Solving heat equation — separation of variables: Worked examples

This page is a shooting gallery. We list every kind of situation the linear-homogeneous heat problem can throw at you, then knock down each one with a full worked example. Keep the parent recipe handy: Solving heat equation — separation of variables.

Recall A word about "every scenario" (scope, honestly stated)

This page exhausts the linear homogeneous rod: all homogeneous boundary types — Dirichlet ( $u = 0$ ), Neumann ( $u_{x} = 0$ ), and a note on mixed / Robin ends — plus every input shape (single/multiple/constant/piecewise/zero) and both time limits. What we do not treat here (they need extra machinery and live on other pages): non-homogeneous BCs (ends held at nonzero or time-dependent temperatures — handled by subtracting a steady state), source terms $u_{t} = α^{2} u_{xx} + Q$ , and full Robin eigenvalue transcendental equations. So "every scenario" means every scenario of the homogeneous separable problem, which is what this chapter builds. The eigenfunction/eigenvalue machinery behind all BC types is Sturm-Liouville Theory.

The scenario matrix

Every worked example below is tagged with the cell it covers.

Cell	What makes it special	Example
A. Single mode	$f (x)$ is already one sine — read off, no integral	Ex 1
B. Sum of modes	$f (x)$ is a few sines added — read off several	Ex 2
C. Constant profile	$f (x) =$ const — must integrate, only odd $n$ survive	Ex 3
D. Triangle / tent	piecewise-linear peak — integration by parts	Ex 4
E. Degenerate input	$f (x) \equiv 0$ — the trivial (dead) rod, uniqueness	Ex 5
F. Limiting behaviour	$t \to \infty$ and $t \to 0^{+}$ — convergence subtleties	Ex 6
G. Neumann + mixed twist	insulated $u_{x} = 0$ (cosines) and one mixed end	Ex 7
H. Word problem	real rod, real numbers, "how long to cool?"	Ex 8

Cases A–H hit: single/multiple/constant/piecewise inputs, the zero-degenerate input, both time limits (with convergence discussed), the Neumann and mixed boundary families, and a physical estimate.

Ex 1 — Cell A: a single sine mode

Worked example Statement

Solve the rod problem on $0 < x < L$ with $u (0, t) = u (L, t) = 0$ and initial shape $f (x) = 5 sin \frac{2 π x}{L} .$

Forecast: Before reading on — do you think you need to compute an integral here? Guess which $B_{n}$ are nonzero.

Step 1 — Match to the template. The general series at $t = 0$ is $f (x) = \sum_{n} B_{n} sin \frac{nπ x}{L}$ . Why this step? Because the $sin \frac{nπ x}{L}$ are orthogonal (each independent), a sine on the left can only match the same sine on the right. If $f$ is already one of them, no integral is needed — we just read coefficients.

Step 2 — Read off. Comparing $5 sin \frac{2 π x}{L}$ with the $n = 2$ term: $B_{2} = 5$ , and every other $B_{n} = 0$ .

Step 3 — Attach the decay. Mode $n = 2$ decays like $e^{- α^{2} (2 π / L)^{2} t} = e^{- 4 α^{2} π^{2} t / L^{2}}$ . Why the 4? The decay rate is $\propto n^{2} = 2^{2} = 4$ . $u (x, t) = 5 sin \frac{2 π x}{L} e^{- 4 α^{2} π^{2} t / L^{2}} .$

Verify: At $t = 0$ , $e^{0} = 1$ , so $u (x, 0) = 5 sin \frac{2 π x}{L} = f (x)$ . ✓ Ends: $sin 0 = 0$ and $sin 2 π = 0$ , so $u (0, t) = u (L, t) = 0$ . ✓

Ex 2 — Cell B: a sum of modes

Worked example Statement

Same rod, initial shape a combination: $f (x) = 3 sin \frac{π x}{L} - 2 sin \frac{4 π x}{L} .$

Forecast: How many terms will the final sum have? Which fades faster, the $3$ -term or the $2$ -term?

