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小木虫论坛-学术科研互动平台 » 计算模拟区 » 程序语言 » MATLAB/Mathematica » 各位大神能不能帮小弟看看动力学编程的问题
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你的CEO

铁虫 (小有名气)

[求助] 各位大神能不能帮小弟看看动力学编程的问题 已有1人参与

**************程序如下，见二楼*******************
各位大神能不能帮小弟看看动力学编程的问题，误差特别大，拟合出来结果有问题，小弟刚学matlab，试着做了动力学参数拟合，还不太会希望大神求助a
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1楼 2015-09-22 15:18:53
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你的CEO

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你的CEO: 回帖置顶 2015-09-22 15:24:43
*****************结果：    *****************


f =
   -0.0873    0.0186    0.0075   -0.0466
   -0.1390   -0.0120   -0.0035    0.0014
   -0.1442   -0.0453    0.0047   -0.0482
   -0.1777   -0.0013   -0.0102    0.0310
   -0.1811   -0.0579   -0.0219    0.0084

Local minimum possible.

lsqnonlin stopped because the size of the current step is less than
the default value of the step size tolerance.

<stopping criteria details>

ci =
  1.0e+003 *
   -8.7033    9.5305
   -0.6189    0.6776
   -0.1243    0.1361
   -0.0808    0.0885


使用函数lsqnonlin()估计得到的参数值为:
        k1 = 413.5893 ± 9116.9019
        k2 = 29.3478 ± 648.2084
        k3 = 5.8897 ± 130.2370
        k4 = 3.8247 ± 84.6506
        The sum of the squares is: 1.2e-001
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3楼2015-09-22 15:20:00
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你的CEO

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你的CEO: 回帖置顶 2015-09-22 15:24:05
你的CEO: 取消置顶 2015-09-22 15:29:56
你的CEO: 回帖置顶 2015-09-22 15:30:06
*******上面那个程序有点小问题，大神们看着个吧！！！！！！！******************


% 动力学参数辨识: 用积分法进行反应速率分析得到速率常数k和反应级数n
% Analysis of kinetic rate data by using the integral method
% Reaction of the type -- rate = kCA^order
% order - reaction order
% rate -- reaction rate vector
% YA -- yield vector for reactant A
% T -- vector of reaction time
% N -- number of data points
% k- reacion rate constant

clear all
clc
format short
global Y_exp Y_sim
tspan = [0,21.065];  % t=w/F(MEOH.0),W:催化剂的质量，F(MEOH.0)：甲醇的初始进料流率
k0 = [2 0.5 0.1 0.1];  
lb = [0 0 0 0];
ub = [+inf  +inf  +inf  +inf ];
Y0 = [0 0 0 0 ];
Y_exp =[0.3901 0.0334 0.0067 0.0095;
        0.4431 0.038  0.0075 0.0045;
        0.4449 0.0488 0.0075 0.01;
        0.4751 0.044  0.0098 0.0026;
        0.482  0.051  0.0101 0.0043];

% 使用函数lsqnonlin()进行参数估计
[k,resnorm,residual,exitflag,output,lambda,jacobian] = lsqnonlin(@ObjFunc,k0,lb,ub,optimset('TolFun',1.0000e-20),tspan,Y0,Y_exp);      
ci = nlparci(k,residual,jacobian)
%[k,resnorm,residual,exitflag,output,lambda,jacobian] = lsqnonlin(@ObjFunc4LNL,k0,lb,ub,optimset('TolFun',1.0000e-6),Y0,Y_exp);      
%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 ± %.4f\n',k(1),ci(1,2)-k(1))
fprintf('\tk2 = %.4f ± %.4f\n',k(2),ci(2,2)-k(2))
fprintf('\tk3 = %.4f ± %.4f\n',k(3),ci(3,2)-k(3))
fprintf('\tk4 = %.4f ± %.4f\n',k(4),ci(4,2)-k(4))

fprintf('\tThe sum of the squares is: %.1e\n\n',resnorm)


