PAGE 1

Wave Equation

models a vibrating string

string length \(L\), held tight, ends fixed

A straight horizontal line representing a string in its natural position, with endpoints labeled x=0 and x=L.

displacement from natural position is \(y(x,t)\)

A curved line representing a displaced string above a dashed horizontal line (y=0). The endpoints are labeled x=0 and x=L. An arrow points down from the curve to the dashed line, labeled y(x,t).

intuitively, the string does not "want" to be displaced from natural position

(like a spring)

so, if displaced like

A small concave-down curve representing a displaced string.

\(y_{xx} < 0\) , the string will have

a downward restoring acceleration \(\rightarrow y_{tt} < 0\)

\(\hookrightarrow\) negative
PAGE 2

similarly, if displaced like

A small sketch of a concave up curve.

\( y_{xx} > 0 \), string supplies an

upward restoring acceleration \( \rightarrow y_{tt} > 0 \)

\( \rightarrow \) acceleration is proportional to concavity (same sign)

\[ \frac{\partial^2 y}{\partial t^2} = a^2 \frac{\partial^2 y}{\partial x^2} \]
\[ y_{tt} = a^2 y_{xx} \]

1-D Wave Equation

\( a^2 \) determines wave/vibration speed

\( a^2 \) in context of string

is
tension
density

Boundary condition:

\( y(0, t) = y(L, t) = 0 \)

ends fixed at zero (Dirichlet)

\( y_x(0, t) = y_x(L, t) = 0 \)

A diagram showing a wave segment with vertical dashed lines at the boundaries. Arrows indicate the ends can move up and down while maintaining a horizontal slope.

ends fixed horizontally (Neumann)

(can move up/down)

PAGE 3

Initial Conditions and the Wave Equation

Initial conditions: two needed since time derivative is order 2

  • \( y(x, 0) = f(x) \) — initial displacement (plucking)
  • \( y_t(x, 0) = g(x) \) — initial velocity (strumming)

Today, we will solve the case w/ ends fixed at 0 and initial displacement only ("Problem A")

\[ y_{tt} = a^2 y_{xx} \]
\( 0 < x < L, \quad t > 0 \)
\[ y(0, t) = y(L, t) = 0 \]
ends fixed at 0
\[ y_t(x, 0) = 0 \]
no initial velocity
\[ y(x, 0) = f(x) \]
initial displacement

Method of Separation of Variables

We will again use the method of separation of variables

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

\[ y_{tt} = X T'' \]

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

PAGE 4

wave eq. becomes

\[ X T'' = a^2 X'' T \]

\[ \frac{X''}{X} = \frac{T''}{a^2 T} = \text{separation constant} = -\lambda \]
just like w/ heat eq.

we get two ODEs out of that:

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

\[ T'' + a^2 \lambda T = 0 \]

Boundary Conditions (BC)

  • \( y(0, t) = 0 \rightarrow X(0)T(t) = 0 \rightarrow \) \( X(0) = 0 \)
  • \( y(L, t) = 0 \rightarrow X(L)T(t) = 0 \rightarrow \) \( X(L) = 0 \)

\( X \) solution is something we've seen before in heat eq.

\[ \lambda_n = \frac{n^2 \pi^2}{L^2} \]
eigenvalues
\[ X_n = \sin\left(\frac{n \pi x}{L}\right) \]
eigenfunctions

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

PAGE 5

Wave Equation: Time Solution and Modes

\[ T'' + a^2 \lambda T = 0 \]

\[ T'' + \frac{a^2 n^2 \pi^2}{L^2} T = 0 \]

\[ T(t) = A \cos\left( \frac{an\pi}{L} t \right) + B \sin\left( \frac{an\pi}{L} t \right) \]

IC: \( y_t(x, 0) = 0 \)

\( X(x) T'(0) = 0 \rightarrow T'(0) = 0 \)

no initial velocity

\( \vdots \)

\( B = 0 \)

\[ T_n = \cos\left( \frac{n\pi a}{L} t \right) \]

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

for wave, the time solution is also periodic

for each \( n \), there is one solution: \( y_n = X_n T_n \)

each of these is called

a "mode" or "harmonic"

