Define the Schwartz space S ( R n ) \mathcal{S}\left(\mathbb{R}^{n}\right) S ( R n )   and the space of tempered distributions S ′ ( R n ) \mathcal{S}^{\prime}\left(\mathbb{R}^{n}\right) S ′ ( R n )  . State the Fourier inversion theorem for the Fourier transform of a Schwartz function.
Consider the initial value problem:
∂ 2 u ∂ t 2 − Δ u + u = 0 , x ∈ R n , 0 < t < ∞ u ( 0 , x ) = f ( x ) , ∂ u ∂ t ( 0 , x ) = 0 \begin{gathered} \frac{\partial^{2} u}{\partial t^{2}}-\Delta u+u=0, x \in \mathbb{R}^{n}, 0<t<\infty \\ u(0, x)=f(x), \quad \frac{\partial u}{\partial t}(0, x)=0 \end{gathered} ∂ t 2 ∂ 2 u  − Δ u + u = 0 , x ∈ R n , 0 < t < ∞ u ( 0 , x ) = f ( x ) , ∂ t ∂ u  ( 0 , x ) = 0  
for f f f   in the Schwartz space S ( R n ) \mathcal{S}\left(\mathbb{R}^{n}\right) S ( R n )  .
Show that the solution can be written as
u ( t , x ) = ( 2 π ) − n / 2 ∫ R n e i x ⋅ ξ u ^ ( t , ξ ) d ξ , u(t, x)=(2 \pi)^{-n / 2} \int_{\mathbb{R}^{n}} e^{i x \cdot \xi} \hat{u}(t, \xi) d \xi, u ( t , x ) = ( 2 π ) − n / 2 ∫ R n  e i x ⋅ ξ u ^ ( t , ξ ) d ξ , 
where
u ^ ( t , ξ ) = cos  ( t 1 + ∣ ξ ∣ 2 ) f ^ ( ξ ) \hat{u}(t, \xi)=\cos \left(t \sqrt{1+|\xi|^{2}}\right) \hat{f}(\xi) u ^ ( t , ξ ) = cos ( t 1 + ∣ ξ ∣ 2  ) f ^  ( ξ ) 
and
f ^ ( ξ ) = ( 2 π ) − n / 2 ∫ R n e − i x ⋅ ξ f ( x ) d x . \hat{f}(\xi)=(2 \pi)^{-n / 2} \int_{\mathbb{R}^{n}} e^{-i x \cdot \xi} f(x) d x . f ^  ( ξ ) = ( 2 π ) − n / 2 ∫ R n  e − i x ⋅ ξ f ( x ) d x . 
State the Plancherel-Parseval theorem and hence deduce that
∫ R n ∣ u ( t , x ) ∣ 2 d x ≤ ∫ R n ∣ f ( x ) ∣ 2 d x . \int_{\mathbb{R}^{n}}|u(t, x)|^{2} d x \leq \int_{\mathbb{R}^{n}}|f(x)|^{2} d x . ∫ R n  ∣ u ( t , x ) ∣ 2 d x ≤ ∫ R n  ∣ f ( x ) ∣ 2 d x .