function f = ObjFunc(k,tspan,Y0,Y_exp)           % 目标函数

[t,Y_sim1] = ode45(@KineticsEqs1,tspan,Y0,[],k);   
f1 = 1*(Y_sim1(end,1)-Y_exp(1,1));
f2= 5*(Y_sim1(end,2)-Y_exp(1,2));
f3= 10*(Y_sim1(end,3)-Y_exp(1,3));
f4= 10*(Y_sim1(end,4)-Y_exp(1,4));
[t,Y_sim2] = ode45(@KineticsEqs2,tspan,Y0,[],k);  
f5 = 1*(Y_sim2(end,1)-Y_exp(2,1));
f6 =5*(Y_sim2(end,2)-Y_exp(2,2));
f7 =10*(Y_sim2(end,3)-Y_exp(2,3));
f8 =10*(Y_sim2(end,4)-Y_exp(2,4));
[t,Y_sim3] = ode45(@KineticsEqs3,tspan,Y0,[],k);  
f9 = 1*(Y_sim3(end,1)-Y_exp(3,1));
f10 =5*(Y_sim3(end,2)-Y_exp(3,2));
f11 =10*(Y_sim3(end,3)-Y_exp(3,3));
f12 =10*(Y_sim3(end,4)-Y_exp(3,4));
[t,Y_sim4] = ode45(@KineticsEqs4,tspan,Y0,[],k);  
f13 = 1*(Y_sim4(end,1)-Y_exp(4,1));
f14 =5*(Y_sim4(end,2)-Y_exp(4,2));
f15 =10*(Y_sim4(end,3)-Y_exp(4,3));
f16 =10*(Y_sim4(end,4)-Y_exp(4,4));
[t,Y_sim5] = ode45(@KineticsEqs5,tspan,Y0,[],k);  
f17 = 1*(Y_sim5(end,1)-Y_exp(5,1));
f18 =5*(Y_sim5(end,2)-Y_exp(5,2));
f19 =10*(Y_sim5(end,3)-Y_exp(5,3));
f20 =10*(Y_sim5(end,4)-Y_exp(5,4));
f=[f1  f2  f3  f4;
   f5  f6  f7  f8;
   f9  f10 f11 f12;
   f13 f14 f15 f16;
   f17 f18 f19 f20]
% ------------------------------------------------------------------
function dYdt = KineticsEqs1(t,Y,k)
%p=[0.9 0.7 0.5 0.6 0.8]; p:压强
%m=[0.6 0.5 0.4 0.25 0.3];  m:氧醇比
%n=[5.4 2 2 2 2.1];    % n:氮醇比
%yO2=(m-0.5*Y(1)-Y(2)-0.5*Y(3))/(m+n+1-0.5*Y(1)+0.5*Y(3));
%yMEOH=(1-3*Y(1)-2*Y(2)-Y(3)-2*Y(4))/(m+n+1-0.5*Y(1)+0.5*Y(3));
yO2=(0.6-0.5*Y(1)-Y(2)-0.5*Y(3))/(0.6+5.4+1-0.5*Y(1)+0.5*Y(3));
yMEOH=(1-3*Y(1)-2*Y(2)-Y(3)-2*Y(4))/(0.6+5.4+1-0.5*Y(1)+0.5*Y(3));
dYdt=...
[k(1)*(0.9^2.2)*yMEOH^1*yO2^1.2;
k(2)*(0.9^2)*yMEOH^1*yO2^1;
k(3)*(0.9^2)*yMEOH^1*yO2^1;
k(4)*(0.9^2)*yMEOH^1.0*yO2^1;
];
function dYdt = KineticsEqs2(t,Y,k)
%p=[0.9 0.7 0.5 0.6 0.8]; p:压强
%m=[0.6 0.5 0.4 0.25 0.3];  m:氧醇比
%n=[5.4 2 2 2 2.1];    % n:氮醇比
%yO2=(m-0.5*Y(1)-Y(2)-0.5*Y(3))/(m+n+1-0.5*Y(1)+0.5*Y(3));
%yMEOH=(1-3*Y(1)-2*Y(2)-Y(3)-2*Y(4))/(m+n+1-0.5*Y(1)+0.5*Y(3));
yO2=(0.5-0.5*Y(1)-Y(2)-0.5*Y(3))/(0.5+2+1-0.5*Y(1)+0.5*Y(3));
yMEOH=(1-3*Y(1)-2*Y(2)-Y(3)-2*Y(4))/(0.5+52+1-0.5*Y(1)+0.5*Y(3));
dYdt=...
[k(1)*(0.7^2.2)*yMEOH^1*yO2^1.2;
k(2)*(0.7^2)*yMEOH^1*yO2^1;
k(3)*(0.7^2)*yMEOH^1*yO2^1;
k(4)*(0.7^2)*yMEOH^1.0*yO2^1;
];


