Writing function based on a vector of roots

1 次查看(过去 30 天)
In the following code, one of lines plotted is a vector of solutions to a nonlinear function (plotted against a parameter). I want to find the intersection point, but I don't know how to do it. Does anyone have any suggestion?
clc;
close all;
clear all;
lambda = 1;
Fprime = @(z) lambda.*exp(-lambda.*z);
F = @(z) 1-lambda.*exp(-lambda.*z);
F2prime = @(z) -lambda^2.*exp(-lambda.*z);
intz = @(z) (1-(lambda.*z+1).*exp(-lambda.*z))./lambda;
funz = @(z) z.*(1-F(z))+intz(z);
ph = 0.8;
pl = 0.4;
t = 0.6;
theta =1.3;
sigma = 0.6;
alpha=0.6;
epsilon = 0.1;
y = @(z,psi) (((ph-pl)/(1-t))*psi*(z.*(1-F(z))+intz(z))).^(1/theta);
u = @(z,psi) ph.*((1+epsilon).*y(z,psi)).^alpha + (1-ph).*y(z,psi).^alpha;
N2 = 200;
psigrid = linspace(.5,5,N2);
zbar2 = zeros(1,N2);
Gfun = @(z,psi) psi.*z;
rhs = @(z,psi) (sigma/.8).*(u(z,psi)-Gfun(z,psi)+(1-ph).*psi.*(1-F(z)).*z);
for k=1:N2
rhstest = @(z) rhs(z,psigrid(1,k));
Gfuntest = @(z) Gfun(z,psigrid(1,k));
zbarfun = @(z) rhstest(z) - Gfuntest(z);
zbar2(1,k) = fsolve(zbarfun,.2);
end
figure
plot(psigrid,zbar2);
hold on;
plot(psigrid,.3.*ones(1,N2));
  3 个评论
John
John 2018-3-1
Strangely enough, it's still not showing...
Thank you very much for your help!
Star Strider
Star Strider 2018-3-1
As always, my pleasure!
I’ve not heard back yet from Rena Berman. I’ll let you know. She might post back here as well, since I included the URL to your Question in my email to her.

请先登录,再进行评论。

采纳的回答

Star Strider
Star Strider 2018-3-1
This will calculate the x-coordinate of the intersection, and plot a green pentagram there:
intx = interp1(zbar2, psigrid, 0.3, 'linear');
figure
plot(psigrid,zbar2)
hold on;
plot(psigrid,.3.*ones(1,N2))
plot(intx, 0.3, 'pg', 'MarkerFaceColor','g', 'MarkerSize',10)
hold off
The rest of your code (before and including the for loop) is unchanged, so I didn’t post it.

更多回答(0 个)

类别

Help CenterFile Exchange 中查找有关 MATLAB 的更多信息

产品

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by