Phase Portraits and Equilibrium Types

One path for each solution

A solution of a coupled pair is two functions, x(t) and y(t). Instead of drawing each against t, plot y against x: as t runs, the point (x, y) moves, and the whole solution becomes one path in the plane. Time is no longer on an axis, so an arrow on the path shows which way it runs.

Every starting point has exactly one path through it, and two paths never cross, because the pair gives one velocity at each point. A drawing of many paths at once is a phase portrait, and it is read by what the paths do near the equilibrium, here the origin, where the velocity is zero.

The straight paths

A start on an eigenvector v stays on the line through v: the solution is $e^{\lambda t}v$e^(λt) v, which only changes length. If $\lambda$λ is negative the point runs straight into the origin along that line; if $\lambda$λ is positive it runs straight out.

For M = (1 1; 4 1), the eigenvalue 3 has eigenvector (1, 2) and −1 has (1, −2), so the lines y = 2x and y = −2x are paths. Every other path is the sum of a part along each line, and each part grows or shrinks at its own rate. The signs of the two eigenvalues decide the shape of everything else.

One of each sign: a saddle

With eigenvalues 3 and −1, a path is $A{e}^{3t}\left(1,2\right)+B{e}^{-t}\left(1,-2\right)$A e^(3t)(1, 2) + B e^(−t)(1, −2). Early on, the $e^{-t}$e^(−t) part is the larger, and the path comes in beside the line y = −2x. As it nears the origin that part shrinks while the $e^{3t}$e^(3t) part grows, so the path turns and leaves beside y = 2x.

Only a start exactly on y = −2x, with A = 0, reaches the origin. Every other start, however close, is carried away, so the equilibrium is unstable. This shape is a saddle point.

Four paths of the pair with M = (1 1; 4 1), drawn from $A{e}^{3t}\left(1,2\right)+B{e}^{-t}\left(1,-2\right)$A e^(3t)(1, 2) + B e^(−t)(1, −2) with $A=\pm 0.02$A = ±0.02 and $B=\pm 1$B = ±1. Each comes in beside y = −2x, where $\lambda =-1$λ = −1, turns short of the origin, and leaves beside y = 2x, where $\lambda =3$λ = 3. The arrows mark each path at t = 0.6.

Both negative: a stable node

Take $\frac{dx}{dt}=-3x+y$dx/dt = −3x + y, $\frac{dy}{dt}=x-3y$dy/dt = x − 3y, so M = (−3 1; 1 −3). Its characteristic polynomial is ${\lambda }^2+6\lambda +8=\left(\lambda +2\right)\left(\lambda +4\right)$λ² + 6λ + 8 = (λ + 2)(λ + 4), so the eigenvalues are −2, with eigenvector (1, 1), and −4, with eigenvector (1, −1). Both parts of every solution shrink, so every path runs into the origin. This is a stable node.

The two parts do not shrink equally. The $e^{-4t}$e^(−4t) part dies faster, so near the origin only the $e^{-2t}$e^(−2t) part is left, and paths arrive tangent to y = x, the direction of the eigenvalue nearer zero. From (2, 0) the solution is $x=e^{-2t}+e^{-4t}$x = e^(−2t) + e^(−4t), $y=e^{-2t}-e^{-4t}$y = e^(−2t) − e^(−4t), and $\frac{y}{x}$y/x is 0.245 at t = 0.25, 0.462 at t = 0.5, 0.762 at t = 1 and 0.964 at t = 2.

Four paths of the pair with M = (−3 1; 1 −3), through (2, 0), (−2, 0), (0, 2) and (0, −2) at t = 0; the gold one starts at the dot (2, 0). All four run into the origin and arrive along y = x, the line of $\lambda =-2$λ = −2.

Both positive: an unstable node

Take M = (3 1; 1 3). Its eigenvalues are 4, with eigenvector (1, 1), and 2, with eigenvector (1, −1). Both parts grow, so every path runs away from the origin. This is an unstable node.

It is the stable node run backward: reversing time changes the sign of every eigenvalue. Near the origin the $e^{2t}$e^(2t) part is the larger, so paths leave tangent to y = −x, the direction of the eigenvalue nearer zero, and farther out the $e^{4t}$e^(4t) part takes over and turns them toward the direction of y = x.

