How to read a PDE line, one symbol at a time
Objective: What does Δu mean when x and y are the spatial variables?
A PDE formula is compressed language. Before manipulating it, expand every symbol into a noun or an action: what is unknown, where it lives, which direction changes, what is supplied, and what must be found.
Read the objects before manipulating the equation: In “L[u]=f in Ω, u=g on ∂Ω”, u is the unknown field; L is a rule that differentiates or combines u; square brackets mean “apply L to u”; f is the interior source; Ω (capital omega) is the region; ∂Ω is the edge of that region; and g supplies values on that edge. The result to establish is: For u_t=k(u_xx+u_yy)+q, the subscript t means one time derivative, xx and yy mean two derivatives in the named spatial direction, k is diffusivity, and q is added heat per unit time. The equation says “time change = diffusion caused by curvature + local supply.” Keep the domain, time interval, forcing, initial data, boundary data, and solution class attached to every step.
The worked question is: On the square 0<x<π, 0<y<π, read and verify u(x,y,t)=e^{-2t}sin x sin y for u_t=u_xx+u_yy with zero boundary values. Begin with “Differentiate in time: u_t=−2e^{-2t}sin x sin y.” and finish with “On x=0, x=π, y=0, or y=π, one sine factor is zero; hence all four edges have value zero.” Then check the interior PDE and every relevant data surface separately. A plot can illustrate the result but cannot replace these checks.
For u_t=k(u_xx+u_yy)+q, the subscript t means one time derivative, xx and yy mean two derivatives in the named spatial direction, k is diffusivity, and q is added heat per unit time. The equation says “time change = diffusion caused by curvature + local supply.”
Read the equality sign as a balance between the left and right sides, not as an instruction to solve immediately.
Translate each derivative: u_t asks how the field changes while position is fixed; u_xx and u_yy measure how its spatial slopes themselves change.
Then attach the region, initial state, and boundary data; without them the displayed differential law still describes many possible fields.
On the square 0<x<π, 0<y<π, read and verify u(x,y,t)=e^{-2t}sin x sin y for u_t=u_xx+u_yy with zero boundary values.
- Differentiate in time: u_t=−2e^{-2t}sin x sin y.
- Differentiate twice in each spatial direction: u_xx=−e^{-2t}sin x sin y and u_yy=−e^{-2t}sin x sin y, so their sum equals u_t.
- On x=0, x=π, y=0, or y=π, one sine factor is zero; hence all four edges have value zero.
Result: The formula satisfies both the interior heat equation and all four boundary edges; at t=0 its initial field is sin x sin y.