# Differential Equations

## Systems of Equations # Stability in the First Approximation

• Consider an autonomous system of $$n$$ differential equations:

${\frac{{d{x_i}}}{{dt}} = {f_i}\left( {{x_1},{x_2}, \ldots ,{x_n}} \right),\;\;}\kern0pt {i = 1,2, \ldots ,n.}$

In vector form it can be written as

${\mathbf{X’} = \mathbf{f}\left( \mathbf{X} \right),\;\;}\kern0pt {\text{where}\;\;\mathbf{X} = \left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}}\\ \vdots \\ {{x_n}} \end{array}} \right],\;\;}\kern0pt {\mathbf{f} = \left[ {\begin{array}{*{20}{c}} {{f_1}}\\ {{f_2}}\\ \vdots \\ {{f_n}} \end{array}} \right].}$

Suppose that the system has the equilibrium position $$\mathbf{X} = \mathbf{0},$$ which we will investigate for stability. It is assumed that the functions $${f_i}\left( \mathbf{X} \right)$$ are twice continuously differentiable in a neighborhood of the origin. Therefore, we can expand these functions in the Maclaurin series in the variables $${x_i},$$ retaining first order terms. Then the system of equations takes the following form:

${\frac{{d{x_i}}}{{dt}} = \frac{{\partial {f_i}}}{{\partial {x_1}}}\left( 0 \right){x_1} + \frac{{\partial {f_i}}}{{\partial {x_2}}}\left( 0 \right){x_2} + \cdots } + {\frac{{\partial {f_i}}}{{\partial {x_n}}}\left( 0 \right){x_n} } + {{R_i}\left( {{x_1},{x_2}, \ldots ,{x_n}} \right).}$

Here, $${R_i}\left( {{x_1},{x_2}, \ldots ,{x_n}} \right)$$ describe the terms of the second order of smallness with respect to the variables $${{x_1},{x_2}, \ldots ,{x_n}}.$$

In vector-matrix notation, the system looks as follows:

$\mathbf{X’} = J\mathbf{X} + \mathbf{R}\left( \mathbf{X} \right),$

where the Jacobian $$J$$ is determined by the matrix

$J = \left[ {\begin{array}{*{20}{c}} {\frac{{\partial {f_1}}}{{\partial {x_1}}}}&{\frac{{\partial {f_1}}}{{\partial {x_2}}}}& \vdots &{\frac{{\partial {f_1}}}{{\partial {x_n}}}}\\ {\frac{{\partial {f_2}}}{{\partial {x_1}}}}&{\frac{{\partial {f_2}}}{{\partial {x_2}}}}& \vdots &{\frac{{\partial {f_2}}}{{\partial {x_n}}}}\\ \cdots & \cdots & \vdots & \cdots \\ {\frac{{\partial {f_n}}}{{\partial {x_1}}}}&{\frac{{\partial {f_n}}}{{\partial {x_2}}}}& \vdots &{\frac{{\partial {f_n}}}{{\partial {x_n}}}} \end{array}} \right].$

The values of the partial derivatives in this matrix are calculated at the series expansion point, i.e. in this case at $$\mathbf{X} = \mathbf{0}.$$

In the study of the stability, we can consider the linearized system

$\mathbf{X’} = J\mathbf{X},$

instead of the original nonlinear one, which is called the system of equations of the first approximation with respect to the original system.

The question in which cases the original nonlinear system and the corresponding system of the first approximation have the same character of stability was resolved by A.M.Lyapunov ($$1857-1918$$). The following theorems hold:

### Lyapunov theorem on stability in the first approximation.

If all eigenvalues $${\lambda _i}$$ of the Jacobian $$J$$ have negative real parts, then the zero solution $$\mathbf{X} = \mathbf{0}$$ of the original and linearized systems is asymptotically stable.

### Lyapunov theorem on instability in the first approximation.

If at least one of the eigenvalues $${\lambda _i}$$ of $$J$$ has a positive real part, then the zero solution $$\mathbf{X} = \mathbf{0}$$ of the original and linearized systems is unstable.

In critical cases, when the real parts of all eigenvalues $${\lambda _i}$$ of the Jacobian $$J$$ are non-positive and there is at least one eigenvalue with zero real part, the zero solution can be stable or unstable. In this case, it’s impossible to determine the character of stability in the first approximation and one should use other methods of stability analysis.

Thus, the Lyapunov theorems given above allow us to study stability of the zero solution of nonlinear systems where the equilibrium position is characterized by the eigenvalues with a nonzero real part. Such equilibrium points are called rough. A more precise definition is formulated as follows:

The equilibrium position $$\mathbf{X} = \mathbf{0}$$ of an $$n$$th order autonomous system

$\mathbf{X’} = \mathbf{f}\left( \mathbf{X} \right)$

is called rough if the corresponding Jacobian $$J$$ has exactly $$n$$ pairwise distinct eigenvalues with a nonzero real part.

Note that in the case of a second order system, only the following three types of equilibrium points are rough:

The above types of points are characterized by the eigenvalues with a nonzero real part. On the contrary, an equilibrium point of the type “center” is not rough since it always has purely imaginary eigenvalues.

Thus, the application of the method of stability analysis in the first approximation is limited by rough (or structurally stable) systems.

• ## Solved Problems

Click a problem to see the solution.

### Example 1

Investigate the stability of the zero equilibrium point of the system using the first approximation method.
${\frac{{dx}}{{dt}} = y + 3{x^2} + 2{y^2},\;\;}\kern0pt{\frac{{dy}}{{dt}} = – 2x – y + xy.}$

### Example 2

Find the equilibrium position of the system and investigate its stability in the first approximation.
$\frac{{dx}}{{dt}} = {x^2} – y,\;\;\frac{{dy}}{{dt}} = x – 1.$

### Example 3

Determine the equilibrium positions of the system and explore their stability. Draw a schematic phase portrait of the corresponding linearized system.
${\frac{{dx}}{{dt}} = {e^{x + y}} – 1,\;\;}\kern0pt{\frac{{dy}}{{dt}} = \ln \left( {1 + x} \right).}$

### Example 4

Using equations of the first approximation investigate the stability of the zero solution of the nonlinear system:
${\frac{{dx}}{{dt}} = \tan \left( {x + y} \right) – y,\;\;}\kern0pt {\frac{{dy}}{{dt}} = 3\sin x + 2{e^y} – 2.}$

### Example 5

Using the first approximation method, investigate the stability of the equilibrium point of the system
${\frac{{dx}}{{dt}} = \ln \left( {x + y} \right),\;\;}\kern0pt{\frac{{dy}}{{dt}} = \arctan \frac{{2x}}{y}.}$

### Example 6

Using the first approximation method, investigate the stability of the zero solution of the system
${\frac{{dx}}{{dt}} = \sin \left( {x + y} \right) – y,\;\;}\kern0pt{\frac{{dy}}{{dt}} = {y^2} + 2x.}$
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