Analysis near singularities

Systems of differential equations: Non-linear differential equations

Analysis near singularities

Consider the system of differential equations

$\left\{\begin{aligned} \dfrac{\dd x}{\dd t} &= f(x,y)\\[0.25cm] \dfrac{\dd y}{\dd t} &= g(x,y)\end{aligned}\right.$ in the plane, where $f$ and $g$ are 'neat' functions (here, 'neat' means that derivatives exist and are continuous functions). Assume that $(x_0,y_0)$ is a singular point, that is, $f(x_0,y_0)=g(x_0,y_0)=0$ . In order to study the behaviour of a solution near $(x_0,y_0)$ we apply linearization: we consider the Jacobian matric $J$ in $(x_0,y_0)$ of the system of differential equations:

$J(x_0,y_0)= \matrix{\dfrac{\partial f}{\partial x}(x_0,y_0) & \dfrac{\partial f}{\partial y}(x_0,y_0) \\ \dfrac{\partial g}{\partial x}(x_0,y_0) & \dfrac{\partial g}{\partial y}(x_0,y_0) }$ We then get the following linear system of differential equations in matrix-vector form:

$\cv{ x'(t)\\ y'(t)} = J\cv{x(t)-x_0\\ y(t)-y_0}$ When we write

$X(t)=x(t)-x_0\quad\text{and}\quad Y(t)=y(t)-y_0$ then the constructed systems becomes

$\cv{ X'(t)\\ Y'(t)}= J\cv{X(t)\\ Y(t)}$ The following theorem gives conditions under which you can relate the stability of the singular point $(0,0)$ in the new system with the stability of $(x_0,y_0)$ in the original system of differential equations.

If the real part of the eigenvalues of $J$ is nonzero, then the behaviour of the non-linear system in the neighbourhood of $(x_0,y_0)$ is qualitatively the same as that of the system

$\cv{ X'(t)\\ Y'(t)} = J\cv{X(t)\\ Y(t)}$ in the neighbourhood of $(0,0)$ . So, if there are two negative eigenvalues, then the solutions $\bigl(x(t),y(t)\bigr)$ starting close to $(x_0,y_0)$ approach the point $(x_0,y_0)$ as $t$ goes to infinity.

Two illustrative examples.

Consider

$\left\{\begin{aligned} x' &= -2x-y^2\\[0.25cm] y' &= -y-x^2\end{aligned}\right.$ Then $(0,0)$ is a singular point. The general shape of the Jacobian matrix $(x,y)$ is

$J(x,y) =\matrix{\dfrac{\partial (-2x-y^2)}{\partial x} & \dfrac{\partial (-2x-y^2)}{\partial y}\\ \dfrac{\partial (-y-x^2)}{\partial x} & \dfrac{\partial (-y-x^2)}{\partial y}}=\matrix{-2 & -2y\\ -2x & -1}$ Linearization in $(0,0)$ gives

$\cv{ x'\\ y'}=\matrix{-2 & 0\\ 0 & -1}\cv{x\\ y}$ The eigenvalues of the Jacobian matrix are negative and thus all solution curves near $(0,0)$ move towards the origin.

We consider the van der Pol equation

$\left\{\begin{aligned} \dfrac{\dd x}{\dd t} & =\frac{1}{\varepsilon}\left(y+x-\frac{x^3}{3}\right)\\[0.25cm] \dfrac{\dd y}{\dd t} & =-\varepsilon x\end{aligned}\right.$ with $0<\varepsilon\ll 1$ . Then $(0,0)$ is a singular point. Linearization gives

$\cv{x'\\ y'}=\matrix{\dfrac{1}{\varepsilon} & \dfrac{1}{\varepsilon}\\ -\varepsilon & 0}\cv{x\\ y}$ The eigenvalues of the Jacobian matrix are

$\lambda_{1,2}=\frac{1}{2\varepsilon}\pm\frac{1}{2}\sqrt{\frac{1}{\varepsilon^2}-4}$ ie

$\lambda_{1,2}=\frac{1}{2\varepsilon}\left(1\pm\sqrt{1-4\varepsilon^2}\right)$ If $0\lt \varepsilon\lt \frac{1}{2}$ we have two positive real eigenvalues and all solution near $(0,0)$ move away from the origin: we have a repelling singular point. If $\frac{1}{2}\lt \varepsilon\lt 1$ we have two complex eigenvalues with positive real part and expanding spirals around $(0,0)$ . According to the theorem of Hartman-Grobman this behaviour is also true for the van der Pol equation. Where solutions spiral to is not covered by the theorem of Hartman-Grobman, but in this case they converge to a limit cycle in the phase plane.