5.17 Phase portraitsIB Maths: Applications and Interpretation HL: Revision notes

Two populations, or two connected quantities, can be described by the coupled linear system $\frac{dx}{dt}=ax+by,\qquad \frac{dy}{dt}=cx+dy,$ which in matrix form is $\frac{d}{dt}\begin{pmatrix} x \\ y \end{pmatrix}=M\begin{pmatrix} x \\ y \end{pmatrix}$ with $M=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$. A phase portrait is a diagram in the $xy$-plane showing trajectories: the paths $(x(t),y(t))$ followed as $t$ increases, with arrows for direction. At each point $(x,y)$ the trajectory has direction $\begin{pmatrix} \frac{dx}{dt} \\ \frac{dy}{dt} \end{pmatrix}=M\begin{pmatrix} x \\ y \end{pmatrix}$. For the systems in this course the origin is the only equilibrium point, where both rates are zero.

Find the eigenvalues $\lambda$ of $M$ from $\det(M-\lambda I)=0$, then an eigenvector $\mathbf{v}$ for each from $(M-\lambda I)\mathbf{v}=\mathbf{0}$ (a GDC will do this too). You are only given systems with distinct, non-zero eigenvalues. If the system starts at a point on an eigenvector $\mathbf{v}$, the solution is $e^{\lambda t}\mathbf{v}$. It stays on that line through the origin, moving away if $\lambda>0$ and towards the origin if $\lambda<0$. These are the straight-line trajectories. Example: $M=\begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}$ has $\lambda^2-7\lambda+10=0$, so $\lambda=2,5$ with eigenvectors $\begin{pmatrix} 1 \\ -2 \end{pmatrix}$ and $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$.

When the eigenvalues $\lambda_1\neq\lambda_2$ are real, the general solution is $\begin{pmatrix} x \\ y \end{pmatrix}=Ae^{\lambda_1t}\mathbf{v}_1+Be^{\lambda_2t}\mathbf{v}_2.$ Use the initial conditions to find $A$ and $B$. Example: $M=\begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}$, with $x=3$ and $y=0$ at $t=0$. Then $A+B=3$ and $-2A+B=0$, so $A=1$ and $B=2$ (using $\lambda_1=2$, $\mathbf{v}_1=\begin{pmatrix} 1 \\ -2 \end{pmatrix}$, $\lambda_2=5$, $\mathbf{v}_2=\begin{pmatrix} 1 \\ 1 \end{pmatrix}$): $x=e^{2t}+2e^{5t},\qquad y=-2e^{2t}+2e^{5t}.$ Exact solutions are only required in this case. For complex or imaginary eigenvalues you describe the behaviour qualitatively.

The signs of the real eigenvalues decide the portrait:

• Both positive: every solution moves away from the origin. The origin is an unstable node (a source).
• Both negative: every solution moves towards the origin. The origin is a stable node (a sink). Trajectories approach along the eigenvector whose eigenvalue is smaller in magnitude.
• Opposite signs: the origin is a saddle point. Solutions on the eigenvector line of the negative eigenvalue move in to the origin, but every other solution is turned away and eventually moves off along the eigenvector of the positive eigenvalue. To sketch: draw the eigenvector lines first with arrows, then add curved trajectories that follow them.

If the eigenvalues are complex, $\lambda=p\pm qi$ with $q\neq0$, the solutions rotate around the origin:

• $p>0$: spiral away from the origin.
• $p<0$: spiral towards the origin.
• $p=0$ (purely imaginary): closed circles or ellipses around the origin, with neither growth nor decay. The real part $p$ decides growth or decay, and the imaginary part $q$ gives the rotation. To find the sense of rotation, test one point: at $(1,0)$ the direction is $(a,c)$, so $c>0$ means anticlockwise and $c<0$ means clockwise. Example: $\frac{dx}{dt}=-x-4y$, $\frac{dy}{dt}=x-y$ has $\lambda=-1\pm2i$, so a spiral towards the origin; at $(1,0)$ the direction is $(-1,1)$, anticlockwise.

A phase portrait answers the question: what happens to the two quantities in the long term?

• Stable populations: a sink or an inward spiral means both variables settle at the equilibrium (here the origin).
• Unstable: a source or an outward spiral means the variables grow without bound.
• Cycles: imaginary eigenvalues give populations that rise and fall repeatedly.
• Saddle: the outcome depends on the starting point, and only one special line leads to equilibrium. Always state the conclusion in the context of the question, using the eigenvalues as the reason.