# Equilibrium Points of Linear Autonomous Systems

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Investigate the equilibrium positions of the linear autonomous system and draw its phase portrait. $\frac{{dx}}{{dt}} = - x,\;\frac{{dy}}{{dt}} = 2x - 2y.$

### Example 2

Investigate the equilibrium positions of the dynamic system and sketch its phase portrait. $\frac{{dx}}{{dt}} = x + 3y,\;\frac{{dy}}{{dt}} = 2x.$

### Example 3

Investigate the equilibrium points and sketch the phase portrait of the following system: $\frac{{dx}}{{dt}} = 3x - 4y,\;\frac{{dy}}{{dt}} = 2x - y$

### Example 4

Investigate the stability of the system depending on the parameter $$a:$$ $\frac{{dx}}{{dt}} = ax + y,\;\frac{{dy}}{{dt}} = x + ay.$

### Example 1.

Investigate the equilibrium positions of the linear autonomous system and draw its phase portrait. $\frac{{dx}}{{dt}} = - x,\;\frac{{dy}}{{dt}} = 2x - 2y.$

Solution.

1. We write the matrix of the system and compute its determinant:
$A = \left[ {\begin{array}{*{20}{r}} { - 1}&0\\ 2&{ - 2} \end{array}} \right],\;\; \det A = \left| {\begin{array}{*{20}{r}} { - 1}&0\\ 2&{ - 2} \end{array}} \right| = 2 \ne 0.$
As $$\det A \ne 0,$$ the system has the unique equilibrium point $$\mathbf{X} = \mathbf{0}.$$ We find the eigenvalues of the matrix $$A:$$
$\det \left( {A - \lambda I} \right) = 0,\;\; \Rightarrow \left| {\begin{array}{*{20}{c}} { - 1 - \lambda }&0\\ 2&{ - 2 - \lambda } \end{array}} \right| = 0,\;\; \Rightarrow \left( {\lambda + 1} \right)\left( {\lambda + 2} \right) = 0,\;\; \Rightarrow {\lambda _1} = - 1,\;{\lambda _2} = - 2.$
2. Both eigenvalues are real and negative, so the equilibrium point $$\mathbf{X} = \mathbf{0}$$ is a stable node.
3. We derive the isocline equations, i.e. the lines which are tangent to the phase trajectories. The vertical isocline is given by
$\frac{{dx}}{{dt}} = - x = 0\;\;\text{or}\;\;x = 0.$
The horizontal isocline is written as As $$\det A \ne 0,$$ the system has the unique equilibrium point $$\mathbf{X} = \mathbf{0}.$$ We find the eigenvalues of the matrix $$A:$$
$\frac{{dy}}{{dt}} = 2x - 2y = 0\;\;\text{or}\;\;y = x.$
4. Find the equations of the asymptotes. This can be done by calculating the eigenvectors $${\mathbf{V}_1},$$ $${\mathbf{V}_2}$$ of the matrix $$A:$$
$\left( {A - {\lambda _1}I} \right){\mathbf{V}_1} = \mathbf{0},\;\; \Rightarrow \left[ {\begin{array}{*{20}{c}} { - 1 + 1}&0\\ 2&{ - 2 + 1} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{V_{11}}}\\ {{V_{21}}} \end{array}} \right] = \mathbf{0},\;\; \Rightarrow \left[ {\begin{array}{*{20}{r}} 0&0\\ 2&{ - 1} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{V_{11}}}\\ {{V_{21}}} \end{array}} \right] = \mathbf{0},\;\; \Rightarrow 2{V_{11}} - {V_{21}} = 0,\;\; \Rightarrow {V_{11}} = 1,\;{V_{21}} = 2,\;\; \Rightarrow {\mathbf{V}_1} = \left[ {\begin{array}{*{20}{c}} {{V_{11}}}\\ {{V_{21}}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1\\ 2 \end{array}} \right];$
$\left( {A - {\lambda _2}I} \right){\mathbf{V}_2} = \mathbf{0},\;\; \Rightarrow \left[ {\begin{array}{*{20}{c}} { - 1 + 2}&0\\ 2&{ - 2 + 2} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{V_{12}}}\\ {{V_{22}}} \end{array}} \right] = \mathbf{0},\;\; \Rightarrow \left[ {\begin{array}{*{20}{c}} 1&0\\ 2&0 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{V_{12}}}\\ {{V_{22}}} \end{array}} \right] = \mathbf{0},\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {1 \cdot {V_{12}} + 0 \cdot {V_{22}} = 0}\\ {2 \cdot {V_{12}} + 0 \cdot {V_{22}} = 0} \end{array}} \right.,\;\; \Rightarrow {V_{12}} = 0,\;{V_{22}} = 1,\;\; \Rightarrow {\mathbf{V}_2} = \left[ {\begin{array}{*{20}{c}} {{V_{12}}}\\ {{V_{22}}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0\\ 1 \end{array}} \right].$
5. Draw an $$xy$$-plane and show the eigenvectors $${\mathbf{V}_1},$$ $${\mathbf{V}_2},$$ the horizontal isocline $$y = x$$ and sketch the phase portrait of the system (Figure $$19$$). The phase trajectories approach zero touching the line directed along the vector $${\mathbf{V}_1},$$ as this eigenvector corresponds to the smallest (in absolute value) eigenvalue: $$\left| {{\lambda _1}} \right| = 1.$$

