Wave Equation

In the mathematical sense, a wave is any function that moves, and the wave equation is a second-order linear PDE (partial differential equation) to illustrate waves. This type of wave equation is also called the two-way wave equation. Before learning in detail about the wave equation, let’s recall a few terms and definitions that help us in deriving wave equations. They are:

PDE: A partial differential equation (PDE) is an equation that includes both a function and its partial derivatives. Most of the equations are written in x, y, z as the usual spatial variables and t for the time variable. Also, we know that some functions will measure various physical quantities, say u = u(x, y, z, t), which could depend on all three spatial variables and time, or some subset. Thus, we can represent the partial derivatives of u as follows:

u_x = ∂u/∂x

u_xx = ∂²u/∂x²

u_t = ∂u/∂t

u_xt = ∂²u/∂x∂t

Some specific partial differential equations that also occur in physics are given below. These are useful in deriving the wave equation.

Laplace’s equation:

∇²u = 0

This is satisfied by the temperature function given by u = u(x, y, z) in a stable body in thermal equilibrium or by the electrostatic potential u = u(x, y, z) in a region without electric charges.

Heat equation:

u_t = k∇²u

This is satisfied by the temperature function given by u = u(x, y, z, t) of a physical object that conducts heat.

k = parameter depending on the object’s conductivity

Wave Equation

A wave equation is a differential equation involving partial derivatives, representing some medium competent in transferring waves. Its solutions provide us with all feasible waves that can propagate. Generally, it includes a second-order derivative with respect to time, which derives from F = ma or something analogous, and a second derivative with respect to the position, which derives from F = -kx or the similar ones, i.e., inertia and elasticity. The one-dimensional wave equation of a vibrating elastic string is given by:

∂²u/∂t² = c² (∂²u/∂x²)

Where c² = T/ρ

The one-dimensional wave equation was discovered by Jean-Baptiste le Rond d’Alembert, a French scientist.

Wave Equation Derivation

We can derive the wave equation, i.e., one-dimensional wave equation using Hooke’s law. Consider the vital forces on a vibrating string proportional to the curvature at a certain point, as shown below.

Let us assume that,

u = u(x, t) = a string’s displacement from the neutral position u ≡ 0

ρ = ρ(x) = The mass density of the string

k = k(x) = elasticity

Assume a small piece of string in the interval [x, x + ∆x], such that its mass being ρ(x)∆x, and velocity ut(x, t).

Therefore, its kinetic energy, one half mass times velocity squared, is given by:

For the string, the total kinetic energy is given by an integral,

From Hooke’s law, we can write the following.

(k/2)y² = potential energy for a string, where

y = length of the string

For the stretched string, the length is given by arc length, i.e., ds = √(1 + u²_x) dx

Thus, the potential energy will be:

The movement for a given function “u” is represented as the integral, as:

By adding δ times a perturbation h = h(x, t) to the function “u” shows a new movement, such as:

According to the principle of least action, for the function “u” to be a physical solution, the first-order term should disappear for any perturbation h. Then, by simplifying integrals by parts, we get:

On further simplifications, we get:

ρ. u_tt = k . u_xx + k_x . u_x

When k is constant, i.e., the elasticity, we obtain:

u_tt = c² u_xx

This can be written as:

∂²u/∂t² = c² (∂²u/∂x²)

Wave Equation Solution

The solution of a wave equation is quite complicated, but it can be obtained using some linear combinations. Thus, we can get the solution for a wave equation by the characteristic curve of the integration by separable variables, which means the variable separable method of integration. However, the general solution for the wave equation of one-dimensional is of the form:

u(x,t) = F(x + ct) + G(x − ct)

Here, F and G are one variable’s differentiable functions.

The solution of this one-dimensional wave equation is uniquely determined by the initial conditions given below:

u(x, 0) = f(x) ….(1)

ut(x, 0) = g(x)….(2)

The domain of u(x,t) will be R = R × [0,∞).

The function u(x,t) satisfies the wave equation on the interior of R and the conditions (1), (2) on the boundary of R.