Recipe

Derive a PDE from a conservation law

Use when: you're given a conserved quantity, a flux law tying its flow rate to a gradient, and asked what PDE it satisfies

Goal: a PDE for u(x,t), built from a segment-level balance carried to the limit \Delta x\to0

Taught in Physical derivation: the heat equation

1. 1

   Pick a short segment of length \Delta x. Its stored quantity changes only by the net flux crossing its two ends, plus a source term if one is present.

   Example

   A segment with c\rho=2, \Delta x=0.5, flux \phi(x,t)=12 entering and \phi(x+\Delta x,t)=8 leaving: c\rho\,\Delta x\,u_t \approx \phi(x,t) - \phi(x+\Delta x,t) = 12-8=4

   Why: energy inside the segment can only change by what actually crosses its two ends, plus any source, since nothing is created or destroyed inside it.

2. 2

   Replace \phi with the flux law at each end of the segment, expressing it through the gradient of u there.

   Example

   Fourier's law gives \phi=-Ku_x, so c\rho\,\Delta x\,u_t \approx -Ku_x(x,t) + Ku_x(x+\Delta x,t) = K\big[u_x(x+\Delta x,t)-u_x(x,t)\big]

   Why: conservation alone says nothing about speed or direction. Fourier's law's own minus sign is what says heat moves from hot toward cold, not the reverse.

3. 3

   Divide both sides by \Delta x, then let \Delta x\to0 so the bracketed fraction turns into a derivative.

   Example

   \frac{u_x(x+\Delta x,t)-u_x(x,t)}{\Delta x} \;\longrightarrow\; u_{xx} \quad\text{as } \Delta x\to0,

   so the source-free balance becomes c\rho\,u_t = Ku_{xx}.

   • no source term present — the limit gives c\rho\,u_t = Ku_{xx} directly.
   • a source term Q(x,t) is present — carry it through unchanged, giving c\rho\,u_t = Ku_{xx} + Q(x,t).

   Why: the segment-level balance is only rough at finite \Delta x; the limit is what makes it an exact pointwise equation.

4. 4

   Divide the whole equation by c\rho, and call the combination k=K/(c\rho) the diffusivity.

   Example

   For K=0.5 and c\rho=2: k=0.5/2=0.25, so the source-free equation reads u_t=0.25\,u_{xx}.

   Why: this puts the PDE in the standard form u_t=ku_{xx} (or u_t=ku_{xx}+Q/(c\rho) with a source), with one constant carrying all the substance's own traits.