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heisekaishui銀蟲 (小有名氣)
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[求助]
關(guān)于1STOPT擬合的小問題 已有2人參與
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各位前輩,我最近用1stopt擬合了一組數(shù)據(jù),但是發(fā)現(xiàn)擬合的數(shù)據(jù)使用不同階的龍格庫塔方程結(jié)果不同,但是精度都差不多,不知道是什么原因,希望大家不吝賜教。 代碼如下: Parameters k1,k2,k3; Variable t,a,b,c; ODEFunction a'=-k1*a-k2*a; b'=k1*a; c'=k2*a; Data; t=0,2,4,6,8,10; a=1.161575096,1.098850041,1.022186084,0.952491579,0.885120223,0.825879893; b=0,0.060153328,0.132977117,0.197793007,0.254614938,0.306825415; c=0,0.002571727,0.006411895,0.01129051,0.021839935,0.028869787; 同時,我在設(shè)定參數(shù)的范圍時,并沒有起到效果,而且1stopt的結(jié)果縱坐標范圍是0-4而不是0-10不知道是什么原因 |

銀蟲 (小有名氣)

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看看這樣能不能解決你的問題 Parameters k1,k2,k3; Variable t,a,b,c; SharedModel; ODEFunction a'=-k1*a-k2*a; b'=k1*a; c'=k2*a; RowData; 0,2,4,6,8,10; 1.161575096,1.098850041,1.022186084,0.952491579,0.885120223,0.825879893; 0,0.060153328,0.132977117,0.197793007,0.254614938,0.306825415; 0,0.002571727,0.006411895,0.01129051,0.021839935,0.028869787; |

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Parameters k1,k2,k3,k4; 修改方程,這個可以得到穩(wěn)定結(jié)果 Variable t,a,b,c; ODEFunction a'=-k1*a-k2*a; b'=k3*a; c'=k4*a; RowData; 0,2,4,6,8,10; 1.161575096,1.098850041,1.022186084,0.952491579,0.885120223,0.825879893; 0,0.060153328,0.132977117,0.197793007,0.254614938,0.306825415; 0,0.002571727,0.006411895,0.01129051,0.021839935,0.028869787; |

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坐標顯示0-4可能是他的作圖功能不好,或不是正版軟件,只看他的計算值就行了(計算值不是有,每時刻的計算值嗎) 2 1.098850041 .060153328 .002571727 1.08633992640023 6.97307002743706E-02 5.60446897778746E-03 For a: 4 1.022186084 .132977117 .006411895 1.01597773545576 .134944946131689 1.08459367441951E-02 DC: 0.99487234139559 6 .952491579 .197793007 .01129051 .950172900633618 .195935268051574 1.57479148657442E-02 R: 0.999959508427279 8 .885120223 .254614938 .021839935 .888630242171105 .25297524940825 2.03323920723293E-02 RMSE: 0.00692414160103415 10 .825879893 .306825415 .028869787 .831073699086233 .306320753592402 2.46199328847424E-02 SSE: 0.000239718684555859 |

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