function dYdt = KineticsEqs3(t,Y,k)
%p=[0.9 0.7 0.5 0.6 0.8]; p:压强
%m=[0.6 0.5 0.4 0.25 0.3];  m:氧醇比
%n=[5.4 2 2 2 2.1];    % n:氮醇比
%yO2=(m-0.5*Y(1)-Y(2)-0.5*Y(3))/(m+n+1-0.5*Y(1)+0.5*Y(3));
%yMEOH=(1-3*Y(1)-2*Y(2)-Y(3)-2*Y(4))/(m+n+1-0.5*Y(1)+0.5*Y(3));
yO2=(0.4-0.5*Y(1)-Y(2)-0.5*Y(3))/(0.4+2+1-0.5*Y(1)+0.5*Y(3));
yMEOH=(1-3*Y(1)-2*Y(2)-Y(3)-2*Y(4))/(0.4+2+1-0.5*Y(1)+0.5*Y(3));
dYdt=...
[k(1)*(0.5^2.2)*yMEOH^1*yO2^1.2;
k(2)*(0.5^2)*yMEOH^1*yO2^1;
k(3)*(0.5^2)*yMEOH^1*yO2^1;
k(4)*(0.5^2)*yMEOH^1.0*yO2^1;
];

function dYdt = KineticsEqs4(t,Y,k)
%p=[0.9 0.7 0.5 0.6 0.8]; p:压强
%m=[0.6 0.5 0.4 0.25 0.3];  m:氧醇比
%n=[5.4 2 2 2 2.1];    % n:氮醇比
%yO2=(m-0.5*Y(1)-Y(2)-0.5*Y(3))/(m+n+1-0.5*Y(1)+0.5*Y(3));
%yMEOH=(1-3*Y(1)-2*Y(2)-Y(3)-2*Y(4))/(m+n+1-0.5*Y(1)+0.5*Y(3));
yO2=(0.25-0.5*Y(1)-Y(2)-0.5*Y(3))/(0.25+2+1-0.5*Y(1)+0.5*Y(3));
yMEOH=(1-3*Y(1)-2*Y(2)-Y(3)-2*Y(4))/(0.25+2+1-0.5*Y(1)+0.5*Y(3));
dYdt=...
[k(1)*(0.6^2.2)*yMEOH^1*yO2^1.2;
k(2)*(0.6^2)*yMEOH^1*yO2^1;
k(3)*(0.6^2)*yMEOH^1*yO2^1;
k(4)*(0.6^2)*yMEOH^1.0*yO2^1;
];
function dYdt = KineticsEqs5(t,Y,k)
%p=[0.9 0.7 0.5 0.6 0.8]; p:压强
%m=[0.6 0.5 0.4 0.25 0.3];  m:氧醇比
%n=[5.4 2 2 2 2.1];    % n:氮醇比
%yO2=(m-0.5*Y(1)-Y(2)-0.5*Y(3))/(m+n+1-0.5*Y(1)+0.5*Y(3));
%yMEOH=(1-3*Y(1)-2*Y(2)-Y(3)-2*Y(4))/(m+n+1-0.5*Y(1)+0.5*Y(3));
yO2=(0.3-0.5*Y(1)-Y(2)-0.5*Y(3))/(0.3+2.1+1-0.5*Y(1)+0.5*Y(3));
yMEOH=(1-3*Y(1)-2*Y(2)-Y(3)-2*Y(4))/(0.3+2.1+1-0.5*Y(1)+0.5*Y(3));
dYdt=...
[k(1)*(0.8^2.2)*yMEOH^1*yO2^1.2;
k(2)*(0.8^2)*yMEOH^1*yO2^1;
k(3)*(0.8^2)*yMEOH^1*yO2^1;
k(4)*(0.8^2)*yMEOH^1.0*yO2^1;
];
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4楼2015-09-22 15:21:56
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function KineticsEst1_int
% 动力学参数辨识: 用积分法进行反应速率分析得到速率常数k和反应级数n
% Analysis of kinetic rate data by using the integral method
% Reaction of the type -- rate = kCA^order
% order - reaction order
% rate -- reaction rate vector
% YA -- yield vector for reactant A
% T -- vector of reaction time
% N -- number of data points
% k- reacion rate constant