Step 1 — Read off each piece. Why this step? Same orthogonality reasoning as Ex 1 — a linear combination of sines maps term-by-term. $B_{1} = 3$ , $B_{4} = - 2$ , all others $0$ .

Step 2 — Give each its own timer. $n = 1$ decays at rate $α^{2} (π / L)^{2}$ ; $n = 4$ decays at rate $16 α^{2} (π / L)^{2}$ (since $4^{2} = 16$ ). $u (x, t) = 3 sin \frac{π x}{L} e^{- α^{2} π^{2} t / L^{2}} - 2 sin \frac{4 π x}{L} e^{- 16 α^{2} π^{2} t / L^{2}} .$

Step 3 — Observe the split fates. The $n = 4$ hump decays $16/1 = 16$ times faster than the $n = 1$ hump. By the Superposition Principle we simply add the independently-fading pieces.

Verify: $t = 0$ gives back $3 sin \frac{π x}{L} - 2 sin \frac{4 π x}{L}$ . ✓ The ratio of decay rates is $16$ . ✓

Ex 3 — Cell C: a constant (flat hot) rod

Worked example Statement

$L = π$ , $α^{2} = 1 m^{2} / s$ , and the whole rod starts at $u (x, 0) = 100$ °, ends dunked to $0$ °.

Forecast: A flat line is nothing like a sine — do you expect all $B_{n}$ , only odd, or only even?

Step 1 — Set up the projection integral. Why this step? A constant is not a single eigenfunction, so we must "project" it onto each sine using the coefficient formula. $B_{n} = \frac{2}{π} \int_{0}^{π} 100 sin (n x) d x .$

Step 2 — Do the integral. $\int_{0}^{π} sin (n x) d x = [- \frac{c o s n x}{n}]_{0}^{π} = \frac{1 - c o s nπ}{n} = \frac{1 - ( - 1 ) ^{n}}{n} .$ Why $(- 1)^{n}$ ? Because $cos nπ$ alternates: $- 1, + 1, - 1, \dots$ So the bracket is $0$ for even $n$ and $2$ for odd $n$ .

Step 3 — Simplify. $B_{n} = \frac{200}{π} \cdot \frac{1 - ( - 1 ) ^{n}}{n} = ⎩ ⎨ ⎧ \frac{400}{π n} 0 n odd n even .$ Why only odd survive? A flat, left–right symmetric hump has the symmetry of odd sines; even sines are antisymmetric about the midpoint and cancel.

Step 4 — Assemble. $u (x, t) = \frac{400}{π} n odd \sum \frac{1}{n} sin (n x) e^{- n^{2} t} .$

Verify: At $t = 0$ , $x = π /2$ : the terms are $\frac{400}{π} \cdot \frac{s i n ( nπ /2 )}{n}$ . Now $sin (nπ /2) = + 1, - 1, + 1, \dots$ for $n = 1, 3, 5, \dots$ , i.e. $(- 1)^{(n - 1) /2}$ , so the sum is $\frac{400}{π} (1 - \frac{1}{3} + \frac{1}{5} - \dots)$ . That alternating sum is the Leibniz / Gregory series for $arctan 1 = π /4$ (a standard Fourier/Taylor result: put $x = 1$ in $arctan x = x - \frac{x ^{3}}{3} + \frac{x ^{5}}{5} - \dots$ ). Hence the total is $\frac{400}{π} \cdot \frac{π}{4} = 100.$ ✓ This is the Fourier series of the constant $100$ evaluated at the midpoint — where it converges to the function value.

Ex 4 — Cell D: a triangular (tent) initial profile

Worked example Statement

$L = 1$ . The rod starts as a symmetric tent peaking at the middle: $f (x) = {x 1 - x 0 \leq x \leq \frac{1}{2} \frac{1}{2} \leq x \leq 1.$ Find $B_{n}$ (with $L = 1$ , so $sin nπ x$ ).

Figure s01 (alt-text/caption): the orange curve is the tent

f

; the plum dashed vertical line at

x = \frac{1}{2}

is the axis of symmetry; the teal segment along the bottom marks the two ends pinned at

0

. The picture's message:

f

is symmetric about

x = \frac{1}{2}

, so only symmetric (odd-

n

) sine humps can build it — even humps, which are antisymmetric about the midpoint, must have zero weight.

