Part 1 · Oscillators and waves

1.3 Linear equations and the complex substitution

Source: Postorino, week 1 slides, pp. 13–14

Objectives

Vocabulary. Linear differential equation, homogeneous, known term, complex amplitude.
Theorem. Superposition: linear combinations of solutions of a homogeneous linear equation are solutions.
Recipe. The complex substitution: make the equation complex, try $\overset{x}{^} e^{iω t}$ , solve the algebra, take the real part.
Picture. The modulus of a complex amplitude is the amplitude, its argument is the phase.
Key. An oscillator’s energy goes as the square of its amplitude.

Linear differential equations

Definition 1.3.1Linear differential equationEssentialslides w1, p. 13

A linear differential equation of order $k$ has the form $y^{(k)} + a_{k - 1} y^{(k - 1)} + \dots + a_{1} y^{'} + a_{0} y = g (t),$ where

$y$ is the unknown function of $t$ ;
$k$ is the highest order of derivative in $t$ ;
$g (t)$ is the known term (termine noto).

It is linear because $y$ and its derivatives appear only to the power 1 (or 0, as in $g$ ): no $y^{2}$ , $y y^{'}$ or $sin y$ . When $g = 0$ it is homogeneous.

Non-example 1.3.2Not linearbeyond the lectures

The exact pendulum, $\ddot{θ} + ω_{0}^{2} sin θ = 0$ , is not linear: $sin θ$ is not $θ$ to the first power. Its small-angle version, $sin θ \approx θ$ , is linear: the harmonic approximation again.

Theorem 1.3.3Superposition principleslides w1, p. 13

For a homogeneous linear differential equation, any linear combination of particular solutions is again a solution of the same equation (principio di sovrapposizione).

Proofbeyond the lectures

Let $y_{1}$ and $y_{2}$ both satisfy the homogeneous equation, and put $y = α y_{1} + β y_{2}$ in. The derivative of a sum is the sum of the derivatives, and constants come out, so $y^{(j)} = α y_{1}^{(j)} + β y_{2}^{(j)}$ for every $j$ . The left side becomes $α (LHS for y_{1}) + β (LHS for y_{2}) = α \cdot 0 + β \cdot 0 = 0.$ For the inhomogeneous equation the same argument shows that any solution plus a solution of the homogeneous equation is again a solution.

The complex substitution

Take a homogeneous second-order equation with real coefficients $a, b, c$ : $a \overset{x}{¨} + b \overset{x}{˙} + c x = 0.$

Recipe 1.3.4Complex substitutionslides w1, p. 13

Replace the real unknown $x$ with a complex one, $x \to x + i x^{'}$ , where $x^{'}$ is a second real function (not a derivative). Look for it in the form $\overset{x}{^} e^{iω t}$ : $a \frac{d ^{2} ( x + i x ^{'} )}{d t ^{2}} + b \frac{d ( x + i x ^{'} )}{d t} + c (x + i x^{'}) = 0.$
Since $a, b, c$ are real, this splits into one equation for the real part and one for the imaginary part: ${a \overset{x}{¨} + b \overset{x}{˙} + c x = 0 i [a \overset{x}{¨}^{'} + b \overset{x}{˙}^{'} + c x^{'}] = 0 identical to the original!$
So turn the real equation into a complex one, find the complex solution, and take its real part. The real part solves the real equation.

Why bother: the derivatives of $e^{iω t}$ are just multiples of it (Key observation 1.2.6), so the equation becomes algebra in $\overset{x}{^}$ . Working with $cos$ directly mixes $cos$ and $sin$ at every derivative.

WarningOnly for linear equations with real coefficientsbeyond the lectures

Taking the real part commutes with derivatives and with multiplying by real numbers, but not with products: $Re (z^{2}) \neq = (Re z)^{2}$ . So the recipe works only because the equation is linear with real coefficients. For energies, which are squares, go back to the real field first, or use $∣ \overset{x}{^} ∣^{2} = \overset{x}{^} \overset{x}{^}^{*}$ .

Definition 1.3.5Complex amplitudeEssentialslides w1, p. 14

In a solution $\overset{x}{^} e^{iω t}$ , the complex number $\overset{x}{^} = x_{0} e^{i φ}$ is the complex amplitude. Its modulus is the amplitude and its argument is the phase: $Re {\overset{x}{^} e^{iω t}} = Re {x_{0} e^{i (ω t + φ)}} = x_{0} cos (ω t + φ) .$

This is the key to reading every complex amplitude in the book: of a driven oscillator (§1.4 Forced and damped oscillators: resonance), of a light wave (§2.2 Electromagnetic waves in vacuum), of the electrons in a dielectric (§3.1 Dielectrics: the Lorentz model).

The free oscillator, by substitution

Example 1.3.6The harmonic oscillator by complex substitutionslides w1, p. 14

Solve $\overset{x}{¨} + ω_{0}^{2} x = 0$ .

Substitute $x \to \overset{x}{^} e^{iω t}$ , with $\overset{x}{^} \in C$ (amplitude and phase together): $\frac{d ^{2} ( x ^ e ^{iω t} )}{d t ^{2}} + ω_{0}^{2} \overset{x}{^} e^{iω t} = 0 \Rightarrow (iω)^{2} \overset{x}{^} e^{iω t} + ω_{0}^{2} \overset{x}{^} e^{iω t} = 0 \Rightarrow - ω^{2} \overset{x}{^} + ω_{0}^{2} \overset{x}{^} = 0.$
For a non-zero $\overset{x}{^}$ , $ω = ω_{0}$ . The complex solution is $\overset{x}{^} e^{i ω_{0} t} = x_{0} e^{i φ} e^{i ω_{0} t}$ , writing $\overset{x}{^} = x_{0} e^{i φ}$ .
Take the real part: $x (t) = Re {x_{0} e^{i (ω_{0} t + φ)}} = x_{0} cos (ω_{0} t + φ) . Q.E.D.$

This is the harmonic oscillator again, with no trigonometry.

