**Proof plan.** The strategy is to reduce the three-dimensional [wave equation](/page/Wave%20Equation) to a one-dimensional problem via spherical means. Given a solution $u(x, t)$, we average it over the sphere $\partial B(x, r)$ to define the spherical mean $U(r, t)$. By exploiting the relation between the Laplacian and the radial derivative of the spherical average — established via the [Gauss-Green Theorem](/theorems/28) — we show $U$ satisfies a $1$-dimensional PDE in $(r, t)$ called the Euler–Poisson–Darboux equation. For $n = 3$, the substitution $\tilde{U} = rU$ converts this into the standard one-dimensional wave equation, which we solve by [D'Alembert's Formula](/theorems/665). Evaluating at $r = 0$ recovers $u(x, t)$.
**Step 1: Define the spherical mean and derive the Euler–Poisson–Darboux equation.**
For fixed $x \in \mathbb{R}^3$ and $r > 0$, the spherical mean of a function $v: \mathbb{R}^3 \to \mathbb{R}$ over $\partial B(x, r)$ is
\begin{align*}
(M_r v)(x) := \frac{1}{\mathcal{H}^2(\partial B(x, r))} \int_{\partial B(x, r)} v(y) \, d\mathcal{H}^2(y).
\end{align*}
By [Volume Of The Ball And Surface Area Of The Sphere](/theorems/871), $\mathcal{H}^2(\partial B(x, r)) = 4\pi r^2$. For computations it is convenient to rewrite this as an integral over the unit sphere, which the following claim justifies.
[claim:Spherical Mean Via The Unit Sphere]
For any $v \in C(\mathbb{R}^3; \mathbb{R})$, $x \in \mathbb{R}^3$, and $r > 0$:
\begin{align*}
(M_r v)(x) = \frac{1}{\mathcal{H}^2(\partial B(0, 1))} \int_{\partial B(0, 1)} v(x + r\omega) \, d\mathcal{H}^2(\omega).
\end{align*}
[/claim]
[proof]
Consider the affine map
\begin{align*}
\Psi: \mathbb{R}^3 &\to \mathbb{R}^3 \\
z &\mapsto x + rz.
\end{align*}
Its restriction to $\partial B(0, 1)$ is an injective $C^1$ map with $\Psi(\partial B(0, 1)) = \partial B(x, r)$. Since $D\Psi = rI \in \mathbb{R}^{3 \times 3}$, the tangent space $T_\omega \partial B(0, 1) = \omega^\perp$ is mapped by $D\Psi$ to $T_{\Psi(\omega)} \partial B(x, r)$ by multiplication by $r$. By the [Substitution Formula for Hausdorff Integrals](/theorems/868), for any Borel measurable $f: \partial B(x, r) \to [0, \infty]$:
\begin{align*}
\int_{\partial B(x, r)} f(y) \, d\mathcal{H}^2(y) = \int_{\partial B(0, 1)} f(x + r\omega) \cdot J_{\partial B(0,1)} \Psi(\omega) \, d\mathcal{H}^2(\omega).
\end{align*}
The tangential Jacobian $J_{\partial B(0,1)} \Psi(\omega)$ is computed as follows. Let $(e_1, e_2)$ be an orthonormal basis for $T_\omega = \omega^\perp$. Then $D\Psi \cdot e_i = r e_i$, so the $2 \times 2$ Gram matrix of $D\Psi|_{T_\omega}$ with respect to this basis is
\begin{align*}
\bigl((D\Psi \cdot e_i) \cdot (D\Psi \cdot e_j)\bigr)_{i,j} = r^2 \delta_{ij} = r^2 I_2.
\end{align*}
Therefore $J_{\partial B(0,1)} \Psi(\omega) = \sqrt{\det(r^2 I_2)} = r^2$. Substituting:
\begin{align*}
\int_{\partial B(x, r)} f(y) \, d\mathcal{H}^2(y) = r^2 \int_{\partial B(0, 1)} f(x + r\omega) \, d\mathcal{H}^2(\omega).
\end{align*}
Dividing by $\mathcal{H}^2(\partial B(x, r)) = r^2 \mathcal{H}^2(\partial B(0, 1))$ (by the same scaling, or by [Volume Of The Ball And Surface Area Of The Sphere](/theorems/871)):
\begin{align*}
(M_r v)(x) = \frac{r^2}{r^2 \mathcal{H}^2(\partial B(0, 1))} \int_{\partial B(0, 1)} v(x + r\omega) \, d\mathcal{H}^2(\omega) = \frac{1}{\mathcal{H}^2(\partial B(0, 1))} \int_{\partial B(0, 1)} v(x + r\omega) \, d\mathcal{H}^2(\omega).
\end{align*}
[/proof]
Define $U(x; r, t) := (M_r u(\cdot, t))(x)$. Since $u$ solves $\partial_t^2 u = \Delta u$, and [differentiation](/page/Derivative) in $t$ commutes with averaging in $\omega$ (the unit-sphere representation shows the integrand depends smoothly on $t$, and $\partial B(0,1)$ is compact):
\begin{align*}
\partial_t^2 U = M_r(\partial_t^2 u) = M_r(\Delta u).
