[proofplan] We take the [Fourier transform](/page/Fourier%20Transform) only in the spatial variables, reducing the [wave equation](/page/Wave%20Equation) to a second-order constant-coefficient ordinary differential equation in $t$ for each frequency $\xi$. Solving that ODE gives the displayed formula for $\hat{u}(t,\xi)$. We then justify that the multipliers are smooth Schwartz multipliers, invert the Fourier transform to construct $u$, and verify the equation and initial data. Uniqueness follows by applying the same Fourier-transformed ODE argument to the difference of two solutions. [/proofplan]

[step:Define the spatial Fourier-side candidate] For each fixed $t \in \mathbb{R}$, define the multiplier functions $c_t:\mathbb{R}^n \to \mathbb{R}$ and $m_t:\mathbb{R}^n \to \mathbb{R}$ by \begin{align*} c_t(\xi)=\cos(t|\xi|) \end{align*} and \begin{align*} m_t(\xi)=\frac{\sin(t|\xi|)}{|\xi|} \end{align*} for $\xi \neq 0$, with $m_t(0)=t$. The removable singularity at $\xi=0$ is removed by the [power series](/page/Power%20Series) \begin{align*} m_t(\xi)=\sum_{k=0}^{\infty}\frac{(-1)^k t^{2k+1}|\xi|^{2k}}{(2k+1)!}. \end{align*} Likewise \begin{align*} c_t(\xi)=\sum_{k=0}^{\infty}\frac{(-1)^k t^{2k}|\xi|^{2k}}{(2k)!}. \end{align*} Thus $c_t,m_t \in C^\infty(\mathbb{R}^n)$. Define the Fourier-side candidate $U:\mathbb{R}\times\mathbb{R}^n \to \mathbb{C}$ by \begin{align*} U(t,\xi)=c_t(\xi)\hat{f}(\xi)+m_t(\xi)\hat{g}(\xi). \end{align*} For every multi-index $\alpha$ and every compact interval $I \subset \mathbb{R}$, the functions $\partial_\xi^\alpha c_t(\xi)$ and $\partial_\xi^\alpha m_t(\xi)$ are bounded by a polynomial in $|\xi|$, uniformly for $t \in I$. Since $\hat{f},\hat{g} \in \mathcal{S}(\mathbb{R}^n)$, it follows that $U(t,\cdot)\in \mathcal{S}(\mathbb{R}^n)$ for every $t \in \mathbb{R}$, and $t \mapsto U(t,\cdot)$ is smooth as a map $\mathbb{R}\to\mathcal{S}(\mathbb{R}^n)$. [/step]

[step:Construct the physical-space solution by inverse Fourier transform]Define $u:\mathbb{R}\times\mathbb{R}^n \to \mathbb{C}$ by the inverse Fourier transform \begin{align*} u(t,x)=\frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^n} U(t,\xi)e^{i x\cdot \xi}\,d\mathcal{L}^n(\xi). \end{align*} Because $U(t,\cdot)\in\mathcal{S}(\mathbb{R}^n)$ and the dependence on $t$ is smooth in the Schwartz topology, differentiation under the integral is valid for every $t$-derivative and every $x$-derivative. Hence $u\in C^\infty(\mathbb{R}\times\mathbb{R}^n)$, $u(t,\cdot)\in\mathcal{S}(\mathbb{R}^n)$ for every $t$, and the spatial Fourier transform of $u(t,\cdot)$ is $U(t,\cdot)$.[/step]

[guided]The construction is made on the Fourier side because the spatial operator $\Delta$ becomes multiplication by $-|\xi|^2$. We have already defined \begin{align*} U(t,\xi)=\cos(t|\xi|)\hat{f}(\xi)+m_t(\xi)\hat{g}(\xi), \end{align*} where $m_t(0)=t$ and $m_t(\xi)=\sin(t|\xi|)/|\xi|$ for $\xi\neq0$. The power-series expansions of $\cos(t|\xi|)$ and $m_t(\xi)$ show that the apparent singularity at $\xi=0$ is removable and that both multipliers are smooth functions of $\xi$. Since $\hat{f}$ and $\hat{g}$ are Schwartz functions, multiplying them by these smooth polynomially bounded multipliers keeps the result in $\mathcal{S}(\mathbb{R}^n)$. Thus $U(t,\cdot)\in\mathcal{S}(\mathbb{R}^n)$ for every $t\in\mathbb{R}$. This is the key regularity point: it ensures that the inverse Fourier integral is absolutely convergent after applying any polynomial weight or derivative. We therefore define \begin{align*} u(t,x)=\frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^n} U(t,\xi)e^{i x\cdot \xi}\,d\mathcal{L}^n(\xi). \end{align*} Because $U$ is smooth in $t$ with values in the [Schwartz space](/page/Schwartz%20Space), every derivative in $t$ remains rapidly decreasing in $\xi$ on compact $t$-intervals. Differentiating in $x_i$ only inserts the factor $i\xi_i$, which also preserves rapid decrease. Hence differentiation under the integral is justified for all mixed derivatives, and $u\in C^\infty(\mathbb{R}\times\mathbb{R}^n)$. Finally, the [Fourier inversion formula](/theorems/528) for Schwartz functions gives that the spatial Fourier transform of $u(t,\cdot)$ is exactly $U(t,\cdot)$. Thus the physical-space function $u$ has the claimed Fourier representation.[/guided]

