Phase portrait

A phase portrait draws the solutions of a system of differential equations as curves in the plane of its variables, one curve through each starting point. This page shows the linear system $x' = Ax$ for a $2 \times 2$ matrix $A$. Moving the point $(\tau, \Delta)$ on the trace–determinant map takes the picture through saddles, nodes, spirals and centers, and the eigenvalue plane beside it shows the two eigenvalues behind each picture.

Every curve follows the direction field. Its arrows all have one length and show only the direction; the speed shows in the equal-time marks, which crowd together where the motion is slow. Some solutions run along a straight line through the origin. If $Av = \lambda v$, then $x(t) = e^{\lambda t} v$ is a solution, because its derivative $\lambda e^{\lambda t} v$ is exactly $A$ applied to it. So the straight lines are the eigen-directions of $A$, and $\lambda$ is the rate along each: the motion goes out when $\lambda > 0$, in when $\lambda < 0$, and not at all when $\lambda = 0$.

The eigenvalues solve $\lambda^2 - \tau\lambda + \Delta = 0$, so $\tau$ is their sum, $\Delta$ their product and $\tau^2 - 4\Delta$ the discriminant. Each boundary on the map is one event in the eigenvalue plane. Below $\Delta = 0$ the eigenvalues have opposite signs and the picture is a saddle; on that line one of them is 0, and a whole line of points stays at rest. On the parabola $\tau^2 = 4\Delta$ the two eigenvalues meet, and above it they are a complex pair. With $\Delta > 0$ the sign of $\tau$ separates decay from growth, and on $\tau = 0$ the pair lies on the imaginary axis. So $\tau$ and $\Delta$ fix the eigenvalues and the type of picture, except at repeated eigenvalues, where the type also depends on whether $A$ is a multiple of the identity.

For complex eigenvalues $\alpha \pm \beta i$ with $\beta > 0$, every nonzero solution goes once around the origin in time $2\pi/\beta$ and, over that time, is scaled by $e^{2\pi\alpha/\beta}$. That is why marks one turn apart lie on one ray from the origin, each scaled by the same factor. When $\alpha = 0$ the factor is 1 and every nonzero orbit closes into an ellipse: a center. With $\Delta = 1$ and $A = \begin{pmatrix} 0 & 1 \\ -1 & \tau \end{pmatrix}$, the system is the damped oscillator $x'' + 2\zeta x' + x = 0$ with $\zeta = -\tau/2$.

Crossing the parabola, the two straight lines close like scissors into one and then disappear as the picture becomes a spiral. Away from the vertex, a typical matrix on the parabola has only that one line, and every other solution enters or leaves the origin tangent to it. At $(\tau, \Delta) = (-2, 1)$, the matrix $-I$ has the same eigenvalues, yet every direction is a straight line. Changing the basis, $A \mapsto PAP^{-1}$, reshapes the picture but keeps $\tau$, $\Delta$, the eigenvalues and the type. A multiple of the identity stays as it is.