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$f'(t)=af(t)(K-f(t))-bf(t)g(t)$ for $a,b,c,d,t,K>0$


This system has 3 fixed points (You can evaluate them if you set the 2 equations = 0). One point is $(\frac{d}{c},\frac{a}{b}(K-\frac{d}{c}))$

I would like to know if this point is asymptotically stable for $K>\frac{d}{c}$, so if the solution converges to this point for $t\to\infty$, correct ?

I have no idea and would really appreciate if someone could show me how to do it so I can use the method for similar equations.

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I guess there is an extra $x$ in your second equation. Tell us what you have tried so far for this homework problem. – Hans Engler Nov 24 '12 at 18:57
The only idea I have is to solve the differential equations, but I think this is the wrong way. – Montaigne Nov 25 '12 at 14:05

Consider a differential equation $X' = F(X)$, where $X(t) = (x_1(t),x_2(t))$. This can also be written as

$$ \begin{align*} x_1' &= F_1(x_1,x_2), \\ x_2' &= F_2(x_1,x_2), \end{align*} $$

where $F = (F_1,F_2)$. Suppose the system has an equilibrium at $X = X_0$. The first step is to compute the linearization about $X_0$, which is given by

$$ Y' = D_XF(X_0)\,Y. $$

To unravel this notation a bit, suppose $X_0 = (\alpha,\beta)$. Then $$D_XF(X_0) = \left( \begin{array}{cc} \frac{\partial F_1}{\partial x_1}(\alpha,\beta) & \frac{\partial F_1}{\partial x_2}(\alpha,\beta) \\ \frac{\partial F_2}{\partial x_1}(\alpha,\beta) & \frac{\partial F_2}{\partial x_2}(\alpha,\beta) \end{array} \right).$$ This is just the matrix of the first partial derivatives of the components of $F$ evaluated at the equilibrium point. In your problem, $X_0 = \left(\frac{d}{c},\frac{a}{b}\left(K-\frac{d}{c}\right)\right)$ and $$F(f,g) = \left(\begin{array}{c} af(K-f)-bfg \\ cfg-dg \end{array}\right)$$ (so that $F_1(f,g) = af(K-f)-bfg$ and $F_2(f,g) = cfg-dg$).

A linear system like this is asymptotically stable at the origin (and hence $X=X_0$ is an asymptotically stable equilibrium of $X' = F(X)$) if both eigenvalues of the matrix $D_XF(X_0)$ have negative real part.

For your problem, it is indeed the case that both eigenvalues have negative real part when $K > d/c$, so the equilibrium in question is indeed asymptotically stable.

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Thank you for your answer, but what ist $D_X F(X_0)$ explicitly in my case? – Montaigne Dec 3 '12 at 20:58
@Montaigne, I have edited my answer to explain that notation. Does this make it clearer? Please post what you're getting for that matrix if you're still unsure. – Antonio Vargas Dec 3 '12 at 21:28
Thank you, it made it clearer, but still I have a problem with $F_1$ and $F_2$. To calcuate the derivative of $F_1$ in $x_1$ I need to know the derivative of $f$ and $g$ in the point $x_1$, but how can this be evaluated not knowing $f$ and $g$ explicitly – Montaigne Dec 3 '12 at 21:37
@Montaigne, $x_1 = f$ here, and $x_2 = g$. The symbol $\frac{\partial F_1}{\partial x_1}$ just means the derivative of $F_1$ with respect to the first argument (since I defined $F_1$ as the map $(x_1,x_2) \mapsto F_1(x_1,x_2)$). – Antonio Vargas Dec 3 '12 at 21:42
Ok, so I understood you correctly, this is my matrix (I think there should be a $x_2$ instead of a $x_1$ in the right down corner of your matrix): \begin{array}{cc} \ -af'(t)f'(t)-bf'(t)g(t) & -bf(t)g'(t) \\ \ cf'(t)g(t) & g'(t)(cf(t)-d) \end{array} – Montaigne Dec 3 '12 at 21:55

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