The issue of treating an inverse Fourier transform in terms of a tempered distribution.

Question

Consider the wave equation $$ u_{tt}=\Delta{u} \quad u(x,0)=f(x) \quad u_t(x,0)=g(x) \tag{*} $$

A solution to this equation is given by $$ u(.,t)=f*\partial_t\Phi_t+g*\Phi_t \tag{**} $$ where $\Phi_t$ is the inverse Fourier transform of the function $\Psi_t(\xi)=\frac{\text{sin}2\pi |\xi|t}{2\pi |\xi|}$. Now for $n=3$, how do we show that $\Phi_t$ as a tempered distribution is given by $$ \displaystyle \langle \Phi_t, h \rangle := \frac{t}{4\pi} \int_{S^2} h(t\omega)\ d\omega \tag{***} $$ That is $$ \displaystyle \Phi_t (h) := \frac{t}{4\pi} \int_{S^2} h(t\omega)\ d\omega $$ for all Schwartz function $h$?

Note that, I am using the following definition of Fourier transform:

$$ F[f]=\hat{f}(\xi)=\!\int\limits_{\mathbb{R}^3}\!f(x)e^{-2\pi ix\cdot\xi}dx $$

and the Fourier transform of a tempered distribution is defined as:

$$ \bigl\langle F[\lambda],\varphi\bigr\rangle =\bigl\langle \lambda,\check{\varphi}\bigr\rangle \quad\forall\,\varphi\in \mathcal{S} $$

OR:

How to verify that $(**)$ is a solution to $(*)$ where $\Phi_{t}$ is a tempered distribution given by $(***)$?

Answer 1

Representation $\Psi_t(\xi)=\frac{\sin{(2\pi |\xi|t)}}{2\pi |\xi|}$ corresponds to Fourier transform written in the form $$ F[f]=\hat{f}(\xi)=\!\int\limits_{\mathbb{R}^3}\!f(x)e^{-2\pi ix\cdot\xi}dx \quad\Rightarrow\quad F[\partial_x^{\alpha}f]=(2\pi i\xi)^{\alpha}\hat{f}(\xi)\tag{1} $$ generally preferred in Harmonic Analysis.  In Partial Differential Equations, another form of Fourier transform is generally preferred $$ F[f]=\hat{f}(\xi)=\!\int\limits_{\mathbb{R}^3}\!f(x)e^{-ix\cdot\xi}dx \quad\Rightarrow\quad F[\partial_x^{\alpha}f]=(i\xi)^{\alpha}\hat{f}(\xi)\tag{2} $$ which looks more habitual in the section tagged pde, with the inverse transform written in the form $$ F^{-1}[g]=\check{g}(x)=\frac{1}{(2\pi)^3}\!\int\limits_{\mathbb{R}^3}\!g(\xi)e^{ix\cdot\xi}d\xi. $$ In version $(1)$, a pde $\,u_{tt}=\Delta u\,$ transforms to $\,\hat{u}_{tt}= -4\pi^2|\xi|^2\hat{u}\,$, while it transforms to $\,\hat{u}_{tt}=-|\xi|^2\hat{u}\,$ in version $(2)$.   In fact, this extra $2\pi$-factor looks just like an absolutely useless stumbling block in Partial Differential Equations, where exactly for this reason version $(1)$ is overwhelmingly disliked.  So, in version $(1)$, solution of the Cauchy problem really looks like $$ \hat{u}(\xi,t)=\hat{f}(\xi)\cos{(2\pi|\xi|t)}+ \hat{g}(\xi)\frac{\sin{(2\pi|\xi|t)}}{2\pi|\xi|}, $$ while it must look like $$ \hat{u}(\xi,t)=\hat{f}(\xi)\cos{(|\xi|t)}+ \hat{g}(\xi)\frac{\sin{(|\xi|t)}}{|\xi|} $$ in version $(2)$.  Namely thereby, in version $(2)$, it must be $\Psi_t(\xi)=\frac{\sin{(|\xi|t)}}{|\xi|}$.

Denote by $\mathcal{S}'$ the Schwartz space of tempered distributions dual to the Schwartz space of rapidly decreasing functions $\mathcal{S}=\mathcal{S} (\mathbb{R}^3)$. Denote by $\delta_R\in \mathcal{S}'$ a delta function supported on a sphere $S_R=\{x\in\mathbb{R}^3\colon |x|=R\}$ of radius $R>0$, otherwise called a spherical delta, defined by the identity $$ \langle \delta_R,\varphi\rangle=\int\limits_{S_R}\varphi(x)\,ds_x\quad \forall\,\varphi\in \mathcal{S}. $$ To find Fourier transform of $\delta_{S_R}$ in $\mathcal{S}'$, recall that a Fourier transform $\hat{f}\in\mathcal{S}'$ of some distribution $f\in\mathcal{S}'$ is traditionally defined as the linear functional $\hat{f}$ subject to the rule $$ \langle\hat{f},\varphi\rangle=\langle f,\hat{\varphi}\rangle\quad\forall\, \varphi\in\mathcal{S}.\tag{3} $$ Note that there can be no definition of the distribution Fourier transform other than $(3)$ according to the fundamental principle underlying the Standard Theory of Distributions. This principle reads something like: A distribution definition must not contradict to its classical prototype. Say, for a function $f\in L^1$, with its classical Fourier transform $\hat{f}$ defined by $(2)$, the distribution Fourier transform $\hat{f}$ is to be defined exactly by the identity $(3)$, since $f$ is to be embedded into $\mathcal{S}'$ as a regular distribution, i.e., as the linear functional defined by the identity $$ \langle f,\varphi\rangle=\int\limits_{\mathbb{R}^3}f(x)\varphi(x)\,dx\quad \forall\,\varphi\in\mathcal{S},\tag{4} $$ whence readily follows $$ \langle\hat{f},\varphi\rangle= \!\int\limits_{\mathbb{R}^3}\!\hat{f}(\xi)\varphi(\xi)\,d\xi= \!\int\limits_{\mathbb{R}^3}\!\varphi(\xi)\,d\xi \!\int\limits_{\mathbb{R}^3}\!f(x)e^{-i x\cdot\xi}dx= \!\int\limits_{\mathbb{R}^3}\!f(x)\,dx \!\int\limits_{\mathbb{R}^3}\!\varphi(\xi) e^{-i x\cdot\xi}\,d\xi= \langle{f},\hat{\varphi}\rangle $$ which implies definition $(3)$.  Note that there is no alternative to $(4)$ for embedding $\,f\in L^1\,$ into $\,\mathcal{S}'\,$ as a linear functional. Say, a sesquilinear form $$ \langle f,\varphi\rangle=\int\limits_{\mathbb{R}^3}f(x)\overline{\varphi(x)}\,dx\quad \forall\,\varphi\in\mathcal{S},\tag{5} $$ with a complex conjugate $\,\overline{\varphi}$, is absolutely unacceptable as an alternative to the bilinear form $(4)$, since $(5)$ defines a functional antilinear in $\varphi$, and hence living outside $\mathcal{S}'$ traditionally defined as a space of linear functionals on $\mathcal{S}$.

