# Differential Equations

## Systems of Equations # Equilibrium Points of Linear Autonomous Systems

### Types of Equilibrium Points

Let a second order linear homogeneous system with constant coefficients be given:

$\left\{ \begin{array}{l} \frac{{dx}}{{dt}} = {a_{11}}x + {a_{12}}y\\ \frac{{dy}}{{dt}} = {a_{21}}x + {a_{22}}y \end{array} \right..$

This system of equations is autonomous since the right hand sides of the equations do not explicitly contain the independent variable $$t.$$

In matrix form, the system of equations can be written as

${\mathbf{X’} = A\mathbf{X},\;\;\text{where}\;\;\mathbf{X} = \left[ {\begin{array}{*{20}{c}} x\\ y \end{array}} \right],\;\;}\kern0pt {A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}\\ {{a_{21}}}&{{a_{22}}} \end{array}} \right].}$

The equilibrium positions can be found by solving the stationary equation

$A\mathbf{X} = \mathbf{0}.$

This equation has the unique solution $$\mathbf{X} = \mathbf{0}$$ if the matrix $$A$$ is nonsingular, i.e. provided that $$\det A \ne 0.$$ In the case of a singular matrix, the system has an infinite number of equilibrium points.

Classification of equilibrium points is determined by the eigenvalues $${\lambda _1},{\lambda _2}$$ of the matrix $$A.$$ The numbers $${\lambda _1},{\lambda _2}$$ can be found by solving the auxiliary equation

${\lambda ^2} – \left( {{a_{11}} + {a_{22}}} \right)\lambda + {a_{11}}{a_{22}} – {a_{12}}{a_{21}} = 0.$

In general, when the matrix $$A$$ is nonsingular, there are $$4$$ different types of equilibrium points:

The stability of equilibrium points is determined by the general theorems on stability. So, if the real eigenvalues (or real parts of complex eigenvalues) are negative, then the equilibrium point is asymptotically stable. Examples of such equilibrium positions are stable node and stable focus.

If the real part of at least one eigenvalue is positive, the corresponding equilibrium point is unstable. For example, it may be a saddle.

Finally, in the case of purely imaginary roots (when the equilibrium point is a center), we are dealing with the classical stability in the sense of Lyapunov.

Our next goal is to study the behavior of solutions near the equilibrium positions. For second order systems, it is convenient to do this graphically using the phase portrait, which is a set of phase trajectories in the coordinate plane. The arrows on the phase trajectories show the direction of movement of the point (i.e., a particular state of the system) over time.

Let’s discuss each type of equilibrium point and the corresponding phase portraits.

### Stable and Unstable Node

The eigenvalues $${{\lambda _1},{\lambda _2}}$$ of the points of type “node” satisfy the conditions:

${\lambda _1},{\lambda _2} \in \Re,\;\;{\lambda _1} \cdot {\lambda _2} \gt 0.$

The following particular cases may arise here.

### The roots $${{\lambda _1},{\lambda _2}}$$ are distinct $$\left( {{\lambda _1} \ne {\lambda _2}} \right)$$ and negative $$\left( {{\lambda _1} \lt 0, {\lambda _2}} \lt 0\right).$$

Draw a schematic phase portrait for this system. Suppose for definiteness that $$\left| {{\lambda _1}} \right| \lt \left| {{\lambda _2}} \right|.$$ The general solution has the form

$\mathbf{X}\left( t \right) = {C_1}{e^{{\lambda _1}t}}{\mathbf{V}_1} + {C_2}{e^{{\lambda _2}t}}{\mathbf{V}_2},$

where $${\mathbf{V}_1} = {\left( {{V_{11}},{V_{21}}} \right)^T},$$ $${\mathbf{V}_2} = {\left( {{V_{12}},{V_{22}}} \right)^T}$$ are eigenvectors corresponding to the eigenvalues $${{\lambda _1},{\lambda _2}}$$ and $${C_1},{C_2}$$ are arbitrary constants.

Since both eigenvalues are negative, then the solution $$\mathbf{X} = \mathbf{0}$$ is asymptotically stable. Such an equilibrium point is called stable node. As $$t \to \infty,$$ the phase curves tend to the origin $$\mathbf{X} = \mathbf{0}.$$

Specify the direction of the phase trajectories. Since

${x\left( t \right) = {C_1}{V_{11}}{e^{{\lambda _1}t}} + {C_2}{V_{12}}{e^{{\lambda _2}t}},\;\;}\kern0pt {y\left( t \right) = {C_1}{V_{21}}{e^{{\lambda _1}t}} + {C_2}{V_{22}}{e^{{\lambda _2}t}},}$

the derivative $$\large\frac{{dy}}{{dx}}\normalsize$$ is

${\frac{{dy}}{{dx}} \text{ = }}\kern0pt{\frac{{{C_1}{V_{21}}{\lambda _1}{e^{{\lambda _1}t}} + {C_2}{V_{22}}{\lambda _2}{e^{{\lambda _2}t}}}}{{{C_1}{V_{11}}{\lambda _1}{e^{{\lambda _1}t}} + {C_2}{V_{12}}{\lambda _2}{e^{{\lambda _2}t}}}}.}$

Divide the numerator and denominator by $${{e^{{\lambda _1}t}}}:$$

${\frac{{dy}}{{dx}} \text{ = }}\kern0pt{\frac{{{C_1}{V_{21}}{\lambda _1} + {C_2}{V_{22}}{\lambda _2}{e^{\left( {{\lambda _2} – {\lambda _1}} \right)t}}}}{{{C_1}{V_{11}}{\lambda _1} + {C_2}{V_{12}}{\lambda _2}{e^{\left( {{\lambda _2} – {\lambda _1}} \right)t}}}}.}$

In this case, $${\lambda _2} – {\lambda _1} \lt 0.$$ Therefore, the terms with the exponential function tend to zero as $$t \to \infty.$$ As a result, at $${C_1} \ne 0,$$ we obtain:

$\lim\limits_{t \to \infty } \frac{{dy}}{{dx}} = \frac{{{V_{21}}}}{{{V_{11}}}}.$

that is the phase trajectories become parallel to the eigenvector $${\mathbf{V}_1}$$ as $$t \to \infty.$$

If $${C_1} = 0,$$ the derivative at any $$t$$ equals

$\frac{{dy}}{{dx}} = \frac{{{V_{22}}}}{{{V_{12}}}},$

i.e. the phase trajectory lies on a line directed along the eigenvector $${\mathbf{V}_2}.$$

Now we consider the behavior of the phase trajectories as $$t \to -\infty.$$ Obviously, the coordinates $$x\left( t \right), y\left( t \right)$$ tend to infinity, and the derivative $$\large\frac{{dy}}{{dx}}\normalsize$$ at $${C_2} \ne 0$$ takes the following form:

${\frac{{dy}}{{dx}} \text{ = }}\kern0pt{\frac{{{C_1}{V_{21}}{\lambda _1}{e^{\left( {{\lambda _1} – {\lambda _2}} \right)t}} + {C_2}{V_{22}}{\lambda _2}}}{{{C_1}{V_{11}}{\lambda _1}{e^{\left( {{\lambda _1} – {\lambda _2}} \right)t}} + {C_2}{V_{12}}{\lambda _2}}} }={ \frac{{{V_{22}}}}{{{V_{12}}}},}$

that is the phase curves at the points at infinity become parallel to the vector $${\mathbf{V}_2}.$$

Accordingly, when $${C_2} = 0,$$ the derivative is

$\frac{{dy}}{{dx}} = \frac{{{V_{21}}}}{{{V_{11}}}}.$

In this case, the phase trajectory is determined by the direction of the eigenvector $${\mathbf{V}_1}.$$

Given the above properties of the phase trajectories, the phase portrait of a stable node is shown schematically in Figure $$2.$$ Figure 2. Figure 3.

Similarly, we can study the behavior of the phase trajectories for other types of equilibrium points. Furthermore, omitting the detailed analysis, we consider basic qualitative characteristics of the other equilibrium points.

