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.
Theorem of Hartman-Grobman 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.
