Worked examples — Heat equation (parabolic) 1D — derivation from Fourier's law

Before we start, three symbols in one place so nothing is used before it is meant:

See the parent derivation if any of this feels new, and Thermal Diffusivity and Material Properties for where $\alpha$ comes from.

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The scenario matrix

Every heat-equation question you'll meet is one (or a blend) of these cells. The examples afterward are tagged with the Cell they exercise.

Cell	Case class	What makes it different	Example
A	Curvature negative (a hill / hot bump)	$u_{xx}<0\Rightarrow u_t<0$: point cools	Ex 1
B	Curvature positive (a valley / cold dip)	$u_{xx}>0\Rightarrow u_t>0$: point warms	Ex 2
C	Curvature zero (straight-line / degenerate)	$u_{xx}=0\Rightarrow u_t=0$: steady state	Ex 3
D	Verify a candidate solution (exponential decay)	Plug into both sides, must match	Ex 4
E	Sign / stability twist (backward heat eq)	wrong sign $\Rightarrow$ blow-up	Ex 5
F	Limiting behaviour as $t\to \infty$ and $\alpha$ scaling	who wins the race to equilibrium	Ex 6
G	Real-world word problem with units	must produce numbers + correct SI	Ex 7
H	Source term $f\ne 0$ (non-homogeneous)	extra $+f$ shifts the balance	Ex 8

Cells A, B, C are the three signs of curvature (all cases covered). D–H are the standard twists. Together they fill the table.

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Example 1 — Cell A: a hot bump cools

Forecast: At $x=0$, $\cos 0=1$, so $u=28$ — this is the top of a bump (a hill). Guess the sign of $u_t$ before reading on. Hills lose heat to lower neighbours…

Step 1 — Compute the curvature. $u_{xx}=\frac{d^2}{d{x}^2}[20+8\cos \pi x]=-8{\pi }^2\cos \pi x.$ Why this step? The equation only cares about $u_{xx}$; the constant $20$ has zero curvature, so it drops out. See the shape in the figure below.

Step 2 — Evaluate at $x=0$. $u_{xx}{|}_{x=0}=-8{\pi }^2(1)=-8{\pi }^2.$ Why this step? At $x=0$ we sit on the peak (red dot in figure); $\cos 0=1$. Negative curvature confirms a hill (∩).

Step 3 — Apply the equation. $u_t=\alpha u_{xx}=3\cdot (-8{\pi }^2)=-24{\pi }^2\approx -236.9\text{K/s}.$ Why this step? $u_t=\alpha u_{xx}$ turns curvature into a time-rate.

Verify: $u_t<0$, so the spot cools — exactly as the hill picture predicts (neighbours are colder, energy leaks out). Units: $\alpha$ in m²/s times $u_{xx}$ in K/m² gives K/s. ✓

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Example 2 — Cell B: a cold dip warms

Forecast: At $x=1$, $\cos \pi =-1$, so $u=12$ — the bottom of the dip. Guess the sign of $u_t$: valleys sit below their neighbours…

Step 1 — Reuse the curvature. $u_{xx}=-8{\pi }^2\cos \pi x.$ Why this step? Same profile, so the same second derivative — we only change the point.

Step 2 — Evaluate at $x=1$. $u_{xx}{|}_{x=1}=-8{\pi }^2\cos \pi =-8{\pi }^2(-1)=+8{\pi }^2.$ Why this step? $\cos \pi =-1$ flips the sign: now curvature is positive, a valley (∪).

Step 3 — Apply the equation. $u_t=3\cdot (8{\pi }^2)=24{\pi }^2\approx 236.9\text{K/s}.$

Verify: $u_t>0$, the dip warms. Notice the numbers in Ex 1 and Ex 2 are exact opposites ($\mp 24{\pi }^2$) — the hill and valley of one $\cos$ curve heat/cool at mirror-image rates. Sanity ✓.

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Example 3 — Cell C: the degenerate straight line

Forecast: A straight line has no bend. What is the curvature of a straight line? Guess $u_t$ before Step 1.

Step 1 — First derivative. $u_x=\frac{d}{dx}[5+7x]=7.$ Why this step? We need $u_{xx}$; get there via $u_x$ first.

