(@(k) SSE_minimization_1parm(tspan,Concentration_M,k,y0),k0,options);
CPUtimex = cputime;
CPUtime_delay = CPUtimex - CPUtime0;
%
%---------------------/plotting calculated data/-------------------------------------
%
xupperlimit = Time_M(length(Time_M));
xval = ([0:1:xupperlimit])';
%
yvector = data4plot_1parm(xval,k,y0);
plot(xval,yvector, 'r');
hold on;
%---------------------/printing calculated data/-------------------------------------
%
disp('RESULTS:');
disp(['CPU time: ',sprintf('%0.5f\t',CPUtime_delay),' sec']);
disp(['k: ',sprintf('%1.3E\t',k')]);
disp(['fval: ',sprintf('%1.3E\t',fval)]);
disp(['exitflag: ',sprintf('%1.3E\t',exitflag)]);
disp(output);
disp(['Output: ',output.message]);
相应的函数，它使用优化的参数k得出计算出的y = f(t)数据:
The corresponding function, which uses the optimized parameter k to yield the calculated y = f(t) data :
function val = data4plot_1parm(tspan_inp,k_inp,y0_inp)
f = @(Tt,Ty) KIN1PARM(Tt,Ty,k_inp);
size_limit = length(y0_inp);
options = odeset('NonNegative',1:size_limit,'RelTol',1e-4,'AbsTol',1e-4);
[ts,val_theo] = ode45(f, tspan_inp, y0_inp, options);
代码运行优化周期时，总是会给出不同的参数k值，这与使用ln(y)vs t计算的值不同(对于该系列的exp.数据应为7.0e-4).
The code runs optimization cycles always giving different values of parameter k, which are different from the value calculated using ln(y) vs t (should be around 7.0e-4 for that series of exp. data).
看一下ode求解器的结果(SSE_minimization_1parm => val_theo)，我发现ode函数为我提供了一个零向量.
Looking at the outcome of the ode solver (SSE_minimization_1parm => val_theo) I found that the ode function gives me a vector of zeroes.
请有人帮我弄清楚ode求解器的功能吗?
Could someone help me , please, to figure out what's going with the ode solver ?
非常感谢！
推荐答案
所以这是我现在能得到的最好的东西.以我的方式，我将ordinatus值表示为时间，将横坐标值表示为要尝试建模的测量量.另外，您似乎已经为求解器设置了很多选项，我都省略了这些选项.首先是您使用ode45()提出的解决方案，但使用了非零的y0 = 100，我只是从看数据(半对数图)中猜测"到了这个问题.
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
结果可以在下面看到.令我惊讶的是，y0 = 100的初始猜测与找到的最佳A0非常吻合.结果可以在下面看到:
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:
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