Matlab plotting direction field, stable or unstable?

조회 수: 2 (최근 30일)
Robin
Robin 2011년 7월 26일
I'm trying to figure out which equilibrium solutions are stable and unstable for this differnetial equation (Gompertz model) y'=y(1-ln(y)(y-3), so here is my code:
%%-----------------------------------------------------------
[y]=solve('y*(1-log(y))*(y-3)=0'); % solving for the equilibrium solution
disp('The Equilibrium solutions are y=')
y
% plotting the direction field
figure, hold on
syms t c
[T Y]= meshgrid(0:.5:12, -5:.5:12);
S= Y.*(1-log(Y)).*(Y-3);
L=sqrt(1+S.^2);
quiver(T, Y, 1./L, S./L, 0.5), axis tight
xlabel 'S', ylabel 'y'
hold off
%%-----------------------------------------------
however the graph it self is sort of ambiguous, i am having difficulty "reading the direction field" from about 2.5 and 3- there appears to be a continuous of equilibrium solutions.
Does anyone have any idea as to why the graph appears to be inconclusive on this interval?
I tried to graph the function and I found the limit as the function went to zero, and I found that it was 0. But I still am not sure why it appears this way. I think maybe the maximum value of f(y) has to something to do with it?
%--------------------------------
% code for graphing f(y) and finding the limit:
clear all
f=@(y) (y.*(1-log(y)).*(y-3));
Y=0:.1:4;
Z=f(Y);
plot(Y,Z)
title 'Plot of y.*(1-log(y)).*(y-3)'
xlabel y
ylabel f(y)
syms y
limit(y*(1-log(y))*(y-3),y,0)

답변 (0개)

카테고리

Help CenterFile Exchange에서 Programming에 대해 자세히 알아보기

Community Treasure Hunt

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

Start Hunting!

Translated by