Chapter 12.RP, Problem 1RP

In Problems 1-6, find all the critical points for the given system, discuss the type and stability of each critical point, and sketch the phase plane diagrams near each of the critical points.

$\begin{array}{l}\frac{dx}{dt}=-x-4y,\\ \frac{dy}{dt}=-2x-7y\end{array}$

To determine

All the critical points for the system, type and stability of each critical points and sketch the phase plane diagram.

$\begin{array}{l}\frac{dx}{dt}=-x-4y,\\ \frac{dy}{dt}=-2x-7y\end{array}$

Answer to Problem 1RP

Solution:

The point $\left(0,0\right)$ is the only critical point. The point $\left(0,0\right)$ is a saddle point. It is always unstable and Figure $\left(1\right)$ shows trajectories with their flow arrows.

Explanation of Solution

Given:

The systems of differential equations: $\begin{array}{l}\frac{dx}{dt}=-x-4y,\\ \frac{dy}{dt}=-2x-7y\end{array}$

Approach:

1) Critical points are the solutions of $x^{\prime }=0$ and $y^{\prime }=0$.

2) Trajectories are the parametric solutions of the system of differential equations.

3) Eigenvalues are the roots of characteristic equation.

4) Classification of critical points:

i) For complex conjugates eigenvalues:

• When the real part $\lambda$ is zero, this type of critical point is called a center. It is neutrally stable.
• When the real part λ is nonzero, this type of critical point is called a spiral point. It is asymptotically stable if real part is less than $0$, it is unstable if real part is greater than $0$.

ii) For real eigenvalues:

• Distinct eigenvalues
  • When both eigenvalues are positive or both are negative, this type of critical point is called a node. It is asymptotically stable if eigenvalues are both negative, unstable if eigenvalues are both positive.
  • When eigenvalues have opposite signs, this type of critical point is called a saddle point. It is always unstable.
• Repeated eigenvalues
  • When there are two linearly independent eigenvectors, this type of critical point is called a proper node. It is asymptotically stable if eigenvalue is less than $0$ and unstable if eigenvalue is greater than $0$.
  • When there is only one linearly independent eigenvector, this type of critical point is called an improper node. It is asymptotically stable if eigenvalue is less than $0$, unstable if eigenvalue is greater than $0$.

Calculation:

Solve $\frac{dx}{dt}=0$ and $\frac{dy}{dt}=0$.

$\begin{array}{rll}-x-4y& =& 0\\ x& =& -4y\end{array}$…$\left(1\right)$

$\begin{array}{rll}-2x-7y& =& 0\\ 2x& =& -7y\end{array}$…$\left(2\right)$

Substitute $-4y$ for $x$ in Equation $\left(2\right)$ and solve.

$\begin{array}{rll}-8y& =& -7y\\ y& =& 0\end{array}$

From Equation $\left(2\right)$, $x=0$.

So, $\left(0,0\right)$ is the only critical point.

Write the given system in matrix form.

$dX=\left[\begin{array}{rr}\hfill -1& \hfill -4\\ \hfill -2& \hfill -7\end{array}\right]X$.

Here, $X=\left[\begin{array}{r}\hfill x\\ \hfill y\end{array}\right]$

Find the eigenvalue of the matrix $\left[\begin{array}{rr}\hfill -1& \hfill -4\\ \hfill -2& \hfill -7\end{array}\right]$.

Eigenvalues are the roots of

$x^2+8x-1=0$…$\left(1\right)$

So, the eigenvalues are $-4\pm \sqrt{17}$.

Since both eigenvalues are distinct with opposite signs, $\left(0,0\right)$ is a saddle point. It is always unstable.

Figure $\left(1\right)$ shows trajectories with their flow arrows.

Figure $\left(1\right)$

Therefore, $\left(0,0\right)$ is the only critical point. The point $\left(0,0\right)$ is a saddle point. It is always unstable and Figure $\left(1\right)$ shows trajectories with their flow arrows.

Conclusion:

Hence, $\left(0,0\right)$ is the only critical point. The point $\left(0,0\right)$ is a saddle point. It is always unstable and Figure $\left(1\right)$ shows trajectories with their flow arrows.