Question

我在SciPy中使用fsolve函数来求解某些非线性方程组，并且我注意到将结果与MATLAB的fsolve进行比较以得到准确的输入和初始条件，结果却有所不同。因此，我想知道如何使用SciPy的fsolve与MATLAB中的结果相同。

目前，我已经在Matlab中配置了fsolve

options = optimoptions('fsolve','Display','iter-detailed','PlotFcn',@optimplotfirstorderopt);
options.StepTolerance = 1e-14;
%options.OptimalityTolerance = 1e-14
options.FunctionTolerance = 1e-14;
options.MaxIterations = 100000;
options.MaxFunctionEvaluations = 400;
%     options.Diagnostics = 'on'
options.Algorithm = 'levenberg-marquardt';%'trust-region'%
fun= @solveTransmissionLineForGAndC;

我已经在SciPy中像这样配置fsolve

fsolve(self.CandGTransmissionLineDistributedE,[g0,c0],Yin,xtol=1e-14,maxfev= 100000,epsfcn=1e-14)

因此，例如，对于相同的四个输入，我从MATLAB和SciPy两者获得的输出是

SciPy：[0.0263397261951065,0.0255996001756125,0.0255996001756125,0.0248335970729323]

MATLAB： [0.036897213114754,0.036897213114754,0.036897213114754,0.036897213114754,0.036897213114754]

如何在SciPy的fsolve中选择求解器算法（或方法）？

编辑：添加MWE

def WandG(GW,Zin):
    G,W = GW # define real variables to be used for the function

    R = (G**2)*(Zin.real - 0.1164) +(( 70e-12 * ( W/100 ) )**2)*( 2.9036155068866304e+16*(Zin.real - 0.1164) ) - G 

    I = (G**2)*( Zin.imag - 18.743998408378143 * (1-W/200) ) + (((W/100)*70e-12)**2)*( 2.9036155068866304e+16*(Zin.imag - 18.743998408378143 * (1-W/200)) ) + 170399985.53071037*(70e-12*(W/100) )

    return R,I

sol = optimize.root(WandG, [0.136879496, 47.04],(12.652884410928804+14.632724423763523j), jac=False, method='lm',tol=1e-14)

使用Scipy的解决方案将是

sol.x[0] = 0.0795283512113496 # G
sol.x[1] = 36.53727146377749  # W

对于具有相同初始条件的MATLAB，求解器和Zin而言，它会返回，我更信任它，因为它与W的最终结果（W=50

相似）

G = 0.0794
W = 44.5234

SciPy的fsolve是否支持不同的求解器算法？

0 个答案: