matlab 解非线性方程只有复数解。

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sawdaga
sawdaga on 19 May 2023
Answered: xexbmk on 19 May 2023
matlab 解非线性方程只有复数解,按道理来说应该是有实数解的,这是什么原因呢,可以通过其他方法得到实数解吗
clear;
clc;
syms fai
a=1.45;n=2.68;m=0.6269; ks = 0.00004821176 ; q=2*10^(-5);
se=(1/(1+(a*fai)^n))^m
krw = se^(1/2)*(1-(1-se^(1/m))^m)^2
kw = ks * krw
vpasolve( kw-q==0, fai )
ans =
- 0.2706702470197805855288345099971 + 0.27709160310623304634861907521709i

Accepted Answer

xexbmk
xexbmk on 19 May 2023
F = matlabFunction( kw - q, 'vars', fai );
Result = fzero( F, rand )
Result = 0.387352732188288
F( Result ) = -6.7762635780344e-21

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