WarningSlide slip

The slide’s last line is $Re {e^{i (ω_{0} t + φ)}}$ , without the factor $x_{0}$ in front.

RemarkWhy

ω = - ω_{0}

is not a new solutionbeyond the lectures

$- ω^{2} + ω_{0}^{2} = 0$ also allows $ω = - ω_{0}$ . Its real part is $x_{0} cos (- ω_{0} t + φ) = x_{0} cos (ω_{0} t - φ)$ : the same family of solutions, with the phase renamed. For real motion one sign is enough.

Energy

The elastic force is conservative, so mechanical energy is conserved.

Theorem 1.3.7Energy of the oscillatorslides w1, p. 14

For $x (t) = x_{0} cos (ω_{0} t + φ)$ , $E = \frac{1}{2} m v^{2} + \frac{1}{2} k x^{2} = \frac{1}{2} k x_{0}^{2} .$

Proof

$v (t) = - ω_{0} x_{0} sin (ω_{0} t + φ)$ . So $E = \frac{1}{2} m ω_{0}^{2} x_{0}^{2} sin^{2} (ω_{0} t + φ) + \frac{1}{2} k x_{0}^{2} cos^{2} (ω_{0} t + φ) = \frac{1}{2} k x_{0}^{2},$ using $m ω_{0}^{2} = k$ and $sin^{2} + cos^{2} = 1$ .

Energy sloshes between kinetic and potential, and the total stays fixed.

Key observation 1.3.8Energy goes as amplitude squaredslides w1, p. 14

The energy of an oscillator is proportional to the square of its maximum amplitude. This holds for essentially every oscillatory or wave phenomenon. It is why the intensity of light goes as $∣ E ∣^{2}$ (Theorem 2.2.10), and why $∣ \overset{x}{^} ∣^{2} = \overset{x}{^} \overset{x}{^}^{*}$ is the quantity to compute.

Exercises

Exercise 1.3.1beyond the lectures

Show that $\overset{x}{^} = x_{0} e^{i φ}$ gives the same solution as $x (t) = A cos ω_{0} t + B sin ω_{0} t$ , and relate $A, B$ to $x_{0}, φ$ .

Solution

$x_{0} cos (ω_{0} t + φ) = x_{0} cos φ cos ω_{0} t - x_{0} sin φ sin ω_{0} t$ . So $A = x_{0} cos φ$ , $B = - x_{0} sin φ$ , $x_{0} = A^{2} + B^{2}$ and $tan φ = - B / A$ .

Exercise 1.3.2beyond the lectures

Which of these are linear? (a) $\overset{y}{¨} + t^{2} y = 0$ ; (b) $\overset{y}{¨} + y \overset{y}{˙} = 0$ ; © $\overset{y}{˙} = sin t$ ; (d) $\overset{y}{˙} + e^{y} = 0$ .

Solution

(a) linear: the coefficient $t^{2}$ may depend on $t$ , $y$ still appears to the first power. (b) not: $y \overset{y}{˙}$ . © linear, inhomogeneous. (d) not: $e^{y}$ .

Exercise 1.3.3beyond the lectures

An oscillator starts at $x (0) = 0$ with velocity $v (0) = v_{0} > 0$ . Find $\overset{x}{^}$ .

Solution

$x (0) = x_{0} cos φ = 0$ and $v (0) = - ω_{0} x_{0} sin φ = v_{0}$ . So $φ = - π /2$ , $x_{0} = v_{0} / ω_{0}$ : $\overset{x}{^} = (v_{0} / ω_{0}) e^{- iπ /2} = - i v_{0} / ω_{0}$ , and $x (t) = (v_{0} / ω_{0}) sin ω_{0} t$ .

Exercise 1.3.4beyond the lectures

Using the complex substitution, solve $\overset{x}{¨} + γ \overset{x}{˙} + ω_{0}^{2} x = 0$ (damped, not driven). Try $x = \overset{x}{^} e^{s t}$ with complex $s$ .

Solution

$s^{2} + γ s + ω_{0}^{2} = 0$ , so $s = - γ /2 \pm i ω_{0}^{2} - γ^{2} /4$ . For $γ < 2 ω_{0}$ , $x (t) = x_{0} e^{- γ t /2} cos (ω_{d} t + φ), ω_{d} = ω_{0}^{2} - γ^{2} /4 .$ The free motion dies out over a time of order $2/ γ$ ; when the oscillator is driven, only the steady state is left (§1.4 Forced and damped oscillators: resonance).

Exercise 1.3.5beyond the lectures

Show that the time average of the kinetic energy of a harmonic oscillator equals that of the potential energy, each $\frac{1}{4} k x_{0}^{2}$ .

Solution

$⟨ sin^{2} ⟩ = ⟨ cos^{2} ⟩ = \frac{1}{2}$ over a period. So $⟨ \frac{1}{2} m v^{2} ⟩ = \frac{1}{2} m ω_{0}^{2} x_{0}^{2} \cdot \frac{1}{2} = \frac{1}{4} k x_{0}^{2}$ , and $⟨ \frac{1}{2} k x^{2} ⟩ = \frac{1}{4} k x_{0}^{2}$ . Their sum is $E = \frac{1}{2} k x_{0}^{2}$ . (This is the root of equipartition, §5.2 Rayleigh–Jeans and Planck.)