clear all
clc
format short
global Y_exp Y_sim
tspan = [0,21.065];  % t=w/F(MEOH.0),W:催化剂的质量，F(MEOH.0)：甲醇的初始进料流率
k0 = [2 0.5 0.1 0.1];  
lb = [0 0 0 0];
ub = [+inf  +inf  +inf  +inf ];
Y0 = [0 0 0 0 ];
Y_exp =[0.3901 0.0334 0.0067 0.0095;
        0.4431 0.038  0.0075 0.0045;
        0.4449 0.0488 0.0075 0.01;
        0.4751 0.044  0.0098 0.0026;
        0.482  0.051  0.0101 0.0043];

% 使用函数lsqnonlin()进行参数估计
[k,resnorm,residual,exitflag,output,lambda,jacobian] = lsqnonlin(@ObjFunc,k0,lb,ub,optimset('TolFun',1.0000e-20),tspan,Y0,Y_exp);      
ci = nlparci(k,residual,jacobian)
%[k,resnorm,residual,exitflag,output,lambda,jacobian] = lsqnonlin(@ObjFunc4LNL,k0,lb,ub,optimset('TolFun',1.0000e-6),Y0,Y_exp);      
%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 ± %.4f\n',k(1),ci(1,2)-k(1))
fprintf('\tk2 = %.4f ± %.4f\n',k(2),ci(2,2)-k(2))
fprintf('\tk3 = %.4f ± %.4f\n',k(3),ci(3,2)-k(3))
fprintf('\tk4 = %.4f ± %.4f\n',k(4),ci(4,2)-k(4))

fprintf('\tThe sum of the squares is: %.1e\n\n',resnorm)

% 残差关于拟合值的残差图



%拟合效果图(实验与拟合的比较)
%[t4plot Y4plot] = ode45(@KineticsEqs1,[tspan(1)  tspan(end)],Y0,[],k0)
%figure
%plot(tspan,Y_exp(:,1),'bo',t4plot,Y4plot(:,1),'r--');
%legend('Exp','Model')
%xlabel('空时t=w/F_A_0, h')
%ylabel('收率Y_D_M_M')
%title('拟合效果图')
%figure
%plot(tspan,Y_exp(:,2),'bo',t4plot,Y4plot(:,2),'r--');
%legend('Exp','Model')
%xlabel('空时t=w/F_A_0, h')
%ylabel('收率Y_M_F')
%title('拟合效果图')
%figure
%plot(tspan,Y_exp(:,3),'bo',t4plot,Y4plot(:,3),'r--');
%legend('Exp','Model')
%xlabel('空时t=w/F_A_0, h')
%ylabel('收率Y_F_A')
%title('拟合效果图')
%figure
%plot(tspan,Y_exp(:,4),'bo',t4plot,Y4plot(:,4),'r--');
%legend('Exp','Model')
%xlabel('空时t=w/F_A_0, h')
%ylabel('收率Y_D_M_E')
%title('拟合效果图')

