The one-dimensional wave equation is a second-order linear partial differential equation (PDE) that describes the propagation of waves, such as sound waves, light waves, and water waves, in a medium along a single spatial dimension. It is a fundamental equation in mathematical physics and engineering and serves as a prototype for many more complex wave-like phenomena. The standard form of the wave equation is:
$$\frac{\partial^2 u}{\partial t^2} = c^2 \frac{\partial^2 u}{\partial x^2}$$
where:
- $u(x,t)$ is the displacement at position $x$ and time $t$,
- $c$ is the speed of wave propagation,
- $\frac{\partial^2 u}{\partial t^2}$ is the second-order time derivative,
- $\frac{\partial^2 u}{\partial x^2}$ is the second-order spatial derivative.

Let us derive the wave equation for a vibrating string under tension.
Assumptions:
- The string is perfectly flexible and homogeneous.
- The motion is only in the vertical plane (transverse).
- No damping or external forces act on the string.
Let:
- $T$ be the constant tension in the string,
- $\mu$ be the mass per unit length,
- $u(x,t)$ be the transverse displacement at point $x$ and time $t$.
Consider an infinitesimal element of the string between $x$ and $x + \Delta x$.
By Newton's second law:
$$\mu \Delta x \cdot \frac{\partial^2 u}{\partial t^2} = T \sin(\theta_2) - T \sin(\theta_1)$$
Using the small-angle approximation: $\sin(\theta) \approx \tan(\theta) = \frac{\partial u}{\partial x}$, we get:
$$\mu \Delta x \cdot \frac{\partial^2 u}{\partial t^2} = T\left(\frac{\partial u}{\partial x}(x + \Delta x) - \frac{\partial u}{\partial x}(x)\right)$$
Dividing by $\Delta x$ and taking the limit as $\Delta x \to 0$:
$$\frac{\partial^2 u}{\partial t^2} = \frac{T}{\mu} \frac{\partial^2 u}{\partial x^2}$$
Thus:
$$\frac{\partial^2 u}{\partial t^2} = \frac{T}{\mu} \frac{\partial^2 u}{\partial x^2}$$
Let $c^2 = \frac{T}{\mu}$, then we obtain the standard wave equation:
$$\frac{\partial^2 u}{\partial t^2} = c^2 \frac{\partial^2 u}{\partial x^2}$$

The general solution is given by D'Alembert’s formula:
$$u(x,t) = f(x - ct) + g(x + ct)$$
Where:
- $f$ and $g$ are arbitrary functions,
- $f(x - ct)$ represents a wave traveling to the right,
- $g(x + ct)$ represents a wave traveling to the left.
This form shows that the wave travels without changing its shape at speed $c$.

Suppose we are given:
- Initial displacement: $u(x,0) = \varphi(x)$
- Initial velocity: $\frac{\partial u}{\partial t}(x,0) = \psi(x)$
Then, D’Alembert’s solution becomes:
$$u(x,t) = \frac{1}{2} \left[ \varphi(x - ct) + \varphi(x + ct) \right] + \frac{1}{2c} \int_{x - ct}^{x + ct} \psi(s) \, ds$$

There are common types of boundary conditions:
1. Fixed End Conditions (Dirichlet BC): $u(0,t) = 0$, $u(L,t) = 0$
2. Free End Conditions (Neumann BC): $\frac{\partial u}{\partial x}(0,t) = 0$, $\frac{\partial u}{\partial x}(L,t) = 0$
3. Mixed Conditions: One end fixed, other free.

To solve using this method:
Assume $u(x,t) = X(x)T(t)$
Substitute into the wave equation:
$$X(x)T''(t) = c^2 X''(x) T(t)$$
Divide both sides:
$$\frac{T''(t)}{c^2 T(t)} = \frac{X''(x)}{X(x)} = -\lambda$$
This gives two ODEs:
- $X'' + \lambda X = 0$
- $T'' + \lambda c^2 T = 0$
The form of solutions depends on boundary conditions.

Let $0 < x < L$, and BCs:
- $u(0,t) = 0$, $u(L,t) = 0$
Solving the spatial part:
$$X'' + \lambda X = 0, \, X(0)= X(L) = 0$$
Solution:
$$X_n(x) = \sin\left(\frac{n \pi x}{L}\right), \; \lambda_n = \left(\frac{n \pi}{L}\right)^2$$
Time part:
$$T_n(t) = A_n \cos\left(\frac{n \pi ct}{L}\right) + B_n \sin\left(\frac{n \pi ct}{L}\right)$$
General solution:
$$u(x,t) = \sum_{n=1}^{\infty} \left[A_n \cos\left(\frac{n \pi ct}{L}\right) + B_n \sin\left(\frac{n \pi ct}{L}\right)\right] \sin\left(\frac{n \pi x}{L}\right)$$
Coefficients $A_n$, $B_n$ determined from initial conditions using Fourier series.