Four paths of the pair with M = (3 1; 1 3), drawn from $a{e}^{4t}\left(1,1\right)+b{e}^{2t}\left(1,-1\right)$a e^(4t)(1, 1) + b e^(2t)(1, −1) with $a=\pm 0.3$a = ±0.3 and $b=\pm 1$b = ±1. Each leaves the origin along y = −x, where $\lambda =2$λ = 2, and bends toward the direction of y = x, where $\lambda =4$λ = 4.

Complex eigenvalues: a spiral

Take $\frac{dx}{dt}=-x-2y$dx/dt = −x − 2y, $\frac{dy}{dt}=2x-y$dy/dt = 2x − y, so M = (−1 −2; 2 −1). Its characteristic polynomial is ${\lambda }^2+2\lambda +5$λ² + 2λ + 5, with discriminant 4 − 20 = −16, so $\lambda =-1\pm 2i$λ = −1 ± 2i. There is no real eigenvector, so no path is straight: every path turns.

The two parts of the eigenvalue do two jobs. The real part, −1, shrinks the distance from the origin like $e^{-t}$e^(−t). The imaginary part, 2, turns the point at 2 radians per unit time. At (1, 0) the velocity is (−1, 2), up and to the left, so the turning is counterclockwise. One full turn takes $\frac{2\pi }{2}=\pi$2π/2 = π units of time, and multiplies the distance by $e^{-\pi }\approx 0.043$e^(−π) ≈ 0.043. The paths spiral in: a stable spiral.

With a positive real part the distance grows instead and the paths spiral out, an unstable spiral.

Two paths of the pair with M = (−1 −2; 2 −1), from (2.5, 0) in gold and from (−2.5, 0) dashed. Each turns counterclockwise and, after one full turn, is back on its own side of the x-axis at $2.5e^{-\pi }\approx 0.108$2.5e^(−π) ≈ 0.108 from the origin.

A real part of zero: a center

When the eigenvalues are purely imaginary the distance neither shrinks nor grows, and every path closes into a loop round the origin. This is a center. The pair $\frac{dx}{dt}=v$dx/dt = v, $\frac{dv}{dt}=-4x$dv/dt = −4x, simple harmonic motion with x for the displacement and v for the velocity, has characteristic polynomial ${\lambda }^2+4$λ² + 4 and eigenvalues $\pm 2i$±2i. Its loops are the ellipses $\frac{1}{2}{v}^2+2x^2=$½v² + 2x² = constant.

The loop of $\frac{dx}{dt}=v$dx/dt = v, $\frac{dv}{dt}=-4x$dv/dt = −4x through (1, 0): an ellipse reaching $x=\pm 1$x = ±1 and $v=\pm 2$v = ±2, with $\frac{1}{2}{v}^2+2x^2=2$½v² + 2x² = 2 all the way round, traveled clockwise. Drag the start out to R = 2 and the loop doubles in each direction, with $\frac{1}{2}{v}^2+2x^2=8$½v² + 2x² = 8; the time round, $T=\pi$T = π, does not change.

Read the type from the eigenvalues

Opposite signs: a saddle point. Both negative: a stable node. Both positive: an unstable node. Complex with a negative real part: a stable spiral; with a positive real part, an unstable spiral; with a real part of zero, a center.

The determinant gives a quick first check, because it is the product of the eigenvalues. For (1 1; 4 1) it is −3, which is negative, so the eigenvalues have opposite signs and the origin is a saddle before either is found.

With constant terms in the pair, the equilibrium is not at the origin. Find it first by setting both rates to zero, then measure from it: the eigenvalues of what is left classify it. The arms race below does exactly this.

The usual mistakes

Calling every path through the origin straight. Only the eigenvector directions are straight paths; the others curve.

Taking the period of a spiral from the real part. The real part sets how fast the distance changes; the turning comes from the imaginary part. For $-1\pm 2i$−1 ± 2i, one turn takes $\frac{2\pi }{2}=\pi$2π/2 = π, not $2\pi$2π.

Reading complex eigenvalues as stable whatever their real part. $-1\pm 2i$−1 ± 2i spirals in, $1\pm 2i$1 ± 2i spirals out, and $\pm 2i$±2i loops round the origin without end.

Calling a saddle stable because some paths reach it. Only the paths that start exactly on one line do.

An arms race, and hares and lynx

In the applications below, the balance point of two countries’ spending is a saddle, and only spending that starts on one line through it settles there. Then a linear model of hares and lynx near their balance has eigenvalues $-0.1\pm 0.7i$−0.1 ± 0.7i, a stable spiral, and its imaginary part gives the time from one peak in the hares to the next.