Skip to content
Part BCSIR NET June 2023eigenvalue-growth-rates

Eigenvalue growth rates

Suppose x(t) is the solution of the initial value problem in R2\mathbb{R}^{2}: ẋ = Ax, x(0)=x0x(0) = x_{0}, where A = [[5, 4], [1, 2]]. Which of the following statements is true?

  1. A.x(t) is a bounded solution for some x0≠0x_{0} \ne 0.
  2. B.e−6te^{-6t}|x(t)| → 0 as t→∞t \to \infty, for all x0≠0x_{0} \ne 0.
  3. C.e−te^{-t}|x(t)| →∞\to \infty as t→∞t \to \infty, for all x0≠0x_{0} \ne 0.
  4. D.e−10te^{-10t}|x(t)| → 0 as t→∞t \to \infty, for all x0≠0x_{0} \ne 0.✓ Correct

The idea, in plain words

The long-run behaviour of a linear system is decided entirely by the two eigenvalues, which you can get from the trace and determinant without solving anything. Both come out positive here, so every non-zero solution grows without bound.

Solution

Eigenvalues of A are 6 and 1 (trace 7, det 6), both positive: every non-zero solution grows like e6te^{6t} or ete^{t}.

So (1) fails, (2) fails along the 6-eigendirection, (3) fails along the 1-eigendirection, and (4) holds since 10 > 6.

Why each option is right or wrong

Checked against 2 symbolic computations
  • A.WrongExecution slip

    False.

    SymPy gives the eigenvalues of A as 6 and 1 — both strictly positive (the characteristic polynomial λ2−7λ+6\lambda^{2} - 7\lambda + 6 has roots 1 and 6, verified).

    Every non-zero solution is a combination of e6te^{6t} and ete^{t} modes

    so all of them grow without bound.

    Bounded solutions would need an eigenvalue with non-positive real part.

  • B.WrongBoundary and endpoint

    False for the initial conditions lying along the dominant eigendirection.

    If x0x_{0} is an eigenvector for λ=6\lambda = 6, then |x(t)| = |x0∣e6tx_{0}|e^{6t} exactly

    so e−6te^{-6t}|x(t)| = |x0x_{0}|, a non-zero constant.

    The claim is 'for all x0≠0′x_{0} \ne 0', and this one starting point defeats it — damping at exactly the top growth rate leaves a constant, not zero.

  • C.WrongBoundary and endpoint

    False, defeated by the other eigendirection.

    Along the λ=1\lambda = 1 eigenvector, |x(t)| = |x0∣etx_{0}|e^{t}

    so e−te^{-t}|x(t)| = |x0x_{0}| — constant, not tending to infinity.

    The slowest mode is what breaks this one, just as the fastest broke (B).

  • D.Correct

    True.

    Every solution satisfies |x(t)| ≤ C(1 + t)e6tt)e^{6t} for some constant

    since 6 is the largest eigenvalue.

    Multiplying by e−10te^{-10t} leaves something bounded by C(1 + t)e−4tt)e^{-4t} → 0.

    The trigger: compare the damping exponent against the largest eigenvalue — strictly greater gives decay for every initial condition, equal gives a constant along that mode.

3 of these 3 wrong options correspond to a named reasoning error — the same errors recur across subjects.

The trap it tests

Numerical convergence

A scheme assumed to converge, or to converge at its advertised rate.

Drill statements like this

Related counterexample: W(f, g) ≡ 0 implies f and g are linearly dependent

More on this topic

The chapter behind this: Linear ODE, Wronskian and systems — free to read

From Ordinary Differential Equations › Linear ODE, Wronskian, variation of parameters, systems

Last revised . Found a mistake? Tell us — corrections are the fastest thing we act on.

ShareWhatsAppTelegram