### The roots $${{\lambda _1},{\lambda _2}}$$ are distinct $$\left( {{\lambda _1} \ne {\lambda _2}} \right)$$ and positive $$\left( {{\lambda _1} \gt 0, {\lambda _2}} \gt 0\right).$$

In this case, the point $$\mathbf{X} = \mathbf{0}$$ is an unstable node. Its phase portrait is shown in Figure $$3.$$

Note that in the case of both stable and unstable node, the phase trajectories touch the line, which is directed along the eigenvector corresponding to the smallest (in absolute value) eigenvalue $$\lambda.$$

### Dicritical Node

Let the auxiliary equation have one zero root of multiplicity $$2,$$ i.e. consider the case $${\lambda _1} = {\lambda _2} = {\lambda} \ne 0.$$ The system has a basis of two eigenvectors, i.e. the geometric multiplicity of the eigenvalue $$\lambda$$ is $$2.$$ In terms of the linear algebra, this means that the dimension of the eigenspace of $$A$$ is equal to $$2:$$ $$\dim \ker A = 2.$$ This situation occurs in systems of the form

${\frac{{dx}}{{dt}} = \lambda x,}\;\; {\frac{{dy}}{{dt}} = \lambda y.}$

The direction of the phase trajectories depends on the sign of $$\lambda.$$ Here the following two cases can arise:

### Case $${\lambda _1} = {\lambda _2} = {\lambda} \lt 0.$$

Such an equilibrium position is called a stable dicritical node (Figure $$4$$).

### Case $${\lambda _1} = {\lambda _2} = {\lambda} \gt 0.$$

This combination of eigenvalues corresponds to an unstable dicritical node (Figure $$5$$). Figure 4. Figure 5.

### Singular Node

Let the eigenvalues of $$A$$ be again coincident: $${\lambda _1} = {\lambda _2} = {\lambda} \ne 0.$$ Unlike the previous case, we assume that the geometric multiplicity of the eigenvalue (or in other words, the dimension of the eigenspace) is now $$1.$$ This means that the matrix $$A$$ has only one eigenvector $${\mathbf{V}_1}.$$ The second linearly independent vector required for the basis is defined as a generalized eigenvector $${\mathbf{W}_1}$$ connected to $${\mathbf{V}_1}.$$

### Case $${\lambda _1} = {\lambda _2} = {\lambda} \lt 0$$.

The equilibrium point is called stable singular node (Figure $$6$$).

### Case $${\lambda _1} = {\lambda _2} = {\lambda} \gt 0$$.

The equilibrium position is called unstable singular node (Figure $$7$$). Figure 6. Figure 7.

The equilibrium point is a saddle under the following condition:

${\lambda _1},{\lambda _2} \in \Re,\;\;{\lambda _1} \cdot {\lambda _2} \lt 0.$

Since one of the eigenvalues is positive, the saddle is an unstable equilibrium point. Suppose, for example, $${\lambda _1} \lt 0,{\lambda _2} \gt 0.$$ The eigenvalues $${\lambda _1}$$ and $${\lambda _2}$$ are associated with the corresponding eigenvectors $${\mathbf{V}_1}$$ and $${\mathbf{V}_2}.$$ The straight lines directed along the eigenvectors $${\mathbf{V}_1},$$ $${\mathbf{V}_2},$$ are called separatrices. These are the asymptotes of other phase trajectories that have the form of a hyperbola. Each of the separatrices can be associated with a certain direction of motion.

If the separatrix is associated with a negative eigenvalue $${\lambda _1} \lt 0,$$ i.e. in our case is directed along the vector $${\mathbf{V}_1},$$ the movement along it occurs towards the equilibrium point $$\mathbf{X} = \mathbf{0}.$$ And conversely, at $${\lambda _2} \gt 0,$$ i.e. for the separatrix associated with the vector $${\mathbf{V}_2},$$ the movement is directed from the origin. The phase portrait of the saddle is shown schematically in Figure $$8.$$

### Stable and Unstable Focus

Now suppose that the eigenvalues $${\lambda _1},{\lambda _2}$$ are complex numbers whose real parts are non-zero. If the matrix $$A$$ is composed of real numbers, the complex roots will be presented in the form of complex conjugate numbers:

${\lambda _{1,2}} = \alpha \pm i\beta .$

Find out what kind of phase trajectories are in the neighborhood of the origin. Construct a complex solution $${\mathbf{X}_1}\left( t \right)$$ corresponding to the eigenvalue $${\lambda _1} = \alpha + i\beta :$$

${{\mathbf{X}_1}\left( t \right) = {e^{{\lambda _1}t}}{\mathbf{V}_1} } = {{e^{\left( {\alpha + i\beta } \right)t}}\left( {\mathbf{U} + i\mathbf{W}} \right),}$

where $${\mathbf{V}_1} = \mathbf{U} + i\mathbf{W}$$ is the complex-valued eigenvector associated with the eigenvalue $${\lambda _1},$$ $$\mathbf{U}$$ and $$\mathbf{W}$$ are real vector functions. As a result, we obtain:

