Worked-out solutions

For the linear system of differential equations $\quad\frac{\dd}{\dd t}\vec{x}=A\, \vec{x}\quad$ the nature of the equilibrium $(0,0)$ and the phase portrait depend on the eigenvalues of the matrix $A$ and their multiplicity.

1. $\quad\frac{\dd}{\dd t}\vec{x}=A\, \vec{x}\quad$ with $A = \matrix{1 & 1\\3 & -1\\}$
   $\quad$The characteristic polynomial of $A$ is equal to $$\begin{aligned}\det(A-\lambda\, I) &=\left|\begin{array}{cc} 1-\lambda & 1 \\ 3 & -1-\lambda \end{array}\right|\\\\ &= (1-\lambda)(-1-\lambda)-3 \\\\ &=\lambda^2 -4\\\\ &=(\lambda-2)(\lambda+2)\end{aligned}$$ $\quad$The eigenvalues of $A$ are the zeros of the characteristic polynomial: $2$ and $-2$.
   $\quad$The equilibrium is a saddle point because there is one positive and one negative eigenvalue.
   
   $\quad$Let $\cv{v_1\\ v_2}$ be an eigenvector corresponding to the eigenvalue $2$.
   $\quad$So: $\matrix{1 & 1\\3 & -1\\}\cdot \cv{v_1\\ v_2}=2\cv{v_1\\ v_2}$.
   
   $\quad$Then: $\left\{\begin{aligned} v_1+v_2 &= 2v_1 \\[0.2cm] 3v_1-v_2 &= 2v_2\end{aligned}\right.$
   
   $\quad$Thus: $v_2=v_1$ and $v_1$ can be freely chosen.
   $\quad$An eigenvector corresponding to the eigenvalue $2$ (with integral coefficients) is $\vec{v}=\cv{1\\1}$.
   $\quad$Similarly, $\vec{v}=\cv{1\\ -3}$ is an eigenvector corresponding to the eigenvalue $-2$.
   $\quad$The general solution is: $$\vec{x}(t)=\alpha e^{2t}\cdot \cv{1\\1}+\beta e^{-2t}\cdot \cv{1\\-3}$$ $\quad$with constants $\alpha$ and $\beta$.
   $\quad$The figure below shows a phase portrait of this linear system of differential equations.
   $\quad$
2. $\quad\frac{\dd}{\dd t}\vec{x}=A\, \vec{x}\quad$ with $A = \matrix{2 & 1\\2 & 3\\}$
   $\quad$The characteristic polynomial of $A$ is equal to $$\begin{aligned}\det(A-\lambda\, I) &=\left|\begin{array}{cc} 2-\lambda & 1 \\ 2 & 3-\lambda \end{array}\right|\\\\ &= (2-\lambda)(3-\lambda)-2 \\\\ &=\lambda^2 -5\lambda+4\\\\ &=(\lambda-1)(\lambda-4)\end{aligned}$$ $\quad$The eigenvalues of $A$ are the zeros of the characteristic polynomial: $1$ and $4$.
   $\quad$The equilibrium is repelling because there are two positive eigenvalues.
   
   $\quad$Let $\cv{v_1\\ v_2}$ be an eigenvector with eigenvalue $1$:
   $\quad$So: $\matrix{2 & 1\\2 & 3\\}\cdot \cv{v_1\\ v_2}=\cv{v_1\\ v_2}$.
   
