How do I evaluate this integral?

[docnet]

Homework Statement

How do I evaluate this integral?

Relevant Equations

Kirchhoff's formula
$$u(t,x)=\frac{1}{4\pi t}\int_{||y-x||=t}h(y)dS(y)+\frac{\partial}{\partial t}\Big[\frac{1}{4\pi t}\int_{||y-x||=t}g(y)dS(y)\Big]$$

The goal is to evaluate the below integrals. Please note $x\in \mathbb{R}^3$

The issue is that I do not understand the meaning of the integration boundary $||y-x||=t$ and the meaning of the notation $dS(y)$. Would someone be kind to explain these notations to me like I am five? are $x$ and $y$ both in $\mathbb{R}^3$? what is the equality with $t$ doing in the integral?
Thank you.

$$u(t,x)=\frac{1}{4\pi t}\int_{||y-x||=t}h(y)dS(y)+\frac{\partial}{\partial t}\Big[\frac{1}{4\pi t}\int_{||y-x||=t}g(y)dS(y)\Big]$$
where
$$g(y)=h(y)=exp(-\frac{1}{1-y^2})$$ over the unit disk
and $0$ else

 

[Office_Shredder]

$x$ is the input of the function, so is constant. $y$ is the variable you are integrating over. $t$ is also a constant, so $||y-x||=t$ is integrating over the region of $y$ which is constant distance from $x$, i.e. a sphere of radius t centered at x.

 

[docnet]

After trying to evaluate this integral last night, I think maybe my professor did not want us to evaluate the integrals, But to understand how the integrals describe a 3D wave solution. Thank you! @Office_Shredder

 

[docnet]

I think that according to the formula, $u(t,x,y,z)$ is a function that describes the amplitude of the wave at position $(x,y,z)$ for all times t. At time $t=0$ the wave is confined to the unit ball and described by the function $exp(−\frac{1}{1−r^2})$. The wave propagates outwards as t increases, and is contained by larger and larger spherical shells centered at the origin. At an arbitrary point $(x_0,y_0,z_0)$ outside of the unit ball, the amplitude of the wave is zero at all times except for one special time the wave passes through.

My professor tells me that it is possible to show that the wave vanishes outside of the spherical shell $\{x\in\mathbb{R}^3|t−1≤||x||≤t+1\}$.

I tried to evaluate Kirchoff's formula, and frankly I feel embarrassed about my work, and hesitated posting this here. But I very much I want to learn and get better at math.

@Orodruin

@Office_Shredder

$$u(t,x)=\frac{1}{4\pi t}\int_{||r-x||=t}h(r)dS(r)+\frac{\partial}{\partial t}\Big[\frac{1}{4\pi t}\int_{||r-x||=t}g(r)dS(r)\Big]$$
Plug in $h(r)$, $g(r)$ and make a change of boundaries $r\Rightarrow r+x$
$$\frac{1}{4\pi t}\int_{||r||=t}exp\Big(-\frac{1}{1-(r+x)^2}\Big)dS(r)+\frac{\partial}{\partial t}\Big[\frac{1}{4\pi t}\int_{||r||=t}exp\Big(-\frac{1}{1-(r+x)^2}\Big)dS(r)\Big]$$
$$=\frac{1}{16\pi^2 t}\int_0^\pi \int_0^{2\pi}exp\Big(-\frac{1}{1-(r+x)^2}\Big)sin\phi d\theta d\phi$$$$+\frac{\partial}{\partial t}\Big[\frac{1}{16\pi^2 t}\int_0^\pi \int_0^{2\pi} exp\Big(-\frac{1}{1-(r+x)^2}\Big)sin\phi d\theta d\phi \Big]$$
$$=exp\Big(-\frac{1}{1-(r+x)^2}\Big)\Big[\frac{1}{4\pi t}+\frac{\partial}{\partial t}\Big(\frac{1}{4\pi t} \Big)\Big]$$
$$=exp\Big(-\frac{1}{1-(r+x)^2}\Big)\frac{t-4\pi}{16\pi^2 t^2}$$

The solution does not make sense because says the wave has negative amplitude at the origin at $t<4π$ and nears $0$ for $t>4π$.

 

[docnet]

I am so sorry about the latex format issues. Post updated.

 

[Orodruin]

You seem to just have inserted the expression for the bump function for when its argument is < 1, but the integration domain is not necessarily restricted to this region.

 

[docnet]

thank you @Orodruin for the clear advice on the boundary. So I re-evaluated the integrals over the unit ball centered at the origin, and got the same answer because I used this method of spherical means (at the bottom of first page in the linked Stanford document) that does not take the radius of the ball into account. That is truly strange to think the spherical average does not depend on the radius, surely the document must contain an error?