Step 2 — Second derivative. $u_{xx}=\frac{d}{dx}[7]=0.$ Why this step? The slope $7$ is constant, so its rate of change (the curvature) is zero — a straight line bends nowhere.

Step 3 — Apply the equation. $u_t=\alpha \cdot 0=0\text{for every }x\text{ and every }\alpha .$

Verify: $u_t=0$ everywhere — the profile is already steady. This matches the parent note: the 1D steady state ($u_{xx}=0$) is exactly a straight line. See Laplace Equation (steady-state heat) for the multi-dimensional cousin. ✓

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Example 4 — Cell D: verify an exponential-decay solution

Forecast: A solution must make the left side equal the right side identically. Guess: the exponent $-4\alpha t$ — where does the $4$ have to come from?

Step 1 — Left side, $u_t$. $u_t=\frac{\partial }{\partial t}\left[e^{-4\alpha t}\cos 2x\right]=-4\alpha e^{-4\alpha t}\cos 2x.$ Why this step? Chain rule on the exponential: the time-derivative pulls down the factor $-4\alpha$. $\cos 2x$ is a constant in $t$.

Step 2 — Right side, first the space derivatives. $u_x=-2e^{-4\alpha t}\sin 2x,u_{xx}=-4e^{-4\alpha t}\cos 2x.$ Why this step? Each $x$-derivative pulls down a factor $2$ from $\cos 2x$; twice gives $2\cdot 2=4$ and returns to $\cos$ with a minus.

Step 3 — Multiply by $\alpha$ and compare. $\alpha u_{xx}=\alpha (-4)e^{-4\alpha t}\cos 2x=-4\alpha e^{-4\alpha t}\cos 2x=u_t.✓$ Why this step? Both sides are now the same expression, so the equation holds for all $x,t,\alpha$.

Verify: They match term-for-term. The general pattern: $e^{-\alpha n^{2}t}\cos (nx)$ solves the equation, decaying faster for larger $n$ (wigglier profiles die quicker). This is the seed of Separation of Variables for the Heat Equation and Fourier Series Solutions. ✓

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Example 5 — Cell E: the sign twist (why the minus matters)

Forecast: In the correct equation heat spreads and dies away. Guess: will the wrong-sign version grow or shrink in time?

Step 1 — Compute the pieces. $u_t=\beta e^{\beta t}\sin \pi x,u_{xx}=-{\pi }^2{e}^{\beta t}\sin \pi x.$ Why this step? Standard derivatives; we solve for the growth rate $\beta$ by matching.

Step 2 — Correct equation $u_t=\alpha u_{xx}$ (α = 1). $\beta =-{\pi }^2\approx -\mathrm{9.87.}$ Why this step? Matching coefficients of $e^{\beta t}\sin \pi x$. Negative $\beta$ ⇒ decay — physically stable, heat smooths out.

Step 3 — Wrong-sign equation $u_t=-\alpha u_{xx}$. $\beta =-(-{\pi }^2)=+{\pi }^2\approx +\mathrm{9.87.}$ Why this step? The flipped sign flips $\beta$. Positive $\beta$ ⇒ explosive growth — the profile amplifies forever.

Verify: Correct: $\beta =-{\pi }^2<0$ (dies). Wrong: $\beta =+{\pi }^2>0$ (blows up). This is precisely the [!mistake] in the parent note — the minus in $q=-k{u}_x$ is what keeps the physics stable. See Classification of PDEs (elliptic, parabolic, hyperbolic). ✓

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Example 6 — Cell F: limiting behaviour and the $\alpha$ race

Forecast: Bigger $\alpha$ = faster diffusion. Guess which rod flattens sooner, and guess the final temperature.

Step 1 — Decay rates. ${\lambda }_P={\alpha }_P{\pi }^2=2{\pi }^2\approx 19.7{\text{s}}^{-1},{\lambda }_Q={\alpha }_Q{\pi }^2=8{\pi }^2\approx 79.0{\text{s}}^{-1}.$ Why this step? The mode $\sin (\pi x)$ decays like $e^{-\alpha {\pi }^{2}t}$; the exponent's magnitude is the decay rate.

