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zhangholmes至尊木虫 (著名写手)
逍遥门主
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[求助]
matlab非线性最优化求解，微分方程组的拟合，参数估计
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如题，用微分方程组拟合4组不同温度下的数据，求理论值a和b，并使a和b的值分别与实验值y1和y2的差方和最小。并求此时的各个参数值(k1,k2,k3,k4,k5,k6)，求各位大侠略施援手。 0<k1<30, 0<k2<10, 0<k3<5, 0<k4<20, 0<k5<20, 0<k6<5 t=[0.494660000000000,0.502760000000000,0.618930000000000,0.621470000000000,0.623470000000000,0.819730000000000,0.825080000000000,0.828350000000000]; T=463 K时， y1= [0.00800000000000000,0.00990000000000000,0.00782000000000000,0.0108400000000000,0.00937000000000000,0.00882000000000000,0.00614000000000000,0.0108200000000000]; y2=[0.0625100000000000,0.0727400000000000,0.0540300000000000,0.0667600000000000,0.0620200000000000,0.0494400000000000,0.0452000000000000,0.0509400000000000]; %y1和y2分别为a和b的实验值 T=473 K时， y1=[0.00829000000000000,0.0107700000000000,0.00938000000000000,0.0172000000000000,0.0128100000000000,0.0139200000000000,0.0108200000000000,0.0179000000000000]; y2=[0.0622600000000000,0.0719200000000000,0.0525700000000000,0.0604900000000000,0.0586200000000000,0.0444300000000000,0.0405900000000000,0.0439200000000000]; T=483 K时， y1=[0.0169300000000000,0.0229900000000000,0.0188900000000000,0.0313900000000000,0.0265800000000000,0.0258100000000000,0.0194200000000000,0.0293300000000000]; y2=[0.0537100000000000,0.0598300000000000,0.0431500000000000,0.0465000000000000,0.0450300000000000,0.0326700000000000,0.0321200000000000,0.0327000000000000]; T=493 K时， y1=[0.0328500000000000,0.0418500000000000,0.0330300000000000,0.0512500000000000,0.0424800000000000,0.0389500000000000,0.0300900000000000,0.0443800000000000]; y2=[0.0381700000000000,0.0415400000000000,0.0292800000000000,0.0273400000000000,0.0295700000000000,0.0197800000000000,0.0216900000000000,0.0180800000000000]; pA = 101*(1 - (2 b + a)); pW = 101*(a+ b); da/dt=k(4).*k(1).*k(2).*k(3).*pA.*pW./(k(1).*k(2).*k(4).*k(5).*pA^2+k(1).*k(2).*k(3).*pA.*pW+k(4)+k(1).*k(2).*k(4).*pA.*pW+k(1).*k(4).*pW; db/dt=k(6).*k(1).*k(2).*k(4).*k(5).*pA^2./(k(1).*k(2).*k(4).*k(5).*pA^2+k(1).*k(2).*k(3).*pA.*pW+k(4)+k(1).*k(2).*k(4).*pA.*pW+k(1).*k(4).*pW; [ 来自科研家族 木虫催化 ] |
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zhangholmes
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谢谢回复，公式最后忘记加了一个括号。不要求统一参数，每组数据都有一组独立的参数，共四组参数。 公式如下： da/dt=k(4).*k(1).*k(2).*k(3).*pA.*pW./(k(1).*k(2).*k(4).*k(5).*pA^2+k(1).*k(2).*k(3).*pA.*pW+k(4)+k(1).*k(2).*k(4).*pA.*pW+k(1).*k(4).*pW）; db/dt=k(6).*k(1).*k(2).*k(4).*k(5).*pA^2./(k(1).*k(2).*k(4).*k(5).*pA^2+k(1).*k(2).*k(3).*pA.*pW+k(4)+k(1).*k(2).*k(4).*pA.*pW+k(1).*k(4).*pW）; |
3楼2013-05-12 15:59:36
