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qs001

新虫 (初入文坛)

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虫号: 2219462
注册: 2013-01-03
专业: 防灾工程

[求助] fsolve计算结果问题

计算结果感觉误差好大，是由于初值取值造成的影响吗，还是要设置option来控制精度，第一次使用matlab一窍不通，请大家帮忙看下。
                                    First-Order                   Norm of
Iteration  Func-count Residual    optimality    Lambda          step
   0          8       39.2026    1.13e+003       0.01
   1       16    0.00103045          0.705       0.001    0.294456
   2       24 3.55079e-005       0.0727    0.0001    0.237055
   3       32 7.28141e-006          0.19    1e-005    0.144512
   4       40 2.91808e-006       0.0123    1e-006    0.0248491
   5       48 2.90763e-006    3.58e-007    1e-007 0.000338087
   6       56 2.90763e-006    1.02e-009    1e-008 1.90914e-006

Equation solved, fsolve stalled.

fsolve stopped because the relative size of the current step is less than the
default value of the step size tolerance and the vector of function values
is near zero as measured by the selected value of the function tolerance.

y =

-2.2413 2.2083 -0.1310 0.1318 1.0596 0.0211 -0.0272

fval =

0.0000
0.0008
-0.0008
0.0003
0.0000
0.0000
-0.0011
0.0006

下面是我要解的方程组
function F=myfun(x)
b=1;a=1;
F=[-0.0575*x(3)/a-x(7)*a*0.5-x(6);
x(3)/a^2+x(4)+17.39*x(7)*log(a)+0.8695652174*b*log(a);
-x(1)/(x(5)^2)-x(2)+x(3)/(x(5)^2)+x(4)-92.49880554*x(7)*log(x(5))+0.8695652174*b*log(x(5));
0.0065*x(1)/x(5)-0.0026*x(2)*x(5)-0.0575*x(3)/x(5)+x(7)*x(5)*((0.7-0.4*log(x(5)))/1.4-0.9347826087);
-2*x(1)/(x(5)^2)+109.8901099*x(7)-b;
2*x(1)/(1.5^2)-109.8901099*x(7)-b;
0.444444444444444*x(1)/a^2+x(2)+109.8901099*x(7)*log(1.5*a);
-x(1)*log(1.5*a/x(5))+0.5*(1.5^2*a^2-x(5)^2)*x(2)-x(3)*log(x(5)/a)+0.5*(x(5)^2-a^2)*x(4)+17.39130435*x(7)*(0.5*x(5)^2*(log(x(5))+0.5)-0.5*a^2*(log(a)+0.5))+109.8901099*x(7)*(0.5*1.5*1.5*a^2*(0.5+log(1.5*a))-0.5*x(5)^2*(log(x(5))+0.5))+0.0209742*1.5^2*a^2*b+0.8695652174*b*(0.5*x(5)^2*(log(x(5))+0.5)-0.5*a^2*(log(a)+0.5))];

选取的初值是x0=[-2,2,-0.5,0.5,1.05,0.01,-0.01]

回复此楼