Forecast: Look at the plum dashed line. Will even modes appear?

Step 1 — Split the coefficient integral at the peak. Why this step? $f$ is defined by two different formulas either side of $x = \frac{1}{2}$ , so the single integral becomes two. $B_{n} = 2 \int_{0}^{1} f (x) sin (nπ x) d x = 2 [I_{1} \int_{0}^{1/2} x sin (nπ x) d x + I_{2} \int_{1/2}^{1} (1 - x) sin (nπ x) d x] .$ (Here $\frac{2}{L} = 2$ since $L = 1$ .)

Step 2 — Do the integration by parts explicitly on $I_{1}$ . Why integration by parts? The integrand is (polynomial) $\times$ (sine); parts lowers the polynomial degree until only sines/cosines remain. Let $u = x, d v = sin (nπ x) d x$ , so $d u = d x, v = - \frac{cos ( nπ x )}{nπ}$ . Then $I_{1} = [- \frac{x c o s ( nπ x )}{nπ}]_{0}^{1/2} + \frac{1}{nπ} \int_{0}^{1/2} cos (nπ x) d x .$ The remaining integral is $\frac{1}{nπ} [\frac{sin ( nπ x )}{nπ}]_{0}^{1/2} = \frac{sin ( nπ /2 )}{( nπ ) ^{2}}$ . Evaluating the bracket at $x = \frac{1}{2}$ (lower limit $x = 0$ gives $0$ ): $I_{1} = - \frac{\frac{1}{2} c o s ( nπ /2 )}{nπ} + \frac{s i n ( nπ /2 )}{( nπ ) ^{2}} .$

Step 3 — Do $I_{2}$ and watch the boundary terms cancel. Write $I_{2} = \int_{1/2}^{1} sin (nπ x) d x - \int_{1/2}^{1} x sin (nπ x) d x$ . The first piece is $[- \frac{c o s ( nπ x )}{nπ}]_{1/2}^{1} = \frac{cos ( nπ /2 ) - cos ( nπ )}{nπ}$ . The second, by the same parts as Step 2, $\int_{1/2}^{1} x sin (nπ x) d x = [- \frac{x c o s ( nπ x )}{nπ}]_{1/2}^{1} + \frac{s i n ( nπ x )}{( nπ ) ^{2}}_{1/2}^{1} = - \frac{c o s ( nπ )}{nπ} + \frac{\frac{1}{2} c o s ( nπ /2 )}{nπ} - \frac{s i n ( nπ /2 )}{( nπ ) ^{2}} .$ Now add $I_{1} + I_{2}$ . Collect the $\frac{\frac{1}{2} cos ( nπ /2 )}{nπ}$ terms: $I_{1}$ contributes $- \frac{\frac{1}{2} cos ( nπ /2 )}{nπ}$ ; from $I_{2}$ the $cos (nπ /2) / nπ$ from the first piece and the $- \frac{1}{2} cos (nπ /2) / nπ$ from the subtracted second piece combine to $+ \frac{\frac{1}{2} cos ( nπ /2 )}{nπ}$ . These two cancel exactly — that is the "boundary terms at $x = \frac{1}{2}$ cancel by symmetry." The $cos (nπ) / nπ$ terms also cancel. What survives is only the sine terms: $I_{1} + I_{2} = \frac{s i n ( nπ /2 )}{( nπ ) ^{2}} + \frac{s i n ( nπ /2 )}{( nπ ) ^{2}} = \frac{2 s i n ( nπ /2 )}{( nπ ) ^{2}} .$ Therefore $B_{n} = 2 (I_{1} + I_{2}) = \frac{4}{( nπ ) ^{2}} sin \frac{nπ}{2} .$ Why $sin \frac{nπ}{2}$ ? It equals $0$ for even $n$ and $\pm 1$ for odd $n$ — the tent, being symmetric, keeps only odd modes, exactly as the figure predicted.