${{\mathbf{X}_1}\left( t \right) = {e^{\alpha t}}{e^{i\beta t}}\left( {\mathbf{U} + i\mathbf{W}} \right) } = {{e^{\alpha t}}\left( {\cos \beta t + i\sin \beta t} \right)\left( {\mathbf{U} + i\mathbf{W}} \right) } = {{e^{\alpha t}}\left( {\mathbf{U}\cos \beta t + i\mathbf{U}\sin \beta t }\right.}+{\left.{ i\mathbf{W}\cos \beta t – \mathbf{W}\sin \beta t} \right) } = {{e^{\alpha t}}\left( {\mathbf{U}\cos \beta t + – \mathbf{W}\sin \beta t} \right) } + {i{e^{\alpha t}}\left( {\mathbf{U}\sin \beta t + \mathbf{W}\cos \beta t} \right).}$

The real and imaginary parts in the last expression form the general solution of the type

${\mathbf{X}\left( t \right) = {C_1}\text{Re}\left[ {{\mathbf{X}_1}\left( t \right)} \right] + {C_2}\text{Im}\left[ {{\mathbf{X}_1}\left( t \right)} \right] } = {{e^{\alpha t}}\left[ {{C_1}\left( {\mathbf{U}\cos \beta t – \mathbf{W}\sin \beta t} \right)} \right. } + {\left. {{C_2}\left( {\mathbf{U}\sin \beta t + \mathbf{W}\cos \beta t} \right)} \right] } = {{e^{\alpha t}}\left[ {\mathbf{U}\left( {{C_1}\cos \beta t + {C_2}\sin \beta t} \right)} \right. } + {\left. {\mathbf{W}\left( {{C_2}\cos \beta t – {C_1}\sin \beta t} \right)} \right].}$

We represent the constant $${C_1},{C_2}$$ as

${C_1} = C\sin \delta ,\;\;{C_2} = C\cos \delta ,$

where $$\delta$$ is an auxiliary angle. Then the solution is written as

${\mathbf{X}\left( t \right) = C{e^{\alpha t}}\left[ {\mathbf{U}\left( {\sin \delta \cos \beta t + \cos \delta \sin \beta t} \right)} \right. } + {\left. {\mathbf{W}\left( {\cos\delta \cos \beta t – \sin \delta \sin \beta t} \right)} \right] } = {C{e^{\alpha t}}\left[ {\mathbf{U}\sin \left( {\beta t + \delta } \right)} \right. + \left. {\mathbf{W}\cos \left( {\beta t + \delta } \right)} \right].}$

Thus, the solution $$\mathbf{X}\left( t \right)$$ can be expanded in the basis of the vectors $$\mathbf{U}$$ and $$\mathbf{W}:$$

$\mathbf{X}\left( t \right) = \mu \left( t \right)\mathbf{U} + \eta \left( t \right)\mathbf{W},$

where the coefficients $$\mu \left( t \right),$$ $$\eta \left( t \right)$$ are given by

${\mu \left( t \right) = C{e^{\alpha t}}\sin \left( {\beta t + \delta } \right),\;\;}\kern0pt {\eta \left( t \right) = C{e^{\alpha t}}\cos\left( {\beta t + \delta } \right).}$

This shows that the phase trajectories are spirals. When $$\alpha \lt 0,$$ the spirals twist approaching the origin. Such an equilibrium position is called stable focus. Accordingly, when $$\alpha \gt 0,$$ we have an unstable focus.

The direction of twist can be identified by the sign of the coefficient $${a_{21}}$$ in the original matrix $$A.$$ Indeed, consider the derivative $${\large\frac{{dy}}{{dt}}\normalsize},$$ for example, at the point $$\left( {1,0} \right):$$

$\frac{{dy}}{{dt}}\left( {1,0} \right) = {a_{21}} \cdot 1 + {a_{22}} \cdot 0 = {a_{21}}$

The positive coefficient $${a_{21}} \gt 0$$ corresponds to the twist counterclockwise as shown in Figure $$9.$$ When $${a_{21}} \lt 0,$$ the spirals will twist in a clockwise direction (Figure $$10$$).

