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- Homework Statement
- please see below

- Relevant Equations
- $$\partial_t^2u(t,r)=\partial^2_ru(t,r)+\frac{n-1}{r}\partial_ru(t,r)$$

Hi all, My question is about the attenuation and delay terms in part (1). ~~what are attenuation and delay terms describing in physical phenomenon? thank you.~~

Consider the wave equation for spherical waves in ##n##-dimensions given by

$$\partial_t^2u(t,r)=\partial^2_ru(t,r)+\frac{n-1}{r}\partial_ru(t,r)$$

for an unknown function ##u:(0,\infty)\times(0,\infty)\rightarrow \mathbb{R}##

(1) Consider twice continuously differentiable functions ##\alpha:(0,\infty)\rightarrow (0,\infty)## (the attenuation) and ##\beta:(0,\infty)\mathbb{R}## (the delay) and ##f:\mathbb{R}\rightarrow \mathbb{R}## and make the ansatz

$$u(t,r)=\alpha(r)f(t-\beta(r))$$

(2) We insert this ansatz into the spherical wave equation

$$\partial_t^2\Big[\alpha(r)f(t-\beta(r))\Big]-\partial_r^2\Big[\alpha(r)f(t-\beta(r))\Big]-\frac{n-1}{r}\partial_r\Big[\alpha(r)f(t-\beta(r))\Big]$$

We compute the first term

$$\partial_t^2\Big[\alpha(r)f(t-\beta(r))\Big]\Rightarrow \boxed{\alpha(r)\partial_t^2f(t-\beta(r))}$$

The first derivative of the second term $$-\partial_r\Big[\alpha(r)f(t-\beta(r))\Big] =-\partial_r\alpha(r)f(t-\beta(r))+\alpha(r)\partial_r\beta(r)\partial_rf(t-\beta(r))$$

The derivative of the first term of the first derivative

$$-\partial_r^2\alpha(r)f(t-\beta(r))+\partial_r\alpha(r)\partial_r\beta(r)\partial_rf(t-\beta(r))$$

The derivative of the second term of the first derivative

$$\Big[\partial_r\alpha(r)\partial_r\beta(r)+\alpha(r)\partial^2_r\beta(r)\Big]\partial_rf(t-\beta(r))-\alpha(r)[\partial_r\beta(r)]^2\partial_r^2f(t-\beta(r))$$

That gives the second derivative of the second term $$\Rightarrow \boxed{-\partial_r^2\alpha(r)f(t-\beta(r))+\partial_r\alpha(r)\partial_r\beta(r)\partial_rf(t-\beta(r))}$$

$$\boxed{+\Big[\partial_r\alpha(r)\partial_r\beta(r)+\alpha(r)\partial^2_r\beta(r)\Big]\partial_rf(t-\beta(r))-\alpha(r)[\partial_r\beta(r)]^2\partial_r^2f(t-\beta(r))}$$

The third term:

$$-\frac{n-1}{r}\partial_r\Big[\alpha(r)f(t-\beta(r))\Big]\Rightarrow \boxed{-\frac{n-1}{r}\Big[\partial_r\alpha(r)f(t-\beta(r))-\alpha(r)\partial_r\beta(r)\partial_rf(t-\beta(r))\Big]}$$

We collect the coefficients of the ##f(t-\beta(r))## terms $$\partial_r^2\alpha(r)+\frac{n-1}{r}\partial_r\alpha(r)$$

and the coefficients of the ##\partial_rf(t-\beta(r))## terms

$$\partial_r\alpha(r)\partial_r\beta(r)\alpha(r)+\Big[\partial_r\alpha(r)\partial_r\beta(r)+\alpha(r)\partial^2_r\beta(r)\Big]+\frac{n-1}{r}\alpha(r)\partial_r\beta(r)$$

and the coefficients of the ##\partial_r^2f(t-\beta(r))## terms

$$\alpha(r)[\partial_r\beta(r)]^2-\alpha(r)$$

To make the brackets banish to zero, we set the three collections of terms equal to zero. Setting each collection of terms equal to zero gives the following system of ODEs

$$\boxed{\begin{cases}

\partial_r^2\alpha(r)+\frac{n-1}{r}\partial_r\alpha(r) =0 \\-2\partial_r\alpha(r)+\frac{n-1}{r}\alpha(r) =0\\

