求助！三个未知数 四个方程用1stopt(Anto2Fit)约束优化
计算小白，新来的没啥金币
用的Anto2Fit 5.5试用版（明明三个parameters为啥说超出四个了）
方程都比较长
不知道为什么三个未知数会出现四个结果
第四个结果好像是我给的常数变量或者变参量（偷梁换柱？）
贴上mathematica里写的表达式（这个软件里运行过三个方程三个变量的求解，方程本身没问题）
Constant cc = 3*10^8;
la = cc/10^9;
k0 = 2*pi/la;
r1 =0.17532;
r2 =0.21216;
t1 =0.77139;
t2 =-0.3507;
Parameter k1[30,60],k2[-2,0],dz[0.5,0.7];
Function a=sin(k1*dz)*(exp(-k2*dz)+exp(k2*dz));
b=cos(k1*dz)*(exp(-k2*dz)-exp(k2*dz));
c=cos(k1*dz)*(exp(-k2*dz)+exp(k2*dz));
d=sin(k1*dz)*(exp(-k2*dz)-exp(k2*dz));
kk=2*(k1^2+k2^2);
(r1*c)-(r2*d)+r1*(b*k1-a*k2)/(2*k0)+r1*(a*k0*k2+b*k0*k1)/kk-r2*(a*k1+b*k2)/(2*k0)-r2*(a*k0*k1-b*k0*k2)/kk-(b*k1-a*k2)/(2*k0)+(a*k0*k2+b*k0*k1)/kk;
(r2*c)+(r1*d)+r1*(a*k1+b*k2)/(2*k0)+r1*(a*k0*k1-b*k0*k2)/kk+r2*(b*k1-a*k2)/(2*k0)+r2*(a*k0*k2+b*k0*k1)/kk-(a*k1+b*k2)/(2*k0)-(b*k0*k2-a*k0*k1)/kk;
(t1*c)-(t2*d)+t1*(b*k1-a*k2)/(2*k0)+t1*(a*k0*k2+b*k0*k1)/kk-t2*(a*k1+b*k2)/(2*k0)-t2*(a*k0*k1-b*k0*k2)/kk+2;
(t2*c)+(t1*d)+t1*(a*k1+b*k2)/(2*k0)+t1*(a*k0*k1-b*k0*k2)/kk+t2*(b*k1-a*k2)/(2*k0)+t2*(a*k0*k2+b*k0*k1)/kk;
//方程我用了两种写法第二种太长了
Function
(r1*cos(k1*dz)*(exp(-k2*dz)+exp(k2*dz)))-(r2*sin(k1*dz)*(exp(-k2*dz)-exp(k2*dz)))+r1*(cos(k1*dz)*(exp(-k2*dz)-exp(k2*dz))*k1-sin(k1*dz)*(exp(-k2*dz)+exp(k2*dz))*k2)/(2*k0)+r1*(sin(k1*dz)*(exp(-k2*dz)+exp(k2*dz))*k0*k2+cos(k1*dz)*(exp(-k2*dz)-exp(k2*dz))*k0*k1)/2*(k1^2+k2^2)-r2*(sin(k1*dz)*(exp(-k2*dz)+exp(k2*dz))*k1+cos(k1*dz)*(exp(-k2*dz)-exp(k2*dz))*k2)/(2*k0)-r2*(sin(k1*dz)*(exp(-k2*dz)+exp(k2*dz))*k0*k1-cos(k1*dz)*(exp(-k2*dz)-exp(k2*dz))*k0*k2)/2*(k1^2+k2^2)-(cos(k1*dz)*(exp(-k2*dz)-exp(k2*dz))*k1-sin(k1*dz)*(exp(-k2*dz)+exp(k2*dz))*k2)/(2*k0)+(sin(k1*dz)*(exp(-k2*dz)+exp(k2*dz))*k0*k2+cos(k1*dz)*(exp(-k2*dz)-exp(k2*dz))*k0*k1)/2*(k1^2+k2^2);
(r2*cos(k1*dz)*(exp(-k2*dz)+exp(k2*dz)))+(r1*sin(k1*dz)*(exp(-k2*dz)-exp(k2*dz)))+r1*(sin(k1*dz)*(exp(-k2*dz)+exp(k2*dz))*k1+cos(k1*dz)*(exp(-k2*dz)-exp(k2*dz))*k2)/(2*k0)+r1*(sin(k1*dz)*(exp(-k2*dz)+exp(k2*dz))*k0*k1-cos(k1*dz)*(exp(-k2*dz)-exp(k2*dz))*k0*k2)/2*(k1^2+k2^2)+r2*(cos(k1*dz)*(exp(-k2*dz)-exp(k2*dz))*k1-sin(k1*dz)*(exp(-k2*dz)+exp(k2*dz))*k2)/(2*k0)+r2*(sin(k1*dz)*(exp(-k2*dz)+exp(k2*dz))*k0*k2+cos(k1*dz)*(exp(-k2*dz)-exp(k2*dz))*k0*k1)/2*(k1^2+k2^2)-(sin(k1*dz)*(exp(-k2*dz)+exp(k2*dz))*k1+cos(k1*dz)*(exp(-k2*dz)-exp(k2*dz))*k2)/(2*k0)-(cos(k1*dz)*(exp(-k2*dz)-exp(k2*dz))*k0*k2-sin(k1*dz)*(exp(-k2*dz)+exp(k2*dz))*k0*k1)/2*(k1^2+k2^2);
