Complex Eigenvalues — Question 8

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Question 8

For a nonzero real frequency ω\omega, consider X′=(−1−ωω−1)X,X(0)=(10).X'=\begin{pmatrix}-1&-\omega\\\omega&-1\end{pmatrix}X, \qquad X(0)=\binom 10. An instrument records only selected times. At every integer n≥0n\ge 0 it reports X(n)=e−n(1,0)TX(n)=e^{-n}(1,0)^T. Assume the measurements are exact.

Tasks

  1. Solve the IVP and determine every nonzero frequency consistent with all the integer-time measurements.

  2. Can these measurements determine rotation direction or the number of revolutions per unit time? Explain why the observed points do not imply a straight-line continuous trajectory.

  3. An additional measurement is X(1/4)=e−1/4(0,1)TX(1/4)=e^{-1/4}(0,1)^T. Find every frequency still allowed, including negative frequencies.

  4. If one also knows 0<ω<4π0<\omega<4\pi, recover the frequency uniquely. Exhibit two different frequencies that fit both sets of measurements when that extra bound is absent.

Original worksheet page 1: question and worked solution for 5-8-008
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Question 8 – Solution

Strategy. Sampling observes angles only modulo complete revolutions; additional information is needed to recover the continuous frequency.

Step 1: Solve and impose integer samples. The eigenvalues are −1±iω-1\pm i\omega, and X(t)=e−t(cos⁡ωtsin⁡ωt).\boxed{X(t)=e^{-t}\binom{\cos\omega t}{\sin\omega t}.} The measurement at n=1n=1 requires ω=2πk\omega=2\pi k, k∈ℤk\in\mathbb Z. Every such value satisfies all integer samples; the nonzero-frequency assumption excludes k=0k=0.

Step 2: Identify what the samples conceal. The sign of kk gives counterclockwise or clockwise rotation, and |k||k| gives revolutions per unit time. Neither is fixed by the integer samples. For every permitted nonzero frequency the continuous angle changes and the radius decays, tracing a spiral. The sampled points all lie on one ray because whole turns occur between them, not because that ray is an invariant trajectory.

Step 3: Use the quarter-time observation. The additional angle condition is ω/4=π/2+2πm\omega/4=\pi/2+2\pi m. Thus ω=2π(1+4m),m∈ℤ.\boxed{\omega=2\pi(1+4m),\qquad m\in\mathbb Z.} These values already satisfy the integer constraints. Negative possibilities remain, for example m=−1m=-1 gives ω=−6π\omega=-6\pi. No zero frequency occurs in this family.

Step 4: Use the external frequency bound. Under 0<ω<4π0<\omega<4\pi, only m=0m=0 remains, giving ω=2π\boxed{\omega=2\pi}. Without that bound, 2π2\pi and −6π-6\pi both fit all stated observations, despite opposite directions and different speeds. The figure plots ety(t)e^t y(t) for these two solutions: they agree at the recorded times but differ between them. The vertical variable is the rescaled component, not the radius or a phase-plane coordinate.

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