### Example 2.

Investigate the equilibrium positions of the dynamic system and sketch its phase portrait. $\frac{{dx}}{{dt}} = x + 3y,\;\frac{{dy}}{{dt}} = 2x.$

Solution.

We first make sure that the determinant is not zero:

$A = \left( {\begin{array}{*{20}{c}} 1&3\\ 2&0 \end{array}} \right),\;\; \det A = \left| {\begin{array}{*{20}{c}} 1&3\\ 2&0 \end{array}} \right| = - 6 \ne 0.$

Hence, the system has the unique equilibrium point at the origin.

1. We solve this problem without computing the eigenvalues and eigenvectors.
2. As the determinant $$\det A \lt 0,$$ then the zero equilibrium point is a saddle. This follows from the bifurcation diagram in Figure $$18.$$
3. Define the equations of isoclines. The vertical isocline is described by the linear function:
$\frac{{dx}}{{dt}} = x + 3y = 0,\;\; \Rightarrow y = - \frac{x}{3}.$
The equation of the horizontal isocline is
$\frac{{dy}}{{dt}} = 2x = 0,\;\; \Rightarrow x = 0\;\left( {y-\text{axis}} \right).$
4. Find the equations of the separatrices which have the form $$y = kx.$$ Substituting this into the original system, we obtain a quadratic equation for the coefficient $$k:$$
$\left\{ \begin{array}{l} \frac{{dx}}{{dt}} = x + 3y\\ \frac{{dy}}{{dt}} = 2x \end{array} \right.,\;\; \Rightarrow \left\{ \begin{array}{l} \frac{{dx}}{{dt}} = x + 3kx\\ \frac{{kdx}}{{dt}} = 2x \end{array} \right.,\;\; \Rightarrow 2x = k\left( {x + 3kx} \right),\;\; \Rightarrow 3{k^2}x + kx - 2x = 0,\;\; \Rightarrow 3{k^2} + k - 2 = 0,\;\; \Rightarrow D = 24,\;{k_{1,2}} = \frac{{ - 1 \pm 5}}{6} = - 1,\;\frac{2}{3}.$
Thus, the equations of the separatrices are as follows:
$y = - x,\;\;y = \frac{2}{3}x.$
5. Draw the isoclines and separatrices on the phase plane and sketch the phase trajectories (Figure $$20$$).
6. Determine the direction of motion along the phase trajectories. Take, for example, the point $$\left( {1,0} \right)$$ and calculate the derivative $$\frac{{dy}}{{dt}}$$ at this point:
$\frac{{dy}}{{dt}}\left( {1,0} \right) = 2 \cdot 1 = 2 \gt 0.$
As the derivative $${\frac{{dy}}{{dt}}} \gt 0,$$ then the point crosses the $$x$$-axis in the upward direction with increasing time $$t.$$ We mark this on the phase plane. Then, using symmetry, we can easy specify the direction of movement for the other phase trajectories (Figure $$20$$).

### Example 3.

Investigate the equilibrium points and sketch the phase portrait of the following system: $\frac{{dx}}{{dt}} = 3x - 4y,\;\frac{{dy}}{{dt}} = 2x - y$

Solution.

