Question

MATLAB中fsolve解非线性方程问题

clear all;clc;close all;
N1=1;%
N2=2;%%
P1=750/1120;
f=50;
%%%%基值定义开始
%定
SacB=1120;%MVA
UacB=550;%kV
IacB=SacB/UacB/(sqrt(3));
ZacB=UacB*UacB/SacB;%ohm
%定义
SdcB=1120;
UdcB=250;
K=UdcB/UacB;
IdcB=sqrt(3)*IacB/K;
ZdcB=K*K*ZacB;
%%%%基值定义完毕
%%
Kb=2;
Kp=1;
Kunit=1；
Dxi=Kb*0.12*3/pi/K/K*SacB/450;
Dxr=Kb*0.12*3/pi/K/K*SacB/450;
Tr=(1.05*525/105.75);
Ti=(1.0375*525/105.75);
Rdc=20/ZdcB;
gam=17/180*pi;
Ids=3/IdcB;
pi=3.14159；
P2=0;
Q2=0;
P3=0;
Q3=0;
v1=1.08;
v4=1;
R12=6.875E-9*16000*500*500/100/ZacB/N2;
X12=1.0E-7*16000*500*500/100/ZacB/N2+0.181/N1;
R34=0;
X34=0.0732*2*pi*f/ZacB;
R22=4.7241/Kunit/ZacB;X22=-340.2973/Kunit/ZacB;
R33=4.7812/Kunit/ZacB;X33=-308.5175/Kunit/ZacB;
%R22=16.6152/Kunit/ZacB;X22=-638.0362/Kunit/ZacB;
%R33=16.611/Kunit/ZacB;X33=-574.8844/Kunit/ZacB;
l=1;
syms deta1 v2 deta2 v3 deta3;
%五个未知数；需要五个方程
eq1=P1-real(v1*(cos(deta1)+j*sin(deta1))*conj((v1*(cos(deta1)+j*sin(deta1))-v2*(cos(deta2)+j*sin(deta2)))/(R12+j*X12)));
eq2=-P2+real(v2*(cos(deta2)+j*sin(deta2))*conj((v2*(cos(deta2)+j*sin(deta2))-v1*(cos(deta1)+j*sin(deta1)))/(R12+j*X12)))+Kp*(3*sqrt(2)*Kb*v3/pi/Ti/K*cos(gam)+(Rdc-Dxi)*Ids)*Ids;
eq3=-Q2+imag(v2*(cos(deta2)+j*sin(deta2))*conj((v2*(cos(deta2)+j*sin(deta2))-v1*(cos(deta1)+j*sin(deta1)))/(R12+j*X12)))+126*3/SacB+Kp*Ids*(sqrt((3*sqrt(2)*Kb*v2/pi/Tr/K)^2-(3*sqrt(2)*Kb*v3/pi/Ti/K*cos(gam)+(Rdc-Dxi)*Ids)^2));
eq4=-P3+real(v3*(cos(deta3)+j*sin(deta3))*conj((v3*(cos(deta3)+j*sin(deta3))-v4)/(R34+j*X34)))-Kp*(3*sqrt(2)*Kb*v3/pi/Ti/K*cos(gam)-Dxi*Ids)*Ids;
eq5=-Q3+imag(v3*(cos(deta3)+j*sin(deta3))*conj((v3*(cos(deta3)+j*sin(deta3))-v4)/(R34+j*X34)))+126*3/SacB-Kp*Ids*(sqrt((3*sqrt(2)*Kb*v3/pi/Ti/K)^2-(3*sqrt(2)*Kb*v3/pi/Ti/K*cos(gam)-Dxi*Ids)^2));

eq=['F=' char('[') eq1 char(';') eq2 char(';') eq3 char(';') eq4 char(';') eq5 char(']') char(';')];
fid=fopen(['pffun' num2str(l) '.m'],'w');
fwrite(fid,'function F = pffun(x)');
fwrite(fid,char(13));
fwrite(fid,eq);
fwrite(fid, char(13));
fwrite(fid,'return');
fclose(fid);

主函数为
icount=1;
fun=str2func(['pffun' num2str(icount)]);
sol=fsolve(str2func(['pffun' num2str(icount)]),[0.48;1.165; 0.3;1.02;0.118]);
%syms deta1 v2 deta2 v3 deta3;

为什么得到以下结果
Optimizer appears to be converging to a point which is not a root.
Norm of relative change in X is less than max(options.TolX^2,eps) but
sum-of-squares of function values is greater than or equal to sqrt(options.TolFun)
Try again with a new starting guess.

有时又是这样的结果Maximum number of function evaluations reached:
increase options.MaxFunEvals.

Answer 1

第一个结果是说，精度options.Tol已经满足要求，但这个精度options.TolFun不满足，让你更改初值试试。
第二个结果是说已经达到最大计算次数，可以通过提高options.MaxFunEvals来增加计算次数。

matlab也是做数值计算的，对计算非线性函数难免受初值选取、精确度满足、步长选择等影响。上面的问题基本也是这样，可以通过设置options来修正，options是个大域名，计算后可以通过在命令窗口输入options来查看各个变量，输入options.变量名 来查看各个变量值，比如options.MaxFunEvals。
x = fsolve(fun,x0,options)你可以多了解下它的用法，网上有很多，第三项是options选项，options各变量有默认值，你可以更改，来控制计算。
如果再不能的话，估计就是函数太过复杂，或太过矫情，不是fsolve所能计算的。

options=optimset('MaxFunEvals',~...)
optimset
Display: [ off | iter | notify | final ]
MaxFunEvals: [ positive scalar ]
MaxIter: [ positive scalar ]
.............