Problems with integral in fsolve

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Oeer
Oeer 2019 年 4 月 9 日
コメント済み: Walter Roberson 2019 年 4 月 9 日
Hey
I have three functions with three unknown variables, but I can't find a solution for them. the features look like this:
functions.PNG
I've tried to do the following code:
m = 0.31;
r = 0.05;
delta = 0.02;
y_bar = 2;
w_1 = 5;
w_2_bar = 5;
sigma = 1;
b_bar = 3;
z_0=[0.5,3,4];
sol=fsolve(@(z)([
w_1*(1-z(1))-z(2)-((1+delta)/((1-m)*(1+r)))*int((b_bar+(1-z(1))*(1-m)*x+(1-m)*(1+r)*z(2))*(1/(2*pi*sigma^2))*exp(-((x-w_2_bar)^2)/(2*sigma^2)),x,0,z(3))
+((1+delta)/(1+r))*int(((1-z(1))*k+(1+r)*z(2))*(1/(2*pi*sigma^2))*exp(-((k-w_2_bar)^2)/(2*sigma^2)),k,z(3),inf);
z(1)*w_1+int(z(1)*h*(1/(2*pi*sigma^2))*exp(-((h-w_2_bar)^2)/(2*sigma^2)),h,0,z(3))
-int((b_bar-m*((1-z(1))*y+(1+r)*z(2)))*(1/(2*pi*sigma^2))*exp(-((y-w_2_bar)^2)/(2*sigma^2)),y,0,z(3));
z(3)-(1/(1-z(1))*y_bar+((1+r)/(1-z(1)))*z(2))]),z_0);
Hope there is one who can correct my code so it works, thank you very much if it is you. :)

採用された回答

Walter Roberson
Walter Roberson 2019 年 4 月 9 日
Code attached.
It is not fast code. You could improve the performance by changing the symbolic integrations into numeric integrations.
Or, since you are using symbolic integration, you could process your function once with symbolic z variables, and simplify. vpasolve() does a good job on what is left, or you can matlabFunction() and fsolve()
  1 件のコメント
Walter Roberson
Walter Roberson 2019 年 4 月 9 日
Having the int() inside the fsolve() is slow, and the int() have closed form representations if you use symbolic z. The performance gains from executing once symbolically and using the result with vpasolve() or fsolve() is substantial.

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その他の回答 (1 件)

Torsten
Torsten 2019 年 4 月 9 日
編集済み: Torsten 2019 年 4 月 9 日
function main
z_0 = [0.5,3,4];
options = optimset('TolFun',1e-8,'TolX',1e-8);
sol = fminsearch(@fun,z_0,options)
end
function res = fun(z)
m = 0.31;
r = 0.05;
delta = 0.02;
y_bar = 2;
w_1 = 5;
w_2_bar = 5;
sigma = 1;
b_bar = 3;
re(1) = w_1*(1-z(1))-z(2)-((1+delta)/((1-m)*(1+r)))*integral(@(x)(b_bar+(1-z(1))*(1-m)*x+(1-m)*(1+r)*z(2))*(1/(2*pi*sigma^2))*exp(-((x-w_2_bar).^2)/(2*sigma^2)),0,z(3),'ArrayValued',1)...
+((1+delta)/(1+r))*integral(@(k)((1-z(1))*k+(1+r)*z(2))*(1/(2*pi*sigma^2)).*exp(-((k-w_2_bar).^2)/(2*sigma^2)),z(3),inf,'ArrayValued',1);
re(2) = z(1)*w_1+integral(@(h)z(1)*h*(1/(2*pi*sigma^2)).*exp(-((h-w_2_bar).^2)/(2*sigma^2)),0,z(3),'ArrayValued',1)...
-integral(@(y)(b_bar-m*((1-z(1))*y+(1+r)*z(2)))*(1/(2*pi*sigma^2)).*exp(-((y-w_2_bar).^2)/(2*sigma^2)),0,z(3),'ArrayValued',1);
re(3) = z(3)-(1/(1-z(1))*y_bar+((1+r)/(1-z(1)))*z(2));
res = norm(re)
end

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