\end{align*}
The following claim relates $M_r(\Delta v)$ to radial derivatives of $M_r v$.
[claim:Spherical Mean Identity For The Laplacian]
For any $v \in C^2(\mathbb{R}^3; \mathbb{R})$, $x \in \mathbb{R}^3$, and $r > 0$:
\begin{align*}
M_r(\Delta v)(x) = \partial_r^2 (M_r v)(x) + \frac{2}{r} \partial_r (M_r v)(x).
\end{align*}
[/claim]
[proof]
**Step 1a.** By the unit-sphere representation (Claim: Spherical Mean Via The Unit Sphere), $(M_r v)(x) = \frac{1}{\mathcal{H}^2(\partial B(0,1))}\int_{\partial B(0,1)} v(x + r\omega) \, d\mathcal{H}^2(\omega)$. Differentiating under the integral sign with respect to $r$ (justified since $v \in C^2$ and $\partial B(0,1)$ is compact):
\begin{align*}
\partial_r (M_r v)(x) = \frac{1}{\mathcal{H}^2(\partial B(0,1))} \int_{\partial B(0, 1)} \omega \cdot (\nabla v)(x + r\omega) \, d\mathcal{H}^2(\omega).
\end{align*}
The integrand is $(\partial_\nu v)(x + r\omega)$, where $\nu(y) = (y - x)/r = \omega$ is the outward unit normal to $\partial B(x, r)$ at $y = x + r\omega$. Applying the change-of-variables identity from Claim: Spherical Mean Via The Unit Sphere (which gives $\int_{\partial B(x,r)} f \, d\mathcal{H}^2 = r^2 \int_{\partial B(0,1)} f(x + r\omega) \, d\mathcal{H}^2(\omega)$) with $f = \partial_\nu v$:
\begin{align*}
\partial_r (M_r v)(x) = \frac{1}{\mathcal{H}^2(\partial B(x, r))} \int_{\partial B(x, r)} \partial_\nu v(y) \, d\mathcal{H}^2(y).
\end{align*}
**Step 1b.** By the [Gauss-Green Theorem](/theorems/28) applied to $B(x, r)$:
\begin{align*}
\int_{\partial B(x, r)} \partial_\nu v \, d\mathcal{H}^2 = \int_{\partial B(x, r)} \nabla v \cdot \nu \, d\mathcal{H}^2 = \int_{B(x, r)} \Delta v \, d\mathcal{L}^3.
\end{align*}
So $\partial_r (M_r v)(x) = \frac{1}{\mathcal{H}^2(\partial B(x, r))} \int_{B(x, r)} \Delta v(y) \, d\mathcal{L}^3(y)$.
**Step 1c.** Write $\sigma := \mathcal{H}^2(\partial B(0, 1))$ for brevity, so that $\mathcal{H}^2(\partial B(x, r)) = \sigma r^2$ by [Volume Of The Ball And Surface Area Of The Sphere](/theorems/871). Differentiating again in $r$ using the Leibniz rule (the domain of [integration](/page/Integral) depends on $r$):
\begin{align*}
\partial_r \left(\frac{1}{\sigma r^2} \int_{B(x, r)} \Delta v \, d\mathcal{L}^3\right) &= -\frac{2}{\sigma r^3} \int_{B(x, r)} \Delta v \, d\mathcal{L}^3 + \frac{1}{\sigma r^2} \frac{d}{dr}\int_{B(x, r)} \Delta v \, d\mathcal{L}^3.
\end{align*}
The derivative of the volume integral is computed by differentiating the substitution $y = x + r\omega$, $d\mathcal{L}^3(y) = r^2 \, d\mathcal{L}^3|_{\text{polar}}$:
\begin{align*}
\frac{d}{dr}\int_{B(x, r)} \Delta v \, d\mathcal{L}^3 = \int_{\partial B(x, r)} \Delta v \, d\mathcal{H}^2 = \sigma r^2 \, M_r(\Delta v)(x).
\end{align*}
Substituting back and using Step 1b:
\begin{align*}
\partial_r^2 (M_r v)(x) &= -\frac{2}{r} \cdot \frac{1}{\sigma r^2}\int_{B(x,r)} \Delta v \, d\mathcal{L}^3 + M_r(\Delta v)(x) = -\frac{2}{r}\partial_r(M_r v)(x) + M_r(\Delta v)(x).
\end{align*}
Rearranging: $M_r(\Delta v)(x) = \partial_r^2 (M_r v)(x) + \frac{2}{r}\partial_r(M_r v)(x)$.
[/proof]
Applying the claim to $v = u(\cdot, t)$ gives the **Euler–Poisson–Darboux equation**:
\begin{align*}
\partial_t^2 U = \partial_r^2 U + \frac{2}{r} \partial_r U.