   $\quad$Then: $\left\{\begin{aligned} 2v_1 + V_2 & v_1 = \\[0.2cm] 2v_1+3v_2 &= v_2\end{aligned}\right.$
   
   $\quad$Thus: $v_2=-v_1$ and $v_1$ can be freely chosen.
   $\quad$An eigenvector corresponding to the eigenvalue $1$ (with integral coefficients) is $\vec{v}=\cv{1\\-1}$.
   $\quad$Similarly, $\vec{v}=\cv{1\\ 2}$ is an eigenvector corresponding to the eigenvalue $4$.
   $\quad$The general solution is: $$\vec{x}(t)=\alpha e^{t}\cdot \cv{1\\-1}+\beta e^{4t}\cdot \cv{1\\2}$$ $\quad$with constants $\alpha$ and $\beta$.
   $\quad$The figure below shows a phase portrait of this linear system of differential equations.
   $\quad$
3. $\quad\frac{\dd}{\dd t}\vec{x}=A\, \vec{x}\quad$ with $A = \matrix{-1 & -2\\2 & -1\\}$
   $\quad$The characteristic polynomial of $A$ is equal to $$\begin{aligned}\det(A-\lambda\, I) &=\left|\begin{array}{cc} -1-\lambda & -2 \\ 2 & -1-\lambda \end{array}\right|\\\\ &= (-1-\lambda)^2+4 \\\\ &= (\lambda+1)^2+4 \\\\&=\lambda^2 +2\lambda+5\end{aligned}$$ $\quad$The eigenvalues of $A$ are the zeros of the characteristic polynomial.
   $\quad$Via the $abc$-formula or by completing the square with $(\lambda+1)^2+4$ we get
   $\quad$the eigenvalues $-1+2\ii$ and $-1-2\ii$.
   $\quad$Complex eigenvalues having a negative real part means
   $\quad$that we are dealing with a shrinking spiral.
   $\quad$The figure below shows a phase portrait of this linear system of differential equations.
   $\quad$
4. $\quad\frac{\dd}{\dd t}\vec{x}=A\, \vec{x}\quad$ with $A = \matrix{1 & 2\\-2 & 1\\}$
   $\quad$The characteristic polynomial of $A$ is equal to $$\begin{aligned}\det(A-\lambda\, I) &=\left|\begin{array}{cc} 1-\lambda & 2 \\ -2 & 1-\lambda \end{array}\right|\\\\ &= (1-\lambda)^2+4 \\\\ &= (\lambda-1)^2+4 \\\\&=\lambda^2 -2\lambda+5\end{aligned}$$ $\quad$The eigenvalues of $A$ are the zeros of the characteristic polynomial.
   $\quad$ Via the $abc$-formula or by completing the square with $\lambda-1)^2+4$ we get
   $\quad$the eigenvalues $1+2\ii$ and $1-2\ii$.
   $\quad$Complex eigenvalues having a negative real part means
   $\quad$that we are dealing with an expanding spiral.
   $\quad$The figure below shows a phase portrait of this linear system of differential equations.
   $\quad$
5. $\quad\frac{\dd}{\dd t}\vec{x}=A\, \vec{x}\quad$ with $A = \matrix{0 & -2\\2 & 0\\}$
   $\quad$The characteristic polynomial of $A$ is equal to $$\begin{aligned}\det(A-\lambda\, I) &=\left|\begin{array}{cc} -\lambda & -2 \\ 2 & \lambda \end{array}\right|\\\\ &= \lambda^2+4 \end{aligned}$$ $\quad$The eigenvalues of $A$ are the zeros of the characteristic polynomial.
   $\quad$Via the $abc$ -formula or by completing the square with $\lambda)^2+4$ we get
   $\quad$the eigenvalues $2\ii$ and $-2\ii$ are.
   $\quad$Complex eigenvalues with a real part equal to $0$ means
   $\quad$that we are dealing with circular solution curve.
   $\quad$The figure below shows a phase portrait in this linear system of differential equations.
   $\quad$The only differences with the exercise set 3 are the direction of the direction field and the curves.
   $\quad$

Worked-out solutions

For the linear system $\quad\frac{\dd}{\dd t}\vec{x}=A\, \vec{x}\quad$ the nature of the equilibrium $(0,0)$ depends on the trace and the determinant of the matrix $A$. We use the scheme below.

