# Stability in the First Approximation

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Investigate the stability of the zero equilibrium point of the system using the first approximation method.

$\frac{{dx}}{{dt}} = y + 3{x^2} + 2{y^2},\;\frac{{dy}}{{dt}} = - 2x - y + xy.$

### 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.

$\frac{{dx}}{{dt}} = {e^{x + y}} - 1,\; \frac{{dy}}{{dt}} = \ln \left( {1 + x} \right).$

### Example 4

Using equations of the first approximation investigate the stability of the zero solution of the nonlinear system:

$\frac{{dx}}{{dt}} = \tan \left( {x + y} \right) - y,\; \frac{{dy}}{{dt}} = 3\sin x + 2{e^y} - 2.$

### Example 5

Using the first approximation method, investigate the stability of the equilibrium point of the system

$\frac{{dx}}{{dt}} = \ln \left( {x + y} \right),\; \frac{{dy}}{{dt}} = \arctan \frac{{2x}}{y}.$

### Example 6

Using the first approximation method, investigate the stability of the zero solution of the system

$\frac{{dx}}{{dt}} = \sin \left( {x + y} \right) - y,\; \frac{{dy}}{{dt}} = {y^2} + 2x.$

### Example 1.

Investigate the stability of the zero equilibrium point of the system using the first approximation method.

$\frac{{dx}}{{dt}} = y + 3{x^2} + 2{y^2},\;\frac{{dy}}{{dt}} = - 2x - y + xy.$

Solution.

In this case, the functions $${f_1},{f_2}$$ have the form:

${f_1}\left( {x,y} \right) = y + 3{x^2} + 2{y^2},\;\; {f_2}\left( {x,y} \right) = - 2x - y + xy.$

Obviously, they are continuous and infinitely differentiable in a neighborhood of the origin and equal to zero at $$\mathbf{X} = \mathbf{0}.$$ Also, the order of the nonlinear terms in both functions is equal to or greater than $$2.$$ Thus, all requirements of the theorem on stability in the first approximation are satisfied. Compute the elements of the Jacobian matrix $$J$$ at the equilibrium point $$\mathbf{X} = \mathbf{0}:$$

$\frac{{\partial {f_1}}}{{\partial x}} = 6x,\;\; \frac{{\partial {f_1}}}{{\partial y}} = 1 + 6y,\;\; \frac{{\partial {f_2}}}{{\partial x}} = - 2 + y,\;\; \frac{{\partial {f_2}}}{{\partial y}} = - 1 + x,$
$\Rightarrow {\left. {\frac{{\partial {f_1}}}{{\partial x}}} \right|_{\substack{ x = 0\\ y = 0}}} = 0,\;\; {\left. {\frac{{\partial {f_1}}}{{\partial y}}} \right|_{\substack{ x = 0\\ y = 0}}} = 1,\;\; {\left. {\frac{{\partial {f_2}}}{{\partial x}}} \right|_{\substack{ x = 0\\ y = 0}}} = - 2,\;\; {\left. {\frac{{\partial {f_2}}}{{\partial y}}} \right|_{\substack{ x = 0\\ y = 0}}} = - 1,\;\; \Rightarrow J = \left[ {\begin{array}{*{20}{r}} 0&1\\ { - 2}&{ - 1} \end{array}} \right].$

Find the eigenvalues:

$\det \left( {J - \lambda I} \right) = 0,\;\; \Rightarrow \left| {\begin{array}{*{20}{c}} {0 - \lambda }&1\\ { - 2}&{ - 1 - \lambda } \end{array}} \right| = 0,\;\; \Rightarrow \lambda \left( {\lambda + 1} \right) + 2 = 0,\;\; \Rightarrow {\lambda ^2} + \lambda + 2 = 0,\;\; \Rightarrow D = - 7,\;\; \Rightarrow {\lambda _{1,2}} = \frac{{ - 1 \pm \sqrt { - 7} }}{2} = - \frac{1}{2} \pm \frac{{i\sqrt 7 }}{2}.$

The auxiliary equation has a pair of complex conjugate roots, the real part of which is negative:

$\text{Re}\left[ {{\lambda _1}} \right] \lt 0,\;\; \text{Re}\left[ {{\lambda _2}} \right] \lt 0.$

Hence, the zero solution of the system is stable by the theorem on stability in the first approximation. This equilibrium point is a focus.

