**Proof plan.** We introduce characteristic coordinates $\xi = x + t$, $\eta = x - t$, in which the [wave equation](/page/Wave%20Equation) reduces to $\partial_\xi \partial_\eta U = 0$. Two successive integrations yield $U(\xi, \eta) = F(\xi) + G(\eta)$ for one-variable profiles $F$ and $G$, so $u(x, t) = F(x + t) + G(x - t)$. Imposing the initial data produces a system of two equations for $F$ and $G$, which we solve by adding and subtracting. The formula follows by evaluating the resulting expressions at $x + t$ and $x - t$.
**Step 1: Change to characteristic coordinates.**
Define the characteristic coordinate map
\begin{align*}
\Psi: \mathbb{R} \times [0, \infty) &\to \mathbb{R}^2 \\
(x, t) &\mapsto (x + t,\, x - t)
\end{align*}
and set $\Omega = \Psi(\mathbb{R} \times [0, \infty)) = \{(\xi, \eta) \in \mathbb{R}^2 : \xi \geq \eta\}$. Define the transformed solution
\begin{align*}
U: \Omega &\to \mathbb{R} \\
(\xi, \eta) &\mapsto u\!\left(\frac{\xi + \eta}{2},\, \frac{\xi - \eta}{2}\right).
\end{align*}
Since $u \in C^2(\mathbb{R} \times [0, \infty); \mathbb{R})$ and $\Psi$ is a smooth diffeomorphism onto $\Omega$, the composition $U$ belongs to $C^2(\Omega; \mathbb{R})$.
[claim:Transformed Equation]
The function $U$ satisfies $\partial_\xi \partial_\eta U(\xi, \eta) = 0$ for all $(\xi, \eta) \in \operatorname{int}(\Omega)$.
[/claim]
[proof]
Write $x = (\xi + \eta)/2$ and $t = (\xi - \eta)/2$. By the chain rule:
\begin{align*}
\partial_\xi U(\xi, \eta) &= \frac{1}{2}\partial_x u(x, t) + \frac{1}{2}\partial_t u(x, t), \\
\partial_\eta U(\xi, \eta) &= \frac{1}{2}\partial_x u(x, t) - \frac{1}{2}\partial_t u(x, t).
\end{align*}
Differentiating the second expression with respect to $\xi$:
\begin{align*}
\partial_\xi \partial_\eta U(\xi, \eta) &= \frac{1}{4}\partial_x^2 u(x, t) - \frac{1}{4}\partial_t^2 u(x, t) = -\frac{1}{4}\bigl(\partial_t^2 u(x, t) - \partial_x^2 u(x, t)\bigr) = 0,
\end{align*}
where the final equality uses $\partial_t^2 u - \partial_x^2 u = 0$ in $\mathbb{R} \times (0, \infty)$.
[/proof]
**Step 2: Integrate the transformed equation.**
[claim:General Solution Form]
There exist [functions](/page/Function) $F \in C^2(\mathbb{R}; \mathbb{R})$ and $G \in C^2(\mathbb{R}; \mathbb{R})$ such that $U(\xi, \eta) = F(\xi) + G(\eta)$ for all $(\xi, \eta) \in \Omega$.
[/claim]
[proof]
Since $\partial_\xi \partial_\eta U = 0$ on $\operatorname{int}(\Omega)$, for each fixed $\eta$ the map $\xi \mapsto \partial_\eta U(\xi, \eta)$ has vanishing $\xi$-[derivative](/page/Derivative) and is therefore constant in $\xi$. That is, there exists a function $\gamma \in C^1(\mathbb{R}; \mathbb{R})$ such that
\begin{align*}
\partial_\eta U(\xi, \eta) = \gamma(\eta) \quad \text{for all } (\xi, \eta) \in \operatorname{int}(\Omega).
\end{align*}
The regularity $\gamma \in C^1$ follows from $U \in C^2$. Define
\begin{align*}
G: \mathbb{R} &\to \mathbb{R} \\
\eta &\mapsto \int_0^{\eta} \gamma(s) \, d\mathcal{L}^1(s).
\end{align*}
Then $G' = \gamma \in C^1(\mathbb{R}; \mathbb{R})$, so $G \in C^2(\mathbb{R}; \mathbb{R})$. The difference $U(\xi, \eta) - G(\eta)$ satisfies $\partial_\eta(U(\xi, \eta) - G(\eta)) = \gamma(\eta) - \gamma(\eta) = 0$, so for each fixed $\xi$ it is constant in $\eta$. Since $(\xi, \xi) \in \Omega$ for every $\xi \in \mathbb{R}$ (corresponding to $t = 0$), define
\begin{align*}
F: \mathbb{R} &\to \mathbb{R} \\
\xi &\mapsto U(\xi, \xi) - G(\xi).
