**Proof plan.** The idea is Hadamard's **method of descent**: treat a function of two spatial variables as a function of three spatial variables that is independent of the third coordinate, apply the known three-dimensional [Kirchhoff's Formula](/theorems/666), and evaluate the resulting surface integrals by projecting the sphere $\partial B(x, t) \subset \mathbb{R}^3$ onto the disk $B(\bar{x}, t) \subset \mathbb{R}^2$. The projection is justified by the [Area Formula](/theorems/25), which converts the $\mathcal{H}^2$-integral over the hemisphere into an $\mathcal{L}^2$-integral over the disk, introducing the Jacobian factor $(t^2 - |y - \bar{x}|^2)^{-1/2}$.

**Step 1: Embed the two-dimensional problem in three dimensions.**

Given $g, h: \mathbb{R}^2 \to \mathbb{R}$, define $\tilde{g}, \tilde{h}: \mathbb{R}^3 \to \mathbb{R}$ by $\tilde{g}(x_1, x_2, x_3) := g(x_1, x_2)$ and $\tilde{h}(x_1, x_2, x_3) := h(x_1, x_2)$. Since $\tilde{g} \in C^3(\mathbb{R}^3; \mathbb{R})$ and $\tilde{h} \in C^2(\mathbb{R}^3; \mathbb{R})$, [Kirchhoff's Formula](/theorems/666) provides a solution $\tilde{u} \in C^2(\mathbb{R}^3 \times [0, \infty); \mathbb{R})$ of the three-dimensional wave equation with data $(\tilde{g}, \tilde{h})$. For any $s \in \mathbb{R}$, define the translated function

\begin{align*} \tilde{u}_s: \mathbb{R}^3 \times [0, \infty) &\to \mathbb{R} \\ (x_1, x_2, x_3, t) &\mapsto \tilde{u}(x_1, x_2, x_3 + s, t). \end{align*}

Since the Laplacian $\Delta = \partial_{x_1}^2 + \partial_{x_2}^2 + \partial_{x_3}^2$ commutes with translations in $x_3$ (i.e. $\Delta(\tilde{u}_s) = (\Delta \tilde{u})_s$), and $\tilde{u}$ satisfies $\partial_t^2 \tilde{u} - \Delta \tilde{u} = 0$, the same identity holds for $\tilde{u}_s$:

\begin{align*} \partial_t^2 \tilde{u}_s - \Delta \tilde{u}_s = (\partial_t^2 \tilde{u} - \Delta \tilde{u})(x_1, x_2, x_3 + s, t) = 0. \end{align*}

Moreover, the initial data of $\tilde{u}_s$ coincide with those of $\tilde{u}$, since $\tilde{g}$ and $\tilde{h}$ are independent of $x_3$:

\begin{align*} \tilde{u}_s(x_1, x_2, x_3, 0) &= \tilde{g}(x_1, x_2, x_3 + s) = g(x_1, x_2) = \tilde{g}(x_1, x_2, x_3), \\ \partial_t \tilde{u}_s(x_1, x_2, x_3, 0) &= \tilde{h}(x_1, x_2, x_3 + s) = h(x_1, x_2) = \tilde{h}(x_1, x_2, x_3). \end{align*}

So $\tilde{u}_s$ and $\tilde{u}$ both solve the three-dimensional wave equation with identical initial data $(\tilde{g}, \tilde{h})$. By [Uniqueness For The Wave Equation](/theorems/669),

\begin{align*} \tilde{u}_s = \tilde{u} \quad \text{for every } s \in \mathbb{R}, \end{align*}

so $\tilde{u}$ is independent of $x_3$. In particular, $\tilde{u}(x_1, x_2, x_3, t) = \tilde{u}(x_1, x_2, 0, t)$ for every $x_3 \in \mathbb{R}$, so we may define

