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PAGE 1

Heat Eq. (part 3)

\[ u_t = k u_{xx} \quad 0 < x < L \quad t > 0 \]

A simple diagram of a horizontal cylindrical rod. The left end is marked with x equals 0 and the right end is marked with x equals L.

now we will insulate the ends (like w/ the lateral surface)

heat cannot leave the rod through the ends anymore.

→ \[ \frac{\partial u}{\partial x} = 0 \quad \text{at } x=0 \text{ and } x=L \]

\[ u_x(0, t) = 0 \]

\[ u_x(L, t) = 0 \]

\( \} \) Neumann boundary conditions

same initial condition as before

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

solve by using separation of variables

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

\[ u_t = X T' \]

\[ u_{xx} = X'' T \]

\[ u_t = k u_{xx} \]

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

separate:

\[ \frac{X''}{X} = \frac{T'}{kT} = -\lambda \]

same separation constant

two ODE's:

\[ X'' + \lambda X = 0 \]

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

BC's:

\[ u_x(0, t) = 0 \rightarrow X'(0) T(t) = 0 \rightarrow X'(0) = 0 \]

\[ u_x(L, t) = 0 \rightarrow X'(L) T(t) = 0 \rightarrow X'(L) = 0 \]

solve for \( X \):

usual solution:

\[ X(x) = A \cos(\sqrt{\lambda} x) + B \sin(\sqrt{\lambda} x) \]

\( (\lambda \neq 0) \)

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\[ X'(x) = -\sqrt{\lambda} A \sin(\sqrt{\lambda} x) + \sqrt{\lambda} B \cos(\sqrt{\lambda} x) \]

\[ X'(0) = 0 = \sqrt{\lambda} B \rightarrow B = 0 \]

\[ X'(L) = 0 = -\sqrt{\lambda} A \sin(\sqrt{\lambda} L) \]
require \( A \neq 0 \)

so, \( \sin(\sqrt{\lambda} L) = 0 \)

so, \( \sqrt{\lambda} L = n\pi \)

\( n = 1, 2, 3, \dots \)

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

eigenvalues

(same as before)

solution for each \( n \) is

\[ X_n = \cos(\sqrt{\lambda} x) \]

(drop A scaling constant)

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

eigenfunctions

cosines this time

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

\( X'(0) = X'(L) = 0 \)

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what if \( \lambda = 0 \)?

\[ X'' = 0 \rightarrow X(x) = Ax + B \]

\[ X'(x) = A \]

\[ X'(0) = X'(L) = 0 \rightarrow A = 0 \]

BC's: \( X'(0) = X'(L) = 0 \)

so, \( X = \text{constant} \) also meets the BC's

\[ X = 1 \]

for the case when \( \lambda = 0 \)

(drop the B scaling constant)

notice \( \lambda = 0 \rightarrow \text{eigenvalue is } 0 \)

notice \( \lambda = 0 \) and \( X = 1 \) are included in

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

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

so,

\[ \lambda_n = \frac{n^2 \pi^2}{L^2} \] \[ X_n = \cos\left(\frac{n\pi}{L} x\right) \]
\[ n = 0, 1, 2, 3, \dots \]
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Solving the Time-Dependent Equation

how we solve \( T' + k\lambda T = 0 \) (same as before)

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

\( n = 0, 1, 2, 3, \dots \)

for each \( n \), solution is \( u_n = X_n T_n = e^{-kn^2\pi^2t/L^2} \cos\left(\frac{n\pi x}{L}\right) \)

general solution is linear combination of all

\[ u(x,t) = \sum_{n=0}^{\infty} A_n e^{-kn^2\pi^2t/L^2} \cos\left(\frac{n\pi x}{L}\right) \]

let's separate out \( n=0 \) (\( A_0 \))

\[ u(x,t) = \frac{1}{2}A_0 + \sum_{n=1}^{\infty} A_n e^{-kn^2\pi^2t/L^2} \cos\left(\frac{n\pi x}{L}\right) \]
looks like a cosine series! let's give \( n=0 \) an extra \( \frac{1}{2} \) factor

Initial Condition (IC): \( (t=0) \)

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

\[ f(x) = \frac{1}{2}A_0 + \sum_{n=1}^{\infty} A_n \cos\left(\frac{n\pi x}{L}\right) \]

cosine series!