\end{align*}
Then $F(\xi) = U(\xi, \eta) - G(\eta)$ for all $(\xi, \eta) \in \Omega$, and hence $U(\xi, \eta) = F(\xi) + G(\eta)$. The regularity $F \in C^2(\mathbb{R}; \mathbb{R})$ follows from $F(\xi) = U(\xi, \xi) - G(\xi)$ with $U \in C^2$ and $G \in C^2$.
[/proof]
**Step 3: Translate back and impose initial conditions.**
The identity $U(\xi, \eta) = F(\xi) + G(\eta)$ in original coordinates reads
\begin{align*}
u(x, t) = F(x + t) + G(x - t) \quad \text{for all } (x, t) \in \mathbb{R} \times [0, \infty).
\end{align*}
Evaluating at $t = 0$:
\begin{align*}
u(x, 0) = F(x) + G(x) = g(x).
\end{align*}
Differentiating $F(x + t) + G(x - t)$ with respect to $t$ by the chain rule gives $F'(x + t) - G'(x - t)$. Evaluating at $t = 0$:
\begin{align*}
\partial_t u(x, 0) = F'(x) - G'(x) = h(x).
\end{align*}
**Step 4: Solve for $F$ and $G$.**
Integrating $F'(x) - G'(x) = h(x)$ over $[0, x]$:
\begin{align*}
F(x) - G(x) = \int_0^x h(y) \, d\mathcal{L}^1(y) + C
\end{align*}
where $C = F(0) - G(0) \in \mathbb{R}$. Adding the relations $F(x) + G(x) = g(x)$ and $F(x) - G(x) = \int_0^x h \, d\mathcal{L}^1 + C$:
\begin{align*}
F(x) &= \frac{1}{2}g(x) + \frac{1}{2}\int_0^x h(y) \, d\mathcal{L}^1(y) + \frac{C}{2}.
\end{align*}
Subtracting the second relation from the first:
\begin{align*}
G(x) &= \frac{1}{2}g(x) - \frac{1}{2}\int_0^x h(y) \, d\mathcal{L}^1(y) - \frac{C}{2}.
\end{align*}
**Step 5: Assemble the formula.**
Evaluating $F$ at $x + t$ and $G$ at $x - t$:
\begin{align*}
u(x, t) &= \frac{1}{2}g(x + t) + \frac{1}{2}\int_0^{x + t} h(y) \, d\mathcal{L}^1(y) + \frac{C}{2} \\
&\quad + \frac{1}{2}g(x - t) - \frac{1}{2}\int_0^{x - t} h(y) \, d\mathcal{L}^1(y) - \frac{C}{2} \\
&= \frac{1}{2}\bigl(g(x + t) + g(x - t)\bigr) + \frac{1}{2}\left(\int_0^{x + t} h(y) \, d\mathcal{L}^1(y) - \int_0^{x - t} h(y) \, d\mathcal{L}^1(y)\right) \\
&= \frac{1}{2}\bigl(g(x + t) + g(x - t)\bigr) + \frac{1}{2}\int_{x - t}^{x + t} h(y) \, d\mathcal{L}^1(y).
\end{align*}
The constants $C$ cancel, and the two [integrals](/page/Integral) combine by the additivity property $\int_0^{a} - \int_0^{b} = \int_b^{a}$ for the [Lebesgue integral](/page/Lebesgue%20Integral).
**Step 6: Verify regularity and uniqueness.**
Since $g \in C^2(\mathbb{R}; \mathbb{R})$ and $h \in C^1(\mathbb{R}; \mathbb{R})$, the [fundamental theorem of calculus](/theorems/632) gives $x \mapsto \int_0^x h(y) \, d\mathcal{L}^1(y) \in C^2(\mathbb{R}; \mathbb{R})$, so $F, G \in C^2(\mathbb{R}; \mathbb{R})$. The chain rule then gives $u(x, t) = F(x + t) + G(x - t) \in C^2(\mathbb{R} \times [0, \infty); \mathbb{R})$. For uniqueness: every $C^2$ solution satisfies $\partial_\xi \partial_\eta U = 0$ by the Transformed Equation claim, hence has the form $F(x + t) + G(x - t)$ by the General Solution Form claim, and Steps 3–4 show the initial data determine $F$ and $G$ uniquely up to the constant $C$, which cancels in the sum.
