I have been solving the wave equation in the upper half-space $\mathbb{R}^{n+1}_+$ $$ D^2_{tt}u(x,t)=\nabla^2{u}, \quad u(x,0)=f(x), \quad D_tu(x,0)=g(x), \quad x\in \mathbb{R}^n, \quad t>0 $$ using Fourier transform and obtained the following:
$$ \hat{u}(\xi,t)=\hat{f}(\xi)\cos{(t|\xi|)}+ \hat{g}(\xi)\frac{\sin{(t|\xi|)}}{|\xi|}, \quad \xi\in \mathbb{R}^n, \quad t>0. $$ Hence, the inversion formula gives for $(x,t)\in \mathbb{R}^{n+1}_+$, $$ u(x,t)=\frac{1}{(2\pi)^n}\int_{\mathbb{R}^n}\hat{f}(\xi)\cos{(|\xi|t)}e^{i\xi\cdot x}d\xi+ \frac{1}{(2\pi)^n}\int_{\mathbb{R}^n}\hat{g}(\xi)\frac{\sin{(|\xi|t)}}{|\xi|}e^{i\xi\cdot x}d\xi=u_1+u_2. $$ I consider first $$u_2=\frac{1}{(2\pi)^n}\int_{\mathbb{R}^n}\hat{g}(\xi)\frac{\sin{(|\xi|t)}}{|\xi|}e^{i\xi\cdot x}d\xi=(g\ast T)(x). $$ Where $$T(x)=(\hat{T(\xi)})^\vee=\left[\left(\frac{\sin{(|\xi|t)}}{|\xi|}\right)^\vee\right]^\wedge.$$
For $n=1$ that is, in one dimension;
I consider $\frac{1}{2}\chi_{[-t,t]}=\left(\frac{\sin{(|\xi|t)}}{|\xi|}\right)^\vee$ from the fact that $\hat{\chi}_{[-t,t]}=\int_{-t}^{t}1\cdot e^{-ix\cdot\xi}dx$.
- My first problem is that I am failing to get an expression of D'Alembert solution
$$u(x,t)=F(x+t)+ G(x-t).$$
- My second problem is that I want to see how the solution for $n=2$, that is, in $2$-dimension is obtained. Especially, on how to evaluate along the boundary of a circle.