Steady-states and non-trivial eigenvalues of equation

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katara
katara 2020-9-11
回答： David Goodmanson 2020-9-11
I have the equation:
And want to find the steady-states and nontrivial eigenvalues of the equation to determine if they are unstable or stable. This is my code so far:
syms y r K theta
eqn = r*y*(y./theta -1)*(1-y./K)==0;
s=solve(eqn)
derivative = diff(r*y*(y./theta -1)*(1-y./K));
eigenvalue1 = subs(derivative,y,theta)
eigenvalue2 = subs(derivative,y,K)
eigenvalue3 = subs(derivative,y,0)
This gives that the steady-states are: 0, theta, K.
And the eigenvalues are:
eigenvalue1 = -r*(theta/K - 1)
eigenvalue2 = -r*(K/theta - 1)
eigenvalue3 = -r
Is this the correct way of finding the steady-states and eigenvalues? I know that the variables don't have any values so it is difficult for me to determine if they are stable or not.

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回答（1 个）

David Goodmanson
David Goodmanson 2020-9-11
Hi katara,
the fixed points are at 0,theta,K as you said. Although you don't have specific values, the derivatives, which you refer to as eig1, eig2, eig3, just depend on the assumptions for [1] the sign of r and [2] whether theta/K is > 0 or < 0. With those four overall possibilites you should find either two stable fixed points and one unstable fixed point, or vise versa. And theta = K is a special case.

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