[step:Verify the wave equation on the Fourier side] For fixed $\xi \in \mathbb{R}^n$, the map $a_\xi:\mathbb{R}\to\mathbb{C}$ defined by \begin{align*} a_\xi(t)=U(t,\xi) \end{align*} satisfies \begin{align*} a_\xi''(t)+|\xi|^2a_\xi(t)=0. \end{align*} Indeed, for $\xi\neq0$ this follows by differentiating $\cos(t|\xi|)$ and $\sin(t|\xi|)/|\xi|$ twice in $t$, and for $\xi=0$ it gives $a_0''(t)=0$ because $c_t(0)=1$ and $m_t(0)=t$. Taking the spatial Fourier transform of $\partial_t^2u-\Delta u$ gives \begin{align*} \widehat{\partial_t^2u-\Delta u}(t,\xi)=\partial_t^2U(t,\xi)+|\xi|^2U(t,\xi)=0. \end{align*} Since the Fourier transform is injective on $\mathcal{S}(\mathbb{R}^n)$, $\partial_t^2u-\Delta u=0$. [/step]

[step:Check the initial position and velocity] At $t=0$, the multiplier values are $c_0(\xi)=1$ and $m_0(\xi)=0$, so \begin{align*} U(0,\xi)=\hat{f}(\xi). \end{align*} Fourier inversion gives $u(0,\cdot)=f$. Differentiating $U$ with respect to $t$ gives \begin{align*} \partial_tU(t,\xi)=-|\xi|\sin(t|\xi|)\hat{f}(\xi)+\cos(t|\xi|)\hat{g}(\xi). \end{align*} At $t=0$ this becomes \begin{align*} \partial_tU(0,\xi)=\hat{g}(\xi). \end{align*} Fourier inversion gives $\partial_tu(0,\cdot)=g$. [/step]

[step:Identify the Fourier multiplier operators] For $h\in\mathcal{S}(\mathbb{R}^n)$, define $\cos(t\sqrt{-\Delta})h$ as the inverse Fourier transform of $c_t\hat{h}$, and define $\frac{\sin(t\sqrt{-\Delta})}{\sqrt{-\Delta}}h$ as the inverse Fourier transform of $m_t\hat{h}$. The Schwartz multiplier bounds established above imply that both maps send $\mathcal{S}(\mathbb{R}^n)$ to itself. Their Fourier transforms are, by definition, \begin{align*} \widehat{\cos(t\sqrt{-\Delta})h}(\xi)=\cos(t|\xi|)\hat{h}(\xi) \end{align*} and \begin{align*} \widehat{\frac{\sin(t\sqrt{-\Delta})}{\sqrt{-\Delta}}h}(\xi)=m_t(\xi)\hat{h}(\xi). \end{align*} Applying this with $h=f$ and $h=g$ gives \begin{align*} \widehat{\cos(t\sqrt{-\Delta})f+\frac{\sin(t\sqrt{-\Delta})}{\sqrt{-\Delta}}g}(\xi)=U(t,\xi). \end{align*} Fourier inversion therefore yields \begin{align*} u(t,\cdot)=\cos(t\sqrt{-\Delta})f+\frac{\sin(t\sqrt{-\Delta})}{\sqrt{-\Delta}}g. \end{align*} [/step]

[step:Prove uniqueness among smooth Schwartz-valued solutions] Let $v:\mathbb{R}\times\mathbb{R}^n\to\mathbb{C}$ be another solution with $v(t,\cdot)\in\mathcal{S}(\mathbb{R}^n)$ for every $t$, smooth dependence $t\mapsto v(t,\cdot)$ in the Schwartz topology, and the same initial data. Define $w:\mathbb{R}\times\mathbb{R}^n\to\mathbb{C}$ by \begin{align*} w(t,x)=u(t,x)-v(t,x). \end{align*} Then $w$ satisfies \begin{align*} \partial_t^2w-\Delta w=0,\qquad w(0,\cdot)=0,\qquad \partial_tw(0,\cdot)=0. \end{align*} Let $W:\mathbb{R}\times\mathbb{R}^n\to\mathbb{C}$ denote the spatial Fourier transform of $w$, so \begin{align*} W(t,\xi)=\widehat{w(t,\cdot)}(\xi). \end{align*} For each fixed $\xi$, $b_\xi:\mathbb{R}\to\mathbb{C}$ defined by $b_\xi(t)=W(t,\xi)$ satisfies \begin{align*} b_\xi''(t)+|\xi|^2b_\xi(t)=0,\qquad b_\xi(0)=0,\qquad b_\xi'(0)=0. \end{align*} The unique solution of this scalar initial-value problem is $b_\xi(t)=0$ for all $t\in\mathbb{R}$. Hence $W(t,\xi)=0$ for every $(t,\xi)\in\mathbb{R}\times\mathbb{R}^n$. By injectivity of the Fourier transform on $\mathcal{S}(\mathbb{R}^n)$, $w(t,\cdot)=0$ for every $t$, so $u=v$. This proves uniqueness and completes the proof. [/step]