(t1*cos(k1*dz)*(exp(-k2*dz)+exp(k2*dz)))-(t2*sin(k1*dz)*(exp(-k2*dz)-exp(k2*dz)))+t1*(cos(k1*dz)*(exp(-k2*dz)-exp(k2*dz))*k1-sin(k1*dz)*(exp(-k2*dz)+exp(k2*dz))*k2)/(2*k0)+t1*(sin(k1*dz)*(exp(-k2*dz)+exp(k2*dz))*k0*k2+cos(k1*dz)*(exp(-k2*dz)-exp(k2*dz))*k0*k1)/2*(k1^2+k2^2)-t2*(sin(k1*dz)*(exp(-k2*dz)+exp(k2*dz))*k1+cos(k1*dz)*(exp(-k2*dz)-exp(k2*dz))*k2)/(2*k0)-t2*(sin(k1*dz)*(exp(-k2*dz)+exp(k2*dz))*k0*k1-cos(k1*dz)*(exp(-k2*dz)-exp(k2*dz))*k0*k2)/2*(k1^2+k2^2)+2;
(t2*cos(k1*dz)*(exp(-k2*dz)+exp(k2*dz)))+(t1*sin(k1*dz)*(exp(-k2*dz)-exp(k2*dz)))+t1*(sin(k1*dz)*(exp(-k2*dz)+exp(k2*dz))*k1+cos(k1*dz)*(exp(-k2*dz)-exp(k2*dz))*k2)/(2*k0)+t1*(sin(k1*dz)*(exp(-k2*dz)+exp(k2*dz))*k0*k1-cos(k1*dz)*(exp(-k2*dz)-exp(k2*dz))*k0*k2)/2*(k1^2+k2^2)+t2*(cos(k1*dz)*(exp(-k2*dz)-exp(k2*dz))*k1-sin(k1*dz)*(exp(-k2*dz)+exp(k2*dz))*k2)/(2*k0)+t2*(sin(k1*dz)*(exp(-k2*dz)+exp(k2*dz))*k0*k2+cos(k1*dz)*(exp(-k2*dz)-exp(k2*dz))*k0*k1)/2*(k1^2+k2^2);
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按你第一种写法, 如果最后四个方程是等于0的约束的话，求得一个如下的解(其中四个方程的误差平方和为2.07045657911154E-05）（注意，下面不是1stopt的结果，所以不保证和1stopt的结果一致）
k1 = 48.5939462992421, k2 = -0.161774554371271, dz = 0.588775267394145
请问求解思路是什么，用matlab还是mathemetica求解的吗？
1. 主要使用自己写的MathSword软件进行求解。（PS：附件可以下载MathSword，在打开程序界面最上端选择【高级优化】，在优化模块右键选择上传的附件 solve.vb文件，然后点击求解，就能看到你问题的答案了。）
2. 至于你说的方程用什么约束比较合适，不太明白你的意思。因为一个方程本身就是一个约束，有约束就可以构建误差，一般的优化器内部会根据你构建的1个或多个误差进行优化。
代码写的有问题。
k1: 48.5939435307918
k2: -0.161774549680272
dz: 0.588775301298235
------------------------------------------------
Constant cc = 3*10^8,
r1 =0.17532,
r2 =0.21216,
t1 =0.77139,
t2 =-0.3507,
la = cc/10^9,
k0 = 2*pi/la;
Parameter k1=[30,60],k2=[-2,0],dz=[0.5,0.7];
ConstStr kk=2*(k1^2+k2^2),
a=sin(k1*dz)*(exp(-k2*dz)+exp(k2*dz)),
b=cos(k1*dz)*(exp(-k2*dz)-exp(k2*dz)),
c=cos(k1*dz)*(exp(-k2*dz)+exp(k2*dz)),
d=sin(k1*dz)*(exp(-k2*dz)-exp(k2*dz));
Function
(r1*c)-(r2*d)+r1*(b*k1-a*k2)/(2*k0)+r1*(a*k0*k2+b*k0*k1)/kk-r2*(a*k1+b*k2)/(2*k0)-r2*(a*k0*k1-b*k0*k2)/kk-(b*k1-a*k2)/(2*k0)+(a*k0*k2+b*k0*k1)/kk=0;
(r2*c)+(r1*d)+r1*(a*k1+b*k2)/(2*k0)+r1*(a*k0*k1-b*k0*k2)/kk+r2*(b*k1-a*k2)/(2*k0)+r2*(a*k0*k2+b*k0*k1)/kk-(a*k1+b*k2)/(2*k0)-(b*k0*k2-a*k0*k1)/kk=0;
(t1*c)-(t2*d)+t1*(b*k1-a*k2)/(2*k0)+t1*(a*k0*k2+b*k0*k1)/kk-t2*(a*k1+b*k2)/(2*k0)-t2*(a*k0*k1-b*k0*k2)/kk+2=0;
(t2*c)+(t1*d)+t1*(a*k1+b*k2)/(2*k0)+t1*(a*k0*k1-b*k0*k2)/kk+t2*(b*k1-a*k2)/(2*k0)+t2*(a*k0*k2+b*k0*k1)/kk=0;
感谢大神
1stOpt正确代码如下，4个方程3个未知数，只能有近似解。
结果：
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