function f = ObjFunc(k,tspan,Y0,Y_exp)           % 目标函数

[t,Y_sim1] = ode45(@KineticsEqs1,tspan,Y0,[],k);   
f1 = 1*(Y_sim1(end,1)-Y_exp(1,1));
f2= 5*(Y_sim1(end,2)-Y_exp(1,2));
f3= 10*(Y_sim1(end,3)-Y_exp(1,3));
f4= 10*(Y_sim1(end,4)-Y_exp(1,4));
[t,Y_sim2] = ode45(@KineticsEqs2,tspan,Y0,[],k);  
f5 = 1*(Y_sim2(end,1)-Y_exp(2,1));
f6 =5*(Y_sim2(end,2)-Y_exp(2,2));
f7 =10*(Y_sim2(end,3)-Y_exp(2,3));
f8 =10*(Y_sim2(end,4)-Y_exp(2,4));
[t,Y_sim3] = ode45(@KineticsEqs3,tspan,Y0,[],k);  
f9 = 1*(Y_sim3(end,1)-Y_exp(3,1));
f10 =5*(Y_sim3(end,2)-Y_exp(3,2));
f11 =10*(Y_sim3(end,3)-Y_exp(3,3));
f12 =10*(Y_sim3(end,4)-Y_exp(3,4));
[t,Y_sim4] = ode45(@KineticsEqs4,tspan,Y0,[],k);  
f13 = 1*(Y_sim4(end,1)-Y_exp(4,1));
f14 =5*(Y_sim4(end,2)-Y_exp(4,2));
f15 =10*(Y_sim4(end,3)-Y_exp(4,3));
f16 =10*(Y_sim4(end,4)-Y_exp(4,4));
[t,Y_sim5] = ode45(@KineticsEqs5,tspan,Y0,[],k);  
f17 = 1*(Y_sim5(end,1)-Y_exp(5,1));
f18 =5*(Y_sim5(end,2)-Y_exp(5,2));
f19 =10*(Y_sim5(end,3)-Y_exp(5,3));
f20 =10*(Y_sim5(end,4)-Y_exp(5,4));
A=[Y_sim1(end,;
    Y_sim2(end,;
    Y_sim3(end,;
    Y_sim4(end,;
    Y_sim5(end,]
f=[f1  f2  f3  f4;
   f5  f6  f7  f8;
   f9  f10 f11 f12;
   f13 f14 f15 f16;
   f17 f18 f19 f20]
% ------------------------------------------------------------------
function dYdt = KineticsEqs1(t,Y,k)
%p=[0.9 0.7 0.5 0.6 0.8]; p:压强
%m=[0.6 0.5 0.4 0.25 0.3];  m:氧醇比
%n=[5.4 2 2 2 2.1];    % n:氮醇比
%yO2=(m-0.5*Y(1)-Y(2)-0.5*Y(3))/(m+n+1-0.5*Y(1)+0.5*Y(3));
%yMEOH=(1-3*Y(1)-2*Y(2)-Y(3)-2*Y(4))/(m+n+1-0.5*Y(1)+0.5*Y(3));
yO2=(0.6-0.5*Y(1)-Y(2)-0.5*Y(3))/(0.6+5.4+1-0.5*Y(1)+0.5*Y(3));
yMEOH=(1-3*Y(1)-2*Y(2)-Y(3)-2*Y(4))/(0.6+5.4+1-0.5*Y(1)+0.5*Y(3));
dYdt=...
[k(1)*(0.9^2.2)*yMEOH^1*yO2^1.2;
k(2)*(0.9^2)*yMEOH^1*yO2^1;
k(3)*(0.9^2)*yMEOH^1*yO2^1;
k(4)*(0.9^2)*yMEOH^1.0*yO2^1;
];
function dYdt = KineticsEqs2(t,Y,k)
%p=[0.9 0.7 0.5 0.6 0.8]; p:压强
%m=[0.6 0.5 0.4 0.25 0.3];  m:氧醇比
%n=[5.4 2 2 2 2.1];    % n:氮醇比
%yO2=(m-0.5*Y(1)-Y(2)-0.5*Y(3))/(m+n+1-0.5*Y(1)+0.5*Y(3));
%yMEOH=(1-3*Y(1)-2*Y(2)-Y(3)-2*Y(4))/(m+n+1-0.5*Y(1)+0.5*Y(3));
yO2=(0.5-0.5*Y(1)-Y(2)-0.5*Y(3))/(0.5+2+1-0.5*Y(1)+0.5*Y(3));
yMEOH=(1-3*Y(1)-2*Y(2)-Y(3)-2*Y(4))/(0.5+52+1-0.5*Y(1)+0.5*Y(3));
dYdt=...
[k(1)*(0.7^2.2)*yMEOH^1*yO2^1.2;
k(2)*(0.7^2)*yMEOH^1*yO2^1;
k(3)*(0.7^2)*yMEOH^1*yO2^1;
k(4)*(0.7^2)*yMEOH^1.0*yO2^1;
];