\end{align*}
**Step 2: Reduce to the one-dimensional wave equation.**
Set $\tilde{U}(x; r, t) := r \, U(x; r, t)$. Compute:
\begin{align*}
\partial_r \tilde{U} &= U + r \, \partial_r U, \\
\partial_r^2 \tilde{U} &= 2\partial_r U + r \, \partial_r^2 U = r\left(\partial_r^2 U + \frac{2}{r}\partial_r U\right) = r \, \partial_t^2 U = \partial_t^2 \tilde{U}.
\end{align*}
So $\tilde{U}$ satisfies the one-dimensional wave equation $\partial_t^2 \tilde{U} = \partial_r^2 \tilde{U}$ for $r > 0$, $t > 0$.
**Step 3: Determine initial data for $\tilde{U}$.**
At $t = 0$:
\begin{align*}
\tilde{U}(x; r, 0) &= r \, (M_r g)(x), \\
\partial_t \tilde{U}(x; r, 0) &= r \, (M_r h)(x).
\end{align*}
As $r \to 0^+$, [continuity](/page/Continuity) of $g$ gives $(M_r g)(x) \to g(x)$, so $\tilde{U}(x; 0, 0) = 0$. Similarly $\partial_t \tilde{U}(x; 0, 0) = 0$.
**Step 4: Apply d'Alembert's formula and extract $u(x, t)$.**
Define $\tilde{U}_0(x; r) := r(M_r g)(x)$ and $\tilde{V}_0(x; r) := r(M_r h)(x)$ for $r \geq 0$. Extend both to $r < 0$ as odd [functions](/page/Function): $\tilde{U}_0(x; -r) := -\tilde{U}_0(x; r)$ and $\tilde{V}_0(x; -r) := -\tilde{V}_0(x; r)$. Since $\tilde{U}(x; 0, t) = 0$ (the spherical mean is bounded as $r \to 0^+$), this odd extension is continuous. By [D'Alembert's Formula](/theorems/665) applied to the odd-extended data on $\mathbb{R}$:
\begin{align*}
\tilde{U}(x; r, t) = \frac{1}{2}\bigl[\tilde{U}_0(x; r + t) + \tilde{U}_0(x; r - t)\bigr] + \frac{1}{2}\int_{r - t}^{r + t} \tilde{V}_0(x; s) \, d\mathcal{L}^1(s).
\end{align*}
For $0 < r < t$, the argument $r - t$ is negative. Applying the odd extension $\tilde{U}_0(x; r - t) = -\tilde{U}_0(x; t - r)$ and splitting the integral at $s = 0$ (using the substitution $s \mapsto -s$ and oddness of $\tilde{V}_0$ to fold the negative part onto $[0, t - r]$) gives the half-line form:
\begin{align*}
\tilde{U}(x; r, t) = \frac{1}{2}\bigl[\tilde{U}_0(x; r + t) - \tilde{U}_0(x; t - r)\bigr] + \frac{1}{2}\int_{t - r}^{t + r} \tilde{V}_0(x; s) \, d\mathcal{L}^1(s).
\end{align*}
Since $\tilde{U}(x; 0, t) = 0$ and $\tilde{U}$ is smooth in $r$, the [limit](/page/Limit) $u(x, t) = \lim_{r \to 0^+} \tilde{U}(x; r, t)/r$ equals $\partial_r \tilde{U}(x; r, t)\big|_{r=0}$ by the definition of the derivative. Differentiating the half-line form with respect to $r$:
\begin{align*}
\partial_r \tilde{U}(x; r, t) = \frac{1}{2}\bigl[\tilde{U}_0'(x; r + t) + \tilde{U}_0'(x; t - r)\bigr] + \frac{1}{2}\bigl[\tilde{V}_0(x; t + r) + \tilde{V}_0(x; t - r)\bigr].
\end{align*}
Evaluating at $r = 0$:
\begin{align*}
u(x, t) = \tilde{U}_0'(x; t) + \tilde{V}_0(x; t) = \partial_t\bigl[t(M_t g)(x)\bigr] + t(M_t h)(x).
\end{align*}
**Step 5: Write out the formula.**
Substituting the definition $(M_t g)(x) = \frac{1}{\mathcal{H}^2(\partial B(x,t))}\int_{\partial B(x,t)} g \, d\mathcal{H}^2$ into Step 4, and writing $\mathcal{H}^2(\partial B(x, t)) = 4\pi t^2$ (by [Volume Of The Ball And Surface Area Of The Sphere](/theorems/871)):
\begin{align*}
u(x, t) = \frac{\partial}{\partial t}\left(\frac{t}{4\pi t^2} \int_{\partial B(x, t)} g \, d\mathcal{H}^2\right) + \frac{t}{4\pi t^2}\int_{\partial B(x, t)} h \, d\mathcal{H}^2 = \frac{\partial}{\partial t}\left(\frac{1}{4\pi t^2}\int_{\partial B(x, t)} t \, g \, d\mathcal{H}^2\right) + \frac{1}{4\pi t^2}\int_{\partial B(x, t)} t \, h \, d\mathcal{H}^2,
\end{align*}
which is the stated formula.