In the figure below $\beta=\mathrm{sp}(A)$ and $\gamma=\det(A)$. Then $\beta^2-4\gamma$ is the discriminant of the characteristic equation of $A$. When the discriminant is equal to 0, then the stability of the balance depends on the sign of $\beta$: repelling if$\beta>0$ and attracting if $\beta<0$. There are six other cases:

1. repelling equilibrium: $\beta>0$ and $\gamma>0$.
2. saddle point (semi-stable equilibrium) $\gamma<0$.
3. attracting equilibrium: $\beta<0$ and $\gamma>0$.
4. expanding spiral: $\beta>0$ and $\beta^2<4\gamma$.
5. periodic solutions around the equilibrium: $\beta=0$ and $\beta^2<4\gamma$.
6. shrinking spiral: $\beta<0$ and $\beta^2<4\gamma$.

1. $\quad\frac{\dd}{\dd t}\vec{x}=A\, \vec{x}\quad$ with $A = \matrix{1 & 5\\3 & 2\\}$
   $\quad\mathrm{sp}(A)=1+2=3\quad\text{and}\quad\det(A)=1\times 2-5\times 3=-13$.
   $\quad$Because $\det(A)<0$ the equilibrium is a saddle point.
   $\quad$The eigenvalues of matrix $A$ are $$\lambda_{1,2}=\frac{\mathrm{sp}(A)\pm\sqrt{\mathrm{sp}(A)-4\det(A)}}{2}=\frac{3\pm\sqrt{61}}{2}$$ $\quad$There is a positive and a negative eigenvalue.
   $\quad$This confirms that the equilibrium $(0,0)$ is a saddle point.
   $\quad$The figure below shows a phase portrait in this linear system of differential equations.
   $\quad$
2. $\quad\frac{\dd}{\dd t}\vec{x}=A\, \vec{x}\quad$ with $A = \matrix{0 & -2\\1 & -3\\}$
   $\quad\mathrm{sp}(A)=0-3=-3\quad\text{and}\quad\det(A)=0\times (-3)-1\times (-2)=2$.
   $\quad$Because $\mathrm{sp}(A)<0$ and $\det(A)>0$ we have an attracting equilibrium.
   $\quad$The eigenvalues of the matrix $A$ are $$\lambda_{1,2}=\frac{\mathrm{sp}(A)\pm\sqrt{\mathrm{sp}(A)-4\det(A)}}{2}=-2\;\;\text{en}\;\; -1$$ $\quad$Thus, there are two negative eigenvalues.
   $\quad$This confirms that the equilibrium $(0,0)$ is attracting.
   $\quad$The figure below shows a phase portrait in this linear system of differential equations.
   $\quad$
3. $\quad\frac{\dd}{\dd t}\vec{x}=A\, \vec{x}\quad$ with $A = \matrix{-2 & 4\\-3 & 4\\}$
   $\quad\mathrm{sp}(A)=-2+4=2\quad\text{and}\quad\det(A)=-2\times 4-4\times (-3)=4$.
   $\quad$Because $\mathrm{sp}(A)>0$ and $\det(A)>0$ we have a repelling equilibrium.
   $\quad$Because $\mathrm{sp}(A)>0$ and $\bigl(\mathrm{sp}(A)\bigr)^2<4\det(A)$ we have an expanding spiral.
   $\quad$The eigenvalues of the matrix $A$ are $$\lambda_{1,2}=\frac{\mathrm{sp}(A)\pm\sqrt{\mathrm{sp}(A)-4\det(A)}}{2}=1\pm 3\ii$$ $\quad$Thus, there are two complex eigenvalues with a positive real part.
   $\quad$This confirms that the equilibrium $(0,0)$ is repelling
   $\quad$in the form of an expanding spiral.
   $\quad$The figure below shows a phase portrait in this linear system of differential equations.
   $\quad$
4. $\quad\frac{\dd}{\dd t}\vec{x}=A\, \vec{x}\quad$ with $A = \matrix{2 & 1\\1 & 3\\}$
   $\quad\mathrm{sp}(A)=2+3=5\quad\text{and}\quad\det(A)=2\times 3-1\times 1=5$.
   $\quad$Because $\mathrm{sp}(A)>0$ and $\det(A)>0$ we have a repelling equilibrium.
   $\quad$The eigenvalues of the matrix $A$ are $$\lambda_{1,2}=\frac{\mathrm{sp}(A)\pm\sqrt{\mathrm{sp}(A)-4\det(A)}}{2}=\frac{5\pm\sqrt{5}}{2}$$ $\quad$Thus, there are two positive eigenvalues.
   $\quad$that the equilibrium $(0,0)$ is repelling.
   $\quad$The figure below shows a phase portrait in this linear system of differential equations.
   $\quad$
5. $\quad\frac{\dd}{\dd t}\vec{x}=A\, \vec{x}\quad$ with $A = \matrix{-2 & -1\\1 & 2\\}$
   $\quad\mathrm{sp}(A)=-2+2=0\quad\text{and}\quad\det(A)=-2\times 2-1\times (-1)=-3$.
   $\quad$Because $\det(A)<0$ we have a saddle point.
   $\quad$The eigenvalues of matrix $A$ are $$\lambda_{1,2}=\frac{\mathrm{sp}(A)\pm\sqrt{\mathrm{sp}(A)-4\det(A)}}{2}=\pm\sqrt{3}$$ $\quad$There is a positive and a negative eigenvalue.
   $\quad$This confirms that the equilibrium $(0,0)$ is a saddle point.
   $\quad$The figure below shows a phase portrait in this linear system of differential equations.
   $\quad$
6. $\quad\frac{\dd}{\dd t}\vec{x}=A\, \vec{x}\quad$ with $A = \matrix{-1 & -2\\2 & 1\\}$
   $\quad\mathrm{sp}(A)=-1+1=0\quad\text{and}\quad\det(A)=-1\times 1-2\times (-2)=3$.
   $\quad$Because $\text{sp(A)=0$ and $\det(A)>0$ we have periodic solutions around the equilibrium.
   $\quad$The eigenvalues of the matrix $A$ are $$\lambda_{1,2}=\frac{\mathrm{sp}(A)\pm\sqrt{\mathrm{sp}(A)-4\det(A)}}{2}=\pm\sqrt{3}\ii$$ $\quad$Thus, there are two complex eigenvalues with real part equal to $0$.
   $\quad$This confirms that we have periodic solutions around the equilibrium $(0,0)$.
   $\quad$The figure below shows a phase portrait in this linear system of differential equations.
   $\quad$

