A pressure disturbance in 3D progates according to 3D equation $u_{tt}=\triangle u $
Suppose at t=0, a bomb explodes creating a disturbance exactly at position x=0 and cause an initial velocity to be
$\phi(x)=0$ and $\psi(x)=1$ for $|x|\leq$1 and 0 otherwise
a) At t=10, what is the value of u at the point (10,0,0) (can leave your answer as an integral)
Answer: By Kirchoff formula:
$u(x_{0},t)=\frac{1}{4\pi t^{2}}\int\int_{dB(x_{0},t)}\phi(x)+\nabla\phi(x)(x-x_{0})+t\psi(x)dS_{x}$
$u((10,0,0),10)=\frac{1}{4\pi 10^{2}}\int\int_{dB((10,0,0),10)}10\psi(x)dS_{x}$ what I am not sure is the boundary of the integral, how can you implement $\psi(x)=1$ for $|x|\leq1$ ?
b) At $t=10$, what is the value of u at the point $x_0=(20,8,17)$? (Give a numerical value)
I think I am supposed to apply the Kirchoff's formula as above but again, not sure about the region of integration. A sphere center at $(20,8,17)$ and radius $t=10$?
c)Suppose I am at the point $x=(20,20,20)$. At what time will I feel the initial disturbance? i.e at what time $t>0$ will $u((20,20,20),t)$ be non zero.
Answer:
From Kirchoff formula, $u((0,0,0),0)$ will only influence the value of $u((20,20,20),t)$ if $|x-x_0|=t$
That is, for $(20,20,20)$ to be on the sphere center at $(0,0,0)$ with radius t, then $t=20$
I dont know if you will find this of interest, but anyway.
Spherically symmetric solutions of the wave equation have the form$$u(r,t)=F(r-t)/r+G(r+t)/r$$ We will take G=0 because that represents an incoming wave and pick $F$ to match the initial conditions. We arrive at the solution $$u(r,t)=\frac{(r-t)(1-H(r-1-t))}{r}$$
where $H$ is the Heaviside function. This only makes $u_t$ zero almost everywhere at $t=0$. I think this is inevitable with discontinuous initial conditions. There is no need to use the Helmholz equation.