The wave equation for a string of length $L$ fixed at both ends:

$\frac{\partial^2 u}{\partial t^2} = c^2\frac{\partial^2 u}{\partial x^2}, \quad u(0,t) = u(L,t) = 0$

Boundary conditions require sine modes:$u_n(x,t) = \sin\left(\frac{n\pi x}{L}\right)\left[A_n\cos(\omega_n t) + B_n\sin(\omega_n t)\right]$

where $\omega_n = n\pi c/L$ are the normal mode frequencies. The general solution is:

$u(x,t) = \sum_{n=1}^{\infty}\sin\left(\frac{n\pi x}{L}\right)\left[A_n\cos(\omega_n t) + B_n\sin(\omega_n t)\right]$

For initial conditions $u(x,0) = f(x)$ and $\partial_t u(x,0) = g(x)$, the coefficients are:

$A_n = \frac{2}{L}\int_0^L f(x)\sin\left(\frac{n\pi x}{L}\right)dx, \quad B_n = \frac{2}{\omega_n L}\int_0^L g(x)\sin\left(\frac{n\pi x}{L}\right)dx$

A Gaussian wave packet in free space:

$E(x,0) = E_0 e^{-x^2/(2\sigma^2)}e^{ik_0 x}$

evolves according to the wave equation $\partial_t^2 E = c^2\partial_x^2 E$. Fourier decomposing:

$\tilde{E}(k,0) = E_0\sigma\sqrt{2\pi}e^{-(k-k_0)^2\sigma^2/2}$

Each component evolves as $e^{i(kx \pm \omega(k)t)}$ with $\omega(k) = c|k|$:

$E(x,t) = \frac{E_0}{2}\int e^{i(kx - ckt)}e^{-(k-k_0)^2\sigma^2/2}dk$

This gives a wave packet moving at speed $c$ with carrier wave at $k_0$ and envelope spread $\sigma$.

White noise has constant power spectral density $S(\omega) = S_0$ for all frequencies, implying:

$R(\tau) = S_0\delta(\tau)$

i.e., the signal is uncorrelated at different times. Physical white noise is band-limited (e.g., Johnson noise in resistors has spectral density $S_0 = 4k_B T R$ up to cutoff frequency $\sim k_B T/\hbar$).

The wave equation $(\partial_t^2 - c^2\nabla^2)G(\mathbf{x},t) = \delta^3(\mathbf{x})\delta(t)$ with retarded boundary conditions (causality) has Fourier transform:

$(-\omega^2 + c^2k^2)\tilde{G}(\mathbf{k},\omega) = 1$

giving:

$\tilde{G}(\mathbf{k},\omega) = \frac{1}{c^2k^2 - \omega^2 - i\epsilon}$

The $i\epsilon$ prescription ensures the retarded solution. Inverting the Fourier transform:

$G(\mathbf{x},t) = \frac{\delta(t - |\mathbf{x}|/c)}{4\pi|\mathbf{x}|}$

This shows that disturbances propagate on the light cone $|\mathbf{x}| = ct$.

For the diffusion equation $\partial_t u = D\nabla^2 u$ in $\mathbb{R}^3$ with $u(\mathbf{x},0) = \delta^3(\mathbf{x})$:

$\partial_t\tilde{u} = -Dk^2\tilde{u} \implies \tilde{u}(\mathbf{k},t) = e^{-Dk^2 t}$

Inverting (a Gaussian integral):

$u(\mathbf{x},t) = \frac{1}{(4\pi Dt)^{3/2}}e^{-|\mathbf{x}|^2/(4Dt)}$

This describes how an initial point concentration spreads out with characteristic radius $\sqrt{Dt}$, fundamental to Brownian motion and heat conduction.

Light passing through an aperture $A(x,y)$ produces a far-field diffraction pattern. The electric field amplitude in the focal plane of a lens is proportional to the Fourier transform of the aperture:

$E(k_x, k_y) \propto \int\int A(x,y)e^{-i(k_x x + k_y y)}dx\,dy$

For a circular aperture of radius $a$, this gives the Airy pattern:

$I(\theta) \propto \left|\frac{J_1(ka\sin\theta)}{ka\sin\theta}\right|^2$

where $J_1$ is the Bessel function. The first dark ring occurs at $\sin\theta = 1.22\lambda/(2a)$, defining the Rayleigh criterion for optical resolution.