### Example 2.

Find the equilibrium position of the system and investigate its stability in the first approximation.

$\frac{{dx}}{{dt}} = {x^2} - y,\;\frac{{dy}}{{dt}} = x - 1.$

Solution.

We determine the equilibrium point from the system of algebraic equations:

$\left\{ \begin{array}{l} \frac{{dx}}{{dt}} = 0\\ \frac{{dy}}{{dt}} = 0 \end{array} \right.,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {{x^2} - y = 0}\\ {x - 1 = 0} \end{array}} \right.,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {x = 1}\\ {y = 1} \end{array}} \right..$

As the equilibrium point $$\left( {1,1} \right)$$ is not a zero solution, we make a coordinate transformation. We introduce the new variables $$u, v:$$

$u = x - 1,\;\; v = y - 1.$

Substituting into the original system, we obtain:

$\left\{ \begin{array}{l} {\frac{{d\left( {u + 1} \right)}}{{dt}} = {\left( {u + 1} \right)^2} - \left( {v + 1} \right) }\\ {\frac{{d\left( {v + 1} \right)}}{{dt}} = u + \cancel{1} - \cancel{1} } \end{array} \right.,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {{\frac{{du}}{{dt}} = {u^2} + 2u + \cancel{1} - v - \cancel{1}} }\\ {\frac{{dv}}{{dt}} = u} \end{array}} \right.,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {\frac{{du}}{{dt}} = {u^2} + 2u - v}\\ {\frac{{dv}}{{dt}} = u} \end{array}} \right..$

In order to determine the stability of the zero solution, we compute the Jacobian $$J.$$ The right hand sides have the form:

${f_1}\left( {u,v} \right) = {u^2} + 2u - v,\;\; {f_2}\left( {u,v} \right) = u.$

Then

$\frac{{\partial {f_1}}}{{\partial u}} = 2u + 2,\;\; \frac{{\partial {f_1}}}{{\partial v}} = - 1,\;\; \frac{{\partial {f_2}}}{{\partial u}} = 1,\;\; \frac{{\partial {f_2}}}{{\partial v}} = 0.$

Accordingly, the partial derivatives at the point $$\left( {u = 0,v = 0} \right)$$ are as follows:

${\left. {\frac{{\partial {f_1}}}{{\partial u}}} \right|_{\substack{ u = 0\\ v = 0}}} = 2,\;\; {\left. {\frac{{\partial {f_1}}}{{\partial v}}} \right|_{\substack{ u = 0\\ v = 0}}} = - 1,\;\; {\left. {\frac{{\partial {f_2}}}{{\partial u}}} \right|_{\substack{ u = 0\\ v = 0}}} = 1,\;\; {\left. {\frac{{\partial {f_2}}}{{\partial v}}} \right|_{\substack{ u = 0\\ v = 0}}} = 0.$

Find the eigenvalues of the Jacobian $$J:$$

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

Thus, the matrix $$J$$ has one eigenvalue $$\lambda = 1$$ with algebraic multiplicity $$2.$$

By definition, a rough equilibrium point should be described by different eigenvalues. As in this case this condition is violated, we cannot judge the character of stability of the system in the first approximation. For the correct analysis of the stability, other approaches such as the method of Lyapunov functions are required.

### Example 3.

Determine the equilibrium positions of the system and explore their stability. Draw a schematic phase portrait of the corresponding linearized system.

$\frac{{dx}}{{dt}} = {e^{x + y}} - 1,\; \frac{{dy}}{{dt}} = \ln \left( {1 + x} \right).$

Solution.