Step 4 — Assemble. $u (x, t) = \sum_{n odd} \frac{4}{( nπ ) ^{2}} sin \frac{nπ}{2} sin (nπ x) e^{- α^{2} n^{2} π^{2} t} .$

Verify: First mode $B_{1} = \frac{4}{π ^{2}} sin \frac{π}{2} = \frac{4}{π ^{2}} \approx 0.4053$ . Check the series recovers the peak: at $x = \frac{1}{2}$ , $f = \frac{1}{2}$ , and $\sum_{n odd} \frac{4}{( nπ ) ^{2}} sin^{2} \frac{nπ}{2} = \frac{4}{π ^{2}} \sum_{n odd} \frac{1}{n ^{2}} = \frac{4}{π ^{2}} \cdot \frac{π ^{2}}{8} = \frac{1}{2}$ . ✓ (The identity $\sum_{n odd} \frac{1}{n ^{2}} = \frac{π ^{2}}{8}$ is a standard Fourier/Basel result.)

Ex 5 — Cell E: the degenerate / zero input

Worked example Statement

$f (x) \equiv 0$ : rod starts everywhere at $0$ °, ends held at $0$ °.

Forecast: Guess the temperature at $t = 5$ seconds, or any time at all.

Step 1 — Compute all coefficients. $B_{n} = \frac{2}{L} \int_{0}^{L} 0 \cdot sin \frac{nπ x}{L} d x = 0 for every n .$ Why bother? To show the machinery is consistent even in the boring case — no hidden nonzero mode sneaks in.

Step 2 — Assemble. $u (x, t) = 0 for all x, t .$ This is the trivial solution.

Step 3 — Why it is the only solution (uniqueness, stated properly). Why can't some other solution exist? Two independent guarantees: (i) Maximum principle. For the heat equation with no source, the maximum and minimum of $u$ over $[0, L] \times [0, T]$ must occur on the "parabolic boundary" — either at $t = 0$ or at the two ends. Here $u = 0$ on all of those, so $0 \leq u \leq 0$ everywhere: $u \equiv 0$ . Applied to the difference of any two solutions with identical data, it forces them equal — hence uniqueness for the whole Dirichlet problem, not just this case. (ii) Sturm–Liouville / energy view. The eigenfunctions $sin \frac{nπ x}{L}$ form a complete orthogonal basis (Sturm-Liouville Theory); a function whose every coefficient is $0$ is the zero function. See also the general uniqueness under Boundary Conditions — Dirichlet vs Neumann.

Verify: $u_{t} = 0$ , $u_{xx} = 0$ , so $u_{t} = α^{2} u_{xx}$ holds ( $0 = 0$ ). BCs and IC all satisfied. ✓

Ex 6 — Cell F: limiting behaviour and convergence

Worked example Statement

Take the tent solution from Ex 4 ( $L = 1$ , $α^{2} = 1$ ). Describe the profile (a) for very large $t$ , and (b) as $t \to 0^{+}$ — and say honestly how the series converges in each case.

Figure s02 (alt-text/caption): orange = the sharp tent recovered as

t \to 0^{+}

; plum dash-dot = an intermediate time where the corner has already rounded; teal = the lone surviving half-sine bump at large

t

. Message: high modes carry the sharp corner and die first, so the profile smooths from the top down before collapsing to a single hump.

Forecast: Which single shape does every Dirichlet rod collapse to before dying?

Step 1 — Large- $t$ limit (and why the limit is legal). Mode $n$ carries $e^{- n^{2} π^{2} t}$ . The ratio of the $n = 3$ term to the $n = 1$ term is $e^{- (9 - 1) π^{2} t} = e^{- 8 π^{2} t} \to 0$ . Convergence note: for any fixed $t_{0} > 0$ the coefficients times $e^{- n^{2} π^{2} t}$ decay faster than any power of $n$ , so the series and all its $x$ - and $t$ -derivatives converge uniformly on $[0, 1] \times [t_{0}, \infty)$ (Weierstrass $M$ -test with $M_{n} = ∣ B_{n} ∣ e^{- n^{2} π^{2} t_{0}}$ ). Uniform convergence is exactly what licenses term-by-term limiting and differentiation — so keeping only $n = 1$ is rigorous: $u (x, t) \approx B_{1} sin (π x) e^{- π^{2} t} = \frac{4}{π ^{2}} sin (π x) e^{- π^{2} t} .$ Universal shape: a single half-sine bump (teal curve), whatever the start looked like.

