Saddles · Differential Equations

Intuition

When the two eigenvalues have opposite signs, one straight line carries solutions out and the other carries them in. Every other solution does some of each: it comes in close to the incoming line, swings past the origin and leaves along the outgoing one, like a ball rolled over a mountain pass. The origin is then a saddle, and it is unstable: only the solutions that start exactly on the incoming line ever reach it.

A mountain pass. Along the ridge the pass is the lowest point, so a ball set on the ridge path rolls down to it; across the ridge the pass is the highest point, so a ball set anywhere else rolls away into one of the valleys.

The saddle of $A=\begin{pmatrix}1&2\\2&1\end{pmatrix}$ , with eigenvalues $3$ and $-1$ . Solutions come in along the stable line $y=-x$ , turn near the origin and leave along the unstable line $y=x$ ; only those on the stable line itself arrive.

Eigenvalues of opposite signs

If $\det A<0$ the eigenvalues are real with $\lambda_{1}<0<\lambda_{2}$ , and the origin is a saddle. The solutions on the stable line approach the origin; those on the unstable line, , leave it; and every other solution, with and , arrives from the direction of and leaves in the direction of . A saddle is always unstable.

\det A<0\ \Longleftrightarrow\ \lambda_{1}<0<\lambda_{2}

What a saddle does

$\det A=\lambda_{1}\lambda_{2}$ is negative only for real eigenvalues of opposite signs: a complex pair $\alpha\pm i\beta$ has product .

A negative determinant makes a saddle

\det A<0\ \Longrightarrow\ \lambda_{1}<0<\lambda_{2}\ \text{real}

The product of the eigenvalues is the determinant. A complex pair multiplies to a sum of squares, which cannot be negative, so the eigenvalues are real, and two real numbers with a negative product have opposite signs. With one eigenvalue of each sign, one term of every solution grows and the other decays, so every solution with a part along the unstable line leaves.

Proof steps

\lambda_{1}\lambda_{2}=\det A<0

The eigenvalues multiply to the determinant.

(\alpha+i\beta)(\alpha-i\beta)=\alpha^{2}+\beta^{2}\ge0

A complex pair cannot have a negative product, so the eigenvalues are real.

\lambda_{1}<0<\lambda_{2}

Two real numbers with a negative product have opposite signs.

c_{2}\ne0\ \Longrightarrow\ \bigl|\mathbf{x}(t)\bigr|\to\infty

The growing term takes over, so every solution with a part along the unstable line leaves.

Applications

Practice

The Determinant Decides

The eigenvalues multiply to the determinant, so a negative determinant means real eigenvalues of opposite signs: a saddle.

\det A<0\ \Longrightarrow\ \lambda_{1}<0<\lambda_{2}

Try it

Which matrix gives a saddle?