Qualitative Theory of ODE

Singular Point Analysis

Linear System of Two dimension

Consider

\[ \begin{equation} \begin{cases} x'=ax+by\\ y'=cx+dy \end{cases}, \quad \hat{\pmb{x}}=\pmb{A}x \label{eq1} \end{equation} \]

with det\((\pmb{A})\neq 0\).

General method for drawing sketch in a plane

For system of two dimension in equation \(\ref{eq1}\), follow steps below.

(i) check \(\lambda_1\), \(\lambda_2\).

To be more specific, directly check det\((\pmb{A})=\lambda_1\lambda_2\) and tr\((\pmb{A})=\lambda_1+\lambda_2\)

See the pictrues below to pinpoint the position for type of Singular Point.

If the position is above the parabola of the above plane, then go to part (iii) directly.

(ii) determine the straight line.

Solve for \(k\) of the straight line.

\[ \frac{y'}{x'}\Bigg|_{y=kx}=k \]

please do not forget \(k=\infty\) would not be included from above.

For deprecated node, we could only solve for one \(k\).

(iii) determine the type of direction.

Genaral method is to check for

\[ xy'-yx'|_{(x_0,y_0)}=r^2\theta'=\begin{cases}>0, \quad \text{anticlockwise}\\ <0,\quad \text{clockwise}\end{cases} \]

But we can choose special point for direction.

Proof for part (ii)Proof for part (iii)Special cases for (iii)

On linear curve, we have \(y(t)=kx(t)\), so \(y'(t)=kx'(t)\), and

\[ \frac{y'}{x'}\Bigg|_{y=kx}=k \]

Solve for \(k\).

we have polar expression

\[ x=r\cos \theta, \quad y=r\sin \theta \]

then

\[ \begin{align} x'&=r'\cos \theta -r\sin \theta\cdot \theta',\quad \label{xprime}\\y'&=r'\sin \theta + r\cos \theta \cdot\theta'\label{yprime} \end{align} \]

\((\ref{xprime})\times \sin\theta\) - \((\ref{yprime})\times \cos\theta\) and get:

\[ x'\sin\theta-y'\cos\theta=-r\cdot\theta' \]

multiply both sides by \(r\) and get

\[ x'y-y'x=-r^2\cdot \theta' \]

So

\[ xy'-yx'=r^2\cdot \theta' \]

both sides have the same sign.

For \(0<\lambda_1<\lambda_2\) or \(\lambda_1<\lambda_2<0\), we can substitute \((1,0)\) into \(y'\) to see the direction.

For \(\lambda_1<0<\lambda_2\), we can inspect a point \((0,1)\) to see the direction.

For \(\lambda_1=\lambda_2\), first we check if there is indefinite solution of \(k\). If so, then ... we can inspect a point on linear curve.

For \(\lambda=\alpha\pm i\beta\), we can also substitute \((1,0)\) into \(y'\) to see the direction.

Example. Draw the craft.

\[ \begin{cases} x'=4x+\sqrt{3}y\\ y'=2x+5y \end{cases} \]

Answer

• check the type of \(\lambda_1\), \(\lambda_2\).

• determine the linear curve.

• determine the type of direction.

Non-Linear System of Two dimension

This part we focus on

\[ \begin{equation} \begin{cases} x'=P(x,y)=ax+by+\varPhi(x,y)\\ y=Q(x,y)=cx+dy+\varPsi(x,y) \end{cases}\label{eq2} \end{equation} \]

where \(\varPhi(x,y)\), \(\varPsi(x,y)\) satisfy

\[ \varPhi(x,y)=o(r),\quad \varPsi(x,y)=o(r), \quad r=\sqrt{x^2+y^2}\rightarrow 0 \]

with its linear appriximation system

\[ \begin{equation} \begin{cases} x'=ax+by\\ y=cx+dy \end{cases}\label{eq3} \end{equation} \]

We can also use its linear spproximation system to test. To be more specific, we have the following theorem.

Theorem for Non-Linear System of Two dimension

(i) If singular point is \(N_-\)(\(N_+\)) of the linear approsimation system \(\ref{eq3}\), then it is also the \(N_-\)(\(N_+\)) of the original non-linear system \(\ref{eq2}\).

(ii) If singular point is \(F_-\)(\(F_+\)) of the linear approsimation system \(\ref{eq3}\), then it is also the \(F_-\)(\(F_+\)) of the original non-linear system \(\ref{eq2}\).

(iii) If singular point is \(S\) of the linear approsimation system \(\ref{eq3}\), then it is also the \(S\) of the original non-linear system \(\ref{eq2}\).

(iv) If singular point is \(S_-\)(\(S_+\)) or \(D_-\)(\(D_+\)) of the linear approsimation system \(\ref{eq3}\), and there exists \(\delta>0\), such that \(\varPhi(x,y)\), \(\varPsi(x,y)\) satisfy

\[ \varPhi(x,y)=o(r^{1+\delta}), \quad\varPsi(x,y)=o(r^{1+\delta}), \quad r=\sqrt{x^2+y^2}\rightarrow 0 \]

then it is also the \(S_-\)(\(S_+\)) or \(D_-\)(\(D_+\)) of the original non-linear system \(\ref{eq2}\).