[交流] matlab求解方程中的参数 已有4人参与

18楼: Originally posted by dingd at 2012-07-04 13:56:45
用新版本的都是自己或单位买的正版。

function parafit
clear all;
t=[0 2 4 8 24 48];
y=[0.69 0.645 0.635 0.62 0.61 0.61];
y0=0.69;

% Nonlinear least square estimate using lsqnonlin()
beta0=0.5;
lb=[0];ub=[inf];
[beta,resnorm,residual,exitflag,output,lambda,jacobian] = ...
    lsqnonlin(@Func,beta0,lb,ub,[],t,y,y0);         
ci = nlparci(beta,residual,jacobian);
beta;
% result
fprintf('\n Estimated Parameters by Lsqnonlin():\n')
fprintf('\t k = %.4f ± %.4f\n',beta(1),ci(1,2)-beta(1))
fprintf('  The sum of the residual squares is: %.1e\n\n',sum(residual.^2))

% plot of fit results
tspan = [0  max(t)];
[tt yc] = ode45(@ModelEqs,tspan,y0,[],beta);
tc=linspace(0,max(t),200);
yca = spline(tt,yc,tc);
plot(t,y,'ro',tc,yca,'r-');
hold on
xlabel('Time');
ylabel('Concentration');
hold off
% =======================================
function f1 = Func(beta,t,y,y0)        % Define objective function
tspan =t;
[tt yy] = ode45(@ModelEqs,tspan,y0,[],beta);
yc= spline(tt,yy,t);
f1=y-yc;
% ==================================
function dydt = ModelEqs(t,y,beta)          % Model equations
dydt = -beta*(0.7-y).^(1/3)*(y-0.61);

3楼: Originally posted by dbb627 at 2012-07-02 17:39:27
结果如下
Local minimum possible.

lsqnonlin stopped because the final change in the sum of squares relative to
its initial value is less than the default value of the function tolerance.

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