function dYdt = KineticsEqs3(t,Y,k)
%p=[0.9 0.7 0.5 0.6 0.8]; p:压强
%m=[0.6 0.5 0.4 0.25 0.3];  m:氧醇比
%n=[5.4 2 2 2 2.1];    % n:氮醇比
%yO2=(m-0.5*Y(1)-Y(2)-0.5*Y(3))/(m+n+1-0.5*Y(1)+0.5*Y(3));
%yMEOH=(1-3*Y(1)-2*Y(2)-Y(3)-2*Y(4))/(m+n+1-0.5*Y(1)+0.5*Y(3));
yO2=(0.4-0.5*Y(1)-Y(2)-0.5*Y(3))/(0.4+2+1-0.5*Y(1)+0.5*Y(3));
yMEOH=(1-3*Y(1)-2*Y(2)-Y(3)-2*Y(4))/(0.4+2+1-0.5*Y(1)+0.5*Y(3));
dYdt=...
[k(1)*(0.5^2.2)*yMEOH^1*yO2^1.2;
k(2)*(0.5^2)*yMEOH^1*yO2^1;
k(3)*(0.5^2)*yMEOH^1*yO2^1;
k(4)*(0.5^2)*yMEOH^1.0*yO2^1;
];

function dYdt = KineticsEqs4(t,Y,k)
%p=[0.9 0.7 0.5 0.6 0.8]; p:压强
%m=[0.6 0.5 0.4 0.25 0.3];  m:氧醇比
%n=[5.4 2 2 2 2.1];    % n:氮醇比
%yO2=(m-0.5*Y(1)-Y(2)-0.5*Y(3))/(m+n+1-0.5*Y(1)+0.5*Y(3));
%yMEOH=(1-3*Y(1)-2*Y(2)-Y(3)-2*Y(4))/(m+n+1-0.5*Y(1)+0.5*Y(3));
yO2=(0.25-0.5*Y(1)-Y(2)-0.5*Y(3))/(0.25+2+1-0.5*Y(1)+0.5*Y(3));
yMEOH=(1-3*Y(1)-2*Y(2)-Y(3)-2*Y(4))/(0.25+2+1-0.5*Y(1)+0.5*Y(3));
dYdt=...
[k(1)*(0.6^2.2)*yMEOH^1*yO2^1.2;
k(2)*(0.6^2)*yMEOH^1*yO2^1;
k(3)*(0.6^2)*yMEOH^1*yO2^1;
k(4)*(0.6^2)*yMEOH^1.0*yO2^1;
];
function dYdt = KineticsEqs5(t,Y,k)
%p=[0.9 0.7 0.5 0.6 0.8]; p:压强
%m=[0.6 0.5 0.4 0.25 0.3];  m:氧醇比
%n=[5.4 2 2 2 2.1];    % n:氮醇比
%yO2=(m-0.5*Y(1)-Y(2)-0.5*Y(3))/(m+n+1-0.5*Y(1)+0.5*Y(3));
%yMEOH=(1-3*Y(1)-2*Y(2)-Y(3)-2*Y(4))/(m+n+1-0.5*Y(1)+0.5*Y(3));
yO2=(0.3-0.5*Y(1)-Y(2)-0.5*Y(3))/(0.3+2.1+1-0.5*Y(1)+0.5*Y(3));
yMEOH=(1-3*Y(1)-2*Y(2)-Y(3)-2*Y(4))/(0.3+2.1+1-0.5*Y(1)+0.5*Y(3));
dYdt=...
[k(1)*(0.8^2.2)*yMEOH^1*yO2^1.2;
k(2)*(0.8^2)*yMEOH^1*yO2^1;
k(3)*(0.8^2)*yMEOH^1*yO2^1;
k(4)*(0.8^2)*yMEOH^1.0*yO2^1;
];
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2楼2015-09-22 15:19:06
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你的CEO

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2楼: Originally posted by 你的CEO at 2015-09-22 15:19:06
function KineticsEst1_int
% 动力学参数辨识: 用积分法进行反应速率分析得到速率常数k和反应级数n
% Analysis of kinetic rate data by using the integral method
% Reaction of the type -- rate = kCA^order ...

这个有问题，请看三楼的程序
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5楼2015-09-22 15:30:53
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asked6188
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6楼2015-09-22 16:05:54
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你的CEO

铁虫 (小有名气)
引用回帖:
6楼: Originally posted by asked6188 at 2015-09-22 16:05:54
顶一下

谢谢，你也搞动力学？
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7楼2015-09-22 20:01:59
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你的CEO

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求大神关注
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8楼2015-09-22 20:10:08
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dingd

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【答案】应助回帖

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感谢参与，应助指数 +1
你的CEO: 金币+10, ★有帮助 2015-10-10 13:53:40
这种微分方程拟合问题用1stOpt吧，比Matlab简单强大多了。搜一下论坛有不少示例。
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9楼2015-09-23 12:08:06
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