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PAGE 1

Heat Eq. (part 2)

last time:

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

\[ u(0, t) = 0 \quad \text{left end temp} = 0 \]

\[ u(L, t) = 0 \quad \text{right end temp} = 0 \]

\[ u(x, 0) = f(x) \quad \text{initial heat profile} \]

A diagram of a horizontal rod with endpoints at x equals 0 and x equals L. Above the rod is a wavy line representing the initial heat profile function f of x.

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

\[ X(x) = \sin\left(\frac{n\pi x}{L}\right) \]
spatial solution
\[ T(t) = e^{-kn^2\pi^2 t / L^2} \]
temporal solution

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

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

PAGE 2

example:

copper slab thickness 4 cm. \( k = 1.15 \text{ cm}^2/s \)

A 3D drawing of a rectangular slab with a small horizontal cylinder being pulled out from its center, labeled as a core sample. The thickness of the slab is indicated as 4 cm.

"core sample"

entire interior is heated uniformly to \( 100^{\circ}\text{C} \) at \( t = 0 \)

then both faces (left/right) are kept at \( 0^{\circ}\text{C} \) for all \( t \)

A horizontal cylinder representing a 1D rod with boundaries at x equals 0 and x equals 4.

entire plate is same material and uniformly heated

→ heat only flows left or right

(1-D Heat Eq.)

solution (from last page)

\[ u(x, t) = \sum_{n=1}^{\infty} C_n e^{-1.15 n^2 \pi^2 t / 16} \sin\left(\frac{n\pi x}{4}\right) \]

at \( t = 0, \) \( f(x) = 100 \)

PAGE 3
\[ C_n = \frac{2}{4} \int_{0}^{4} 100 \sin\left(\frac{n\pi x}{4}\right) dx = \frac{200 [1 - (-1)^n]}{n\pi} \]

the time solution: \( t \to \infty, u \to 0 \) (heat flows out since both ends are cold)

space solution: periodic because left end and right end are \( 0^{\circ}C \)

what is the temp. at mid point (x=2) 3 seconds later?

\[ u(2,3) = \sum_{n=1}^{\infty} \frac{200 [1 - (-1)^n]}{n\pi} e^{-1.15 n^2 \pi^2 \cdot 3 / 16} \sin\left(\frac{n\pi \cdot 2}{4}\right) \]
infinite sum

in practice, negative exponential \( \to \) fast convergence \( \to \) few terms needed for good approx.

1-term approx (n=1 only) \( \to 15.16^{\circ}C \)

PAGE 4

Surface Plot of Heat Equation Solution \( u(x, t) \)

A 3D surface plot showing the temperature distribution u(x, t) over position x from 0 to 4 and time t from 0 to 5. The surface starts as a high sine-like curve at t=0 and decays toward zero as time increases, with the ends fixed at zero temperature.
PAGE 5

Heat Equation Visualization: Spatial Profiles

The following graph illustrates the temperature profile \( u(x, t) \) across a spatial domain \( x \in [0, 4] \) at several discrete time intervals. The initial condition at \( t = 0 \) is a uniform temperature of 100, which then decays over time as the boundaries are held at 0.

A line graph showing temperature u(x, t) on the y-axis from 0 to 100 versus position x on the x-axis from 0 to 4.0. At t=0, the profile is a rectangular pulse at u=100. As time t increases from 0.1 to 5.0, the curves become increasingly smooth, bell-shaped, and lower in amplitude, converging toward zero.

Key Observations

  • At \( t = 0 \), the temperature is constant at 100 across the domain, except at the boundaries.
  • For \( t > 0 \), the temperature at the boundaries \( x=0 \) and \( x=4 \) is fixed at 0.
  • As time progresses, the peak temperature at the center (\( x=2.0 \)) decreases, and the profile flattens.
PAGE 6

Heat Equation Visualization: Temporal Decay

This graph depicts the temperature \( u(x, t) \) as a function of time \( t \) at specific fixed positions \( x \) within the domain. It highlights the rate of cooling at different points relative to the boundaries.

A line graph showing temperature u(x, t) on the y-axis from 0 to 100 versus time t on the x-axis from 0 to 5. Three curves represent positions x=0.5, x=1.0, and x=2.0. All curves start at u=100 and decay exponentially toward zero, with the position closest to the boundary (x=0.5) decaying the fastest.

Analysis of Decay Rates

  • The curve for \( x = 0.5 \) (closest to the boundary) shows the most rapid initial drop in temperature.
  • The curve for \( x = 2.0 \) (the center of the domain) maintains a higher temperature for longer, as it is furthest from the cooling boundaries.
  • All positions eventually converge to the steady-state temperature of 0 as \( t \to \infty \).
PAGE 7

Heat Conduction in Two Slabs

Two slabs stuck together, both outer faces at \( 0^\circ\text{C} \)

Left slab heated to \( 50^\circ\text{C} \) at \( t=0 \), right slab at \( 100^\circ\text{C} \)

A 3D diagram showing two rectangular slabs joined together. A cylindrical rod is highlighted passing through the center of the slabs. Labels indicate the outer faces are at 0 degrees Celsius.

Left: \( 50^\circ\text{C} \)

Right: \( 100^\circ\text{C} \)

\( 4 \text{ cm} \) thick each

A horizontal cylinder representing a rod of length L equals 8, with endpoints marked x equals 0 and x equals 8.