Step 2 — Small- $t$ limit (the delicate end). As $t \to 0^{+}$ every $e^{- n^{2} π^{2} t} \to 1$ , so formally $u (x, 0^{+}) = \sum B_{n} sin (nπ x) = f (x)$ . Convergence note: at $t = 0$ the smoothing exponentials are gone. Here $B_{n} \sim 1/ n^{2}$ , which is absolutely summable, so the tent's series still converges uniformly to the continuous tent (the tent has no jumps). Contrast the constant- $100$ profile of Ex 3: there $B_{n} \sim 1/ n$ , the series converges only pointwise (not uniformly) inside $(0, π)$ and at the ends the sum is $0 \neq = 100$ — the endpoints and any jump converge to the midpoint value (Gibbs behaviour). So " $u (x, 0) = f (x)$ " holds in the interior but must be read carefully at discontinuities and boundaries.

Step 3 — The smoothing story. For any $t > 0$ the solution is infinitely smooth (the exponentials kill all roughness): high- $n$ wiggles die first, so the tent's corner rounds off instantly, then the whole bump fades. This instant smoothing is why heat flow erases sharp features.

Verify: Amplitude ratio $n = 3$ vs $n = 1$ at $t = 0.1$ is $\frac{B _{3}}{B _{1}} e^{- 8 π^{2} (0.1)} = - \frac{1}{9} e^{- 0.8 π^{2}} \approx - 4.15 \times 1 0^{- 5}$ , negligible — the single-mode approximation is excellent. ✓

Ex 7 — Cell G: the Neumann and mixed twists (cosines, and one of each end)

Worked example Statement

(a) Neumann: same rod, but ends insulated: $u_{x} (0, t) = u_{x} (L, t) = 0$ , initial $f (x)$ . Find the building blocks. (b) Mixed: ends $u (0, t) = 0$ (fixed) and $u_{x} (L, t) = 0$ (insulated). What eigenfunctions now?

Forecast: With sines forbidden in (a), what functions have zero slope at both ends? In (b), which quarter-waves fit?

Step 1 — Redo the space ODE with Neumann BCs. Why redo it? Eigenfunctions are chosen to satisfy the BCs; different BCs ⇒ different eigenfunctions. See Boundary Conditions — Dirichlet vs Neumann. $X^{''} + λ X = 0, X^{'} (0) = 0, X^{'} (L) = 0.$ For $λ = k^{2} > 0$ : $X = A cos k x + B sin k x$ , $X^{'} = - A k sin k x + B k cos k x$ . $X^{'} (0) = B k = 0 \Rightarrow B = 0$ ; $X^{'} (L) = - A k sin k L = 0 \Rightarrow sin k L = 0 \Rightarrow k L = nπ$ . Why cosines now? Zero-slope ends want a curve flat at both ends — $cos \frac{nπ x}{L}$ has horizontal tangents there. Also $λ = 0$ now works ( $X_{0} = 1$ , slope zero everywhere) — a mode Dirichlet rejected! $X_{n} (x) = cos \frac{nπ x}{L}, n = 0, 1, 2, \dots$

Step 2 — Assemble the Neumann solution. $u (x, t) = \frac{A _{0}}{2} + n = 1 \sum \infty A_{n} cos \frac{nπ x}{L} e^{- α^{2} (nπ / L)^{2} t}, A_{n} = \frac{2}{L} \int_{0}^{L} f (x) cos \frac{nπ x}{L} d x .$ This is a Fourier cosine series (Fourier Series). The $n = 0$ term $\frac{A _{0}}{2} = \frac{1}{L} \int_{0}^{L} f d x$ has decay rate $0$ — it never fades. Why one mode survives forever? Insulated ends let no heat escape, so total heat is conserved; the rod settles to its average temperature, not to zero. Contrast Ex 6, where Dirichlet ends drained all heat to $0$ .

