Solving heat equation — separation of variables: Question bank

The machinery being probed: the guess $u = X (x) T (t)$ , the separation constant $- λ$ , the sine eigenfunctions from Sturm-Liouville Theory, the Fourier Series fit, and the Superposition Principle that glues modes together.

True or false — justify

The claim "separation of variables always works for any PDE with any

f (x)

" is true.

False — it needs a linear, homogeneous PDE with homogeneous boundary conditions so that superposition can rebuild any

f

; a nonlinear equation or nonzero (

u (0, t) = 5

) BC breaks the method as stated.

Every solution of

u_{t} = α^{2} u_{xx}

must have the product form

u = X (x) T (t)

False — the building blocks

u_{n} = X_{n} T_{n}

have that form, but the actual solution is a sum

\sum B_{n} X_{n} T_{n}

, which is generally not a single product. Separation finds a basis, not the whole answer.

The separation constant could just as well be written

+ λ

with

λ > 0

and give the same physics.

False —

+ λ

forces

cosh, sinh

in space (can't hit two zero ends) and

e^{+ λ α^{2} t}

in time, which grows without bound. Zero BCs plus physical decay force

- λ

λ > 0

λ = 0

is a valid eigenvalue for the Dirichlet rod.

False —

λ = 0

gives

X = A + B x

; both zero BCs force

A = B = 0

, only the trivial solution. It contributes nothing, so it is excluded.

Higher-frequency modes (

n

large) decay faster than lower ones.

True — the decay rate is

α^{2} (nπ / L)^{2} \propto n^{2}

, so wiggly modes are crushed exponentially quicker; this is why heat smooths sharp features first.

If the ends were insulated (

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

) instead of frozen, the eigenfunctions would still be sines.

False — Neumann (insulated) BCs select cosines

cos (nπ x / L)

and add an

n = 0

constant term; see Boundary Conditions — Dirichlet vs Neumann. Always match the eigenfunction to the BC.

Doubling the diffusivity

α^{2}

changes the shape of the eigenfunctions.

False —

α^{2}

appears only in the time factor

e^{- α^{2} (nπ / L)^{2} t}

, so it changes how fast modes fade, not the spatial sine shapes.

The steady-state (

t \to \infty

) of this problem is a non-trivial temperature profile.

False — with both ends at

0

and no source, every mode

\to 0

, so

u \to 0

everywhere. The rod ends up uniformly cold.

Spot the error

u = X T

, so

u_{t} = X^{'} T^{'}

." — find the mistake.

u_{t}

differentiates in time only:

u_{t} = X T^{'}

(the space factor

X

is a constant with respect to

t

). Similarly

u_{xx} = X^{''} T

. Differentiating both factors at once is the error.

"After dividing,

\frac{T ^{'}}{α ^{2} T} = \frac{X ^{''}}{X}

— since both sides are equal, they're each some function of

x

and

t

." — where's the flaw?

Wrong conclusion: the left depends only on

t

, the right only on

x

; equality for all

x, t

forces both to be the same constant, not a function. That constant is the whole point.

X (0) = 0

because we're told

u (0, t) = 0

, so

X (0) T (t) = 0

, so

T (t) = 0

." — what's wrong?

T \equiv 0

the solution is trivially zero everywhere. To keep a nontrivial solution we instead demand the other factor vanish:

X (0) = 0

sin k L = 0

k L = nπ

, giving

n = 0, 1, 2, \dots

" — is

n = 0

allowed?

No —

n = 0

gives

X = sin 0 = 0

, the trivial eigenfunction. We start at

n = 1

. (Negative

n

just repeat the same sines up to sign, so they add nothing new.)

"Fit

f (x)

with a full Fourier series

\frac{a _{0}}{2} + \sum (a_{n} cos + b_{n} sin)

." — why is that wrong here?

Cosines (and the constant) are nonzero at

x = 0

, violating

u (0, t) = 0

. Dirichlet zero BCs select a pure sine series only.

B_{n} = \int_{0}^{L} f (x) sin \frac{nπ x}{L} d x

." — what's missing?

