Question

哪位高手帮我把下面这个matlab的程序转换成mathematica的啊？急用急用 感激不尽！

>> clc;clear
close all;
%输入部分
% 基本参数
alfa=1.08;
beta=0.05;
N=30.58;
fc=0.01;
fF=1e-4;
fD=1e-4;
lamda=1.54e-4;
%
T1=10;
T2=1;
T3=100;
T4=100;
T5=0.1;
T6=0.1;
T7=100;
T8=100;
T9=100;
T10=100;
T11=1;
%Ei
E1=0;
E2=0.001;
E3=0.001;
E4=0.001;
E5=0;
E6=0;
E7=0;
E8=0.001;
E9=0.001;
E10=0.001;
E11=0.001;
% 时间控制
Time=10000;
Nstep=10000e1;
dt=Time/Nstep;
day=(1:Nstep)*dt;
% T mass in compartment
I1=zeros(1,Nstep);
I2=zeros(1,Nstep);
I3=zeros(1,Nstep);
I4=zeros(1,Nstep);
I5=zeros(1,Nstep);
I6=zeros(1,Nstep);
I7=zeros(1,Nstep);
I8=zeros(1,Nstep);
I9=zeros(1,Nstep);
I10=zeros(1,Nstep);
I11=zeros(1,Nstep);
I1(1)=0;
I2(1)=0;
I3(1)=0;
I4(1)=0;
I5(1)=0;
I6(1)=0;
I7(1)=4000;
I8(1)=0;
I9(1)=0;
I10(1)=0;
I11(1)=0;
% 计算部分
for i=1:Nstep-1
I1(i+1)=I1(i)+dt*(alfa*N-(1+E1)*I1(i)/T1-lamda*I1(i));
I2(i+1)=I2(i)+dt*((1-fc)*I1(i)/T1-(1+E2)*I2(i)/T2-lamda*I2(i));
I3(i+1)=I3(i)+dt*(fc*I1(i)/T1-(1+E3)*I3(i)/T3-lamda*I3(i));
I4(i+1)=I4(i)+dt*(I3(i)/T3-(1+E4)*I4(i)/T4-lamda*I4(i));
I5(i+1)=I5(i)+dt*(I2(i)/T2+I11(i)/T11-(1+E5)*I5(i)/T5-lamda*I5(i));
I6(i+1)=I6(i)+dt*(I5(i)/T5+I10(i)/T10-(1+E6)*I6(i)/T6-lamda*I6(i));
I7(i+1)=I7(i)+dt*(I6(i)/T6-N/beta-E7*I7(i)/T7-lamda*I7(i));
I8(i+1)=I8(i)+dt*(fF*N/beta-(1+E8)*I8(i)/T8-lamda*I8(i));
I9(i+1)=I9(i)+dt*(fD*N/beta-(1+E9)*I9(i)/T9-lamda*I9(i));
I10(i+1)=I10(i)+dt*(E1*I1(i)/T1+E2*I2(i)/T2+E3*I3(i)/T3+E4*I4(i)/T4+...
E5*I5(i)/T5+E6*I6(i)/T6+E7*I7(i)/T7+E8*I8(i)/T8+E9*I9(i)/T9+E11*I11(i)/T11+I4(i)/T4+I8(i)/T8+I9(i)/T9-E10*I10(i)/T10-lamda*I10(i));
I11(i+1)=I11(i)+dt*((1-beta-fc-fD)*N/beta-(1+E11)*I11(i)/T11-lamda*I11(i));
end
% 输出部分
loglog(day,I1,'gv');
hold on;
loglog(day,I2,'bh');
%loglog(day,I3,'y+');
%loglog(day,I4,'g');
loglog(day,I5,'md');
loglog(day,I6,'cx');
loglog(day,I7,'kp');
%loglog(day,I8,'r');
%loglog(day,I9,'b');
%loglog(day,I10,'k');
loglog(day,I11,'rx');
title('数值分析');
xlabel('时间(天)');
ylabel('系统');
legend('I1','I2','I5','I6','I7');
axis([1e-1 1e4 1e0 1e5]);

Answer 1

既然是急用那想来现在是用不着了吧。你这代码似乎就是个常微分方程初值问题求解啊？那我建议你把原方程直接扔进NDSolve解还比较快。