Second-order linear equations

Why the characteristic equation works: it is an eigenvalue problem.

about 8 min

Start from a problem

Problem

A damped oscillator behaves in three sharply different ways: it oscillates forever, it oscillates and decays, or it slides back without oscillating. What decides whether it oscillates?

We need a method that turns a second-order differential equation into an algebra problem, and explains why the answer splits into exactly three cases.

Observe

Choose "Damped oscillator": $x''+2\gamma x'+\omega^2x=0$ . With $\gamma=0$ the solution is an undying sine wave; for small $\gamma>0$ the oscillation decays; once $\gamma$ exceeds $\omega$ it stops oscillating and simply slides back to zero. In the phase portrait these are an ellipse, an inward spiral, and a curve heading straight for the origin.

Interactive experimentNumerical ODE explorer

Equation

x'' + 2\gamma x' + \omega^2 x = 0

Time series

Phase portrait

ω1.00

γ0.00

Initial value x(0)1.50

Initial value x′(0)0.00

step h0.250 · 48 steps

Method Euler RK4 exactEuler error at end: 3.1775RK4 error at end: 4.10e-4

Conjecture

The shape of the solution (oscillating or not) depends on how $\gamma$ compares with $\omega$ , like the sign of a discriminant of some quadratic. And solutions are always exponentials or exponentials times trigonometric functions, all relatives of $e^{rt}$ . Perhaps substituting $y=e^{rt}$ turns the differential equation into an algebraic one.

Definition

Definition 3.1Second-order linear homogeneous equation with constant coefficients

$y''+py'+qy=0,\qquad p,q\in\mathbb R.$

Derivation: the characteristic equation

Substitute $y=e^{rt}$ : $y'=re^{rt}$ , $y''=r^2e^{rt}$ , so $(r^2+pr+q)e^{rt}=0\iff r^2+pr+q=0.$

Definition 3.2Characteristic equation

$r^2+pr+q=0$ is the characteristic equation of the ODE.

Why "characteristic"? Let $v=\begin{pmatrix}y\\y'\end{pmatrix}$ . Then $v'=\begin{pmatrix}y'\\y''\end{pmatrix}=\begin{pmatrix}0&1\\-q&-p\end{pmatrix}v=:Av$ . The characteristic polynomial of $A$ is $\det(A-\lambda I)=\lambda^2+p\lambda+q.$ The characteristic equation of the ODE is literally the characteristic equation of $A$ . And the eigenvector $v=(1,\lambda)$ of $Av=\lambda v$ corresponds exactly to the solution $y=e^{\lambda t}$ (for which $y'=\lambda y$ ). Solving a second-order linear ODE is finding the eigenvalues of a $2\times2$ matrix.

Theorems and proofs

Theorem 3.1The solution space is a 2-dimensional vector space

The solutions of $y''+py'+qy=0$ form a two-dimensional real vector space.

Proof

Linearity: if $y_1,y_2$ are solutions, $(ay_1+by_2)''+p(ay_1+by_2)'+q(ay_1+by_2)=a(\cdots)+b(\cdots)=0$ , so the solution set is a subspace of the space of functions.

Dimension: by existence and uniqueness (for the system $v'=Av$ ), the map $y\mapsto(y(0),y'(0))\in\mathbb R^2$ is a linear bijection from the solution space onto $\mathbb R^2$ : injective because equal initial data give equal solutions, surjective because every initial datum has a solution. So the solution space is isomorphic to $\mathbb R^2$ .

Hence it suffices to find two independent solutions $y_1,y_2$ ; the general solution is $C_1y_1+C_2y_2$ .

Theorem 3.2General solution in three cases

Let the characteristic roots be $r_1,r_2$ and $\Delta=p^2-4q$ .

$\Delta>0$ , distinct real $r_1\ne r_2$ : $y=C_1e^{r_1t}+C_2e^{r_2t}$ .
$\Delta=0$ , $r_1=r_2=r$ : $y=(C_1+C_2t)e^{rt}$ .
$\Delta<0$ , $r=\alpha\pm i\beta$ : $y=e^{\alpha t}(C_1\cos\beta t+C_2\sin\beta t)$ .

Proof

Case 1. $e^{r_1t},e^{r_2t}$ are solutions. Independence: if $ae^{r_1t}+be^{r_2t}\equiv0$ , divide by $e^{r_1t}$ to get $a+be^{(r_2-r_1)t}\equiv0$ ; differentiating gives $b(r_2-r_1)e^{(r_2-r_1)t}\equiv0$ , so $b=0$ and then $a=0$ .

Case 2. $r=-p/2$ and $q=r^2$ . Check $te^{rt}$ : $(te^{rt})'=(1+rt)e^{rt}$ , $(te^{rt})''=(2r+r^2t)e^{rt}$ ; substituting, $e^{rt}\bigl[(2r+p)+(r^2+pr+q)t\bigr]=0$ since $2r+p=0$ and $r$ is a root. $e^{rt}$ and $te^{rt}$ are clearly independent. This is the case where $A$ has only one eigen-direction (like a shear): the missing second solution is supplied by the factor $t$ .

Case 3. $e^{(\alpha+i\beta)t}=e^{\alpha t}(\cos\beta t+i\sin\beta t)$ (Euler's formula, see Taylor expansion) is a complex solution. The coefficients are real, so its real and imaginary parts are real solutions: $e^{\alpha t}\cos\beta t$ and $e^{\alpha t}\sin\beta t$ . They are independent since $\cos\beta t,\sin\beta t$ are ( $\beta\ne0$ ).

For the damped oscillator $p=2\gamma$ , $q=\omega^2$ , $\Delta=4(\gamma^2-\omega^2)$ : $\gamma<\omega$ underdamped (decaying oscillation), $\gamma=\omega$ critical, $\gamma>\omega$ overdamped. Exactly the three behaviours in the experiment.

Example 3.1

$y''+4y=0$ : $r^2+4=0$ , $r=\pm2i$ , $y=C_1\cos2t+C_2\sin2t$ .

$y''-3y'+2y=0$ : $r^2-3r+2=(r-1)(r-2)$ , $y=C_1e^t+C_2e^{2t}$ .

Common mistake

For a double root $r$ the second solution is $te^{rt}$ , not another $e^{rt}$ : two proportional solutions span only one dimension and cannot fill a two-dimensional solution space. For complex roots the general solution is $e^{\alpha t}(C_1\cos\beta t+C_2\sin\beta t)$ with $\beta$ the imaginary part, not some combination like $\alpha^2+\beta^2$ .

Applications

ApplicationInhomogeneous equations

The general solution of $y''+py'+qy=g(t)$ is the homogeneous general solution plus one particular solution; the proof is identical to the first-order case. For forced vibration $x''+\omega^2x=F\cos\Omega t$ the particular solution's amplitude blows up as $\Omega\to\omega$ : resonance.

ApplicationWhy linear algebra reaches so far

An $n$ -th order constant-coefficient linear ODE $\iff$ an eigenvalue problem for an $n\times n$ matrix; the solution space is an $n$ -dimensional vector space; inhomogeneous solutions form its translate. Vibration modes in mechanics and RLC responses in circuits are diagonalisation in disguise.

Remark

Textbooks usually give the name "characteristic equation" first and then ask you to memorise three cases. Here the order is reversed: first see the three behaviours, then guess $e^{rt}$ , then discover that the "characteristic equation" really is the characteristic equation of a matrix. The name is no coincidence; it is one mathematical object seen from two fields.

Exercises

What is the larger root of the characteristic equation of

y''-3y'+2y=0

The general solution of

y''+4y=0