# Phase Portraits of Ordinary Differential Equations

Some differential equations are relatively easy to solve – even though at A-Level this may seem to be the norm, in reality it is very rare for a differential equation to have an explicit solution. The types of differential equations at A-Level can be solved by direct integration or by using a range of nifty techniques like making a substitution to change the differential equation into one that can be solved directly, separation of the variables can be used to solve some equations of the form $f(y)y’=g(x)$; sometimes the integrating factor $\mathrm{e}^{\int P(x) \mathrm{d}x}$ can be used to solve equations of the form $y^{\prime}+P(x)y=Q(x)$ and second-order equations of the form $ay^{\prime \prime}+by’+cy=f(x)$ where $a,b$ and $c$ are constants, can be solved by finding the complementary function and particular integral and combining the two.

However, these are very special cases. Many differential equations that come up in real-life situations are very non-linear in nature. Non-linear differential equations are generally very difficult to solve, but quite often, it is the behaviour of the solutions that interests us more than the actual solution. We may not be really interested in a particular solution but a family of solutions so that we can easily compare their behaviours particularly for equations that evolve over time. In some cases – even though it may be very difficult or even impossible to find the solution to a differential equation analytically, we can still analyse the behaviour through use of a phase-diagram.

A phase-diagram is a vector field that we can use to visually present the solutions to a differential equation. For example here is a second-order differential equation – (this is an example that I did that appears in the book by D. W. Jordan and P. Smith titled Nonlinear Ordinary Differential Equations – An Introduction for Scientists and Engineers Fourth Edition)
$$\ddot{x} = x-x^{2}$$

This second order-differential equation can be separated into a system of first-order differential equations given by $$\dot{x}=y$$ $$\dot{y}=x-x^{2}$$

This is a very common technique for solving differential equations since first-order ODEs are generally easier to solve than second-order ODEs. Note that $$\frac{\mathrm{d}y}{\mathrm{d}x}= \frac{\dot{y}}{\dot{x}}=\frac{x-x^{2}}{y}$$

The phase-portrait of this looks as follows

A phase portrait drawn in SAGE Math for a second-order ODE

At a point on the phase-diagram there is attached a vector which are represented here by the arrows – the vector tells you which direction to move in. So given a starting point it is simply a case of moving from point to point in the direction that the vector tells you to move in. So different starting points will give different paths through the phase-portrait called trajectories, some of which are shown in the following diagram

Phase portrait of a second-order ODE with trajectories

Each trajectory is described by the equation $3y^{2}=-2x^{3}+3x^{2}+C$ for different values of $C$. The trajectories above are for $C=-8$, $-4$, $-0.5$, $0$, $1$, $2$, $4$ and $6$. Here the horizontal $x$-axis represents the position of a particle and the vertical $y$-axis represents the velocity of the particle – remember that $y=\dot{x}$.

So now if we start in the top left hand corner of the diagram on the curve corresponding to when $C=8$ (the right-most blue line) the arrows tell us to move to the right and down – so our displacement increases as we move to the right but our velocity decreases as we move down. Eventually we cross the $x$-axis and here our velocity is instantaneously zero; then we start moving backwards since our velocity becomes negative as we continue moving downwards, and our displacement decreases as we move to the left. So I have not even calculated the actual solution to the ODE but I still know how the solution behaves.