The normalizing factor

\frac{2}{L}

. Since

\int_{0}^{L} sin^{2} \frac{nπ x}{L} d x = \frac{L}{2}

, orthogonality gives

B_{n} = \frac{2}{L} \int_{0}^{L} f sin d x

. Forgetting it scales every coefficient wrong.

"Since each

u_{n}

solves the PDE, their sum solves it only if we're lucky." — correct the reasoning.

Not luck — the equation is linear and homogeneous, so by the Superposition Principle any (even infinite) sum of solutions is automatically a solution. That's why we're allowed to superpose.

Why questions

Why do we guess

u = X T

instead of deriving it?

Because it's an ansatz (trial form): if it produces solutions we lose nothing, and if no product solution existed we'd discover that when the algebra failed. It's a licensed gamble that pays off for linear PDEs.

Why does splitting one hard PDE into two ODEs count as progress?

ODEs in a single variable are solvable with standard techniques (characteristic roots, integrating factors); the PDE mixed

x

and

t

derivatives inseparably. Separation trades one hard problem for two easy ones.

Why must the space problem be an eigenvalue problem rather than a plain ODE?

The zero BCs at both ends over-constrain

X^{''} + λ X = 0

; only special

λ

values (the eigenvalues) admit nonzero solutions. This is exactly the setup of Sturm-Liouville Theory.

Why does the initial condition get handled last?

Because it's the only non-homogeneous condition. We first build all solutions satisfying the PDE and zero BCs, then choose the coefficients

B_{n}

to match

f (x)

— the last free knobs.

Why does the long-time profile always approach a half-sine bump?

Every mode

n \geq 2

decays

n^{2}

times faster than

n = 1

, so after enough time only the

n = 1

term

sin \frac{π x}{L}

survives meaningfully — a universal shape independent of the starting

f

Why does a flat initial profile (

f =

const) excite only odd modes?

A constant is symmetric about the rod's midpoint; the even sines

sin \frac{2 π x}{L}, sin \frac{4 π x}{L}, \dots

are antisymmetric there, so their projection integrals vanish. Only odd sines share the symmetry.

Why is the same

sin \frac{nπ x}{L}

used both as eigenfunction and as the Fourier Series basis?

Because the eigenfunctions of a self-adjoint boundary problem are orthogonal and complete — they double as the basis in which we expand any

f

. Sturm-Liouville guarantees this.

Edge cases

What happens if

f (x)

is already a single sine, say

sin \frac{3 π x}{L}

No integral needed — read off

B_{3} = 1

, all others

0

, and attach its decay:

u = sin \frac{3 π x}{L} e^{- 9 α^{2} π^{2} t / L^{2}}

. A single mode evolves alone.

What if

f (x)

is discontinuous or has a corner (e.g. a step)?

The sine series still converges (in the mean) and the solution is valid for

t > 0

; the diffusion instantly smooths the discontinuity because high-

n

modes are damped, so

u

becomes smooth immediately after

t = 0

What is

u (x, 0)

predicted by the series exactly at the endpoints if

f (0) \neq = 0

The sine series is pinned to

0

x = 0, L

, so it can only converge to

0

there regardless of

f

; if

f (0) \neq = 0

you get a jump (Gibbs) at the ends — a sign the IC is incompatible with the BCs.

What if the initial condition is

f (x) = 0

Every

B_{n} = 0

, so

u \equiv 0

for all time — the trivial solution, and correctly so: a rod starting cold with cold ends never heats.

α^{2} \to 0

(a perfect insulator), what does the solution do?

The time factors

e^{- α^{2} (nπ / L)^{2} t} \to 1

, so

u \to f (x)

frozen in time — no heat conducts, the profile never changes. The math matches the physics.

What if

L \to \infty

(very long rod)?

The eigenvalue spacing

\frac{nπ}{L} \to 0

, the discrete sum becomes an integral, and the sine series turns into a sine transform — the same idea, continuum version, for an unbounded rod.

Recall One-line self-test

Ends frozen at 0, no source, wiggly start — after a long time the rod looks like... ::: a single half-sine hump $sin \frac{π x}{L}$ , fading to cold.

Related equations that reuse this exact machinery: Wave Equation (second time derivative → oscillating not decaying time factor) and Laplace Equation (steady state, no time at all).