Differential Equations

Systems of Equations

Systems Diff Equations Logo

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: \(\left| {{\lambda _1}} \right| = 1.\)
  6. The phase portrait of the system in Example 1
    Figure 19.

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 \(20\)).
  7. The phase portrait of the system in Example 2
    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 \(21\)).
  6. The phase portrait of the system in Example 3
    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 \(a:\)

Page 1 Page 2