We find the equilibrium points by solving the system of algebraic equations:

$\left\{ \begin{array}{l} \frac{{dx}}{{dt}} = 0\\ \frac{{dy}}{{dt}} = 0 \end{array} \right.,\;\; \Rightarrow \left\{ \begin{array}{l} {e^{x + y}} - 1 = 0\\ \ln \left( {1 + x} \right) = 0 \end{array} \right..$

The second equation implies that $$x = 0.$$ Substituting this into the first equation, we have:

${e^y} - 1 = 0,\;\; \Rightarrow {e^y} = 1,\;\; \Rightarrow {y = 0.}$

Thus, the system has the unique equilibrium position $$\left( {x = 0, y = 0} \right).$$ To investigate the stability of the zero solution, we expand the right hand sides in the Maclaurin series:

${f_1}\left( {x,y} \right) = {e^{x + y}} - 1,\;\; {f_2}\left( {x,y} \right) = \ln \left( {1 + x} \right),$
$\frac{{\partial {f_1}}}{{\partial x}} = {e^{x + y}},\;\; \frac{{\partial {f_1}}}{{\partial y}} = {e^{x + y}},\;\; \frac{{\partial {f_2}}}{{\partial x}} = \frac{1}{{1 + x}},\;\; \frac{{\partial {f_2}}}{{\partial y}} = 0.$

The partial derivatives at the point $$\left( {x = 0,y = 0} \right)$$ are equal to

${\left. {\frac{{\partial {f_1}}}{{\partial x}}} \right|_{\substack{ x = 0\\ y = 0}}} = 1,\;\; {\left. {\frac{{\partial {f_1}}}{{\partial y}}} \right|_{\substack{ x = 0\\ y = 0}}} = 1,\;\; {\left. {\frac{{\partial {f_2}}}{{\partial x}}} \right|_{\substack{ x = 0\\ y = 0}}} = 1,\;\; {\left. {\frac{{\partial {f_2}}}{{\partial y}}} \right|_{\substack{ x = 0\\ y = 0}}} = 0.$

We find the eigenvalues of the Jacobian matrix $$J:$$

$J = \left[ {\begin{array}{*{20}{c}} 1&1\\ 1&0 \end{array}} \right],\;\; \det \left( {J - \lambda I} \right) = 0,\;\; \Rightarrow \left| {\begin{array}{*{20}{c}} {1 - \lambda }&1\\ 1&{0 - \lambda } \end{array}} \right| = 0,\;\; \Rightarrow \lambda \left( {\lambda - 1} \right) - 1 = 0,\;\; \Rightarrow {\lambda ^2} - \lambda - 1 = 0,\;\; \Rightarrow D = 5,\;\; \Rightarrow {\lambda _{1,2}} = \frac{{1 \pm \sqrt 5 }}{2} = \frac{1}{2} \pm \frac{{\sqrt 5 }}{2}.$

As you can see, the eigenvalues are real numbers with different signs and approximately equal to

${\lambda _1} = \frac{1}{2} + \frac{{\sqrt 5 }}{2} \approx 0.5 + 1.12 = 1.62,\;\;\; {\lambda _2} = \frac{1}{2} - \frac{{\sqrt 5 }}{2} \approx 0.5 - 1.12 = - 0.62$

Therefore, the linearized system has the zero equilibrium point of the saddle type. A similar conclusion is true with respect to the original nonlinear system.

Construct a schematic phase portrait of the linearized system. Compute the eigenvectors $${\mathbf{V}_1},$$ $${\mathbf{V}_2},$$ associated with the eigenvalues $${\lambda _1}$$ and $${\lambda _2}.$$ For the first eigenvalue $${\lambda _1},$$ we obtain:

$\left( {J - {\lambda _1}I} \right){\mathbf{V}_1} = \mathbf{0},\;\; \Rightarrow \left[ {\begin{array}{*{20}{c}} {\frac{1}{2} - \frac{{\sqrt 5 }}{2}}&1\\ 1&{ - \frac{1}{2} - \frac{{\sqrt 5 }}{2}} \end{array}} \right] \left[ {\begin{array}{*{20}{c}} {{V_{11}}}\\ {{V_{21}}} \end{array}} \right] = \mathbf{0},\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {\left( {\frac{1}{2} - \frac{{\sqrt 5 }}{2}} \right){V_{11}} + {V_{21}} = 0}\\ {{V_{11}} - \left( {\frac{1}{2} + \frac{{\sqrt 5 }}{2}} \right){V_{21}} = 0} \end{array}} \right.,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {\left( {\frac{1}{2} - \frac{{\sqrt 5 }}{2}} \right){V_{11}} + {V_{21}} = 0}\\ {\left( {\frac{1}{2} - \frac{{\sqrt 5 }}{2}} \right){V_{11}} + {V_{21}} = 0} \end{array}} \right.,\;\; \Rightarrow \left( {\frac{1}{2} - \frac{{\sqrt 5 }}{2}} \right){V_{11}} + {V_{21}} = 0.$

