Chain rule helper: if $h$ is differentiable at $a + b$, then the map $s \mapsto h(a + s)$ has derivative $h'(a + b)$ at $b$.
Chain rule helper: if $h$ is differentiable at $a - b$, then the map $s \mapsto h(a - s)$ has derivative $-h'(a - b)$ at $b$.
Chain rule helper: if $h$ is differentiable at $b + a$, then $s \mapsto h(s + a)$ has derivative $h'(b + a)$ at $b$.
Chain rule helper: if $h$ is differentiable at $b - a$, then $s \mapsto h(s - a)$ has derivative $h'(b - a)$ at $b$.
Uniqueness for the wave equation with zero initial data: if $w \in C^2(\mathbb{R}^2)$ satisfies $\partial_t^2 w = \partial_x^2 w$, $w(0, x) = 0$, and $\partial_t w(0, x) = 0$ for all $x$, then $w \equiv 0$.
The difference of two solutions to the wave equation is itself a solution: if $u_1, u_2$ both satisfy $\partial_t^2 u = \partial_x^2 u$, then so does $u_1 - u_2$.
If two $C^2$ functions $u_1, u_2$ have equal initial $t$-derivatives at $t = 0$, then the $t$-derivative of their difference vanishes at $t = 0$.
Uniqueness for the wave equation: two $C^2$ solutions of the 1D wave equation that share the same initial position $u(0, x)$ and the same initial velocity $\partial_t u(0, x)$ must agree everywhere.
Null decomposition for the wave equation: every $C^2$ solution $u$ of the 1D wave equation can be written as $u(t, x) = F(x + t) + G(x - t)$ for some differentiable $F, G : \mathbb{R} \to \mathbb{R}$ (travelling waves).
In a null decomposition $u(t, x) = F(x + t) + G(x - t)$ matching initial data $f$ and $g$, the right-moving wave $F$ satisfies $F'(x) = (f'(x) + g(x))/2$.
Any null decomposition $F(x + t) + G(x - t)$ that matches the initial position $f$ and initial velocity $g$ must coincide with d'Alembert's solution.
Theorem 4.1 (d'Alembert's formula, uniqueness part). The unique $C^2$ solution $u$ of the 1D wave equation $-\partial_t^2 u + \partial_x^2 u = 0$ with initial data $u(0, x) = f(x)$ and $\partial_t u(0, x) = g(x)$ is given by d'Alembert's formula.
If $f(0) = 0$, then the odd extension of $f$ is an odd function: $\tilde{f}(-y) = -\tilde{f}(y)$.
If $g : \mathbb{R} \to \mathbb{R}$ is continuous with $g(0) = 0$, then its odd extension is continuous.
If $g$ is continuous on $[0, \infty)$ with $g(0) = 0$, then its odd extension is continuous on all of $\mathbb{R}$.
Derivative of the odd extension (from $C^1$ on $[0, \infty)$): for $g \in C^1([0, \infty))$ with $g(0) = 0$, the odd extension is differentiable everywhere, with derivative $g'(\lvert x \rvert)$ at $x$.
If $g \in C^1([0, \infty))$ with $g(0) = 0$, then the odd extension of $g$ is in $C^1(\mathbb{R})$.
Derivative of an even extension: for $h \in C^1([0, \infty))$ with $h'(0) = 0$, the function $y \mapsto h(\lvert y \rvert)$ is differentiable everywhere with derivative given by the odd extension of $h'$.
If $f \in C^2([0, \infty))$ with $f(0) = 0$ and $f''(0) = 0$, then the odd extension of $f$ is in $C^2(\mathbb{R})$. The compatibility condition $f''(0) = 0$ is necessary for smoothness across the origin.
Derivative of the odd extension (from $C^1$ on all of $\mathbb{R}$): if $g \in C^1(\mathbb{R})$ and $g(0) = 0$, then the odd extension is differentiable everywhere with derivative $g'(\lvert x \rvert)$ at $x$.