The determinant of this system is

$A = \left[ {\begin{array}{*{20}{c}} 3&{ - 4}\\ 2&{ - 1} \end{array}} \right],\;\; \det A = \left| {\begin{array}{*{20}{c}} 3&{ - 4}\\ 2&{ - 1} \end{array}} \right| = 5 \ne 0.$

Hence, the system has the unique equilibrium point $$\left( {0,0} \right).$$

1. We calculate the eigenvalues of $$A:$$
$\left| {\begin{array}{*{20}{c}} {3 - \lambda }&{ - 4}\\ 2&{ - 1 - \lambda } \end{array}} \right| = 0,\;\; \Rightarrow \left( {\lambda - 3} \right)\left( {\lambda + 1} \right) + 8 = 0,\;\; \Rightarrow {\lambda ^2} - 2\lambda + 5 = 0,\;D = - 16,\;\; \Rightarrow {\lambda _{1,2}} = \frac{{2 \pm 4i}}{2} = 1 \pm 2i.$
2. The eigenvalues $${\lambda_1},$$ $${\lambda_2}$$ are complex conjugate numbers with a positive real part. Therefore, the equilibrium position at the origin is an unstable focus.
3. Find the equations of isoclines. The vertical isocline is described by the following equation:
$\frac{{dx}}{{dt}} = 3x - 4y = 0,\;\; \Rightarrow y = \frac{3}{4}x.$
The horizontal isocline is defined by the equation:
$\frac{{dy}}{{dt}} = 2x - y = 0,\;\; \Rightarrow y = 2x.$
4. Find out the direction of twisting calculating the derivative $$\frac{{dy}}{{dt}}$$ at the point $$\left( {1,0} \right):$$
$\frac{{dy}}{{dt}}\left( {1,0} \right) = 2 \cdot 1 - 0 = 2 \gt 0.$
Thus, the spirals twist counterclockwise.
5. Given the data found, we can construct a schematic phase portrait of the system (Figure $$21$$).

### Example 4.

Investigate the stability of the system depending on the parameter $$a:$$ $\frac{{dx}}{{dt}} = ax + y,\;\frac{{dy}}{{dt}} = x + ay.$

Solution.

Compute the trace and the determinant of the matrix $$A:$$

$A = \left[ {\begin{array}{*{20}{c}} a&1\\ 1&a \end{array}} \right],\;\; \text{tr}\,A = 2a,\;\; \det A = {a^2} - 1.$

The auxiliary equation of the system can be written as

${\lambda ^2} - \text{tr}\,A \cdot \lambda + \det A = 0,\;\; \Rightarrow {\lambda ^2} - 2a\lambda + {a^2} - 1 = 0.$

The discriminant of this equation is

$D = {\left( { - 2a} \right)^2} - 4\left( {{a^2} - 1} \right) = \cancel{4{a^2}} - \cancel{4{a^2}} + 4 = 4 \gt 0.$

As the discriminant is always positive, then the eigenvalues are real for any $$a.$$ This means that if $$\det A \ne 0,$$ the system has the unique equilibrium point at the origin, which will be either a saddle or a node.

Determine for which values of $$a$$ the zero point is a saddle. According to the bifurcation diagram, the saddle exists when $$\det A \lt 0.$$ Then we have the inequality:

$\det A = {a^2} - 1 \lt 0,\;\; \Rightarrow a \in \left( { - 1,1} \right).$

Thus, for $$a \in \left( { - 1,1} \right)$$ there will be a saddle at the origin.

It is clear that the equilibrium point of the type "node" exists for the following values of the parameter $$a:$$

$a \in \left( { - \infty , - 1} \right) \cup \left( {1,\infty } \right).$

Moreover, if $$\text{tr}\,A = 2a \gt 0$$ or $$a \gt 0,$$ then the node will be unstable, and if $$\text{tr}\,A = 2a \lt 0$$ or $$a \lt 0$$ we get a stable node.

As a result, we have:

$a \in \left( {-\infty,-1 } \right) - \text{stable node},$
$a \in \left( {1,\infty } \right) - \text{unstable node}.$

Note that at no values of $$a$$ the equilibrium points lie on the parabola on the bifurcation diagram. Indeed, the equation of this parabola has the form

$\det A = {\left( {\frac{\text{tr}\,A}{2}} \right)^2}.$

When substituting our values, we obtain

${a^2} - 1 = {\left( {\frac{{2a}}{2}} \right)^2},\;\; \Rightarrow {a^2} - 1 = {a^2},\;\; \Rightarrow - 1 = 0,$

which is not true.

