Radial solutions to the Laplace equation

[docnet]

Homework Statement

please see below

Relevant Equations

please see below

Part 1
$$\Delta u(x)=\Delta v(|x|)$$
Substitute $$|x|=r=\sqrt{\sum_{i=1}^n{x^2_i}}$$
$$u'(x)= v'(r)\frac{\sum_{i=1}^nx_i}{\sqrt{\sum_{i=1}^n{x^2_i}}}$$
$$u''(x)=v''(r)\frac{\sum_{i=1}^nx_i}{\sqrt{\sum_{i=1}^n{x^2_i}}}+v'(r)f(x)=v''(r)+v'(r)f(x)$$
$$f(x)=\frac{n\sqrt{\sum_{i=1}^nx_i^2}-\frac{\sum_{i=1}^nx_i \sum_{i=1}^nx_i}{\sqrt{\sum_{i=1}^nx_i^2}}}{\sum_{i=1}^nx_i^2}=\frac{nr-(\frac{r^2}{r})}{r^2}=\frac{n-1}{r}$$
$$u''(x)=v''(r)+\frac{n-1}{r}v'(r)$$part 2
make the substitution $$v'=a$$
$$0=a'+\frac{n-1}{r}a$$
$$\frac{a'}{a}=\frac{1-n}{r}$$
$$\frac{d}{da} ln(a)=\frac{1-n}{r}$$
$$ln(a)=\int\frac{1-n}{r}dr$$
$$ln(a)=(1-n)ln(r)+C$$
$$a=Cr^{1-n}$$
$$v'=Cr^{1-n}$$
$$v=Cr^{-n}+D$$

 

[BvU]

$v'=Cr^{1-n}$ is $C\over r$ for $n = 2$. How do you make the last step ?

 

[docnet]

Hint 2 confused me. It is because I differentiated the $Cr^{1-n}$ term and absorbed (1−n) into C instead of integrating.

$$v′=Cr^{1−n}⇒v=Cr^{2−n}+D$$
Where C and D are constants of integration, giving me a two-parameter family of solutions to the ODE.

when $n=2$, $v=C+D$. when $n=3$, $v=\frac{C}{r}+D$ so the solutions look different for $n=2$ and for $n≥3$

 

[docnet]

I am unsure whether I did this step correctly with respect to the negative term in the numerator.$$f(x)=\frac{n\sqrt{\sum_{i=1}^nx_i^2}-\frac{\sum_{i=1}^nx_i \sum_{i=1}^nx_i}{\sqrt{\sum_{i=1}^nx_ii^2}}}{\sum_{i=1}^nx_i^2}=\frac{nr-(\frac{r^2}{r})}{r^2}$$

 

[docnet]

The reason I think I erred is because

$$\sum_{i=1}^n x_{i} \times \sum_{i=1}^n x_i = (x_1 + x_2 ... x_n)(x_1+x_2...x_n) = (x^2_1+x^2_2+...x^2_n) + crossterms=r^2+crossterms$$

also I think I erred here

$$\frac{\sum_{i=1}^nx_i}{\sqrt{\sum_{i=1}^n{x^2_i}}}=\frac{(x_1 + x_2 ... x_n)}{r}$$

 

[BvU]

when $n=2$, $\ v=C+D$.

I don't think that with your answer the equation $v^′=Cr^{1−n}$ is satisfied .

 

[docnet]

I don't think that with your answer the equation $v^′=Cr^{1−n}$ is satisfied .

oops.. yes you're so right.

$$v=C\frac{r^{2-n}}{2-n}+D$$

 

[BvU]

With $n=2$ ?

 

[docnet]

Sorry sir, the solution is $v=Cln(r)+D$ with $n=2$

If this is correct, then the solution to the ODE $0=v''+\frac{n-1}{r}v$ is

$v=Cln(r)+D$ with $n=2$
$v=C\frac{r^{2-n}}{2-n}+D$ with $n≥3$

 

[BvU]

No need to apologize !

Initially I did not check al the $\Sigma$ stuff -- it basically always confuses me no end.
However, here it's an essential part of the exercise, so:
(Note: $\sum x_i$ isn't a very useful thing normally, so when I see that I start becoming suspicious).

