楼: Originally posted by dingd at 2011-12-17 09:52:26:
用1stOpt试试：
[code]
Variable t,c;
ODEFunction c'=(4.41/96485-(4.41*k+L*4.41/96485)*c)/(1+4.41*k*t);
Data;
0        0
10        0.23211
30        0.45906
50        0.68601
70        0.92328
90        1.21213
110        1.32561
130        1. ...

2楼: Originally posted by 258190169 at 2011-12-16 16:49:37:
本人编写的程序如下但是无法运行，由于是新手，希望高手帮忙调试一下：
function PenicilliumEst
clear all;
t=[0,10,30,50,70,90,110,130,150,160];
y=[0,0.23211,0.45906,0.68601,0.92328,1.21213,1.32561, ...

function PenicilliumEst
clear all;
t=[0,10,30,50,70,90,110,130,150,160];
y=[0,0.23211,0.45906,0.68601,0.92328,1.21213,1.32561,1.34624,1.39782,1.398];
y0=0;

% Nonlinear least square estimate using lsqnonlin()
beta0=[0.001 0.001];
lb=[0 0];ub=[inf 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 k1 = %.4f ± %.4f\n',beta(1),ci(1,2)-beta(1))
fprintf('\t k2 = %.4f ± %.4f\n',beta(2),ci(2,2)-beta(2))
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 = (4.41/96485-(4.41*beta(1)+beta(2)*4.41/96485)*y)/(1+4.41*beta(1)*t);

Variable t,c;
ODEFunction c'=(4.41/96485-(4.41*k+L*4.41/96485)*c)/(1+4.41*k*t);
Data;
0        0
10        0.23211
30        0.45906
50        0.68601
70        0.92328
90        1.21213
110        1.32561
130        1.34624
150        1.39782
160        1.398