Let $${V_{11}} = t.$$ Then

${V_{21}} = - \left( {\frac{1}{2} - \frac{{\sqrt 5 }}{2}} \right){V_{11}} = - \left( {\frac{1}{2} - \frac{{\sqrt 5 }}{2}} \right)t \approx 0.62t,\;\; \Rightarrow {\mathbf{V}_1} = \left[ {\begin{array}{*{20}{c}} {{V_{11}}}\\ {{V_{21}}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} t\\ {0.62t} \end{array}} \right] = t\left[ {\begin{array}{*{20}{c}} 1\\ {0.62} \end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}} 1\\ {0.62} \end{array}} \right].$

Similarly, we find the eigenvector $${\mathbf{V}_2}$$ corresponding to the eigenvalue $${\lambda _2}:$$

$\left( {J - {\lambda _2}I} \right){\mathbf{V}_2} = \mathbf{0},\;\; \Rightarrow \left[ {\begin{array}{*{20}{c}} {\frac{1}{2} + \frac{{\sqrt 5 }}{2}}&1\\ 1&{ - \left( {\frac{1}{2} - \frac{{\sqrt 5 }}{2}} \right)} \end{array}} \right] \left[ {\begin{array}{*{20}{c}} {{V_{12}}}\\ {{V_{22}}} \end{array}} \right] = \mathbf{0},\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {\left( {\frac{1}{2} + \frac{{\sqrt 5 }}{2}} \right){V_{12}} + {V_{22}} = 0}\\ {{V_{12}} - \left( {\frac{1}{2} - \frac{{\sqrt 5 }}{2}} \right){V_{22}} = 0} \end{array}} \right.,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {\left( {\frac{1}{2} + \frac{{\sqrt 5 }}{2}} \right){V_{12}} + {V_{22}} = 0}\\ {\left( {\frac{1}{2} + \frac{{\sqrt 5 }}{2}} \right){V_{12}} + {V_{22}} = 0} \end{array}} \right.,\;\; \Rightarrow \left( {\frac{1}{2} + \frac{{\sqrt 5 }}{2}} \right){V_{12}} + {V_{22}} = 0.$

Suppose that $${V_{12}} = t.$$ Hence,

${V_{22}} = - \left( {\frac{1}{2} + \frac{{\sqrt 5 }}{2}} \right){V_{12}} = - \left( {\frac{1}{2} + \frac{{\sqrt 5 }}{2}} \right)t \approx -1.62t,\;\; \Rightarrow {\mathbf{V}_2} = \left[ {\begin{array}{*{20}{c}} {{V_{12}}}\\ {{V_{22}}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} t\\ {-1.62t} \end{array}} \right] = t\left[ {\begin{array}{*{20}{c}} 1\\ {-1.62} \end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}} 1\\ {-1.62} \end{array}} \right].$

Draw the straight lines passing through the origin and directed along the vectors $${\mathbf{V}_1}$$ and $${\mathbf{V}_2}$$ (Figure $$2$$).

These lines are separatrices of the saddle. Now we can depict the phase trajectories.

Now we specify the direction of the trajectories by selecting any point, for example, the point $$\left( {1,0} \right).$$ The velocity of motion at this point is given by

${\left. {\frac{{dx}}{{dt}}} \right|_{\left( {1,0} \right)}} = {e^{1 + 0}} - 1 \approx 1.72,\;\;\; \kern-0.3pt {\left. {\frac{{dy}}{{dt}}} \right|_{\left( {1,0} \right)}} = \ln \left( {1 + 1} \right) = \ln 2 \approx 0.69$

It is clear that the velocity vector is directed to the upper right side. On this basis, we can specify the direction of the phase trajectories.

This equilibrium point is rough. Therefore, the phase portrait of the original nonlinear system has the same shape in a neighborhood of the zero equilibrium point as shown in Figure $$2$$ for the linearized system.