We consider separately the cases at the boundary points $$a = \pm 1.$$ At these values of $$a$$ the system becomes singular:

$\det A = {a^2} - 1 = 0.$

Let $$a = 1.$$ Calculate the eigenvalues of the matrix:

$A = \left[ {\begin{array}{*{20}{c}} 1&1\\ 1&1 \end{array}} \right],\;\; \det \left( {A - \lambda I} \right) = 0,\;\; \Rightarrow \left| {\begin{array}{*{20}{c}} {1 - \lambda }&1\\ 1&{1 - \lambda } \end{array}} \right| = 0,\;\; \Rightarrow {\left( {\lambda - 1} \right)^2} - 1 = 0,\;\; \Rightarrow \left| {\lambda - 1} \right| = 1,\;\; \Rightarrow {\lambda _{1,2}} = 1 \pm 1 = 0,\;2.$

Find the eigenvector $${\mathbf{V}_1}$$ corresponding to the eigenvalue $${\lambda_1} = 0:$$

$\left[ {\begin{array}{*{20}{c}} {1 - 0}&1\\ 1&{1 - 0} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{V_{11}}}\\ {{V_{21}}} \end{array}} \right] = \mathbf{0},\;\; \Rightarrow \left[ {\begin{array}{*{20}{c}} 1&1\\ 1&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{V_{11}}}\\ {{V_{21}}} \end{array}} \right] = \mathbf{0},\;\; \Rightarrow {V_{11}} + {V_{21}} = 0,\;\; \Rightarrow {V_{21}} = 1,\;{V_{11}} = - 1,\;\; \Rightarrow {\mathbf{V}_1} = \left[ {\begin{array}{*{20}{r}} { - 1}\\ 1 \end{array}} \right].$

Hence, the straight line parallel to the vector $${\mathbf{V}_1}$$ and passing through the origin has the equation $$y = -x.$$ Any point on this line is an equilibrium point, and these equilibrium points are unstable, because the second eigenvalue is positive: $${\lambda_2} = 2 \gt 0.$$

Now let $$a = -1.$$ Similarly, we find the eigenvalues at this $$a:$$

$A = \left[ {\begin{array}{*{20}{r}} { - 1}&1\\ 1&{ - 1} \end{array}} \right],\;\; \det \left( {A - \lambda I} \right) = 0,\;\; \Rightarrow \left| {\begin{array}{*{20}{c}} { - 1 - \lambda }&1\\ 1&{ - 1 - \lambda } \end{array}} \right| = 0,\;\; \Rightarrow {\left( {\lambda + 1} \right)^2} - 1 = 0,\;\; \Rightarrow \left| {\lambda + 1} \right| = 1,\;\; \Rightarrow {\lambda _{1,2}} = - 1 \pm 1 = 0,\; - 2.$

Calculate the eigenvector $${\mathbf{U}_1} = {\left( {{U_{11}},{U_{21}}} \right)^T}$$ associated with the eigenvalue $${\lambda_1} = 0:$$

$\left[ {\begin{array}{*{20}{c}} { - 1 - 0}&1\\ 1&{ - 1 - 0} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{U_{11}}}\\ {{U_{21}}} \end{array}} \right] = \mathbf{0},\;\; \Rightarrow \left[ {\begin{array}{*{20}{r}} { - 1}&1\\ 1&{ - 1} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{U_{11}}}\\ {{U_{21}}} \end{array}} \right] = \mathbf{0},\;\; \Rightarrow - {U_{11}} + {U_{21}} = 0,\;\; \Rightarrow {U_{21}} = 1,\;{U_{11}} = 1,\;\; \Rightarrow {\mathbf{U}_1} = \left[ {\begin{array}{*{20}{c}} {{U_{11}}}\\ {{U_{21}}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1\\ 1 \end{array}} \right].$

This vector corresponds to the straight line $$y = x$$ passing through the origin. All points of this line are equilibrium points. These equilibrium positions will be stable because $${\lambda_2} = -2 \lt 0.$$

Combining the results, we can write the final answer. The system has the following equilibrium positions depending on the parameter $$a:$$

• When $$a \in \left( { - \infty , - 1} \right),$$ there is a stable node at the origin;
• When $$a = -1,$$ any point on the line $$y = x$$ is a stable equilibrium position;
• When $$a \in \left( { - 1, 1} \right),$$ there is a saddle at the origin;
• When $$a = 1,$$ any point on the line $$y = -x$$ is an unstable equilibrium position;
• When $$a \in \left( {1, \infty} \right),$$ there is an unstable node at the origin.