Thus, taking into account the direction of twist, there are only $$4$$ different types of focus. Schematically, they are shown in Figures $$9-12.$$ Figure 9. Figure 10. Figure 11. Figure 12.

### Center

If the eigenvalues of the matrix $$A$$ are purely imaginary numbers, then this equilibrium point is called a center. For a matrix with real elements, the imaginary eigenvalues are complex conjugate pairs. In the case of a center, the phase trajectories are formally obtained from the equation of spirals at $$\alpha = 0$$ and are ellipses, i.e. they describe periodic motion of a point in the phase space. A center equilibrium position is stable in the sense of Lyapunov.

There are two types of centers, which differ in the direction of movement of the points (Figures $$13, 14$$). As in the case of focus, the direction of movement can be determined by the sign of the derivative $$\large\frac{{dy}}{{dt}}\normalsize$$ at some point. If we take the point $$\left( {1,0} \right),$$ then

$\frac{{dy}}{{dt}}\left( {1,0} \right) = {a_{21}}.$

that is the direction of rotation is determined by the sign of the coefficient $${a_{21}}.$$ Figure 13. Figure 14.

Thus, we have considered different types of equilibrium points in the case of a non-singular matrix $$A$$ $$\left( {\det A \ne 0} \right).$$ Taking into account the direction of phase trajectories, there are total $$13$$ different phase portraits (shown, respectively, in Figures $$2-14$$).

We now turn to the case of a singular matrix $$A.$$

### Singular Matrix

If the matrix is singular, then it has one or both eigenvalues equal to zero. In this case, there are the following special cases:

### Case $${\lambda _1} \ne 0, {\lambda _2} = 0$$.

Here, the general solution has the form

$\mathbf{X}\left( t \right) = {C_1}{e^{{\lambda _1}t}}{\mathbf{V}_1} + {C_2}{\mathbf{V}_2},$

where $${\mathbf{V}_1} = {\left( {{V_{11}},{V_{21}}} \right)^T},$$ $${\mathbf{V}_2} = {\left( {{V_{12}},{V_{22}}} \right)^T},$$ are the eigenvectors corresponding to the eigenvalues $${\lambda _1}$$ and $${\lambda _2}.$$ It turns out that in this case the whole line passing through the origin and directed along the vector $${\mathbf{V}_2}$$ consists of the equilibrium points (these points do not have a special name). The phase trajectories are rays parallel to the other eigenvector $${\mathbf{V}_1}.$$ Depending on the sign of $${\lambda _1},$$ the motion at $$t \to \infty$$ occurs either in the direction of the line $${\mathbf{V}_2}$$ (Figure $$15$$), or away from it (Figure $$16$$). Figure 15. Figure 16.

### Case $${\lambda _1} = {\lambda _2} = 0, \dim \ker A = 2.$$

In this case, the dimension of the eigenspace of the matrix is equal to $$2$$ and, therefore, there are two eigenvectors $${\mathbf{V}_1}$$ and $${\mathbf{V}_2}.$$ This may happen when $$A$$ is the zero matrix. The general solution is given by

$\mathbf{X}\left( t \right) = {C_1}{\mathbf{V}_1} + {C_2}{\mathbf{V}_2}.$

It follows that every point in the plane is an equilibrium position of the system.

### Case $${\lambda _1} = {\lambda _2} = 0, \dim \ker A = 1.$$

This case is different from the previous one in that there is only one eigenvector (the matrix $$A$$ will then be non-zero). To construct a basis, we can take the generalized eigenvector $${\mathbf{W}_1}$$ connected to $${\mathbf{V}_1}$$ as a second linearly independent vector. The general solution can be written as

$\mathbf{X}\left( t \right) = \left( {{C_1} + {C_2}t} \right){\mathbf{V}_1} + {C_2}{\mathbf{W}_1}.$

Here, all points of the straight line passing through the origin and directed along the eigenvector $${\mathbf{V}_1}$$ are unstable equilibrium positions. The phase trajectories are straight lines parallel to $${\mathbf{V}_1}.$$ The direction of movement along these lines as $$t \to \infty$$ depends on the constant $${C_2}:$$ with $${C_2} \lt 0,$$ the motion is from left to right, and with $${C_2} \gt 0$$ − in the opposite direction (Figure $$17$$).

As seen, there are $$4$$ different phase portraits in the case of a singular matrix. Therefore, the linear second order autonomous system allows total $$17$$ different phase portraits.

