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258190169

铜虫 (小有名气)

[求助] matlab-常微分方程参数估计

初始数据浓度和时间
t=[0,10,30,50,70,90,110,130,150,160];
c=[0,0.23211,0.45906,0.68601,0.92328,1.21213,1.32561,1.34624,1.39782,1.398];
微分方程,dc/dt=[4.41/96485-(4.41*k+L*4.41/96485)c]/(1+4.41*k*t)

要求: 1.得到拟合参数:k 和L 以及相对偏差
       2.得到拟合曲线和数据点的图
       3.最好附上院程序
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dingd

铁杆木虫 (职业作家)

现在有点明白国产软件发展艰难的原因了,起始Origin的创始人好像也是中国人,但在美国发展,大家有用正版Origin的吗?
9楼2011-12-20 14:10:09
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258190169

铜虫 (小有名气)

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=[0,0.23211,0.45906,0.68601,0.92328,1.21213,1.32561,1.34624,1.39782,1.398];
y0=0;

% Nonlinear least square estimate using lsqnonlin()
beta0=[0.005 0.001];
lb=[0 0];ub=[inf inf];
[beta,resnorm,residual,exitflag,output,lambda,jacobian] = ...
    lsqnonlin(@Func,beta0,lb,ub,[],t,y);         
ci = nlparci(beta,residual,jacobian);

% =======================================
function f = Func(beta,t,y,y0)        % Define objective function
tspan = [0  max(x)];
[tt yy] = ode45(@ModelEqs,tspan,y0,[],beta);
yc= spline(tt,yy,x);
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)

% 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,c0,[],beta);
tc=linspace(0,max(t),200);
yc = spline(tt,yc,tc);
plot(t,c,'ro',tc,yca,'r-');
hold on
xlabel('Time');
ylabel('Concentration');
hold off
2楼2011-12-16 16:49:37
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dbb627

荣誉版主 (著名写手)

【答案】应助回帖

★ ★ ★ ★ ★
感谢参与,应助指数 +1
258190169(金币+20): 多谢大侠 你的QQ是多少可以和你联系一下嘛 2011-12-16 22:12:03
cenwanglai(金币+5, 计算强帖+1): 谢谢给予帮助~ 2011-12-20 09:07:46
引用回帖:
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, ...

CODE:
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);

改的可以运行了,但是初值不合适
The more you learn, the more you know, the more you know, and the more you forget. The more you forget, the less you know. So why bother to learn.
3楼2011-12-16 17:31:51
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dingd

铁杆木虫 (职业作家)

★
dbb627(金币+1): 感谢参与 2011-12-17 11:40:34
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        1.34624
150        1.39782
160        1.398

均方差(RMSE): 0.272257391318038
残差平方和(SSE): 0.66711678414573
相关系数(R): 0.932733204157737
相关系数之平方(R^2): 0.869991230138359
决定系数(DC): 0.576237758605965

参数                  最佳估算
--------------------        -------------
k        -2.32798002359073
l        514537.340812352


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4楼2011-12-17 09:52:26
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