Consider the Cauchy problem: $u_{tt} = c^2 u_{xx}, \quad u(x,0) = f(x), \quad u_t(x,0) = g(x)$

Step 1: Change of Variables

Introduce characteristic coordinates: $\xi = x - ct, \quad \eta = x + ct$

By the chain rule: $\frac{\partial}{\partial x} = \frac{\partial}{\partial \xi} + \frac{\partial}{\partial \eta}, \quad \frac{\partial}{\partial t} = -c\frac{\partial}{\partial \xi} + c\frac{\partial}{\partial \eta}$

Step 2: Transform the PDE

Computing second derivatives: $u_{tt} = c^2\left(\frac{\partial^2 u}{\partial \xi^2} - 2\frac{\partial^2 u}{\partial \xi \partial \eta} + \frac{\partial^2 u}{\partial \eta^2}\right)$ $u_{xx} = \frac{\partial^2 u}{\partial \xi^2} + 2\frac{\partial^2 u}{\partial \xi \partial \eta} + \frac{\partial^2 u}{\partial \eta^2}$

Substituting into $u_{tt} = c^2 u_{xx}$: $c^2\left(\frac{\partial^2 u}{\partial \xi^2} - 2\frac{\partial^2 u}{\partial \xi \partial \eta} + \frac{\partial^2 u}{\partial \eta^2}\right) = c^2\left(\frac{\partial^2 u}{\partial \xi^2} + 2\frac{\partial^2 u}{\partial \xi \partial \eta} + \frac{\partial^2 u}{\partial \eta^2}\right)$

Simplifying: $\frac{\partial^2 u}{\partial \xi \partial \eta} = 0$

Step 3: General Solution

Since $\frac{\partial^2 u}{\partial \xi \partial \eta} = 0$, we have $\frac{\partial u}{\partial \xi} = h(\xi)$ for some function $h$.

Integrating: $u(\xi, \eta) = \int h(\xi)\,d\xi + k(\eta) = F(\xi) + G(\eta)$

In original coordinates: $u(x,t) = F(x - ct) + G(x + ct)$

Step 4: Apply Initial Conditions

From $u(x,0) = f(x)$: $F(x) + G(x) = f(x) \quad \text{...(1)}$

From $u_t(x,0) = g(x)$: $-cF'(x) + cG'(x) = g(x)$

Integrating: $-F(x) + G(x) = \frac{1}{c}\int_0^x g(s)\,ds + C \quad \text{...(2)}$

Step 5: Solve for F and G

Adding equations (1) and (2): $2G(x) = f(x) + \frac{1}{c}\int_0^x g(s)\,ds + C$

Subtracting equation (2) from (1): $2F(x) = f(x) - \frac{1}{c}\int_0^x g(s)\,ds - C$

Step 6: Final Formula

Substituting $F(x-ct)$ and $G(x+ct)$ (the constant $C$ cancels): $u(x,t) = \frac{1}{2}[f(x-ct) + f(x+ct)] + \frac{1}{2c}\int_{x-ct}^{x+ct} g(s)\,ds$

This is D'Alembert's formula.