### Bifurcation Diagram

In the above, we have reviewed the classification of equilibrium points of a linear system based on the eigenvalues. However, the type of an equilibrium point can be determined without computing the eigenvalues $${\lambda _1},{\lambda _2},$$ knowing only the determinant of the matrix $$\det A$$ and its trace $$\text{tr}\,A.$$

Recall that the trace of the matrix is the number equal to the sum of the diagonal elements:

${A = \left( {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}\\ {{a_{21}}}&{{a_{22}}} \end{array}} \right),}\;\; {\text{tr}\,A = {a_{11}} + {a_{22}},}\;\; {\det A = {a_{11}}{a_{22}} – {a_{12}}{a_{21}}.}$

Indeed, the auxiliary equation of the matrix is

${\lambda ^2} – \left( {{a_{11}} + {a_{22}}} \right)\lambda + {a_{11}}{a_{22}} – {a_{12}}{a_{21}} = 0.$

It can be written in terms of the determinant and the trace of the matrix:

${\lambda ^2} – \text{tr}\,A \cdot \lambda + \det A = 0.$

The discriminant of this quadratic equation is given by

$D = {\left( {\text{tr}\,A} \right)^2} – 4\det A.$

Thus, the bifurcation curve delineating the different stability modes is a parabola in the plane $$\left( {\text{tr}\,A,\det A} \right)$$ (Figure $$18$$):

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

The equilibrium points of the type “focus” and “center” are above the parabola. The points of the type “center” are located on the positive $$y$$-axis, i.e. provided that $$\text{tr}\,A = 0.$$ The “nodes” and “saddles” are below the parabola. The parabola itself contains dicritical or singular nodes.

Stable modes of motion exist in the upper left quadrant of the bifurcation diagram. The other three quadrants correspond to unstable equilibrium positions.

### How to Sketch a Phase Portrait

To draw the phase portrait of a second order linear autonomous system with constant coefficients

${\mathbf{X’} = A\mathbf{X},}\;\; {A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}\\ {{a_{21}}}&{{a_{22}}} \end{array}} \right],\;\;}\kern0pt {\mathbf{X} = \left[ {\begin{array}{*{20}{c}} x\\ y \end{array}} \right],}$

it is necessary to do the following steps:

1. Find the eigenvalues of the matrix by solving the auxiliary equation
${\lambda ^2} – \left( {{a_{11}} + {a_{22}}} \right)\lambda + {a_{11}}{a_{22}} – {a_{12}}{a_{21}} = 0.$
2. Determine the type of the equilibrium point and the character of stability.
3. #### Hint:

The type of the equilibrium position can also be determined based on the bifurcation diagram (Figure $$18$$), knowing the trace and the determinant of the matrix:
${\text{tr}\,A = {a_{11}} + {a_{22}},}\;\; {\det A = \left| {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}\\ {{a_{21}}}&{{a_{22}}} \end{array}} \right| } = {{a_{11}}{a_{22}} – {a_{12}}{a_{21}}.}$
4. Find the equations of the isoclines:
${\frac{{dx}}{{dt}} = {a_{11}}x + {a_{12}}y}\;\; {\left( \text{vertical isocline} \right),}$
${\frac{{dy}}{{dt}} = {a_{21}}x + {a_{22}}y}\;\; {\left( \text{horizontal isocline} \right).}$
5. If the equilibrium position is a node or a saddle, it is necessary to compute the eigenvectors and draw the asymptotes parallel to the eigenvectors and passing through the origin.
6. Schematically draw the phase portrait.
7. Show the direction of motion along the phase trajectories (this depends on the stability or instability of the equilibrium point).In the case of a focus, one should determine the direction of trajectories twisting. This can be done by calculating the velocity vector $$\left( {{\large\frac{{dx}}{{dt}}\normalsize}, {\large\frac{{dy}}{{dt}}\normalsize}} \right)$$ at any point, for example, at the point $$\left( {1,0} \right).$$ Similarly, we can determine the direction of movement if the equilibrium position is a center.

The algorithm described here is not a rigid scheme. In the study of a particular system, other tricks and techniques are acceptable in order to draw up the phase portrait.

## 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 $$a:$$
$\frac{{dx}}{{dt}} = ax + y,\;\;\frac{{dy}}{{dt}} = x + ay.$
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Concept
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Problems 1-4