Step 2 — Time for rod Q to reach $1/e$. Peak value is $e^{-{\lambda }_{Q}t}$; set it to $e^{-1}$: ${\lambda }_{Q}t=1\Rightarrow t=\frac{1}{8{\pi }^2}\approx 0.0127\text{s}.$ Why this step? "$1/e$ time" (the time constant) is $1/\lambda$ — a clean way to compare speeds.

Step 3 — Long-time limit. ${\lim }_{t\to \infty }{e}^{-\alpha {\pi }^{2}t}\sin \pi x=0\text{for both rods.}$ Why this step? Any positive exponent times $t$ sends $e^{-\lambda t}\to 0$; the ends are held at $0$, so the whole rod relaxes to $0$.

Verify: ${\lambda }_Q=4{\lambda }_P$ (since $8=4\times 2$), so rod Q equilibrates four times faster — doubling-then-doubling $\alpha$ compounds. Final temperature $0$ for both, matching "diffusion erases structure." ✓

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Example 7 — Cell G: real-world word problem with units

Forecast: Copper conducts superbly ($k$ huge) but also stores a fair amount of energy. Guess whether $\alpha$ is "large" ($\sim 10^{-4}$) or "tiny" ($\sim 10^{-7}$).

Step 1 — Diffusivity. $\alpha =\frac{k}{\rho c}=\frac{400}{8960\times 385}=\frac{400}{3449600}\approx 1.16\times 10^{-4}{\text{m}}^2/\text{s}.$ Why this step? The PDE's time scale is set by $\alpha =k/(\rho c)$, not by $k$ alone (parent [!mistake]). Units: $\frac{\text{W/m⋅K}}{({\text{kg/m}}^3)(\text{J/kg⋅K})}={\text{m}}^2/\text{s}$. ✓

Step 2 — Curvature of the given profile. $u_x=-100x,u_{xx}=-100{\text{K/m}}^2.$ Why this step? $u=300-50x^2$ is a downward parabola (a hill), so constant negative curvature $-100$ everywhere.

Step 3 — Time-rate. $u_t=\alpha u_{xx}=(1.16\times 10^{-4})(-100)\approx -1.16\times 10^{-2}\text{K/s}.$ Why this step? Same $u_t=\alpha u_{xx}$; the negative sign says the rod's warm crest cools, slowly (copper's modest $\alpha$).

Verify: $\alpha \approx 1.16\times 10^{-4}$ m²/s — the "large" $\sim 10^{-4}$ guess, correct for copper. $u_t\approx -0.0116$ K/s, units K/s. ✓ See Thermal Diffusivity and Material Properties.

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Example 8 — Cell H: a heat source (non-homogeneous)

Forecast: Without the source, this hill would cool. The source pumps energy in. Guess: does the source help it warm, or just slow the cooling — and can they exactly cancel?

Step 1 — Curvature. $u_x=-x,u_{xx}=-1{\text{K/m}}^2.$ Why this step? We need the diffusion term $k{u}_{xx}$; the profile is a shallow hill, curvature $-1$.

Step 2 — Assemble the source equation. $\rho c{u}_t=k{u}_{xx}+f=20(-1)+30=-20+30=10.$ Why this step? Diffusion contributes $-20$ (cooling, it's a hill), the heater adds $+30$. Their sum is the net drive on stored energy.

Step 3 — Solve for $u_t$. $u_t=\frac{10}{\rho c}=\frac{10}{10}=1\text{K/s}.$ Why this step? Divide the energy balance by $\rho c$ to isolate the temperature rate.

Verify: Net $u_t=+1$ K/s: the $+30$ source outweighs the $-20$ diffusive cooling, so the spot warms despite being a hill. If the source had been exactly $f=20$, the two would cancel and $u_t=0$ — a source-sustained steady state. ✓

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Recap of the whole matrix

Recall Did every cell get covered?

A hill cools (Ex 1), a valley warms (Ex 2), a straight line is steady (Ex 3), an exponential mode is verified (Ex 4), a wrong sign blows up (Ex 5), bigger $\alpha$ wins the equilibrium race (Ex 6), a real copper rod gives SI numbers (Ex 7), and a source can flip cooling into warming (Ex 8). Every sign of curvature and every standard twist is now something you have seen worked.

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Connections

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