\begin{align*} u(x_1, x_2, t) := \tilde{u}(x_1, x_2, 0, t). \end{align*}

Since $\tilde{u}$ solves $\partial_t^2 \tilde{u} - (\partial_{x_1}^2 + \partial_{x_2}^2 + \partial_{x_3}^2)\tilde{u} = 0$ and $\partial_{x_3}^2 \tilde{u} = 0$ (as $\tilde{u}$ is independent of $x_3$), restricting to $x_3 = 0$ gives $\partial_t^2 u - (\partial_{x_1}^2 + \partial_{x_2}^2) u = 0$. The initial conditions $u(x_1, x_2, 0) = g(x_1, x_2)$ and $\partial_t u(x_1, x_2, 0) = h(x_1, x_2)$ follow from those of $\tilde{u}$, so $u$ solves the two-dimensional wave equation with data $(g, h)$.

**Step 2: Apply Kirchhoff's formula to $\tilde{u}$.**

By [Kirchhoff's Formula](/theorems/666) with $x = (\bar{x}, 0) \in \mathbb{R}^3$:

\begin{align*} u(\bar{x}, t) = \tilde{u}((\bar{x}, 0), t) = \partial_t\bigl[t \, (M_t \tilde{g})((\bar{x}, 0))\bigr] + t \, (M_t \tilde{h})((\bar{x}, 0)), \end{align*}

where

\begin{align*} (M_t \tilde{g})((\bar{x}, 0)) := \frac{1}{\mathcal{H}^2(\partial B((\bar{x}, 0), t))}\int_{\partial B((\bar{x}, 0), t)} \tilde{g} \, d\mathcal{H}^2 \end{align*}

is the spherical mean of $\tilde{g}$ over $\partial B((\bar{x}, 0), t) \subset \mathbb{R}^3$, and likewise for $\tilde{h}$. By [Volume Of The Ball And Surface Area Of The Sphere](/theorems/871), $\mathcal{H}^2(\partial B((\bar{x}, 0), t)) = 4\pi t^2$.

**Step 3: Project the sphere onto the disk.**

The spherical mean involves integrating $\tilde{g}$ over $\partial B((\bar{x}, 0), t) \subset \mathbb{R}^3$ against $\mathcal{H}^2$. Since $\tilde{g}$ is independent of $x_3$, the upper and lower hemispheres contribute equally, so it suffices to compute the integral over the upper hemisphere $S^+$ and double it. The following claim converts this $\mathcal{H}^2$-integral into an $\mathcal{L}^2$-integral over the disk $B(\bar{x}, t) \subset \mathbb{R}^2$.

[claim:Surface Element Of The Upper Hemisphere]

Let $t > 0$ and $\bar{x} \in \mathbb{R}^2$. Let $S^+ := \{(y_1, y_2, y_3) \in \partial B((\bar{x}, 0), t) : y_3 > 0\}$ denote the open upper hemisphere. For every Borel measurable function $f: S^+ \to [0, \infty]$:

\begin{align*} \int_{S^+} f \, d\mathcal{H}^2 = \int_{B(\bar{x}, t)} f\!\left(y_1, y_2, \sqrt{t^2 - |y - \bar{x}|^2}\right) \frac{t}{\sqrt{t^2 - |y - \bar{x}|^2}} \, d\mathcal{L}^2(y), \end{align*}

where $y = (y_1, y_2)$.

[/claim]

[proof]

Define the parametrisation

\begin{align*} \phi: B(\bar{x}, t) &\to \mathbb{R}^3 \\ y = (y_1, y_2) &\mapsto \left(y_1, \, y_2, \, \sqrt{t^2 - |y - \bar{x}|^2}\right). \end{align*}

