Writing function based on a vector of roots

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John
John 2018년 2월 28일
댓글: Star Strider 2018년 3월 1일
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));
  댓글 수: 3
John
John 2018년 3월 1일
Strangely enough, it's still not showing...
Thank you very much for your help!
Star Strider
Star Strider 2018년 3월 1일
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.

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Star Strider
Star Strider 2018년 3월 1일
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.

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