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

\( n = 0, 1, 2, 3, \dots \)

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Example: Heat Equation with Insulated Ends

example \( L=30 \), \( k=1 \), insulated ends

initial heating profile

\[ f(x) = \begin{cases} 0 & 0 < x < 5 \\ 25 & 5 < x < 10 \\ 0 & 10 < x < 30 \end{cases} \]
A graph of the initial heating profile f(x). It shows a rectangular pulse of height 25 between x=5 and x=10, and zero elsewhere on the interval from 0 to 30.
\[ u(x,t) = \frac{25}{6} + \sum_{n=1}^{\infty} \frac{50}{n\pi} \left[ \sin\left(\frac{n\pi}{3}\right) - \sin\left(\frac{n\pi}{6}\right) \right] e^{-n^2\pi^2t/900} \cos\left(\frac{n\pi x}{30}\right) \]

Steady-state solution \( (t \to \infty) \)

\[ u(x,t) = \frac{25}{6} \quad (\text{constant}) \]
A graph of the steady-state solution. It shows a horizontal line at f(x) = 25/6 across the entire interval from x=0 to x=30.

note: the area under the initial profile = area under the steady-state solution

PAGE 7

Heat Equation Analysis

because heat cannot leave the rod, and heat doesn't want any spot to be hotter/colder than the avg. value near it and no concavity

PAGE 8

Heat Equation 3D Surface: \(u(x,t)\)

A 3D surface plot representing the heat equation solution u(x,t). The x-axis ranges from 0 to 30, the t-axis from 0 to 500, and the vertical u(x,t) axis from 0 to 25. An initial high-temperature spike at x=5 quickly dissipates over time, spreading heat across the spatial domain until the surface becomes nearly flat at later times.
PAGE 9

Numerical Solution Visualization: Spatial Distribution

The following graph illustrates the numerical solution \( u(x, t) \) as a function of position \( x \) for various time steps \( t \). This visualization demonstrates the diffusion and smoothing of an initial rectangular pulse over time.

A line graph showing u(x,t) versus x for various time values. At t=0, there is a sharp rectangular pulse between x=5 and x=10 with a height of 25. As time increases (t=5, 20, 100, 500), the pulse spreads out, decreases in peak height, and eventually flattens toward a constant value across the domain.

Key Observations

  • At \( t = 0 \), the solution is a discontinuous top-hat function with high-frequency oscillations (Gibbs phenomenon) at the edges.
  • As \( t \) increases, the sharp edges are smoothed out, representing the diffusive nature of the underlying partial differential equation.
  • For very large \( t \) (e.g., \( t = 500 \)), the solution approaches a steady-state uniform distribution.
PAGE 10

Numerical Solution Visualization: Temporal Evolution

This graph depicts the temporal evolution of the solution \( u(x, t) \) at specific spatial locations \( x \). It shows how the value at different points in the domain converges toward a common equilibrium value over time.

A line graph showing u(x,t) versus time t for various spatial positions x. The curve for x=7.5 starts at a high value of 25 and decays rapidly. Other curves for x=2.5, 12.5, and 25 start at 0, rise to a peak, and then decay. All curves converge to a value of approximately 4.2 as t approaches 500.

Analysis of Temporal Decay

  • The point \( x = 7.5 \) is inside the initial pulse, starting at the maximum value and decaying immediately.
  • Points outside the initial pulse (e.g., \( x = 2.5, 12.5, 25 \)) initially see an increase in \( u \) as the pulse diffuses toward them, followed by a slow decay.
  • All curves asymptotically approach the same constant value, indicating the system is reaching thermal or concentration equilibrium.