Visual walkthrough — Heat equation (parabolic) 1D — derivation from Fourier's law

Before any symbol, here is our whole world: a thin rod, and a wiggly line above it that says how hot each spot is.

---

Step 1 — What is a temperature profile?

WHAT. Picture a thin metal rod lying flat. We measure position along it with a number $x$ (how far from the left end, in metres). At each spot we can measure a temperature. We draw that temperature as a height above the rod. That curve is $u(x,t)$.

WHY. Every symbol we use later is a fact about this curve. Before we can talk about "how curved" or "how it changes," we must first agree that temperature is just a height above each position — a shape we can look at.

• $u$ ::: the temperature (a height), measured in kelvin $\text{K}$
• $x$ ::: position along the rod, in metres
• $t$ ::: time; the curve can wobble as time passes, so we write $u(x,t)$

PICTURE. The rod is the horizontal line; the coloured curve floats above it. Tall = hot, low = cold.

---

Step 2 — The slope $u_x$: how steep is the profile?

WHAT. Zoom into one point on the curve and lay a straight ruler tangent to it. The slope of that ruler is written $u_x$ (the little $x$ means "how fast $u$ changes as you step in the $x$-direction"). Steep uphill = large positive $u_x$; steep downhill = large negative $u_x$; flat = zero.

WHY. Heat is going to care about which way is downhill, because heat rolls downhill from hot to cold. To say "downhill" precisely we need slope. That is the only reason the derivative enters here — it is the machine that turns "hotter over there" into a single number with a sign.

$u_x=\underset{\text{slope of the temperature curve}}{\underbrace{\frac{\partial u}{\partial x}}}$

• $\partial u$ ::: a tiny change in temperature
• $\partial x$ ::: the tiny step in position that caused it
• their ratio ::: rise over run = steepness, exactly like a hill's grade

PICTURE. Green tangent lines at three spots: uphill (positive), flat (zero), downhill (negative).

---

Step 3 — Fourier's Law: heat rolls downhill

WHAT. Define the heat flux $q$ = how much thermal energy crosses a point each second, per unit area, moving in the $+x$ direction. Fourier's law says $q$ is proportional to the negative slope:

$q=\underset{\text{material \& sign}}{\underbrace{-k}}\underset{\text{steepness}}{\underbrace{u_x}}$

WHY the minus sign. Look at the red arrows in the figure. If the curve goes uphill to the right ($u_x>0$, hotter on the right), heat must flow left — toward the cold — so $q$ is negative. The minus sign is what forces "hot → cold." Without it, heat would climb toward the hot spot, which never happens.

• $k$ ::: thermal conductivity ($\text{W/m⋅K}$) — how easily this material passes heat along
• $-$ ::: flips direction so flow is always down the slope
• $q>0$ ::: energy moving right; $q<0$ ::: energy moving left

PICTURE. On a hill sloping up-right, the flux arrow points left (downhill). On a hill sloping up-left, it points right. Always toward the cold.

---

Step 4 — The control slab: bookkeeping for energy

WHAT. Cut out a thin slice of the rod between $x$ and $x+\Delta x$, with face area $A$. This is our control slab. All the physics is now just: energy in − energy out = energy that piled up inside. This is conservation of energy.

WHY. We cannot track every atom. Instead we watch one small box and count energy crossing its two faces. A small box is exactly what lets us later shrink $\Delta x\to 0$ and get a derivative — the geometry does the calculus for us.

Energy stored inside the slab: $E(t)={\int }_x^{x+\Delta x}\underset{\text{energy per volume}}{\underbrace{\rho cu}}\underset{\text{volume slice}}{\underbrace{Ads}}$

• $\rho$ ::: density (${\text{kg/m}}^3$)
• $c$ ::: specific heat ($\text{J/kg⋅K}$) — energy to warm 1 kg by 1 K
• $\rho cu$ ::: therefore energy per cubic metre stored at temperature $u$
• $Ads$ ::: volume of a paper-thin sub-slice at position $s$

PICTURE. The slab, its left face at $x$ with flux $q(x)$ entering, its right face at $x+\Delta x$ with flux $q(x+\Delta x)$ leaving.

---

Step 5 — In minus out: the flux imbalance

WHAT. Energy entering the left face per second is $Aq(x)$. Energy leaving the right face is $Aq(x+\Delta x)$. So the net energy gained by the slab per second is

$\frac{dE}{dt}=\underset{\text{in at left}}{\underbrace{Aq(x)}}-\underset{\text{out at right}}{\underbrace{Aq(x+\Delta x)}}$

WHY. Nothing creates or destroys energy inside (no source yet), so the only way the slab's stored energy changes is by leaking across its two faces. If more comes in the left than leaves the right, the slab warms. This is the whole heart of the derivation: a mismatch of flux is what heats things up.

• $Aq(x)$ ::: rate energy walks in through the left wall
• $Aq(x+\Delta x)$ ::: rate it walks out the right wall
• their difference ::: the net pile-up (positive = warming)

PICTURE. Two arrows of different lengths at the two faces; the leftover (shaded) is what stays behind and heats the slab.

