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MATH
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Examples from Stability of systems

All examples from Stability of systems.

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A nonlinear non-example: logistic iteration

For the parameterised logistic recurrence

xn+1=rxn(1−xn),x0=13,x_{n+1}=r x_n(1-x_n),\qquad x_0=\frac13,

the first terms for r=2r=2 are

x1=49,x2=2(49)(1−49)=4081.x_1=\frac49,\qquad x_2=2\left(\frac49\right)\left(1-\frac49\right) =\frac{40}{81}.

The factor 1−xn1-x_n depends on the current value, so this recurrence is not linear and has no characteristic polynomial of the kind used below. The delayed logistic recurrence

xn+1=rxn(1−xn−1),x0=15,x1=14x_{n+1}=rx_n(1-x_{n-1}),\qquad x_0=\frac15,\quad x_1=\frac14

requires both x0x_0 and x1x_1 before it can be iterated.

Test a recurrence

Consider

xn+1=xn+xn−1.x_{n+1}=x_n+x_{n-1}.

The characteristic equation is

r2−r−1=0,r^2-r-1=0,

whose roots are

r1=1+52,r2=1−52.r_1=\frac{1+\sqrt5}{2}, \qquad r_2=\frac{1-\sqrt5}{2}.

Since ∣r1∣>1|r_1|>1, the corresponding mode grows. The system is unstable. The second root has modulus less than 11, but one growing mode is enough: choosing initial data with a non-zero coefficient of r1nr_1^n produces an unbounded solution.

Classify a damped oscillator

For

mx′′+cx′+kx=0,m x''+c x'+kx=0,

the roots satisfy

mr2+cr+k=0.mr^2+cr+k=0.

Take m=k=1m=k=1 and vary the damping cc.

  • If c>2c>2, the roots are distinct negative real numbers: the motion is overdamped and decays without oscillating.
  • If c=2c=2, the repeated root is r=−1r=-1, so (K1+K2t)e−t→0(K_1+K_2t)e^{-t}\to0: the motion is critically damped and decays.
  • If 0<c<20<c<2, the roots are a complex conjugate pair with negative real part: the motion is underdamped and oscillates while its amplitude decays.
  • If c=0c=0, the roots are r=±ir=\pm i, so the motion is periodic rather than decaying.
  • If −2<c<0-2<c<0, the conjugate roots have positive real part −c/2-c/2, so oscillations grow and the system is unstable.

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