### Example 4.

Using equations of the first approximation investigate the stability of the zero solution of the nonlinear system:

$\frac{{dx}}{{dt}} = \tan \left( {x + y} \right) - y,\; \frac{{dy}}{{dt}} = 3\sin x + 2{e^y} - 2.$

Solution.

We verify that the point $$\left( {x = 0, y = 0} \right)$$ is an equilibrium position for the given system:

${f_1}\left( {0,0} \right) = \tan 0 - 0 = 0,\;\; {f_2}\left( {0,0} \right) = 3\sin 0 + 2{e^0} - 2 = 0 + 2 \cdot 1 - 2 = 0.$

Expand the functions $${f_1},{f_2}$$ (which are continuously differentiable in a neighborhood of the zero point) in the Maclaurin series. The first order partial derivatives have the form

$\frac{{\partial {f_1}}}{{\partial x}} = \frac{1}{{{{\cos }^2}\left( {x + y} \right)}},\;\; \frac{{\partial {f_1}}}{{\partial y}} = \frac{1}{{{{\cos }^2}\left( {x + y} \right)}} - 1,\;\; \frac{{\partial {f_2}}}{{\partial x}} = 3\cos x,\;\; \frac{{\partial {f_2}}}{{\partial y}} = 2{e^y}.$

The values of the derivatives at the point $$\left( {x = 0,y = 0} \right)$$ are

${\left. {\frac{{\partial {f_1}}}{{\partial x}}} \right|_{\left( {0,0} \right)}} = 1,\;\; {\left. {\frac{{\partial {f_1}}}{{\partial y}}} \right|_{\left( {0,0} \right)}} = 0,\;\; {\left. {\frac{{\partial {f_2}}}{{\partial x}}} \right|_{\left( {0,0} \right)}} = 3,\;\; {\left. {\frac{{\partial {f_2}}}{{\partial y}}} \right|_{\left( {0,0} \right)}} = 2.$

We've got the Jacobian matrix $$J$$ for the linearized system in the form:

$J = \left[ {\begin{array}{*{20}{c}} 1&0\\ 3&2 \end{array}} \right].$

Compute its eigenvalues:

$\det \left( {J - \lambda I} \right) = 0,\;\; \Rightarrow \left| {\begin{array}{*{20}{c}} {1 - \lambda }&0\\ 3&{2 - \lambda } \end{array}} \right| = 0,\;\; \Rightarrow \left( {\lambda - 1} \right)\left( {\lambda - 2} \right) = 0,\;\; \Rightarrow {\lambda _1} = 1,\;{\lambda _2} = 2.$

Thus, according to the Lyapunov theorem on stability in the first approximation, the zero solution of the system is unstable. The zero equilibrium point is an unstable node.

We find the eigenvectors and draw a schematic phase portrait of the linearized system near zero. For the eigenvalue $${\lambda _1} = 1,$$ the vector $${\mathbf{V}_1}$$ has the following coordinates:

$\left( {J - {\lambda _1}I} \right){\mathbf{V}_1} = \mathbf{0},\;\; \Rightarrow \left[ {\begin{array}{*{20}{c}} {1 - 1}&0\\ 3&{2 - 1} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{V_{11}}}\\ {{V_{21}}} \end{array}} \right] = \mathbf{0},\;\; \Rightarrow \left[ {\begin{array}{*{20}{c}} 0&0\\ 3&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{V_{11}}}\\ {{V_{21}}} \end{array}} \right] = \mathbf{0},\;\; \Rightarrow 3{V_{11}} + {V_{21}} = 0.$

Let $${V_{11}} = t.$$ Then

${V_{21}} = - 3{V_{11}} = - 3t,\;\; \Rightarrow {\mathbf{V}_1} = \left[ {\begin{array}{*{20}{c}} {{V_{11}}}\\ {{V_{21}}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} t\\ { - 3t} \end{array}} \right] = t\left[ {\begin{array}{*{20}{r}} 1\\ { - 3} \end{array}} \right] \sim \left[ {\begin{array}{*{20}{r}} 1\\ { - 3} \end{array}} \right].$

Determine the eigenvector $${\mathbf{V}_2}$$ for the eigenvalue $${\lambda _2} = 2:$$

