Heat equation (parabolic) 1D — derivation from Fourier's law: Question bank

True or false — justify

A point where the temperature profile is concave up (

u_{xx} > 0

) is currently cooling.

False. Concave up is a valley; its neighbours are hotter, so heat flows in and

u_{t} = α u_{xx} > 0

— it warms. Hills (

u_{xx} < 0

) cool.

The heat equation is symmetric under reversing time (

t \to - t

), like Newton's laws.

False. Replacing

t \to - t

turns

u_{t} = α u_{xx}

into

- u_{t} = α u_{xx}

, a different (backward, unstable) equation. Heat spreads but never spontaneously un-spreads — that broken symmetry is the arrow of time.

If the temperature profile is a straight line everywhere, no point changes temperature.

True. A straight line has

u_{xx} = 0

at every point, so

u_{t} = α \cdot 0 = 0

. The linear profile is the 1D steady state.

Doubling the thermal conductivity

k

always exactly doubles how fast the rod equilibrates.

Not in general. The time-scale is set by

α = k / (ρ c)

; doubling

k

doubles

α

only if

ρ c

is held fixed. A material with large

ρ c

can have modest

α

despite large

k

The flux

q = - k u_{x}

can be nonzero even where the temperature

u

itself is zero.

True. Flux depends on the slope

u_{x}

, not the value

u

. A point at

0 K

sitting on a steep gradient still passes heat through it.

A disturbance at one end of the rod takes some finite time to be felt at the far end.

False (mathematically). The heat equation is parabolic with infinite signal speed: any change is instantly felt everywhere, though its magnitude far away is exponentially tiny. (The wave equation is the finite-speed one.)

At a local maximum of temperature, the heat equation forces the temperature there to be non-increasing.

True. A local max has

u_{xx} \leq 0

, so

u_{t} = α u_{xx} \leq 0

. This is the seed of the maximum principle: interior hot spots can only fade.

The equation

ρ c u_{t} = k u_{xx} + f

with

f > 0

can make even a straight-line profile heat up.

True. With

u_{xx} = 0

the source alone gives

u_{t} = f / (ρ c) > 0

. A heat source warms a flat profile that would otherwise be steady.

Spot the error

"Fourier's law is

q = + k u_{x}

because more gradient means more flux."

The sign is wrong. Heat flows from hot to cold, i.e. down the gradient, so

q = - k u_{x}

. The magnitude does grow with the gradient — but the direction is opposite to

u_{x}

"Since energy is conserved and reversible, the heat equation must be

u_{tt} = α u_{xx}

, second order in time."

Energy conservation gives

ρ c u_{t} = - q_{x}

, which is first order in

t

— storage

ρ c u

carries one time derivative, Fourier's flux carries none. Only

u_{tt}

would need a second physical law (like Newton's for waves). Diffusion is not reversible.

"The diffusivity is just the conductivity,

α = k

Missing the storage term.

α = k / (ρ c)

has units

m^{2} / s

, whereas

k

has units

W/m\cdotK

— they aren't even dimensionally the same, so

α = k

is impossible.

"In Step 5 the right-hand side becomes

+ \partial q / \partial x

It becomes $- \partial q / \partial x$ . Net-in

= q (x) - q (x + Δ x)

, so dividing by

Δ x

gives the negative of the difference quotient, i.e.

- q_{x}

. Dropping that minus flips the sign of the whole PDE.

u_{xx}

measures the slope of the temperature, so

u_{xx} = 0

means the temperature is constant."

u_{xx}

is the curvature, not the slope;

u_{x}

is the slope.

u_{xx} = 0

means the profile is straight (constant slope), which can still rise across the rod — it need not be constant.

"Because

q (x, t) - q (x + Δ x, t)

is the net heat in, and heat leaving is bad, we should write net-in

= q (x + Δ x) - q (x)

No. Both fluxes point in

+ x

. Energy enters the left face at rate

q (x)

and leaves the right face at rate

q (x + Δ x)

, so net gain

= q (x) - q (x + Δ x)

. The proposed version reverses in and out.

