用Scipy/Numpy-NDSolve等价物在Python中求解矩阵微分方程?

2024-10-01 15:28:18 发布

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我有两个numpy数组:9x9和9x1。我想在离散时间点求解微分方程,但在使ODEInt工作时遇到困难。我真的不确定我是否做了正确的事。在

使用Mathematica,方程式为:

Solution = {A[t]} /. NDSolve[{A'[t] == Ab.A[t] && A[0] == A0}, {A[t]}, {t, 0, .5}, MaxSteps -> \[Infinity]];

time = 0.25;
increment = 0.05;
MA = Table[Solution, {t, 0, time, increment}];

其中Ab是9x9矩阵,A0是9x1矩阵(初始)。在这里,我解决时间和生活是美好的。在

在Python实现中,我有以下代码给出了错误的答案:

^{pr2}$

输出为:

[[ -1.07158538e-003  -8.91531917e-001  -3.84315167e-004   0.00000000e+000
    0.00000000e+000   0.00000000e+000   5.40097001e-004   4.49347036e-001
    1.93701289e-004]
 [  3.09311322e+019   9.45061860e+022   2.35327270e+019   2.11901406e+020
    1.63784238e+023   7.60569684e+019   2.29098804e+020   1.89766602e+023
    8.18752241e+019]
 [  9.84409730e+042   3.00774018e+046   7.48949158e+042   6.74394343e+043
    5.21257342e+046   2.42057805e+043   7.29126532e+043   6.03948436e+046
    2.60574901e+043]
 [  3.13296814e+066   9.57239028e+069   2.38359473e+066   2.14631766e+067
    1.65894606e+070   7.70369662e+066   2.32050753e+067   1.92211754e+070
    8.29301904e+066]
 [  9.97093898e+089   3.04649506e+093   7.58599405e+089   6.83083947e+090
    5.27973769e+093   2.45176732e+090   7.38521364e+090   6.11730342e+093
    2.63932422e+090]
 [  3.17333659e+113   9.69573101e+116   2.41430747e+113   2.17397307e+114
    1.68032166e+117   7.80295913e+113   2.35040739e+114   1.94688412e+117
    8.39987500e+113]]

但正确的答案应该是:

{-0.0010733126291998989, -0.8929689437405254, -0.0003849346301906338, 0., 0., 0., 0.0005366563145999495, 0.4464844718702628, 0.00019246731509531696}
{-0.000591095648651598, -0.570032546156741, -0.00023381082725213798, -0.00024790706920038567, 0.00010389803046880286, -0.00005361569187144767, 0.0003273277204077012, 0.2870035216110215, 0.00012300339326137006}
{-0.0003770535829276868, -0.364106358478121, -0.0001492324135668267, -0.0001596072774600538, -0.0011479989178276948, -0.000034744485507007025, 0.00020965172928479557, 0.18378613639965447, 0.00007876820247280559}
{-0.00024100792803807562, -0.23298939195213314, -0.00009543704274825206, -0.00010271831380730501, -0.0013205519868311284, -0.000022472380871477824, 0.00013326471695185768, 0.11685506361394844, 0.00005008078740423007}
{-0.00015437993249587976, -0.1491438843823813, -0.00006111736454518403, -0.00006545797627466387, -0.0005705018939767294, -0.000014272382451480663, 0.00008455890984798549, 0.0741820536557778, 0.00003179071165818503}
{-0.00009882799610556456, -0.09529950309336405, -0.00003909275555213336, -0.00004138741286392128, 0.00006303116741431477, -8.944610716890746*^-6, 0.00005406263888971806, 0.04743157303933772, 0.00002032674776723143}

有人能告诉我我做错了什么吗?在


Tags: 答案numpyabtime时间矩阵数组a0
1条回答
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1楼 · 发布于 2024-10-01 15:28:18

在对odeint的调用中,尝试将tuple(array[Ab])更改为(array(Ab),),甚至只是(Ab,)。也就是说,使用

MA = odeint(deriv, A0, time, (Ab,))

{{cd6>如果没有这个变量,你肯定会解决这个问题。我用的是3x3数组而不是9x9。在

^{pr2}$

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