You have $$u(x_1, x_2, ... , x_n) = v(|\vec x |) = v(r) = v\left (\sqrt{\sum x_i^2\ } \ \right )$$Now we need $$\Delta u \equiv \vec \nabla\cdot\vec\nabla u\ = \vec\nabla\cdot\left ({\partial u\over \partial x_1} , {\partial u\over \partial x_2}, ... ,{\partial u\over \partial x_n}\right ) $$ for which the most useful form here is $$\Delta u(\vec x) = {\partial^2 u\over \partial x_1^2} + {\partial^2 u\over \partial x_2^2} + ... + {\partial^2 u\over \partial x_n^2} = \sum {\partial^2 u\over \partial x_i^2} $$
and this is where we are going to throw in the chain rule. Want to try ?

$\$

 

[docnet]

No need to apologize !

Initially I did not check al the Σ stuff -- it basically always confuses me no end.
However, here it's an essential part of the exercise, so:
(Note: ∑xi isn't a very useful thing normally, so when I see that I start becoming suspicious).

You have u(x1,x2,...,xn)=v(|x→|)=v(r)=v(∑xi2 )Now we need Δu≡∇→⋅∇→u =∇→⋅(∂u∂x1,∂u∂x2,...,∂u∂xn) for which the most useful form here is Δu(x→)=∂2u∂x12+∂2u∂x22+...+∂2u∂xn2=∑∂2u∂xi2
and this is where we are going to throw in the chain rule. Want to try ?

I evaluate the first derivative of $v$ using the chain rule
$$∂_{x_i}\left[v(|x|)\right]=∂_{x_i}v(|x|)\frac{x_i}{\sqrt{x_1^2+...+x_n^2}}$$
Summing over $i$ gives
$$v'(|x|)\frac{x_1+...+x_n}{\sqrt{x_1^2+...+x_n^2}}$$
I evaluate the second derivative of $v$ using the chain rule
$$∂_{x_i}\left[v'(|x|)\frac{x_1+...+x_n}{\sqrt{x_1^2+...+x_n^2}}\right]$$
$$\partial_{x_i} v'(|x|) \frac{x_i}{\sqrt{x_1^2+...+x_n^2}} \frac{x_1+...+x_n}{\sqrt{x_1^2+...+x_n^2}}+ \partial_{x_i} v(|x|) \frac{\sqrt{x_1^2+...+x_n^2}-\frac{x_i(x_1+...+x_n)}{\sqrt{x_1^2+...+x_n^2}}}{x_1^2+...+x_n^2}$$
summing over $i$ gives
$$v''(|x|) \frac{(x_1+...+x_n)^2}{x_1^2+...+x_n^2} +v'(|x|) \frac{n\sqrt{x_1^2+...+x_n^2}-\frac{(x_1+...+x_n)^2}{\sqrt{x_1^2+...+x_n^2}}}{x_1^2+...+x_n^2}$$
$$v''(|x|)\frac{r^2+crossterms}{r^2}+v'(|x|)\frac{nr-\frac{r^2+crossterms}{r}}{r^2}$$

The problem is $(x_1+...+x_n)^2=r^2+crossterms$. It would be nice for the crossterms to become $0$. then we would have
$$v''(|x|)+v'(|x|)\frac{n-1}{r}$$

 

[BvU]

Summing over i gives

What I wanted to bring across is that you do not want to do that: the gradient (the first derivative) is a vector. It is only in the second step (the divergence of a vector) that the summing pops up. That is why I said

(Note: $\sum x_i\$ isn't a very useful thing normally, so when I see that I start becoming suspicious).

And now my eyes are numb from typing in $\LaTeX$ .

I believe you. And that much isn't even necessary !

if that is okay with the homework forum guidelines.

Of course that is okay -- but you miss out on a bit of satisfaction of cracking the exercise, learning, and perhaps avoiding future errors ...