Step 3 — The mixed case (one fixed, one insulated). $X (0) = 0 \Rightarrow A = 0$ , so $X = B sin k x$ ; then $X^{'} (L) = B k cos k L = 0 \Rightarrow cos k L = 0 \Rightarrow k L = (n - \frac{1}{2}) π$ . $X_{n} (x) = sin \frac{( 2 n - 1 ) π x}{2 L}, n = 1, 2, \dots$ Why "quarter-waves"? $sin$ starts at $0$ (fixed end) and must reach a flat top (zero slope) at $x = L$ — that is a quarter, three-quarters, … of a full sine wave. Eigenvalues $λ_{n} = (\frac{( 2 n - 1 ) π}{2 L})^{2}$ . This still fits the Sturm-Liouville Theory framework; only the boundary equation changed.

Verify: Neumann with constant $f = 100$ , $L = π$ : $\frac{A _{0}}{2} = \frac{1}{π} \int_{0}^{π} 100 d x = 100$ , and $A_{n} = \frac{2}{π} \int_{0}^{π} 100 cos (n x) d x = 0$ for $n \geq 1$ . Steady state $= 100$ (the start average). ✓ Mixed check: smallest eigenvalue $k_{1} = \frac{π}{2 L}$ gives $cos (k_{1} L) = cos \frac{π}{2} = 0$ . ✓

Ex 8 — Cell H: a real-world word problem

Worked example Statement

A metal rod $L = π$ metres, thermal diffusivity $α^{2} = 1 m^{2} / s$ , starts at the single mode $u (x, 0) = 80 sin x$ °C, ends held at $0$ °C. How long until the midpoint temperature drops to $10$ °C?

Forecast: Estimate — a few tenths of a second, or several seconds?

Step 1 — Write the solution. Why this step? $f = 80 sin x$ is a pure $n = 1$ mode ( $L = π \Rightarrow sin \frac{nπ x}{π} = sin n x$ ), so $B_{1} = 80$ , decay rate $α^{2} (1)^{2} = 1$ . $u (x, t) = 80 sin x e^{- t} .$

Step 2 — Evaluate at the midpoint $x = π /2$ . $sin \frac{π}{2} = 1$ , so $u (π /2, t) = 80 e^{- t}$ . Why the midpoint? $sin x$ peaks there, so it's the hottest, longest-lasting point — the "worst case" for cooling.

Step 3 — Solve for the time. Set $80 e^{- t} = 10 \Rightarrow e^{- t} = \frac{1}{8} \Rightarrow t = ln 8 \approx 2.079 s .$ Why $ln$ ? Exponential decay is undone by the logarithm — the natural tool for "how long until it reaches value $V$ ."

$t = ln 8 \approx 2.08 seconds .$

Verify (units + value): rate $α^{2} (nπ / L)^{2} = 1 \cdot (1/1)^{2} = 1 s^{- 1}$ , so exponent $- t$ is dimensionless. ✓ Plug back: $80 e^{- l n 8} = 80/8 = 10$ °C. ✓

Recall Quick self-test on the matrix

A single-sine $f$ needs an integral for $B_{n}$ ? ::: No — read it off (Cell A). Constant $f$ keeps which modes? ::: Only odd $n$ (Cell C). As $t \to \infty$ a Dirichlet rod approaches what shape? ::: A half-sine bump, then $0$ (Cell F). Why is keeping only $n = 1$ at large $t$ rigorous? ::: For fixed $t_{0} > 0$ the series converges uniformly (Weierstrass $M$ -test), licensing term-by-term limits. Insulated (Neumann) ends give which eigenfunctions? ::: Cosines, incl. a constant $n = 0$ mode that never fades — rod settles to its average (Cell G). One fixed + one insulated end gives? ::: Quarter-wave sines $sin \frac{( 2 n - 1 ) π x}{2 L}$ . The zero initial rod evolves to? ::: Stays $0$ forever — trivial and unique by the maximum principle (Cell E).