Phase Portraits of Linear Systems

Drop a marble into a round bowl and it rolls to the bottom and stays: that resting point is stable. Balance one on top of a smooth hill and the tiniest nudge sends it rolling away: unstable. Now picture a Pringle chip, or a horse's saddle — from front to back it curves up like a bowl, but side to side it curves down like a hill. A marble placed at the center will sit there in theory, but almost any push sends it sliding off sideways. Those three pictures — bowl, hill, saddle — are exactly the behaviors this page is about.

Instead of one marble, imagine an arrow at every point of the plane telling a particle which way to drift next. For a linear system the rule is simple: $x' = ax + by$ and $y' = cx + dy$. The four numbers $a,b,c,d$ form a matrix $A$, and the whole motion is just $\mathbf{x}' = A\mathbf{x}$. The origin never moves (plug in $x=y=0$ and both rates are zero), so it is the fixed point. Two summary numbers tell you almost everything: the trace $T = a+d$ and the determinant $D = ad - bc$. As a first taste, the one-line system $x' = -x$ just decays smoothly to zero, the simplest stable behavior there is.

The honest answer to "what happens?" comes from the eigenvalues $\lambda = \dfrac{T \pm \sqrt{T^2 - 4D}}{2}$. Their real part says grow or shrink; any imaginary part says rotate. Two real same-sign eigenvalues give a node (the bowl or the hill); opposite signs give a saddle; a complex pair gives a spiral; a purely imaginary pair ($T=0$) gives a center of closed loops. The sliders are literally the matrix: $a$ and $d$ are the self-feedback of $x$ and $y$, while $b$ and $c$ are the cross-coupling that tilts and twists the flow.

Load the Center preset and watch trajectories close into perfect loops; then nudge slider $a$ a hair away from zero and see the loops unwind into a spiral — that is how fragile a center is. Switch to Saddle and notice the two straight eigenvector lines: one feeds into the origin, the other shoots away. Finally take any stable case and flip the sign of $a$ and $d$ together: the trace changes sign and every arrow reverses, turning a sink into a source.

From the eigenvalue product and sum, $D=\lambda_1\lambda_2$ and $T=\lambda_1+\lambda_2$. So a negative determinant forces eigenvalues of opposite sign (a saddle), while a positive determinant with negative trace forces both real parts negative (stable). This is why two numbers classify the whole system.

$\Delta=T^2-4D$. When $\Delta>0$ the square root is real, giving two real eigenvalues (node or saddle). When $\Delta<0$ the eigenvalues are a complex conjugate pair $\alpha\pm i\beta$ with $\alpha=T/2$ and $\beta=\tfrac{1}{2}\sqrt{4D-T^2}$ — rotation, hence a spiral or center. The boundary $\Delta=0$ is the repeated-root star.

$A=\begin{pmatrix}-0.5&-1\\1&-0.5\end{pmatrix}$: $T=-1$, $D=(-0.5)(-0.5)-(-1)(1)=0.25+1=1.25$, $\Delta=1-5=-4<0$. Eigenvalues $\lambda=\tfrac{1}{2}(-1\pm\sqrt{-4})=-0.5\pm i$. Negative real part and nonzero imaginary part $\Rightarrow$ a stable spiral winding into the origin. This is the default system on the page.

For a real eigenvalue $\lambda$, an eigenvector $\mathbf{v}$ solves $(A-\lambda I)\mathbf{v}=0$; a convenient choice is $\mathbf{v}=(b,\ \lambda-a)$ when $b\neq0$. Trajectories starting on that line stay on it, moving out if $\lambda>0$ and in if $\lambda<0$. In a node, generic trajectories enter tangent to the eigendirection of the smaller-magnitude eigenvalue (the "slow" direction).

The origin is asymptotically stable iff both eigenvalues have negative real part, i.e. $T<0$ and $D>0$. It is unstable if any eigenvalue has positive real part ($T>0$, or $D<0$). The borderline cases — centers ($T=0,D>0$) and zero-eigenvalue lines ($D=0$) — are not asymptotically stable and are destroyed by small nonlinear or parameter perturbations.