Download original notes
PAGE 1

Heat Eq. (part 3)

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

now we will insulate the ends

heat cannot flow through the ends

A horizontal cylinder representing a rod. An arrow points to the top surface labeled 'lateral surface insulated'. Arrows point to the left end at x=0 and the right end at x=L, both labeled 'insulate'.

this changes the boundary conditions :

\[ \begin{cases} u_x(0, t) = 0 \\ u_x(L, t) = 0 \end{cases} \]

Neumann BC's

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

same method: separation of variables

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

\[ u_t = X T' \]

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

\[ u_t = k u_{xx} \rightarrow X T' = k X'' T \]

PAGE 2
\[ \frac{\bar{X}''}{\bar{X}} = \frac{T'}{kT} = -\lambda \]
separation constant

two ODE's result:

\[ \bar{X}'' + \lambda \bar{X} = 0 \] \[ T' + k\lambda T = 0 \]

BC's:

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

let's solve

\[ \bar{X}'' + \lambda \bar{X} = 0 \] \[ \bar{X}'(0) = \bar{X}'(L) = 0 \]
\[ \bar{X} = A \cos(\sqrt{\lambda} x) + B \sin(\sqrt{\lambda} x) \] \((\lambda \neq 0)\)
\[ \bar{X}'(x) = -\sqrt{\lambda} A \sin(\sqrt{\lambda} x) + \sqrt{\lambda} B \cos(\sqrt{\lambda} x) \] \[ \bar{X}'(0) = 0 = \sqrt{\lambda} B \quad \rightarrow \quad B = 0 \]
\[ \bar{X}'(L) = 0 = -\sqrt{\lambda} A \sin(\sqrt{\lambda} L) \] require \(A \neq 0\)
PAGE 3

\[ \sin(\sqrt{\lambda} L) = 0 \]

for each \( n \),

\[ \sqrt{\lambda} L = n\pi \]

\( n = 1, 2, 3, \dots \)
\[ \lambda_n = \frac{n^2 \pi^2}{L^2} \]

eigenvalue

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

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

eigenfunctions

(cosine instead of sine)

(drop the scaling constant A)

what if \( \lambda = 0 \) in \( X'' + \lambda X = 0 \)

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

\[ X = Ax + B \]

\[ X' = A \]

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

so, the solution is multiple of 1 (drop B the scaling constant)

\[ X = 1 \]

if

\[ \lambda = 0 \]
notice these are included in the previous case for \( n = 0 \)
PAGE 4

therefore solutions are

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

now \( T' + k\lambda T = 0 \) (exactly the same as in the non-insulated case)

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

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

general solution is linear combination of all

\[ u_n = X_n T_n \]

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

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

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

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

cosine series (missing \( 1/2 \))

PAGE 5

we will artificially include a factor of \(\frac{1}{2}\) for \(n=0\)

\[ u(x,t) = \frac{1}{2} A_0 + \sum_{n=1}^{\infty} A_n e^{-kn^2\pi^2t/L^2} \cos\left(\frac{n\pi x}{L}\right) \]

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

\[ f(x) = \frac{1}{2} A_0 + \sum_{n=1}^{\infty} A_n \cos\left(\frac{n\pi x}{L}\right) \]
Standard cosine series
\[ A_n = \frac{2}{L} \int_{0}^{L} f(x) \cos\left(\frac{n\pi x}{L}\right) dx \]

Example

\(L=30\)    \(K=1\)    insulated ends

initial condition: \[ f(x) = \begin{cases} 0 & 0 < x < 5 \\ 25 & 5 < x < 10 \\ 0 & 10 < x < 20 \end{cases} \]

A graph of the initial condition function f(x). It shows a rectangular pulse of height 25 between x=5 and x=10, and is zero elsewhere on the interval from 0 to 30.

\(\vdots\)

\[ u(x,t) = \frac{25}{6} + \sum_{n=1}^{\infty} \frac{50}{n\pi} \left[ \sin\left(\frac{n\pi}{3}\right) - \sin\left(\frac{n\pi}{6}\right) \right] e^{-n^2\pi^2t/900} \cos\left(\frac{n\pi x}{30}\right) \]
PAGE 6

steady-state (\(t \to \infty\))

\(u = \frac{25}{6}\)

A graph showing the steady-state solution u as a horizontal line at the value 25/6 across the interval from x=0 to x=30.

this is the average initial value

(area under the first graph is the same as area under the second)

  • heat cannot leave or enter after \(t=0\)
  • heat doesn't want to be uneven (no one spot hotter/colder than nearby average)
  • heat also doesn't want any concavity (\(u_{xx} = 0\) as \(t \to \infty\))
PAGE 7

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

A 3D surface plot representing the solution to the heat equation. The x-axis ranges from 0 to 30, the t-axis from 0 to 500, and the vertical axis u(x,t) from 0 to 25. A sharp rectangular pulse at t=0 quickly diffuses and flattens as time t increases, spreading across the x-domain.
PAGE 8

\(u(x,t)\) vs \(x\) for various \(t\)

A 2D line graph showing the evolution of the heat distribution over time. The x-axis ranges from 0 to 30 and the y-axis u(x,t) from 0 to 25. Five curves are shown: t=0 is a sharp blue rectangular pulse between x=5 and x=10; t=5 is a rounded orange peak; t=20 is a lower green curve; t=100 is a flatter red curve; and t=500 is a nearly horizontal purple line, illustrating the diffusion process.
PAGE 9

Analysis of Temperature Distribution Over Time

Temporal Evolution of \( u(x,t) \)

The following graph illustrates the behavior of the function \( u(x,t) \) as a function of time \( t \) for several fixed spatial positions \( x \). This visualization demonstrates how the system approaches a steady state over a long time interval.

A line graph showing u(x,t) versus t for four different values of x: 2.5, 7.5, 12.5, and 25. The horizontal axis represents time t from 0 to 500, and the vertical axis represents u(x,t) from 0 to 25. For x = 7.5, the curve starts very high and decays rapidly. For x = 2.5 and 12.5, the curves rise to a peak before decaying. For x = 25, the curve rises slowly from zero. All curves eventually converge toward a steady-state value of approximately 4.2 as t approaches 500.

Observations

  • For small values of \( x \) (e.g., \( x = 7.5 \)), the initial value of \( u \) is significantly higher, followed by a rapid exponential-like decay.
  • At intermediate positions (e.g., \( x = 2.5 \) and \( x = 12.5 \)), the function exhibits a transient increase before settling into a decay phase.
  • At the furthest position shown (\( x = 25 \)), the value starts at zero and gradually increases toward the equilibrium value.
  • As \( t \to \infty \), all curves converge to a constant value, indicating the system reaches thermal or chemical equilibrium regardless of the spatial coordinate \( x \).