$\left( {J - {\lambda _2}I} \right){\mathbf{V}_2} = \mathbf{0},\;\; \Rightarrow \left[ {\begin{array}{*{20}{c}} {1 - 2}&0\\ 3&{2 - 2} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{V_{12}}}\\ {{V_{22}}} \end{array}} \right] = \mathbf{0},\;\; \Rightarrow \left[ {\begin{array}{*{20}{r}} { - 1}&0\\ 3&0 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{V_{12}}}\\ {{V_{22}}} \end{array}} \right] = \mathbf{0},\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} { - {V_{12}} + 0 \cdot {V_{22}} = 0}\\ {3{V_{12}} + 0 \cdot {V_{22}} = 0} \end{array}} \right..$

Let $${V_{12}} = 0,$$ $${V_{22}} = 1,$$ that is

${\mathbf{V}_2} = \left[ {\begin{array}{*{20}{c}} {{V_{12}}}\\ {{V_{22}}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0\\ 1 \end{array}} \right].$

Now we can plot the asymptotic straight lines passing through the origin and parallel to the vectors $${\mathbf{V}_1}$$ and $${\mathbf{V}_2}$$ (Figure $$3$$).

We take into account that the phase trajectories asymptotically approach the line directed along the vector $${\mathbf{V}_1}$$ with the smallest (in absolute value) eigenvalue $${\lambda _1} = 1.$$ At a distance from the origin, the phase trajectories become parallel to the eigenvector $${\mathbf{V}_2},$$ which is directed along the $$y$$-axis.

The original nonlinear system has the same phase portrait near the origin. This follows from the roughness (structural stability) of the zero equilibrium position.

### Example 5.

Using the first approximation method, investigate the stability of the equilibrium point of the system

$\frac{{dx}}{{dt}} = \ln \left( {x + y} \right),\; \frac{{dy}}{{dt}} = \arctan \frac{{2x}}{y}.$

Solution.

We first determine the equilibrium point:

$\left\{ \begin{array}{l} \frac{{dx}}{{dt}} = 0\\ \frac{{dy}}{{dt}} = 0 \end{array} \right.,\;\; \Rightarrow \left\{ \begin{array}{l} \ln \left( {x + y} \right) = 0\\ \arctan \frac{{2x}}{y} = 0 \end{array} \right.,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {x + y = 1}\\ {x + y \gt 0}\\ {\frac{{2x}}{y} = 0}\\ {y \ne 0} \end{array}} \right.,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {x + y = 1}\\ {x = 0}\\ {x + y \gt 0}\\ {y \ne 0} \end{array}} \right.,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {x = 0}\\ {y = 1} \end{array}} \right..$

Thus, the system has a single equilibrium point at $$\left( {x = 0,y = 1} \right).$$ We study its stability using the first approximation techniques. Compute the partial derivatives of the right hand sides $${f_1}\left( {x,y} \right)$$ and $${f_2}\left( {x,y} \right).$$

$\frac{{\partial {f_1}}}{{\partial x}} = \frac{\partial }{{\partial x}}\left[ {\ln \left( {x + y} \right)} \right] = \frac{1}{{x + y}},$
$\frac{{\partial {f_1}}}{{\partial y}} = \frac{\partial }{{\partial y}}\left[ {\ln \left( {x + y} \right)} \right] = \frac{1}{{x + y}},$
$\frac{{\partial {f_2}}}{{\partial x}} = \frac{\partial }{{\partial x}}\left[ {\arctan\frac{{2x}}{y}} \right] = \frac{1}{{1 + {{\left( {\frac{{2x}}{y}} \right)}^2}}} \cdot \frac{2}{y} = \frac{{{y^2}}}{{{y^2} + 4{x^2}}} \cdot \frac{2}{y} = \frac{{2y}}{{4{x^2} + {y^2}}},$
$\frac{{\partial {f_2}}}{{\partial y}} = \frac{\partial }{{\partial y}}\left[ {\arctan\frac{{2x}}{y}} \right] = \frac{1}{{1 + {{\left( {\frac{{2x}}{y}} \right)}^2}}} \cdot \left( { - \frac{{2x}}{{{y^2}}}} \right) = \frac{{{y^2}}}{{{y^2} + 4{x^2}}} \cdot \left( { - \frac{{2x}}{{{y^2}}}} \right) = - \frac{{2x}}{{4{x^2} + {y^2}}}.$

