7.1.2. Application to Standard PDEs

The Heat Equation ∂𝑢/∂𝑡 = 𝑘∂²𝑢/∂𝑥²

Step 1: Assume 𝑢(𝑥,𝑡) = 𝑋(𝑥)𝑇(𝑡) Substitute into the PDE: 𝑑𝑇/𝑑𝑡 = 𝑘𝑇 ⇒ 𝑑²𝑋/𝑑𝑥² = −𝜆𝑋 Both sides are set equal to a separation constant −𝜆, yielding two ODEs: • Time equation: 𝑑𝑇/𝑑𝑡 + 𝜆𝑘𝑇 = 0 ⇒ 𝑇(𝑡) = 𝐴𝑒^{−𝜆𝑘𝑡} • Spatial equation: 𝑑²𝑋/𝑑𝑥² + 𝜆𝑋 = 0 ⇒ 𝑋(𝑥)= 𝐵sin(√𝜆𝑥)+ 𝐶cos(√𝜆𝑥) Use boundary and initial conditions to determine constants and form the final solution.

This chunk introduces the heat equation, which describes how heat is distributed over time. We start by making an assumption that the solution can be separated into a space function, X, and a time function, T. By substituting this assumption into the heat equation, we separate the variables and introduce a separation constant, which leads us to two ordinary differential equations (ODEs), one for time and one for space. The solutions of these equations provide us with expressions for T(t) and X(x). We can find specific solutions based on the initial and boundary conditions of the problem.

Think of heating a metal rod. If you heat one end, the heat will flow along the rod to the other end over time. The heat equation mathematically describes this process, allowing us to calculate how fast the heat spreads, depending on conditions like material properties and surrounding temperature.

The Wave Equation ∂²𝑢/∂𝑡² = 𝑐²∂²𝑢/∂𝑥²

Assume 𝑢(𝑥,𝑡) = 𝑋(𝑥)𝑇(𝑡). Then, 𝑑²𝑇/𝑑𝑡² = 𝑐²𝑇 ⇒ 𝑑²𝑋/𝑑𝑥² = −𝜆𝑋 Results in two ODEs: • 𝑑²𝑇/𝑑𝑡² + 𝜆𝑐²𝑇 = 0 • 𝑑²𝑋/𝑑𝑥² + 𝜆𝑋 = 0 Solutions will again be sinusoidal, depending on boundary conditions.

The wave equation models how waves propagate through a medium. Similar to the heat equation, we start by assuming a separable solution. By substituting this assumption into the wave equation, we derive two ordinary differential equations. Each ODE corresponds to a different aspect of the wave: one describes how the wave changes over time (T(t)), and the other describes its shape in space (X(x)). The solutions to these equations are sinusoidal, meaning they can oscillate, which is characteristic of wave motion. The specific form of these solutions is influenced by the boundary conditions applied to the problem.

Imagine standing by a lake and throwing a stone into the water. The ripples, or waves, that spread out demonstrate how the wave equation works. Just like the stone causes a disturbance that travels outward, the wave equation predicts how that disturbance moves through the water over time, creating patterns reliant on the initial throw and the environment.

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