So, the only definition we can have is $$ \bigl\langle F[\delta_{S_R}],\varphi\bigr\rangle =\bigl\langle \delta_{S_R},\hat{\varphi}\bigr\rangle =\int\limits_{S_R}\hat{\varphi}(x)\,ds_x =\int\limits_{\mathbb{R}^3}\Bigl(\int\limits_{S_R} e^{-ix\cdot \xi}ds_x\Bigr) \varphi (\xi)\,d\xi\quad \forall\,\varphi\in \mathcal{S} $$ where the order of integration has been interchanged. Hence we find $$ F[\delta_{S_R}]=\int\limits_{S_R} e^{-ix\cdot\xi}ds_x=\int\limits_{S_R} e^{-ix_3|\xi|}ds_x $$ with the RHS integral invariant with respect to rotations of the sphere $S_R$. To establish the rather obvious identity $$ \int\limits_{S_R} e^{-ix\cdot\xi}ds_x=\int\limits_{S_R} e^{-ix_3|\xi|}ds_x \quad\forall\, x\in\mathbb{R}^3,\tag{6} $$ take a rotation $\,T\,$ of the sphere $\,S_R\,$ in $\,\mathbb{R}^3$.  We have $\,T^{-1}=T^{\ast}\,$ and $\,TT^{\ast}=T^{\ast}T=I\,$ while $\,TS_R=S_R\,$ and $\,|{\rm det\,}T|=1$. Hence, for the inner product in $\mathbb{R}^3$, we have $$ (Tz)\cdot\xi=z\cdot T^{\ast}\xi\;\; \forall\,z\in S_R\,. $$ Now fix arbitarary $\,\xi\in\mathbb{R}^3\,$ and choose the rotation $\,T=T_{\xi}\,$ such that $\,T_{\xi}^{\ast}\xi=(0,0,|\xi|)$. Change of variables $\,x=T_{\xi}z\,$ yields $$ \int\limits_{S_R}e^{-ix\cdot\xi}ds_x= \int\limits_{S_R}e^{-i(T_{\xi}z)\cdot\xi}ds_z= \int\limits_{S_R}e^{-iz\cdot (T_{\xi}^{\ast}\xi)}ds_z= \int\limits_{S_R}e^{-iz_3|\xi|}ds_z\,, $$ which implies $(6)$.  Finally, changing to spherical coordinates, we get $$ \begin{align*} \int\limits_{S_R} e^{-ix_3|\xi|}ds_x= \int\limits_{0}^{2\pi}\int\limits_0^\pi e^{-i |\xi| R \cos \theta} R^2 \sin\theta\, d\theta d\varphi =2\pi R^2 \int\limits_0^\pi e^{-i|\xi|R \cos \theta}\sin\theta\, d\theta\\ =4\pi R^2 \int\limits_0^{\pi/2} \cos (|\xi|R\cos \theta)\sin \theta\, d\theta = 4\pi R\frac{\sin (|\xi|R)}{|\xi|}. \end{align*} $$ Thus we have found the Fourier transform of $\delta_{S_R}$ in $\mathcal{S}'$, namely, $$ F[\delta_{S_R}]=4\pi R\frac{\sin (|\xi|R)}{|\xi|}\quad \forall\,R>0, $$ whence follows $$ \Phi_t\overset{\rm def}{=}F^{-1}\Bigl[\frac{\sin (|\xi|t)}{|\xi|}\Bigr]= \frac{1}{4\pi t}\delta_{S_t}\quad \forall\, t>0 $$ which implies that $$ \langle \Phi_t,h\rangle=\frac{1}{4\pi t}\langle\delta_{S_t},h\rangle= \frac{1}{4\pi t}\!\int\limits_{S_t}\,h(x)\,ds_x= \frac{t}{4\pi}\!\int\limits_{S_1}\,h(ty)\,ds_y= \frac{t}{4\pi} \int\limits_{S^2} h(t\omega)\, d\omega\quad \forall\,h\in \mathcal{S} $$ where $S^2\overset{\rm def}{=}S_1\,$.   Q.E.D.