Wave Equation

If a harmonic oscillator is coupled with another one, the disturbance will be transferred to its neighbor. In a long chain of oscillators, the oscillation propagates through all of them in the form of a wave. In this case, the parameter $c$ usually denotes the propagation speed. The momentary elongation at the position $x$ at the given time $t$ can then be written as: $$\psi(x,t) = A\sin\left(\omega\left(t-\frac{x}{c}\right)\right)$$ By keeping the time constant, the only remaining variable is the position $x$, i.e. the elongation along the $x$-axis can also be described with the help of the sine function. The distance between two points with the same elongation is called wavelength and is usually denoted as $\lambda$. The wave has traveled the distance $\lambda$ in the given period time $T$. The speed of propagation can therefore be written according to $v=s/t$ as follows: $$c = \frac{\lambda}{T}$$ If $T$ is replaced by $1/f$, the following very important correlation between the wavelength and the period can be obtained: $$\boxed{c = f\lambda}$$ Similar to the angular frequency $\omega = 2\pi/T$, a wavenumber $k$ can be introduced that is defined as $$\boxed{k = \frac{2\pi}{\lambda}}$$ The wave function for harmonic waves can then be written as: $$\boxed{\psi(x,t) = A\sin(\omega t - kx)}$$ In order to find the so-called wave equation, we can compute the derivation of the wave function with respect to $x$ and $t$. $$\frac{\partial^2 \psi}{\partial t^2} = \omega^2 A \sin(\omega t - kx)$$ $$\frac{\partial^2 \psi}{\partial x^2} = k^2 A \sin(\omega t - kx)$$ In both cases, the chain rule of differentiation has been applied. Dividing this equation system by $\omega$ and using $$c = f\lambda = \omega k$$ we obtain the difference between both functions $$\boxed{\frac{\partial^2 \psi}{\partial x^2} - \frac{1}{c^2}\frac{\partial^2 \psi}{\partial t^2} = 0}$$ For the previous derivation, harmonic oscillations have been assumed. Now we want to show that the same equation can be used for arbitrary waves. For that purpose, we create a disturbance at a position $x_0$. After the time $t_1$, it will then reach the position $x_1 = x_0 + vt_1$. This condition leads to $$\psi(x_1,t_1) = \psi(x_1 - vt_1,0) = \psi(x_0,0)$$ Since the momentary elongation $\psi$ gets independent of $t$, it equals the following function: $$\psi(x,t) = f(x - vt)$$ A wave that is independent of the other two coordinates $y$ and $z$ is called a plane wave. In the following, the argument $x-vt$ will be replaced by $u$. Calculating the derivation of $\psi$ with respect to $x$ and $t$ then leads to $$\frac{\mathrm{d}\psi}{\mathrm{d}x} = \frac{\mathrm{d}f}{\mathrm{d}u}\frac{\mathrm{d}u}{\mathrm{d}x} = \frac{\mathrm{d}f}{\mathrm{d}u}$$ $$\frac{\mathrm{d}\psi}{\mathrm{d}t} = \frac{\mathrm{d}f}{\mathrm{d}u}\frac{\mathrm{d}u}{\mathrm{d}t} = -v\frac{\mathrm{d}f}{\mathrm{d}u}$$ The second derivation then results in $$\frac{\mathrm{d}^2\psi}{\mathrm{d}x^2} = \frac{\mathrm{d}f^2}{\mathrm{d}u^2}$$ $$\frac{\mathrm{d}^2\psi}{\mathrm{d}t^2} = v^2\frac{\mathrm{d}f^2}{\mathrm{d}u^2}$$ The difference of both functions is then exactly equal to the previously derived wave equation for harmonic waves: $$\boxed{\frac{\partial^2 \psi}{\partial x^2} - \frac{1}{v^2}\frac{\partial^2 \psi}{\partial t^2} = 0}$$