PAGE 6

General Solution and Initial Conditions

general solution:

\[ y(x, t) = \sum_{n=1}^{\infty} A_n \cos\left( \frac{n\pi a}{L} t \right) \sin\left( \frac{n\pi}{L} x \right) \]

IC: \( y(x, 0) = f(x) \) initial displacement

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

sine series

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

example

\( L = 20, \quad a = 1, \quad g(x) = 0 \) (no initial velocity)

initial displacement \( f(x) = \)

\[ \begin{cases} \frac{1}{5}x & 0 < x < 5 \\ 1 & 5 < x < 15 \\ \frac{20-x}{5} & 15 < x < 20 \end{cases} \]

A graph of the initial displacement function f(x) over the interval 0 to 20. The function starts at (0,0), rises linearly to (5,1), remains constant at y=1 until x=15, and then decreases linearly back to (20,0), forming a trapezoidal shape.
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Wave Equation Solution and Harmonics

\[ y(x,t) = \sum_{n=1}^{\infty} \frac{8}{n^2\pi^2} \left[ \sin\left(\frac{n\pi}{4}\right) + \sin\left(\frac{3n\pi}{4}\right) \right] \cos\left(\frac{n\pi t}{20}\right) \sin\left(\frac{n\pi x}{20}\right) \]

\( n=1 \):

\[ y_1(x,t) = \frac{8\sqrt{2}}{\pi^2} \cos\left( \underbrace{\frac{\pi}{20}}_{\text{freq.}} t \right) \sin\left( \frac{\pi}{20} x \right) \]

\(\rightarrow\) freq. of vibration of the fundamental mode (\(n=1\))

\[ \frac{\pi}{20} \text{ rad/s} = \frac{\pi/20}{2\pi} = \frac{1}{40} \text{ Hz} \]

\( n=2 \):

Second harmonic

freq. is \( 2 \cdot \frac{\pi}{20} = \frac{1}{20} \text{ Hz} \)

(double of 1st harmonic)

one octave higher than fundamental

\( n=3 \):

Third harmonic

(an octave and a perfect fifth above the fundamental)

the sound we hear is ALL n's put together

PAGE 8

Surface Plot: \( y(x,t) \)

A 3D surface plot showing the displacement y as a function of position x and time t. The surface has a wave-like form with peaks and valleys, colored with a gradient from blue for negative displacement to yellow for positive displacement. The x-axis ranges from 0 to 20, the t-axis from 0 to 40, and the y-axis from -1.00 to 1.00.
PAGE 9

String Displacement Analysis

String Displacement at Snapshots in Time (̲y vs ̲x)

The following graph illustrates the displacement of a string at various discrete time intervals. The horizontal axis represents the position along the string (̲x), ranging from 0.0 to 20.0 units. The vertical axis represents the displacement (̲y), ranging from -1.00 to 1.00 units.

A line graph showing string displacement y versus position x at six different time snapshots: t=0, t=4, t=8, t=12, t=16, and t=20. The curves show a trapezoidal wave pulse that evolves over time, reflecting across the x-axis and changing amplitude.

Legend of Time Snapshots

  • t = 0: Initial state with maximum positive displacement.
  • t = 4: Transitioning state.
  • t = 8: Reduced positive amplitude.
  • t = 12: Transitioning to negative displacement.
  • t = 16: Increased negative amplitude.
  • t = 20: Maximum negative displacement.
PAGE 10

Temporal Oscillation Analysis

Oscillation of Specific Points (̲y vs ̲t)

This graph tracks the displacement (̲y) over time (̲t) for three specific points along the string: \(x = 5\), \(x = 10\), and \(x = 15\). The time axis ranges from 0 to 40 units.

A line graph showing displacement y versus time t for three positions: x=5, x=10, and x=15. The curves for x=5 and x=15 are identical, showing a V-shaped oscillation, while x=10 shows a wider trapezoidal oscillation reaching -1.00 at t=20.

Position Legend

  • x = 5: Blue line showing periodic oscillation.
  • x = 10: Orange line showing a broader dwell at peak displacements.
  • x = 15: Green line, overlapping with the x=5 trajectory.