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PAGE 1

Heat Equation

Diagram of a horizontal conducting rod with endpoints at x=0 and x=L.

conducting rod length L

heat can flow along the rod

but not across lateral surface ("up" or "down")

temperature is u(x, t)

x: position

t: time

the governing equation is the Heat Equation

\[ \frac{\partial u}{\partial t} = k \frac{\partial^2 u}{\partial x^2} \]

(1-D Heat eq.)

or \[ u_t = k u_{xx} \]

(\( k > 0 \))

this is a partial differential equation (PDE)

the heat eq. comes from conservation of energy and Fourier's law

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what does \( u_t = k u_{xx} \) say?

uxx : Concavity of \( u(x,t) \) in space (position)

uxx: Concave up \( \rightarrow u_t > 0 \rightarrow \) temp. rises over time

(\( u_{xx} > 0 \))

uxx: Concave down \( \rightarrow u_t < 0 \rightarrow \) temp. falls over time

(\( u_{xx} < 0 \))

A graph of u vs x showing a wave-like curve with points of local maxima and minima labeled with concavity and rate of change.

Concave Down Region

  • \( u_{xx} < 0 \): hotter than avg. temp. nearby
  • \( u_t < 0 \): temp. drop.

Concave Up Region

  • \( u_{xx} > 0 \)
  • \( u_t > 0 \): temp. increases.
PAGE 3

Solving the Heat Equation: Separation of Variables

How to solve \( u_t = k u_{xx} \)?

Set up:

\( 0 < x < L \)

\( t > 0 \)

A horizontal cylinder representing a rod from x=0 to x=L with temperature labels u(0,t)=T1 and u(L,t)=T2.

Boundary and Initial Conditions

\( u(0, t) = T_1 \)

(left end temp. for all \( t \))

\( u(L, t) = T_2 \)

(right end temp. for all \( t \))

Boundary conditions (BC)

\( u(x, 0) = f(x) \)

(initial temp. profile)

Initial condition (IC)

Let's solve the case where \( T_1 = T_2 = 0 \) (homogeneous BC's)

Method of Separation of Variables

\[ u(x, t) = X(x) T(t) \]

Product of two functions:

  • \( X \): function of \( x \) only
  • \( T \): function of \( t \) only
\[ u_t = \frac{\partial}{\partial t} (X(x) T(t)) = X T' \]

"constant" (with respect to t)

Likewise,

\[ u_{xx} = X'' T \]
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Sub into \( u_t = k u_{xx} \):

\[ X T' = k X'' T \]

separate the functions

\[ \frac{X''}{X} = \frac{T'}{kT} = \text{constant} = -\lambda \quad (\lambda > 0) \]
Where:
\( \frac{X''}{X} \) is a function of \( x \) only
\( \frac{T'}{kT} \) is a function of \( t \) only
\( -\lambda \) is the separation constant

Two ODEs

\( \frac{X''}{X} = -\lambda \Rightarrow \)

\( X'' + \lambda X = 0 \)

\( \frac{T'}{kT} = -\lambda \Rightarrow \)

\( T' + \lambda k T = 0 \)

Boundary Conditions (BC's)

\( u(0, t) = 0 \rightarrow X(0) T(t) = 0 \rightarrow \)

\( X(0) = 0 \)

\( u(L, t) = 0 \rightarrow X(L) T(t) = 0 \rightarrow \)

\( X(L) = 0 \)
PAGE 5

Solving the Spatial Part of the Heat Equation

Let's solve:

\[ X'' + \lambda X = 0, \quad X(0) = X(L) = 0 \]
\[ X(x) = A \cos(\sqrt{\lambda} x) + B \sin(\sqrt{\lambda} x) \]

Applying boundary conditions:

\[ X(0) = A = 0 \]
\[ X(L) = B \sin(\sqrt{\lambda} L) = 0 \]
require \( B \neq 0 \)
\[ \sin(\sqrt{\lambda} L) = 0 \]
\[ \sqrt{\lambda} L = n\pi \]
\( n = 1, 2, 3, \dots \)

Eigenvalues

\[ \lambda_n = \frac{n^2 \pi^2}{L^2} \]

for each \( n \) there is one solution

eigenvalues

so for each \( n \), there is one solution

Eigenfunctions

\[ X_n(x) = \sin\left( \frac{n\pi}{L} x \right) \]

(drop \( B \) scaling constant)

eigenfunctions

PAGE 6

Solving the Temporal Part and General Solution

now

\[ T' + k\lambda T = 0 \]

use

\[ \lambda = \frac{n^2 \pi^2}{L^2} \]
\[ T' + \frac{kn^2\pi^2}{L^2} T = 0 \]
\[ T(t) = C e^{-kn^2\pi^2/L^2 t} \]

one for each \( n = 1, 2, 3, \dots \)

\[ T_n(t) = e^{-kn^2\pi^2/L^2 t} \]

Since \( u(x,t) = X(x) T(t) \)

for each \( n \), there is one solution

\[ u_n(x,t) = X_n T_n \]

general solution is linear combo of all

\[ u(x,t) = \sum_{n=1}^{\infty} C_n e^{-kn^2\pi^2/L^2 t} \sin\left( \frac{n\pi x}{L} \right) \]
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Last unused condition: u(x, 0) = f(x) (initial condition)

\[ f(x) = \sum_{n=1}^{\infty} C_n \sin\left(\frac{n\pi x}{L}\right) \]

sine series half period L

\[ C_n = \frac{2}{L} \int_{0}^{L} f(x) \sin\left(\frac{n\pi x}{L}\right) dx \]