请问 fsolve 得到的解如何判断是否有意义
Optimizer appears to be converging to a point which is not a root.Relative function value changing by less than max(options.TolFun^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.
function ff=jifen3(x)
r1=9.5*10^(-3);r2=9.6*10^(-3);
r3=11.5*10^(-3);r4=11.6*10^(-3);P4=600;
C10=115100;C01=101300;P0=1.01325*10^5;h1=0.003;%h1是铜块的厚度;%令P0=0才能得到实际结果。
rad1=[];
rad2=[];
rad3=[];
rad4=[];
for n=1000:50:1400;
w=2*pi*n/60;
midu=8920;
P1=midu*w.^2*((x(1)^2-(r2^2-r1^2))^(3/2)-((x(1)^2-(r2^2-r1^2))^(1/2)-h1)^3)/(3*(x(1)^2-(r2^2-r1^2))^(1/2));
P3=1.0*10^3*w.^2*((x(2)^2-(r4^2-r2^2))^(3/2)-x(1)^3)/(3*(x(2)^2-(r4^2-r3^2))^(1/2));
%-----------------------------
ff(1)=2*(C10+C01)*((r1^2-(x(1)^2-(r2^2-r1^2)))*(-1/2)*((x(1)^2-(r2^2-r1^2))-x(1)^2)/((x(1)^2-(r2^2-r1^2))*x(1)^2)+log(x(1)/(x(1)^2-(r2^2-r1^2))^(1/2))-(1/2)*log((r1^2-(x(1)^2-(r2^2-r1^2))+x(1)^2)/r1^2))+P0+P1+2*(C10+C01)*((r3^2-(x(2)^2-(r4^2-r3^2)))*(-1/2)*((x(2)^2-(r4^2-r3^2))-x(2)^2)/((x(2)^2-(r4^2-r3^2))*x(2)^2)+log(x(2)/(x(2)^2-(r4^2-r3^2))^(1/2))-(1/2)*log((r3^2-(x(2)^2-(r4^2-r3^2))+x(2)^2)/r3^2))+P3-P4;
s1=(((x(1)^2-(r2^2-r1^2))^(1/2)-r1)/sqrt(2)+r1)*((x(1)^2-(r2^2-r1^2))^(1/2)-r1)/sqrt(2)*2-asin(((x(1)^2-(r2^2-r1^2))^(1/2)-r1)/sqrt(2)/r1)*r1.^2;
s2=pi/4*r1^2-((pi/4-asin(((x(1)^2-(r2^2-r1^2))^(1/2)-r1)/sqrt(2)/r1))*r1^2-((x(1)^2-(r2^2-r1^2))^(1/2)-r1)*r1*sin(pi/4-asin(((x(1)^2-(r2^2-r1^2))^(1/2)-r1)/sqrt(2)/r1)));
%-----------------------------
ff(2)=4*(s1+s2)-pi*((x(2)^2-(r4^2-r3^2))-r3^2);
rad1((1000-950)/50)=(x(1)^2-(r2^2-r1^2))^(1/2)*1000
rad2((1000-950)/50)=x(1)*1000
rad3((1000-950)/50)=(x(2)^2-(r4^2-r3^2))^(1/2)*1000
rad4((1000-950)/50)=x(2)*1000
end
fsolve(@jifen3,)
这样得到的解完全没有意义吗？如何改进才能收敛？
[ 本帖最后由 eight 于 2007-10-15 19:19 编辑 ] 可能这种方式不适合该方程，去数学版区看看解方程的相关帖子吧 一个办法是与实际数据(如果有的话)比较;
另一个办法就是换一种软件试试---例如用1stOpt求解，也正好发挥其求解此方面问题的优势． 不会1stOpt
会的人麻烦帮忙运行一下源程序
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