- #1
Dustinsfl
- 2,281
- 5
Something doesn't seem right in regards to my analysis of the Jacobian. What about when $a=0$ at the second fixed point?
\begin{alignat*}{3}
x' & = & y - ax\\
y' & = & -y + \frac{x}{1 + x}
\end{alignat*}
First, we need to determine the fixed points in the system. So let
\begin{alignat*}{3}
y - ax & = & 0\\
-y + \frac{x}{1 + x} & = & 0
\end{alignat*}
Solving for $y$ in the first equation, we have $y = ax$.
$$
-ax + \frac{x}{1 + x} = 0\Rightarrow ax(1 + x) - x = x(ax + a - 1) = 0
$$
One fixed point is $(0,0)$.
$$
ax + a - 1 = 0\Rightarrow x = \frac{1 - a}{a}
$$
Another fixed point is $\left(\frac{1 - a}{a},a - 1\right)$. The Jacobian for the dynamical system is
$$
\mathcal{J} = \begin{pmatrix}
-a & 1\\
\frac{1}{(1 + x)^2} & -1
\end{pmatrix}.
$$
For the fixed point $(0,0)$, the Jacobian is
$$
\mathcal{J} = \begin{pmatrix}
-a & 1\\
1 & -1
\end{pmatrix}.
$$
Then the trace of the Jacobian is $\text{tr}(\mathcal{J}) = -(a + 1)$ and the determinant is $\det(\mathcal{J}) = a - 1$. When $a = 1$, the determinant is 0 and the two fixed points merge together. Therefore, when $a$ crosses 1, we have a change in stability at $(0,0)$ and a change in the number of fixed points. We will have a saddle when $a < 1$ at the fixed point $(0,0)$. When $a > 1$, the determinant is positive and the trace is negative.
\begin{alignat*}{3}
\left(\text{tr}(\mathcal{J})\right)^2 - 4\det(\mathcal{J}) & = & (a + 1)^2 - 4(a - 1)\\
& = & a^2 - 2a + 5 > 0
\end{alignat*}
Therefore, we have a stable node at the origin. Next, let $\frac{1}{1 + x} = a$. Then we can re-write the Jacobian as
$$
\mathcal{J} = \begin{pmatrix}
-a & 1\\
a^2 & -1
\end{pmatrix}.
$$
The trace of the Jacobian is $\text{tr}(\mathcal{J}) = -(a + 1)$ and the determinant is $\det(\mathcal{J}) = a(1 - a)$. When $a > 1\cup a < 0$, the determinant is negative. Thus, we will have a saddle at $\left(\frac{1 - a}{a},a - 1\right)$. When $0 < a < 1$, the determinant is positive and the trace is negative.
\begin{alignat*}{3}
x' & = & y - ax\\
y' & = & -y + \frac{x}{1 + x}
\end{alignat*}
First, we need to determine the fixed points in the system. So let
\begin{alignat*}{3}
y - ax & = & 0\\
-y + \frac{x}{1 + x} & = & 0
\end{alignat*}
Solving for $y$ in the first equation, we have $y = ax$.
$$
-ax + \frac{x}{1 + x} = 0\Rightarrow ax(1 + x) - x = x(ax + a - 1) = 0
$$
One fixed point is $(0,0)$.
$$
ax + a - 1 = 0\Rightarrow x = \frac{1 - a}{a}
$$
Another fixed point is $\left(\frac{1 - a}{a},a - 1\right)$. The Jacobian for the dynamical system is
$$
\mathcal{J} = \begin{pmatrix}
-a & 1\\
\frac{1}{(1 + x)^2} & -1
\end{pmatrix}.
$$
For the fixed point $(0,0)$, the Jacobian is
$$
\mathcal{J} = \begin{pmatrix}
-a & 1\\
1 & -1
\end{pmatrix}.
$$
Then the trace of the Jacobian is $\text{tr}(\mathcal{J}) = -(a + 1)$ and the determinant is $\det(\mathcal{J}) = a - 1$. When $a = 1$, the determinant is 0 and the two fixed points merge together. Therefore, when $a$ crosses 1, we have a change in stability at $(0,0)$ and a change in the number of fixed points. We will have a saddle when $a < 1$ at the fixed point $(0,0)$. When $a > 1$, the determinant is positive and the trace is negative.
\begin{alignat*}{3}
\left(\text{tr}(\mathcal{J})\right)^2 - 4\det(\mathcal{J}) & = & (a + 1)^2 - 4(a - 1)\\
& = & a^2 - 2a + 5 > 0
\end{alignat*}
Therefore, we have a stable node at the origin. Next, let $\frac{1}{1 + x} = a$. Then we can re-write the Jacobian as
$$
\mathcal{J} = \begin{pmatrix}
-a & 1\\
a^2 & -1
\end{pmatrix}.
$$
The trace of the Jacobian is $\text{tr}(\mathcal{J}) = -(a + 1)$ and the determinant is $\det(\mathcal{J}) = a(1 - a)$. When $a > 1\cup a < 0$, the determinant is negative. Thus, we will have a saddle at $\left(\frac{1 - a}{a},a - 1\right)$. When $0 < a < 1$, the determinant is positive and the trace is negative.
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