Worked-out solution

We consider the linear system of differential equations with the matrix-vector form

$$\frac{\dd}{\dd t}\cv{x\\y}=A\cv{x\\y },\quad\text{where}\quad A=\matrix{-2 & -\alpha\\ 3 & -1}$$ Then $$\mathrm{sp}(A)=-1-2=-3\quad\text{and}\quad\det(A)=-(-2)\times (-1)-3\times (-\alpha)=2+3\alpha$$ When $\alpha<-\frac{2}{3}$, then $\det(A)<0$ and we are dealing with a saddle point.

When $\alpha>-\frac{2}{3}$ then $\det(A)>0$ and we are dealing with an attracting equilibrium.
This can be more specific if you look at the expression $\bigl(\mathrm{sp}(A)\bigr)^2-4\det(A) =9-4(2+3\alpha)=1-12\alpha$. This expression is negative when $\alpha>\frac{1}{12}$ and this implies according to the stability scheme that we then deal with a shrinking spiral.

When $-\frac{2}{3}<\alpha\le \frac{1}{12}$ we have a n attracting equilibrium but it is not a shrinking spiral.

The figure below shows phase portraits of this linear system of differential equations for various values of $\alpha$ that illustrate the three types.

When $\alpha=-\frac{2}{3}$, we have a special case in which each point on the line $y=3x$ is an equilibrium. Each solution goes ion a straight line to a point on this line, as in shown in the phase portrait below.