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AI HL · AHL 9

Differential Equations

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AHL 9.15

Stability of second order equations

AHL 9.15

Stability of second order equations

Target: I can write x″ + ax′ + bx = 0 as a first order system and give its matrix. (+3 more)

By the end of this section

Key points

The system and its matrix

  • For \(\frac{d^2x}{dt^2} + a\frac{dx}{dt} + bx = 0\), let \(y = \frac{dx}{dt}\). Then \(\frac{dx}{dt} = 0x + 1y\) and \(\frac{dy}{dt} = -bx - ay\).
  • The matrix of the system is \(\begin{pmatrix} 0 & 1 \\ -b & -a \end{pmatrix}\). Read it row by row from the two equations. Signs change when you move terms across.
  • Its characteristic equation is \(\lambda^2 + a\lambda + b = 0\). The numbers are the same as the coefficients of the differential equation, so you can check your matrix against this.

What stable means

  • The solution is stable if x tends to 0 as t increases. It is unstable if x grows without limit.
  • Real eigenvalues: stable if both are negative. Unstable if any one is positive, including a saddle (one positive and one negative).
  • Complex eigenvalues \(p \pm qi\): stable if the real part p is negative, unstable if p is positive.
  • If the real part is 0 the solution oscillates for ever. x does not tend to 0, so it is not stable.

Method and check

  • Write the matrix, solve the characteristic equation (or find the eigenvalues) on the calculator, then say real or complex, and give the sign.
  • Always give the reason in terms of the eigenvalues: “both negative”, “positive real part”.
  • Quick check: for \(\lambda^2 + a\lambda + b = 0\), the solution is stable exactly when a and b are both positive.

On the calculator

Eigenvalues of the matrix

  • Enter the 2 by 2 matrix.
  • Use the calculator's eigenvalue function on it.
  • Read both eigenvalues. For a complex pair, read the real and imaginary parts.

Screenshot and video to come

Solving the characteristic equation

  • Use the polynomial solver for λ² + aλ + b = 0, with the degree set to 2.
  • Enter the three coefficients 1, a and b.
  • Read the roots. Complex roots are shown with i.

Screenshot and video to come

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
Writing the system and its matrix

Write \(\frac{d^2x}{dt^2} + 8\frac{dx}{dt} + 15x = 0\) as a system of coupled first order equations and give the matrix of the system.

Working
  1. Let y = dx/dt

    \(\frac{dx}{dt} = y,\quad \frac{dy}{dt} = -8y - 15x\)

    dy/dt = d²x/dt² = −8dx/dt − 15x.

  2. Write each equation with both letters

    \(\frac{dx}{dt} = 0x + 1y,\quad \frac{dy}{dt} = -15x - 8y\)

    Put in the zero coefficient so the pattern shows.

  3. The matrix

    \(\begin{pmatrix} 0 & 1 \\ -15 & -8 \end{pmatrix}\)

    Each row holds the coefficients of x and y in one equation.

  4. Check

    \(\lambda^2 + 8\lambda + 15 = 0\)

    The characteristic equation has the same numbers as the differential equation.

Video walkthrough coming
Real eigenvalues: a saddle

Write \(\frac{d^2x}{dt^2} + 2\frac{dx}{dt} - 8x = 0\) as a system and decide whether the solution is stable.

Working
  1. The system and matrix

    \(\frac{dx}{dt} = y,\quad \frac{dy}{dt} = -2y + 8x;\quad \begin{pmatrix} 0 & 1 \\ 8 & -2 \end{pmatrix}\)

    dy/dt = 8x − 2y.

  2. Characteristic equation

    \(\lambda^2 + 2\lambda - 8 = 0\)

    Same coefficients as the equation.

  3. From the calculator

    \(\lambda = 2 \text{ or } -4\)

    Polynomial solver, or the eigenvalues of the matrix.

  4. Decide

    unstable

    One eigenvalue is positive and one is negative, so this is a saddle and x does not tend to 0.

Video walkthrough coming
Complex eigenvalues: positive real part

(a) Write \(\frac{d^2x}{dt^2} - 4\frac{dx}{dt} + 13x = 0\) as a system of coupled first order equations. (b) Hence decide whether the solution is stable.

Working
  1. The system

    \(\frac{dx}{dt} = y,\quad \frac{dy}{dt} = 4y - 13x\)

    dy/dt = 4y − 13x.

  2. The matrix

    \(\begin{pmatrix} 0 & 1 \\ -13 & 4 \end{pmatrix}\)

  3. Eigenvalues

    \(\lambda^2 - 4\lambda + 13 = 0,\quad \lambda = 2 \pm 3i\)

    From the calculator: complex, because the discriminant is negative.

  4. Decide

    unstable

    The real part 2 is positive, so x grows with larger and larger oscillations.

Video walkthrough coming
Complex eigenvalues: negative real part

Decide whether the solution of \(\frac{d^2x}{dt^2} + 2\frac{dx}{dt} + 10x = 0\) is stable.

Working
  1. Matrix

    \(\begin{pmatrix} 0 & 1 \\ -10 & -2 \end{pmatrix}\)

  2. Eigenvalues

    \(\lambda^2 + 2\lambda + 10 = 0,\quad \lambda = -1 \pm 3i\)

    From the calculator.

  3. Decide

    stable

    The real part is negative, so x tends to 0 as t increases, with oscillations that die away.

Video walkthrough coming
An undamped motion

Decide whether the solution of \(\frac{d^2x}{dt^2} + 9x = 0\) is stable.

Working
  1. Matrix

    \(\begin{pmatrix} 0 & 1 \\ -9 & 0 \end{pmatrix}\)

    There is no x′ term, so a = 0.

  2. Eigenvalues

    \(\lambda^2 + 9 = 0,\quad \lambda = \pm 3i\)

    The real part is 0.

  3. Decide

    not stable

    x oscillates for ever with the same size. It does not tend to 0, so the solution is not stable.

Video walkthrough coming