At the point $$\left( {x = 0,y = 1} \right),$$ the derivatives are as follows:

${\left. {\frac{{\partial {f_1}}}{{\partial x}}} \right|_{\left( {0,1} \right)}} = 1,\;\; {\left. {\frac{{\partial {f_1}}}{{\partial y}}} \right|_{\left( {0,1} \right)}} = 1,\;\; {\left. {\frac{{\partial {f_2}}}{{\partial x}}} \right|_{\left( {0,1} \right)}} = 2,\;\; {\left. {\frac{{\partial {f_2}}}{{\partial y}}} \right|_{\left( {0,1} \right)}} = 0.$

Therefore, the linearized matrix (Jacobian) at the equilibrium point is given by

$J = \left[ {\begin{array}{*{20}{c}} 1&1\\ 2&0 \end{array}} \right].$

Find its eigenvalues $$\lambda:$$

$\det \left( {J - \lambda I} \right) = 0,\;\; \Rightarrow \left| {\begin{array}{*{20}{c}} {1 - \lambda }&1\\ 2&{0 - \lambda } \end{array}} \right| = 0,\;\; \Rightarrow \lambda \left( {\lambda - 1} \right) - 2 = 0,\;\; \Rightarrow {\lambda ^2} - \lambda - 2 = 0,\;\; \Rightarrow D = 9,\;\; \Rightarrow {\lambda _{1,2}} = \frac{{1 \pm 3}}{2} = - 1,2.$

Thus, the equilibrium position $$\left( {0,1} \right)$$ of the linearized system is characterized by two real eigenvalues of opposite sign. According to the Lyapunov theorem on instability in the first approximation, the point $$\left( {0,1} \right)$$ is unstable. This equilibrium point is a saddle.

### Example 6.

Using the first approximation method, investigate the stability of the zero solution of the system

$\frac{{dx}}{{dt}} = \sin \left( {x + y} \right) - y,\; \frac{{dy}}{{dt}} = {y^2} + 2x.$

Solution.

Obviously, the right hand sides of the differential equations are continuous and infinitely differentiable. Therefore, we can expand them in a Maclaurin series:

${f_1}\left( {x,y} \right) = \sin \left( {x + y} \right) - y,\;\;\; {f_2}\left( {x,y} \right) = {y^2} + 2x,$
$\frac{{\partial {f_1}}}{{\partial x}} = \cos \left( {x + y} \right),\;\; \frac{{\partial {f_1}}}{{\partial y}} = \cos \left( {x + y} \right) - 1,\;\; \frac{{\partial {f_2}}}{{\partial x}} = 2,\;\; \frac{{\partial {f_2}}}{{\partial y}} = 2y.$

Substituting $$x = 0, y = 0,$$ we obtain the following values of the derivatives:

${\left. {\frac{{\partial {f_1}}}{{\partial x}}} \right|_{\left( {0,0} \right)}} = 1,\;\; {\left. {\frac{{\partial {f_1}}}{{\partial y}}} \right|_{\left( {0,0} \right)}} = 0,\;\; {\left. {\frac{{\partial {f_2}}}{{\partial x}}} \right|_{\left( {0,0} \right)}} = 2,\;\; {\left. {\frac{{\partial {f_2}}}{{\partial y}}} \right|_{\left( {0,0} \right)}} = 0.$

Compose the Jacobian matrix $$J$$ and calculate its eigenvalues:

$J = \left[ {\begin{array}{*{20}{c}} 1&0\\ 2&0 \end{array}} \right],\;\; \det \left( {J - \lambda I} \right) = 0,\;\; \Rightarrow \left| {\begin{array}{*{20}{c}} {1 - \lambda }&0\\ 2&{ - \lambda } \end{array}} \right| = 0,\;\; \Rightarrow \lambda \left( {\lambda - 1} \right) = 0,\;\; \Rightarrow {\lambda _1} = 0,\;{\lambda _2} = 1.$

We have two eigenvalues, one of which is zero. This means that we have a critical case, for which it is impossible to determine the type of stability in the first approximation.