[\partial_r\beta(r)]^2=1

\end{cases}}$$

(4) The ODE that involves ##α##only is $$\partial_r^2\alpha(r)+\frac{n-1}{r}\partial_r\alpha(r) =0$$

Case: ##n=1##

$$\partial_r^2\alpha(r) =0$$

The roots of the characteristic polynomial ##\lambda^2=0## is just ##0##. The solution is a polynomial of the form

$$\Rightarrow \boxed{\alpha(r)=cr+d} \quad \text{where }\quad c,d\in\mathbb{R}$$

setting ##n=2## gives a second order euler homegenuous ODE.

$$\partial_r^2\alpha(r)+\frac{1}{r}\partial_r\alpha(r) =0$$

The equation has a solution of the form ##r^x##. plugging ##r^x## into the ODE gives the general solution

$$\boxed{\alpha(r)=cln(r)+d}$$

For the case ##n\geq 3## the ODE is

$$\partial_r^2\alpha(r)+\frac{2}{r}\partial_r\alpha(r) =0$$

By similar methods to the case ##n=2##,

$$\boxed{\alpha(r)=\frac{c}{t}+d}$$

(5) The ODE for ##β(r)## is

$$[\partial_r\beta(r)]^2=1\Rightarrow \boxed{\partial_r\beta(r)=\pm 1}$$

(6) The following system of ODEs in ##\alpha(r)##

$$\begin{cases}

\partial_r^2\alpha(r)+\frac{n-1}{r}\partial_r\alpha(r)=0\\

-2\partial_r\alpha(r)+\frac{n-1}{r}\alpha(r) =0\end{cases}$$ shows that there are no solutions, unless ##n=1## or ##n=3##. We assume the solutions is of form $$\alpha(r)=cr^a$$ for some constants ##c,a\in\mathbb{R}##.

Plugging in the solution to the system of ODEs gives

$$\begin{cases}

a(a-1)+(n-1)c=0 \\ -2a+(n-1)=0

\end{cases}$$

whose solution exists for ##a## only if ##n=1## or ##n=3##.

(7) When we plug in ##n=1##, we get ##a=0## which means ##\alpha{r}=c## for some ##c\in\mathbb{R}##.

**What do "attenuation" and "delay" mean in terms of real-life physical phenomena?**Consider the wave equation for spherical waves in ##n##-dimensions given by

$$\partial_t^2u(t,r)=\partial^2_ru(t,r)+\frac{n-1}{r}\partial_ru(t,r)$$

for an unknown function ##u:(0,\infty)\times(0,\infty)\rightarrow \mathbb{R}##

(1) Consider twice continuously differentiable functions ##\alpha:(0,\infty)\rightarrow (0,\infty)## (the attenuation) and ##\beta:(0,\infty)\mathbb{R}## (the delay) and ##f:\mathbb{R}\rightarrow \mathbb{R}## and make the ansatz

$$u(t,r)=\alpha(r)f(t-\beta(r))$$

(2) We insert this ansatz into the spherical wave equation

$$\partial_t^2\Big[\alpha(r)f(t-\beta(r))\Big]-\partial_r^2\Big[\alpha(r)f(t-\beta(r))\Big]-\frac{n-1}{r}\partial_r\Big[\alpha(r)f(t-\beta(r))\Big]$$

We compute the first term

$$\partial_t^2\Big[\alpha(r)f(t-\beta(r))\Big]\Rightarrow \boxed{\alpha(r)\partial_t^2f(t-\beta(r))}$$

The first derivative of the second term $$-\partial_r\Big[\alpha(r)f(t-\beta(r))\Big] =-\partial_r\alpha(r)f(t-\beta(r))+\alpha(r)\partial_r\beta(r)\partial_rf(t-\beta(r))$$

The derivative of the first term of the first derivative

$$-\partial_r^2\alpha(r)f(t-\beta(r))+\partial_r\alpha(r)\partial_r\beta(r)\partial_rf(t-\beta(r))$$

The derivative of the second term of the first derivative

$$\Big[\partial_r\alpha(r)\partial_r\beta(r)+\alpha(r)\partial^2_r\beta(r)\Big]\partial_rf(t-\beta(r))-\alpha(r)[\partial_r\beta(r)]^2\partial_r^2f(t-\beta(r))$$

