Wave equation (hyperbolic) 1D — derivation: Question bank

Before we start, a one-line reminder of the cast so no symbol is unearned:

$u (x, t)$ = the sideways displacement of the string at position $x$ , time $t$ .
$u_{x}$ = slope (how tilted the string is); $u_{xx}$ = curvature (how the slope bends).
$u_{tt}$ = vertical acceleration of a point on the string.
$c = T / ρ$ = wave speed, built from tension $T$ and linear density $ρ$ .
$θ$ = the angle the string makes with the horizontal at a point; its slope is $tan θ = u_{x}$ .
$f$ and $g$ = the two initial conditions: $f (x) = u (x, 0)$ is the starting shape, and $g (x) = u_{t} (x, 0)$ is the starting velocity of every point.

True or false — justify

Speed of a wave grows if you pluck the string harder.

False. In the linear model

c = T / ρ

depends only on the medium (tension and density); a louder pluck raises the amplitude

u

, not the speed.

The wave equation is derived from energy conservation.

False. It comes straight from Newton's second law

F = ma

applied to a tiny element — vertical force from tension equals mass times vertical acceleration. See Newton's second law.

u_{xx} > 0

at a point means the string is steep there.

False.

u_{xx}

is curvature, not steepness.

u_{xx} > 0

means concave-up (a

\cup

dip), and by the equation that point accelerates upward.

If a point on the string has zero slope (

u_{x} = 0

) it must have zero acceleration.

False. Acceleration is set by

u_{xx}

, not

u_{x}

. The crest of a hump has

u_{x} = 0

but large negative

u_{xx}

, so it accelerates strongly downward.

Every function of the form

F (x - c t)

solves the wave equation.

True. The operator factors as

(\partial_{t} - c \partial_{x}) (\partial_{t} + c \partial_{x})

, and

F (x - c t)

is annihilated by the right factor, so it's automatically a rightward-travelling solution.

The wave equation is the same type of PDE as the heat equation.

False. With

A = 1, B = 0, C = - c^{2}

the wave equation is hyperbolic (

B^{2} - 4 A C = 4 c^{2} > 0

); the heat equation

u_{t} = α u_{xx}

has no

u_{tt}

, so

A = 0

and it is parabolic (

B^{2} - 4 A C = 0

). Different type, different behaviour — see Classification of second-order PDEs.

Doubling the tension doubles the wave speed.

False.

c = T / ρ

, so doubling

T

multiplies

c

2 \approx 1.41

, not

2

A rightward pulse

F (x - c t)

keeps its exact shape forever.

True (in this ideal model). The argument

x - c t

just shifts; with no dispersion or damping in the pure wave equation, the profile is rigidly transported.

The horizontal component of tension does work moving the string up and down.

False. Only the vertical component

T sin θ

(with

θ

the string's angle to the horizontal) drives transverse motion; the horizontal components cancel (that's why

T

stays constant along

x

The factorisation

\partial_{t}^{2} - c^{2} \partial_{x}^{2} = (\partial_{t} - c \partial_{x}) (\partial_{t} + c \partial_{x})

is just a formal trick with no justification.

False. Expanding the product term by term,

(\partial_{t} - c \partial_{x}) (\partial_{t} + c \partial_{x}) = \partial_{t}^{2} + c \partial_{t} \partial_{x} - c \partial_{x} \partial_{t} - c^{2} \partial_{x}^{2}

; the two mixed terms cancel (partial derivatives commute,

\partial_{t} \partial_{x} = \partial_{x} \partial_{t}

), leaving exactly

\partial_{t}^{2} - c^{2} \partial_{x}^{2}

Why questions

Why must we assume small slopes (

∣ u_{x} ∣ ≪ 1

) for the derivation to work?

Small slopes let us replace

sin θ

and

tan θ

(with

θ

the angle to the horizontal) by the same

u_{x}

and treat

cos θ \approx 1

, linearising the geometry so tension stays constant and the PDE stays linear.

Why does the equation predict a travelling shape rather than a static one?

Curvature drives each point toward straightness (

u_{tt} \propto u_{xx}

); as a point overshoots, its neighbour is now the most curved, so the "bulge of curvature" — and hence the shape — moves along.

Why is the discriminant test

B^{2} - 4 A C

used to classify this PDE?

With

A, B, C

the coefficients of

u_{tt}, u_{x t}, u_{xx}

, it mirrors the conic-section test: the sign decides whether the characteristic directions are real and distinct. Here

B^{2} - 4 A C = 0 - 4 (1) (- c^{2}) = 4 c^{2} > 0

gives two real characteristic families, the hallmark of wave propagation — see Method of characteristics.

Why do we need two initial conditions (

f = u

and

g = u_{t}

t = 0

) but only for time?

The equation is second order in

t

(an

u_{tt}

), so like

F = ma

it needs both an initial position

f

and an initial velocity

g

to fix the motion; boundary conditions handle the

x

direction separately.

Why does a heavier string (bigger

ρ

) carry waves more slowly?

More mass per length means the same tension force produces less acceleration (

a = F / m

), so disturbances propagate slower — exactly what

c = T / ρ

says.

Why can separation of variables and d'Alembert both solve the same equation?

They're two representations of the same solution space — d'Alembert gives travelling waves

F (x - c t) + G (x + c t)

, separation gives standing-wave sums; a standing wave is just two opposite travelling waves superposed.

Edge cases

What happens to

c

if the tension

T \to 0

(a slack string)?

c = T / ρ \to 0

: with no restoring tension nothing pulls the element back, so no wave can travel — the model degenerates.

If the initial shape is a flat line (

f \equiv 0

, where

f (x) = u (x, 0)

) with zero velocity (

g \equiv 0

, where

g (x) = u_{t} (x, 0)

), what does the string do?

It stays flat forever. d'Alembert gives

u \equiv 0

; with no curvature there is no acceleration, so nothing ever starts.

At a sharp corner (kink) in the initial shape

f

u_{xx}

is undefined — is the wave equation still meaningful?

The strong (pointwise) form breaks at the kink, but d'Alembert's

\frac{1}{2} [f (x - c t) + f (x + c t)]

still transports the corner as a weak/travelling solution; the corner just moves at speed

c

What does the solution look like at the exact instant a rightward and leftward pulse fully overlap?

They superpose linearly (add), possibly cancelling or doubling momentarily, then pass through each other unchanged — a direct consequence of the equation being linear.

In the limit

c \to \infty

(infinitely stiff/light string), how does information travel?

The characteristics

x \pm c t

flatten to instantaneous horizontal lines, so a disturbance would reach all points at once — a physically forbidden idealisation showing why finite

c

matters.

If you set the initial velocity

g (x) = u_{t} (x, 0)

so that

g

integrates to a nonzero constant over all space, is that a problem?

Not for the local wave equation — d'Alembert only ever integrates

g

over the finite interval

[x - c t, x + c t]

, so the value stays finite for any finite

t

For a semi-infinite string fixed at

x = 0

, does a pulse simply vanish when it reaches the wall?

No — it reflects inverted. The fixed end forces

u (0, t) = 0

, which (via an odd image extension) sends back a flipped pulse; energy is conserved, not lost.

Recall One-line self-test

If someone says "curvature = steepness," what's your instant correction? ::: Steepness is $u_{x}$ (slope); curvature is $u_{xx}$ (how the slope changes) — and it's curvature, not steepness, that drives the acceleration.

Question bank — Wave equation (hyperbolic) 1D — derivation

True or false — justify

Spot the error

Why questions

Edge cases