"A steady state means

u = 0

everywhere."

Steady state means

u_{t} = 0

, i.e.

u_{xx} = 0

, giving a linear profile

u = a + b x

. The temperature can be large and varying in space; it just doesn't change in time.

Why questions

Why does a single

α

multiply

u_{xx}

instead of separate constants

k

ρ

c

Dividing the balance

ρ c u_{t} = k u_{xx}

ρ c

collapses all three material constants into the single ratio

α = k / (ρ c)

, the only combination that sets the diffusion time-scale.

Why must Fourier's law carry a minus sign for the final PDE to be physically stable?

The minus turns

- \partial_{x} (- k u_{x})

into

+ k u_{xx}

, giving

u_{t} = + α u_{xx}

— smoothing and stable. A missing minus gives

u_{t} = - α u_{xx}

, the backward heat equation, which amplifies tiny wiggles explosively.

Why can we differentiate under the integral in Step 2 of the parent derivation?

The integration limits

x

and

x + Δ x

are fixed in space; only the integrand depends on

t

, so

d / d t

passes through the integral and hits only

u

, giving

\int ρ c u_{t} A d s

Why does the curvature (not the slope) decide whether a point warms or cools?

What matters is whether a point is hotter or colder than the average of its neighbours. Curvature

u_{xx}

measures exactly that gap; slope

u_{x}

only says which way temperature tilts, which by itself moves no net energy into the point.

Why is the heat equation classified as parabolic and not hyperbolic?

Written as

0 \cdot u_{tt} + 0 \cdot u_{x t} - α u_{xx} + u_{t} = 0

, the second-order part has

B^{2} - A C = 0

(since

A = 0

B = 0

). Zero discriminant is the parabolic case — diffusive, one real characteristic direction.

Why does a solution like

e^{- α π^{2} t} sin (π x)

decay rather than grow?

The mode

sin (π x)

has curvature

- π^{2} sin (π x)

, so its peaks are hills that cool and its troughs are valleys that warm — diffusion flattens it, and the math packages this into the shrinking factor

e^{- α π^{2} t}

Edge cases

What is

u_{t}

everywhere if the initial temperature is uniform (constant) along the whole rod?

Both

u_{x} = 0

and

u_{xx} = 0

, so

u_{t} = 0

: a uniform rod is already in equilibrium and stays there (with no source).

What happens to the flux

q

at a smooth interior peak of the temperature profile?

At the peak

u_{x} = 0

, so

q = - k u_{x} = 0

at that single point — but

q

is nonzero on either side, flowing away from the peak, which is why the peak still cools.

α = 0

(a perfect insulator,

k = 0

), what does the equation say?

u_{t} = 0

everywhere: with no conductivity, no heat moves, so the temperature is frozen at its initial profile forever, however lumpy.

For the very first instant of a sharp square-wave initial temperature, is the corner point's behaviour well-defined?

No — a corner has undefined (infinite) curvature

u_{xx}

, so

u_{t}

isn't finite there at

t = 0

. The equation instantly smooths the corner; for any

t > 0

the profile is smooth and everything is well-defined.

At an insulated boundary (no heat crosses), what condition does Fourier's law force on the profile?

Zero flux means

q = - k u_{x} = 0

, so

u_{x} = 0

at that end — the temperature profile meets an insulated wall with zero slope (flat against the wall).

If the source

f

is negative (a heat sink) on a flat profile, which way does temperature go?

ρ c u_{t} = k \cdot 0 + f = f < 0

, so

u_{t} < 0

: even a perfectly flat, otherwise-steady profile cools where energy is being drained out.

Connections

Heat equation (parabolic) 1D — derivation from Fourier's law
Fourier's Law of Conduction
Conservation Laws and Continuity Equation
Classification of PDEs (elliptic, parabolic, hyperbolic)
Wave Equation (hyperbolic) 1D
Laplace Equation (steady-state heat)
Thermal Diffusivity and Material Properties
Separation of Variables for the Heat Equation
Fourier Series Solutions