If $g \in C^1(\mathbb{R})$ with $g(0) = 0$, then the odd extension of $g$ is in $C^1(\mathbb{R})$.
Derivative of an even extension (from $C^1$ on all of $\mathbb{R}$): if $h \in C^1(\mathbb{R})$ and $h'(0) = 0$, then $y \mapsto h(\lvert y \rvert)$ is differentiable everywhere with derivative the odd extension of $h'$.
Half-line d'Alembert formula in the case $0 \le t \le x$ (Corollary 4.0.1, first branch): $u(t, x) = \frac{1}{2}(f(x + t) + f(x - t)) + \frac{1}{2}\int_{x-t}^{x+t} g(z)\,dz$.
Half-line d'Alembert formula in the case $0 \le x \le t$ (Corollary 4.0.1, second branch): $u(t, x) = \frac{1}{2}(f(x + t) - f(t - x)) + \frac{1}{2}\int_{t - x}^{x + t} g(z)\,dz$.
Regularity for the half-line d'Alembert solution: under the standard assumptions plus the compatibility condition $f''(0) = 0$, the half-line d'Alembert solution is in $C^2$ on $[0, \infty) \times [0, \infty)$.
The half-line d'Alembert solution is jointly $C^2$ on $[0, \infty) \times [0, \infty)$ when $f \in C^2$, $g \in C^1$, $f(0) = g(0) = 0$, and $f''(0) = 0$.
For a solution $u$ of the half-line wave problem with boundary condition $u(t, 0) = 0$, the second spatial derivative of the odd extension (in $x$) vanishes at $x = 0$, for every fixed $t \ge 0$.
$C^2$ regularity of the joint odd extension in $x$: if $u$ is $C^2$ on $[0, \infty) \times [0, \infty)$ with $u(t, 0) = 0$ and the second spatial derivative vanishing at the boundary, then the extension $(t, x) \mapsto u(t, x)$ for $x \ge 0$ and $-u(t, -x)$ for $x < 0$ is jointly $C^2$ on $\mathbb{R}^2$.
The joint odd extension (in $x$) of a half-line wave solution is jointly $C^2$ on $\mathbb{R}^2$, under the standard regularity and compatibility hypotheses.
The joint odd extension (in $x$) of a half-line wave solution continues to satisfy the wave equation $\partial_t^2 = \partial_x^2$ also at negative times $t < 0$.
The joint odd extension (in $x$) of a half-line wave solution is jointly $C^2$ on $\mathbb{R}^2$ and satisfies the wave equation $\partial_t^2 = \partial_x^2$ on all of $\mathbb{R}^2$.
The joint odd extension $\tilde{u}$ (in $x$) of a half-line wave solution $u$ solves the full-line wave equation on $\mathbb{R}^2$ with initial position $\tilde{f}$ (the odd extension of $f$) and initial velocity $\tilde{g}$ (the odd extension of $g$).
Uniqueness for the half-line wave equation: any $C^2$ solution $u$ on $[0, \infty) \times [0, \infty)$ with $u(t, 0) = 0$, $u(0, x) = f(x)$ and $\partial_t u(0, x) = g(x)$ must coincide with the half-line d'Alembert solution.
Corollary 4.0.1: the unique $C^2$ solution to the $1+1$ dimensional initial + boundary value problem $-\partial_t^2 u + \partial_x^2 u = 0$ on $[0, \infty) \times [0, \infty)$ with $u(t, 0) = 0$, $u(0, x) = f(x)$, and $\partial_t u(0, x) = g(x)$ is given by the half-line d'Alembert formula. It coincides with $\frac{1}{2}(f(x+t) + f(x-t)) + \frac{1}{2} \int_{x-t}^{x+t} g(z)\,dz$ for $0 \le t \le x$, and with $\frac{1}{2}(f(x+t) - f(t-x)) + \frac{1}{2}\int_{t-x}^{x+t} g(z)\,dz$ for $0 \le x \le t$.