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E-TRADE TOGETHER GLOBAL ACADEMY

Introduction to Partial Differential Equations

Level 4 · Prerequisites: Courses 2.2 and 4.1 · Connects to Physics and Energy Engineering

ACADEMICALLY REVIEWED · DESKTOP RUNTIME VERIFIED · MOBILE QA ENVIRONMENT BLOCKED

6 canonical classes · 2 laboratories · 3 module assessments · project · final

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MODULE 1

PDE Foundations & Classification

Identify variables, order, type, assumptions and physical interpretation.

Dependencies: Multivariable derivatives, integral calculus and ordinary differential equations.

Next-module connection: Conditions & Analytical Methods

4.3.1 · v1 · 90 min

Heat equation

Why it matters

Diffusion models explain how temperature or concentration spreads through space and time.

Learning objectives

  • Distinguish an ODE from a PDE and identify dependent and independent variables.
  • Interpret every term, unit, domain and condition in the heat equation.

Prerequisites

Courses 2.2 and 4.1; partial derivatives from Course 2.3 are reviewed where needed.

Concept, key terms and notation

For temperature u(x,t), the one-dimensional heat model is u_t=αu_xx on 0<x<L. The diffusivity α has units length²/time. An initial profile u(x,0)=f(x) describes the whole rod at t=0; boundary data describe what happens at x=0 and x=L over time. The model assumes a homogeneous medium and no internal source.

A PDE contains partial derivatives of a multivariable unknown. Heat flow is first order in time and second order in space; α>0 makes it parabolic. Energy moves down temperature gradients, while total energy is conserved only under appropriate insulated boundaries.

Mathematical development and worked examples

Worked example 1

Problem / PDE · domain / conditions · goal: u_t=0.5u_xx, u(x,0)=20+5sin(πx), u(0,t)=u(1,t)=20.

Method, development and calculation: Let v=u−20. Since v_xx=−π²v, v(x,t)=5e^(−0.5π²t)sin(πx). Thus u=20+v.

Solution, verification, interpretation and limitations: Substitution gives u_t=0.5u_xx; both boundaries equal 20 and t=0 recovers the initial profile.

Worked example 2

Problem / PDE · domain / conditions · goal: Check whether u=e^(−4t)cos(2x) solves u_t=u_xx.

Method, development and calculation: u_t=−4e^(−4t)cos(2x) and u_xx=−4e^(−4t)cos(2x).

Solution, verification, interpretation and limitations: The derivatives match; any proposed boundaries still must be checked separately.

Common mistake and counterexample

Incorrect: ‘u_t=αu_xx is an ODE because t is the changing variable.’ It is a PDE because u depends on x and t and partial derivatives occur. Correct classification uses the unknown’s independent variables, not a preferred plot axis.

Knowledge check and summary

  1. A PDE uses partial derivatives of a multivariable unknown.
  2. Initial data cover the spatial domain at one time.
  3. Dirichlet data specify boundary values.
  4. A candidate must satisfy equation and conditions.

Mastery criterion: 4/4 correct with equation, conditions and verification evidence; correct any miss and complete a fresh equivalent check.

Related laboratory and next class

Heat Diffusion Simulation → Wave equation