Then $\phi$ is injective, $C^1$ on $B(\bar{x}, t)$, and $\phi(B(\bar{x}, t)) = S^+$. Writing $r(y) := \sqrt{t^2 - |y - \bar{x}|^2}$ for brevity, the Jacobian matrix is

\begin{align*}

J\phi(y) = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ -\dfrac{y_1 - \bar{x}_1}{r(y)} & -\dfrac{y_2 - \bar{x}_2}{r(y)} \end{pmatrix} \in \mathbb{R}^{3 \times 2}. \end{align*} The Gram matrix $J\phi(y)^T J\phi(y) \in \mathbb{R}^{2 \times 2}$ has entries \begin{align*} (J\phi^T J\phi)_{ij} = \delta_{ij} + \frac{(y_i - \bar{x}_i)(y_j - \bar{x}_j)}{r(y)^2}, \qquad i, j \in \{1, 2\}. \end{align*} Its determinant is computed by the matrix determinant lemma (or direct expansion): \begin{align*} \det(J\phi^T J\phi) &= 1 + \frac{(y_1 - \bar{x}_1)^2 + (y_2 - \bar{x}_2)^2}{r(y)^2} = 1 + \frac{|y - \bar{x}|^2}{t^2 - |y - \bar{x}|^2} = \frac{t^2}{t^2 - |y - \bar{x}|^2}. \end{align*} Therefore $\sqrt{\det(J\phi^T J\phi)} = \dfrac{t}{\sqrt{t^2 - |y - \bar{x}|^2}}$. Since $\phi$ is an injective Lipschitz map from $B(\bar{x}, t) \subseteq \mathbb{R}^2$ into $\mathbb{R}^3$, the [Area Formula](/theorems/25) gives \begin{align*} \int_{S^+} f \, d\mathcal{H}^2 = \int_{\phi(B(\bar{x},t))} f \, d\mathcal{H}^2 = \int_{B(\bar{x}, t)} f(\phi(y)) \sqrt{\det(J\phi(y)^T J\phi(y))} \, d\mathcal{L}^2(y) = \int_{B(\bar{x}, t)} f\!\left(y, \sqrt{t^2 - |y - \bar{x}|^2}\right) \frac{t}{\sqrt{t^2 - |y - \bar{x}|^2}} \, d\mathcal{L}^2(y). \end{align*} [/proof] Applying the claim to $f = \tilde{g}|_{S^+}$, using that $\tilde{g}(y_1, y_2, y_3) = g(y_1, y_2)$, and doubling to account for the lower hemisphere (where $\tilde{g}$ takes the same values by $x_3$-independence): \begin{align*} \int_{\partial B((\bar{x}, 0), t)} \tilde{g} \, d\mathcal{H}^2 = 2 \int_{B(\bar{x}, t)} \frac{t \, g(y)}{\sqrt{t^2 - |y - \bar{x}|^2}} \, d\mathcal{L}^2(y). \end{align*} Since $\mathcal{H}^2(\partial B(0, t)) = 4\pi t^2$ (by [Volume Of The Ball And Surface Area Of The Sphere](/theorems/871)), the spherical mean becomes \begin{align*} (M_t \tilde{g})((\bar{x}, 0)) = \frac{1}{4\pi t^2} \cdot 2 \int_{B(\bar{x}, t)} \frac{t \, g(y)}{\sqrt{t^2 - |y - \bar{x}|^2}} \, d\mathcal{L}^2(y) = \frac{1}{2\pi t}\int_{B(\bar{x}, t)} \frac{g(y)}{\sqrt{t^2 - |y - \bar{x}|^2}} \, d\mathcal{L}^2(y). \end{align*} The same identity holds with $h$ in place of $g$. **Step 4: Assemble the formula.** Substituting back into the expression from Step 2: \begin{align*} u(\bar{x}, t) &= \partial_t\left(\frac{1}{2\pi}\int_{B(\bar{x}, t)} \frac{g(y)}{\sqrt{t^2 - |y - \bar{x}|^2}} \, d\mathcal{L}^2(y)\right) + \frac{1}{2\pi}\int_{B(\bar{x}, t)} \frac{h(y)}{\sqrt{t^2 - |y - \bar{x}|^2}} \, d\mathcal{L}^2(y), \end{align*} which is the stated formula.