matlab 拟合反应动力学参数结果很差。大家帮忙看一下

作者 楚天湘水: 来源: 小木虫 700 14 举报帖子

已知动力学模型为：
dA1dt = r1 + r2;
dA2dt =  - r1-r2;
dA3dt = -r1；
dA4dt = r1 - r2;
dA5dt = r2;

r1 = k(1)*((a(3)*a(2) - (1/2.20)*a(4)*a(1)));
r2 = k(2)*(a(4)*a(2) -  (1/0.40)*a(5)*a(1)));

已知的参数：
时间：tspan = [0 60 120 180 240 300 360 420 600 1200 6900 7500],
活度系数：KineticsData1
% 动力学数据
%  a1 a2 a3 a4 a5
ExpData=...
[0.0000 0.8122 0.3894 0.0000 0.0000
   0.1827 0.6353 0.1599 0.2163 0.0444
   0.2547 0.5668 0.0979 0.2485 0.0805
   0.2854 0.5390 0.0788 0.2415 0.1061
   0.2978 0.5258 0.0738 0.2359 0.1192
   0.3019 0.5210 0.0710 0.2309 0.1288
   0.3092 0.5143 0.0691 0.2264 0.1349
   0.3178 0.5062 0.0680 0.2223 0.1349
   0.3178 0.5062 0.0680 0.2186 0.1433
   0.3211 0.5030 0.0674 0.2154 0.1478
   0.3209 0.5036 0.0673 0.2140 0.1482
   0.3205 0.5045 0.0674 0.2136 0.1476 ]

下面是写的matlab程序：
function Kinetic2
% 动力学ODE方程模型的参数估计
clear all
clc

k0 = [900000 50000];       % 参数初值
lb = [0  0];                % 参数下限
ub = [+inf  +inf ]; % 参数上限
a0 = [0.0000 0.8122 0.3894 0.0000 0.0000];
KineticsData1;
yexp = ExpData(:,1:5); % yexp: 实验数据[a1 a2 a3 a4 a5]

% 使用函数fmincon()进行参数估计
[k,fval,flag] = fmincon(@ObjFunc4Fmincon,k0,[],[],[],[],lb,ub,[],[],a0,yexp);
fprintf('\n使用函数fmincon()估计得到的参数值为:\n')
fprintf('\tk1 = %.4f\n',k(1))
fprintf('\tk2 = %.4f\n',k(2))
fprintf('  The sum of the squares is: %.1e\n\n',fval)
k_fmincon = k;

% 使用函数lsqnonlin()进行参数估计
[k,resnorm,residual,exitflag,output,lambda,jacobian] = ...
lsqnonlin(@ObjFunc4LNL,k0,lb,ub,[],a0,yexp);
ci = nlparci(k,residual,jacobian);
fprintf('\n\n使用函数lsqnonlin()估计得到的参数值为:\n')
fprintf('\tk1 = %.4f\n',k(1))
fprintf('\tk2 = %.4f\n',k(2))
fprintf('  The sum of the squares is: %.1e\n\n',resnorm)
exitflag

% 以函数fmincon()估计得到的结果为初值，使用函数lsqnonlin()进行参数估计
k0 = k_fmincon;
[k,resnorm,residual,exitflag,output,lambda,jacobian] = ...
lsqnonlin(@ObjFunc4LNL,k0,lb,ub,[1e-15],a0,yexp);
ci = nlparci(k,residual,jacobian);
fprintf('\n\n以fmincon()的结果为初值，使用函数lsqnonlin()估计得到的参数值为:\n')
fprintf('\tk1 = %.4f\n',k(1))
fprintf('\tk2 = %.4f\n',k(2))
fprintf('  The sum of the squares is: %.1e\n\n',resnorm)
exitflag

function f = ObjFunc4Fmincon(k,a0,yexp)
tspan = [0 60 120 180 240 300 360 420 600 1200 6900 7500];
[t a] = ode23s(@KineticEqs,tspan,a0,[],k);
y(:,1:5) = a(:,1:5);
f = sum(((y(:,1)-yexp(:,1))/0.01).^2) + sum(((y(:,2)-yexp(:,2))/0.01).^2) ...
+ sum(((y(:,3)-yexp(:,3))/0.005).^2) + sum(((y(:,4)-yexp(:,4))/0.01).^2) ...
+ sum(((y(:,5)-yexp(:,5))/0.01).^2);

function f = ObjFunc4LNL(k,a0,yexp)
tspan = [0 60 120 180 240 300 360 420 600 1200 6900 7500];
[t a] = ode23s(@KineticEqs,tspan,a0,[],k);
y(:,1:5) = a(:,1:5);
f1 = (y(:,1) - yexp(:,1))/0.01;
f2 = (y(:,2) - yexp(:,2))/0.01;
f3 = (y(:,3) - yexp(:,3))/0.005;
f4 = (y(:,4) - yexp(:,4))/0.01;
f5 = (y(:,5) - yexp(:,5))/0.01;
f = [f1; f2; f3; f4; f5];
plot(tspan,yexp(:,1), 'b^',tspan,yexp(:,2), 'ro',tspan,yexp(:,3),'y*',...
tspan,yexp(:,4), 'go',tspan,yexp(:,5), 'cs')
hold on
plot(tspan,y(:,1), 'b-',tspan,yexp(:,2), 'r-',tspan,yexp(:,3),'y-', ...
tspan,yexp(:,4), 'g-',tspan,yexp(:,5), 'c-')
axis([0 800 0 1])

%动力学模型-----------------------------------------------------------------------
function dadt = KineticEqs(t,a ,k)
Keq=[2.20 0.40];
dadt=...
[((k(1)*(a(3)*a(2)-(1/Keq(1))*a(4)*a(1))) + (k(2)*(a(4)*a(2)-(1/Keq(2))*a(5)*a(1))));
-((k(1)*(a(3)*a(2)-(1/Keq(1))*a(4)*a(1))) + (k(2)*(a(4)*a(2)-(1/Keq(2))*a(5)*a(1))));
-(k(1)*(a(3)*a(2)-(1/Keq(1))*a(4)*a(1)));
  ((k(1)*(a(3)*a(2)-(1/Keq(1))*a(4)*a(1))) - (k(2)*(a(4)*a(2)-(1/Keq(2))*a(5)*a(1))));
(k(2)*(a(4)*a(2)-(1/Keq(2))*a(5)*a(1)))];

大家帮忙看一看，哪里有问题，从做出的图可以看出，处理的结果很差啊。
无论K初值给出怎么样，都是拟合出来的a1和实验值a1相差很远，而其他值都拟合很好。

[ Last edited by 楚天湘水 on 2012-5-1 at 00:44 ] 返回小木虫查看更多

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