从解决到Fsolve
问题描述：
我有三个方程中有三个未知数，我想解决。 我用symbolic toolbox指定方程。我知道我可以使用solve函数来要求matlab找到我一个数字解决方案。然而，在3个未知数的3个方程中，matlab应该能够找到解析解（fsolve）。我只是不知道如何更改代码，以便我可以使用fsolve而不是solve。从解决到Fsolve
下面我的代码：
清除所有
syms Kl Kh alpha nu w phi delta P beta zh zl Ezh Ezl
nu1 = (1/(1-nu));
f1 = ((zl * (Kl^alpha))^nu1 + (zh * (Kh^alpha))^nu1) * nu^(nu*nu1) * (w^(-nu*nu1)) - w/phi + delta*(Kl + Kh)*P
f2 = Kh - (((1-beta*(1-delta))*P * (w^(nu1*nu))*(nu^(nu*nu1)))/(beta*alpha* (Ezh)^nu1))^((1-nu)/(alpha+nu-1))
f3 = Kl - (((1-beta*(1-delta))*P * (w^(nu1*nu))*(nu^(nu*nu1)))/(beta*alpha* (Ezl)^nu1))^((1-nu)/(alpha+nu-1))
f1 = subs(f1, {alpha, beta, nu, phi,delta, zh, zl, Ezh, Ezl, P}, {0.27, 0.96, 0.60, 2.15,0.065,1.11687642219068,0.895354204038589,1.07811003137331,0.934120594855956, 0.95})
f2 = subs(f2, {alpha, beta, nu, phi,delta, zh, zl, Ezh, Ezl, P}, {0.27, 0.96, 0.60, 2.15,0.065,1.11687642219068,0.895354204038589,1.07811003137331,0.934120594855956, 0.95})
f3 = subs(f3, {alpha, beta, nu, phi,delta, zh, zl, Ezh, Ezl, P}, {0.27, 0.96, 0.60, 2.15,0.065,1.11687642219068,0.895354204038589,1.07811003137331,0.934120594855956, 0.95})
S = solve([f1 == 0, f2 == 0, f3 == 0],...
[w, Kh, Kl], 'ReturnConditions', true);
答
在此期间我找到了解决办法。
下面是完成代码：
function SSfunction = SSfunction(x)
syms alphaa nu phi delta p betaa zh zl ezh ezl
nu1 = (1/(1-nu));
f1 = ((zl * (x(3)^alphaa))^nu1 + (zh * (x(2)^alphaa))^nu1) * nu^(nu*nu1) * (x(1)^(-nu*nu1)) - x(1)/phi - delta*(x(3) + x(2))*p;
f2 = x(2) - ((betaa*alphaa*(ezh^(nu1)) * (nu^(nu*nu1)))/((1-betaa*(1-delta))*p* (x(1)^(nu*nu1))))^((1-nu)/(1-alphaa-nu));
f3 = x(3) - ((betaa*alphaa*(ezl^(nu1)) * (nu^(nu*nu1)))/((1-betaa*(1-delta))*p* (x(1)^(nu*nu1))))^((1-nu)/(1-alphaa-nu));
f1 = subs(f1, {alphaa, betaa, nu, phi,delta, zh, zl, ezh, ezl, p}, {0.27, 0.96, 0.60, 2.15,0.065,1.11687642219068,0.895354204038589,1.07811003137331,0.934120594855956, 0.950});
f3 = subs(f1, {alphaa, betaa, nu, phi,delta, zh, zl, ezh, ezl, p}, {0.27, 0.96, 0.60, 2.15,0.065,1.11687642219068,0.895354204038589,1.07811003137331,0.934120594855956, 0.950});
f2 = subs(f1, {alphaa, betaa, nu, phi,delta, zh, zl, ezh, ezl, p}, {0.27, 0.96, 0.60, 2.15,0.065,1.11687642219068,0.895354204038589,1.07811003137331,0.934120594855956, 0.950});
SSfunction(1) = eval(f1)
SSfunction(2) = eval(f2)
SSfunction(3) = eval(f3)
end
x0 = [1,2,0.7];
fun = @SSfunction;
x = fsolve(fun,x0)
+0
两条评论。首先，命名与函数本身相同的函数输出（'SSfunction'）通常是一种不好的做法。而且，这种解决方案效率低下，因为每次将fsolve优化称为目标函数时，符号表达式都会转换为浮点。在非常特殊的情况下，如果方程中存在重要性或其他数值问题，则此方法可能会有用。这似乎并非如此。 – horchler
答
使用matlabFunction你的符号表达式转换为可以直接使用fsolve一个量化的数字功能：
...
f = matlabFunction([f1;f2;f3],'Vars',{[w;Kh;Kl]});
w0 = 1;
Kh0 = 1;
Kl0 = 1;
x0 = [w0;Kh0;Kl0];
x = fsolve(f,x0)
这将是一个数量级幅度快于使用fsolve中的符号表达式本身。为了获得更多的速度，你也可以完全通过手动矢量化你的函数来摆脱象征性的数学运算：
alpha = 0.27;
beta = 0.96;
nu = 0.6;
phi = 2.15;
delta = 0.065;
zh = 1.11687642219068;
zl = 0.895354204038589;
Ezh = 1.07811003137331;
Ezl = 0.934120594855956;
P = 0.95;
nu1 = (1/(1-nu));
f = @(w,Kh,Kl)[((zl * (Kl.^alpha))^nu1 + (zh * (Kh.^alpha))^nu1) * nu^(nu*nu1) .* (w.^(-nu*nu1)) - w/phi + delta*(Kl + Kh)*P;
Kh - (((1-beta*(1-delta))*P * (w.^(nu1*nu))*(nu^(nu*nu1)))/(beta*alpha* (Ezh)^nu1))^((1-nu)/(alpha+nu-1));
Kl - (((1-beta*(1-delta))*P * (w.^(nu1*nu))*(nu^(nu*nu1)))/(beta*alpha* (Ezl)^nu1))^((1-nu)/(alpha+nu-1))];
w0 = 1;
Kh0 = 1;
Kl0 = 1;
x0 = [w0;Kh0;Kl0];
x = fsolve(@(x)f(x(1,:),x(2,:),x(3,:)),x0)
'fsolve'用于找到*数值解，而不是分析解。在符号数学工具箱中的求解可以在某些情况下找到解析解（如果它们存在的话），但其他方法则会使用可变精度数学来找到数值解。你在寻找一个数字解决方案吗？ – horchler
是的，我的歉意。我的意思是数字解决方案。 – phdstudent