\( L = 8 \)

Initial temp:

\[ f(x) = \begin{cases} 50 & 0 < x < 4 \\ 100 & 4 < x < 8 \end{cases} \]

The solution for temperature distribution \( u(x, t) \) is given by:

\[ u(x, t) = \sum_{n=1}^{\infty} \frac{100}{n\pi} \left[ 1 + \cos\left(\frac{n\pi}{2}\right) - 2(-1)^n \right] e^{-1.15 n^2 \pi^2 t / 64} \sin\left(\frac{n\pi x}{8}\right) \]
PAGE 8

Surface Plot of Temperature Distribution

Surface Plot \( u(x, t) \) for \( L = 8 \), \( k = 1.15 \), with non-uniform initial conditions.

A 3D surface plot showing temperature u as a function of position x and time t. The initial state at t=0 shows a step function with values at 50 and 100. As time increases, the temperature profile smooths out and decays toward zero at the boundaries.

Plot Observations

  • The vertical axis represents temperature \( u \) from 0 to 100.
  • The horizontal axes represent position \( x \) (from 0 to 8) and time \( t \) (from 0 to 10).
  • The color gradient indicates temperature intensity, with yellow representing higher temperatures and dark purple representing lower temperatures.
PAGE 9

Temperature Profile \( u(x, t) \)

Parameters: \( L = 8, k = 1.15 \)

The following graph illustrates the spatial distribution of temperature \( u \) along a rod of length \( L = 8 \) at various time intervals \( t \). The initial state at \( t = 0 \) shows a step function, which gradually smooths out and decays toward zero over time due to thermal diffusion.

A line graph showing temperature u(x, t) versus position x from 0 to 8. At t=0, the profile is a step function with u=50 for x between 0 and 4, and u=100 for x between 4 and 8. As time t increases from 0.1 to 10.0, the sharp corners smooth out and the overall temperature across the rod decreases toward zero.

Legend and Observations

  • \( t = 0 \): Initial step distribution.
  • \( t = 0.1, 0.5, 1.0, 2.0, 5.0, 10.0 \): Successive profiles showing the diffusion process.
  • The boundary conditions are fixed at \( u(0, t) = 0 \) and \( u(8, t) = 0 \), causing the temperature to drop at the ends.
PAGE 10

Temperature \( u(x, t) \) Over Time

Temporal evolution at specific positions \( x \)

This graph depicts how the temperature \( u \) changes over time \( t \) at three fixed positions along the rod: \( x = 2.0 \), \( x = 4.0 \), and \( x = 6.0 \). Each curve shows an initial value corresponding to the step function at \( t = 0 \) followed by a decay as heat dissipates.

A line graph showing temperature u versus time t from 0 to 10 for three positions. The curve for x=6.0 starts at u=100 and decays rapidly. The curve for x=4.0 starts at u=75 and decays. The curve for x=2.0 starts at u=50 and decays more slowly. All curves converge toward zero as time increases.

Key Data Points

  • \( x = 2.0 \): Starts at \( u \approx 50 \).
  • \( x = 4.0 \): Starts at the midpoint of the step, \( u \approx 75 \).
  • \( x = 6.0 \): Starts at \( u \approx 100 \).
  • All positions exhibit exponential-like decay toward the steady-state temperature of 0.
PAGE 11

Non-homogeneous Boundary Conditions

now let's remove the ends at 0 temp constraint

A simple diagram of a horizontal rod with its left end at x=0 and its right end at x=L.

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

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

if \( T_1, T_2 \) are not zero, the BC's are nonhomogeneous

\[ u_t = k u_{xx} \]

Steady-state solution: \( t \to \infty \) or \( u_t = 0 \) (time no longer changes solution)

\[ u_t = 0, \quad u_{xx} = 0 \to u = C_1 + C_2 x \]

w/ \( u(0) = T_1, \quad u(L) = T_2 \)

\[ u = \frac{T_2 - T_1}{L} x + T_1 = v(x) \]

steady-state solution

(a straight line)

PAGE 12
A coordinate graph showing v(x) on the vertical axis and x on the horizontal axis. A straight line connects the point (0, T1) to the point (L, T2).

what about the transient solution?

when time still matters

\[ u_t = k u_{xx} \]

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

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

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

\[ v(x) = \frac{T_2 - T_1}{L} x + T_1 \]

define \( w(x, t) = u(x, t) - v(x) \)

(original minus steady-state)

\[ w(0, t) = u(0, t) - v(0) = 0 \]

\[ w(L, t) = u(L, t) - v(L) = 0 \]

\[ w_t = u_t \quad w_{xx} = u_{xx} \]

\[ u_t = k u_{xx} \longleftrightarrow w_t = k w_{xx} \]

\( w(x, t) \) has homogeneous BC's and \( w_t = k w_{xx} \)

use original solution for \( w \)

\[ w(x, t) = \sum_{n=1}^{\infty} B_n e^{-k n^2 \pi^2 t / L^2} \sin\left(\frac{n \pi x}{L}\right) \]

PAGE 13

\[ u(x,t) = v(x) + \sum_{n=1}^{\infty} B_n e^{-k n^2 \pi^2 t / L^2} \sin\left(\frac{n \pi x}{L}\right) \]

\[ u(x,t) = \left( \frac{T_2 - T_1}{L} x + T_2 \right) + \sum_{n=1}^{\infty} B_n e^{-k n^2 \pi^2 t / L^2} \sin\left(\frac{n \pi x}{L}\right) \]

initial condition: \( u(x,0) = f(x) \)

\[ f(x) = \left( \frac{T_2 - T_1}{L} x + T_2 \right) + \sum_{n=1}^{\infty} B_n \sin\left(\frac{n \pi x}{L}\right) \]

\[ [f(x) - v(x)] = \sum_{n=1}^{\infty} B_n \sin\left(\frac{n \pi x}{L}\right) \]
Sine series
\[ B_n = \frac{2}{L} \int_{0}^{L} [f(x) - v(x)] \sin\left(\frac{n \pi x}{L}\right) dx \]