That gives the second derivative of the second term $$\Rightarrow \boxed{-\partial_r^2\alpha(r)f(t-\beta(r))+\partial_r\alpha(r)\partial_r\beta(r)\partial_rf(t-\beta(r))}$$

$$\boxed{+\Big[\partial_r\alpha(r)\partial_r\beta(r)+\alpha(r)\partial^2_r\beta(r)\Big]\partial_rf(t-\beta(r))-\alpha(r)[\partial_r\beta(r)]^2\partial_r^2f(t-\beta(r))}$$

The third term:

$$-\frac{n-1}{r}\partial_r\Big[\alpha(r)f(t-\beta(r))\Big]\Rightarrow \boxed{-\frac{n-1}{r}\Big[\partial_r\alpha(r)f(t-\beta(r))-\alpha(r)\partial_r\beta(r)\partial_rf(t-\beta(r))\Big]}$$

We collect the coefficients of the ##f(t-\beta(r))## terms $$\partial_r^2\alpha(r)+\frac{n-1}{r}\partial_r\alpha(r)$$

and the coefficients of the ##\partial_rf(t-\beta(r))## terms

$$\partial_r\alpha(r)\partial_r\beta(r)\alpha(r)+\Big[\partial_r\alpha(r)\partial_r\beta(r)+\alpha(r)\partial^2_r\beta(r)\Big]+\frac{n-1}{r}\alpha(r)\partial_r\beta(r)$$

and the coefficients of the ##\partial_r^2f(t-\beta(r))## terms

$$\alpha(r)[\partial_r\beta(r)]^2-\alpha(r)$$

To make the brackets banish to zero, we set the three collections of terms equal to zero. Setting each collection of terms equal to zero gives the following system of ODEs

$$\boxed{\begin{cases}

\partial_r^2\alpha(r)+\frac{n-1}{r}\partial_r\alpha(r) =0 \\-2\partial_r\alpha(r)+\frac{n-1}{r}\alpha(r) =0\\

[\partial_r\beta(r)]^2=1

\end{cases}}$$

(4) The ODE that involves ##α##only is $$\partial_r^2\alpha(r)+\frac{n-1}{r}\partial_r\alpha(r) =0$$

Case: ##n=1##

$$\partial_r^2\alpha(r) =0$$

The roots of the characteristic polynomial ##\lambda^2=0## is just ##0##. The solution is a polynomial of the form

$$\Rightarrow \boxed{\alpha(r)=cr+d} \quad \text{where }\quad c,d\in\mathbb{R}$$

setting ##n=2## gives a second order euler homegenuous ODE.

$$\partial_r^2\alpha(r)+\frac{1}{r}\partial_r\alpha(r) =0$$

The equation has a solution of the form ##r^x##. plugging ##r^x## into the ODE gives the general solution

$$\boxed{\alpha(r)=cln(r)+d}$$

For the case ##n\geq 3## the ODE is

$$\partial_r^2\alpha(r)+\frac{2}{r}\partial_r\alpha(r) =0$$

By similar methods to the case ##n=2##,

$$\boxed{\alpha(r)=\frac{c}{t}+d}$$

(5) The ODE for ##β(r)## is

$$[\partial_r\beta(r)]^2=1\Rightarrow \boxed{\partial_r\beta(r)=\pm 1}$$

(6) The following system of ODEs in ##\alpha(r)##

$$\begin{cases}

\partial_r^2\alpha(r)+\frac{n-1}{r}\partial_r\alpha(r)=0\\

-2\partial_r\alpha(r)+\frac{n-1}{r}\alpha(r) =0\end{cases}$$ shows that there are no solutions, unless ##n=1## or ##n=3##. We assume the solutions is of form $$\alpha(r)=cr^a$$ for some constants ##c,a\in\mathbb{R}##.

Plugging in the solution to the system of ODEs gives

$$\begin{cases}

a(a-1)+(n-1)c=0 \\ -2a+(n-1)=0

\end{cases}$$

whose solution exists for ##a## only if ##n=1## or ##n=3##.

(7) When we plug in ##n=1##, we get ##a=0## which means ##\alpha{r}=c## for some ##c\in\mathbb{R}##.

**edited: grammar**#### Attachments

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