---

Step 6 — Shrink the slab: the flux difference becomes a derivative

WHAT. Differentiating the stored energy $E$ (its limits are fixed in space, so the $\frac{d}{dt}$ dives inside) and matching Step 5:

${\int }_x^{x+\Delta x}\rho c{u}_{t}Ads=A(q(x)-q(x+\Delta x))$

Divide both sides by $A\Delta x$ and let $\Delta x\to 0$:

$\rho c{u}_t=\underset{=-q_x}{\underbrace{\underset{\Delta x\to 0}{\lim }\frac{q(x)-q(x+\Delta x)}{\Delta x}}}=-\frac{\partial q}{\partial x}$

WHY. Look closely at that fraction. It is exactly (minus) the definition of a derivative: (value here − value a step over) ÷ (size of step). As the box shrinks, the average temperature-change inside becomes the value right at the point, $u_t$, and the flux difference becomes the slope of the flux, $q_x$. The whole reason we chose a tiny slab was to summon this derivative.

• $u_t$ ::: how fast temperature at this point rises in time ($\text{K/s}$)
• $-q_x$ ::: how fast the flux is dropping as you move right = net inflow rate

PICTURE. The slab collapses to a point; the two arrows merge into one number — the slope of $q$.

---

Step 7 — Substitute Fourier's law and meet the equation

WHAT. We now have two facts: $\rho c{u}_t=-q_x$ (Step 6) and $q=-k{u}_x$ (Step 3). Plug the second into the first:

$\rho c{u}_t=-\frac{\partial }{\partial x}\left(\underset{q}{\underbrace{-k{u}_x}}\right)=k{u}_{xx}$

Divide by $\rho c$ and name the constant $\alpha =\frac{k}{\rho c}$:

$\boxed{u_t=\underset{\text{diffusivity}}{\underbrace{\alpha }}\underset{\text{curvature}}{\underbrace{u_{xx}}}}$

WHY. The two minus signs (one from Fourier, one from "in minus out") multiply to a plus. That plus is why heat spreads and smooths instead of blowing up. The second derivative $u_{xx}$ appeared because we differentiated a slope ($u_x$) once more — so the equation is governed by curvature, not slope. See why $\alpha$, not $k$, sets the pace.

• $u_{xx}$ ::: curvature — how much the curve bends (cup up vs bump)
• $\alpha =k/(\rho c)$ ::: turns "how curved" into "how fast" (${\text{m}}^2/\text{s}$)

PICTURE. A dip (cup-up, $u_{xx}>0$) with an up-arrow (warming); a bump (cup-down, $u_{xx}<0$) with a down-arrow (cooling).

---

Step 8 — The degenerate cases: read every sign

WHAT. The equation must make sense in every situation, so let us test the extremes.

Profile shape	$u_{xx}$	$u_t=\alpha u_{xx}$	What happens
Bump / hill	$<0$	$<0$	point cools
Dip / valley	$>0$	$>0$	point warms
Straight line	$=0$	$=0$	no change — steady
Flat (constant)	$=0$	$=0$	already uniform, stays

WHY. Two "no change" rows deserve care. A straight-line profile has slope but no curvature: heat pours out the downhill side exactly as fast as it pours in the uphill side, so the point breaks even — this is the steady state in 1D. A constant profile has no slope at all: nothing flows anywhere. Both give $u_t=0$, but for subtly different reasons — and the equation captures both automatically because it watches curvature, not slope.

PICTURE. Three panels: bump (down-arrow), valley (up-arrow), straight line (flat arrows in = out, net zero).

---

The one-picture summary

Everything above, on one frame: a curve, its slope giving flux (down the hill), the flux mismatch across a shrinking slab giving $u_t$, and the two minus signs collapsing into $u_t=\alpha u_{xx}$.

Recall Feynman retelling of the whole walkthrough

Picture a hot rod as a hilly landscape of temperature. Slope tells you which way is downhill (Step 2). Heat is lazy — it always rolls down the slope, and the minus sign in Fourier's law bakes that in (Step 3). Now cut out a tiny box (Step 4) and just count: energy in the left face, out the right face (Step 5). If more comes in than leaves, the box warms up. Shrink the box and "in minus out" turns into "how fast the flux is changing," which is a derivative (Step 6). Feed Fourier's law in, and the two minus signs cheerfully cancel to a plus (Step 7). What survives is beautiful: the second derivative — the curvature — because we took the slope of a slope. Hills (curving down) lose heat and cool; valleys (curving up) gain heat and warm; straight ramps break even (Step 8). That single sentence — "time-change equals diffusivity times curvature" — is $u_t=\alpha u_{xx}$.

---

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)
• Separation of Variables for the Heat Equation
• Thermal Diffusivity and Material Properties
• Laplace Equation (steady-state heat)
• Wave Equation (hyperbolic) 1D