- - - - -

For later reference and because I seem to enjoy this kind of exercise (is it that obvious ) and somewhat against the PF rules (well,

Barbossa : First, your return to shore was not part of our negotiations nor our agreement so I must do nothing. And secondly, you must be a pirate for the pirate's code to apply and you're not. And thirdly, the code is more what you'd call "guidelines" than actual rules. Welcome aboard the Black Pearl, Miss Turner .)​

​

$$∂_{x_i}\left[v(|x|)\right]=∂_{x_i}v(|x|)\frac{x_i}{\sqrt{x_1^2+...+x_n^2}}$$

Right. Or$$r^2=\sum x_i^2\Rightarrow 2r\,dr = 2\sum x_i\,dx_i
\Rightarrow {\partial r\over \partial x_i} = {x_i\over r}$$Chain rule:
$${\partial u(\vec x)\over \partial x_i}={\partial v(r)\over \partial x_i}
= v' \,{\partial r\over \partial x_i}= v'\,{x_i\over r}

$$as you had. Chain rule again:$$
{\partial^2\, u(\vec x)\over \partial x_i^2}={\partial \over \partial x_i}\left ( v'\,{x_i\over r}\right )=
{\partial v'\over \partial x_i}\, {x_i\over r} + v'\,{\partial x_i\over \partial x_i}\,{1\over r} +
v'\, {x_i}\, {\partial \over \partial x_i}\left({1\over r}\right )

$$Three terms. First one:$$
{\partial v'\over \partial x_i}\, {x_i\over r} = {\partial v'\over \partial r} \, {\partial r \over \partial x_i}\, {x_i\over r}= v'' \, \left ( {x_i^2\over r^2} \right )

$$Second one: $$
v'\,{\partial x_i\over \partial x_i}\,{1\over r} = v'\,{1\over r}

$$ Third:$$
v'\, {x_i}\, {\partial \over \partial x_i}\left({1\over r}\right ) = v'\, {x_i}\, \left ( -1\over r^2 \right ) {\partial r \over \partial x_i} = v'\, {x_i}\, \left ( -1\over r^2 \right ) \, {x_i\over r} = - v'\, {x_i^2 \over r^3}

$$Collect the three again: $$
{\partial^2\, u(\vec x)\over \partial x_i^2}=
v'' \, \left ( {x_i^2\over r^2} \right ) +
v'\,{1\over r} -
v'\, {x_i^2 \over r^3}

$$ and now we do some summing up to write down $\Delta u$ (see post #10) : $$
\begin{align*}
\Delta u(\vec x) = \sum {\partial^2 u\over \partial x_i^2} &=
\sum v'' \, \left ( {x_i^2\over r^2} \right ) +
\sum v'\,{1\over r} -
\sum v'\, {x_i^2 \over r^3} \\ \mathstrut \\ &=
v'' + v'\,{n\over r} - v'\,{1\over r} \\ \mathstrut \\ &= v'' + {n-1\over r}\,v'
\end{align*}

$$as dangled in front of us in the exercise.

Now it's me looking cross-eyed from all the cut&paste $\TeX$. It was fun, though.
I promise to look at your post #11 when I recover tomorrow, and point out where things derail (if at all ).
Although it seems your only issue was with the cross terms, and I hope I have shown that they don't pop up at all.

$\$

 

[docnet]

What I wanted to bring across is that you do not want to do that: the gradient (the first derivative) is a vector. It is only in the second step (the divergence of a vector) that the summing pops up. That is why I saidI believe you. And that much isn't even necessary !

Of course that is okay -- but you miss out on a bit of satisfaction of cracking the exercise, learning, and perhaps avoiding future errors ...

- - - - -

For later reference and because I seem to enjoy this kind of exercise (is it that obvious ) and somewhat against the PF rules (well,

Barbossa : First, your return to shore was not part of our negotiations nor our agreement so I must do nothing. And secondly, you must be a pirate for the pirate's code to apply and you're not. And thirdly, the code is more what you'd call "guidelines" than actual rules. Welcome aboard the Black Pearl, Miss Turner .)​

​

Right. Or$$r^2=\sum x_i^2\Rightarrow 2r\,dr = 2\sum x_i\,dx_i
\Rightarrow {\partial r\over \partial x_i} = {x_i\over r}$$Chain rule:
$${\partial u(\vec x)\over \partial x_i}={\partial v(r)\over \partial x_i}
= v' \,{\partial r\over \partial x_i}= v'\,{x_i\over r}

$$as you had. Chain rule again:$$
{\partial^2\, u(\vec x)\over \partial x_i^2}={\partial \over \partial x_i}\left ( v'\,{x_i\over r}\right )=
{\partial v'\over \partial x_i}\, {x_i\over r} + v'\,{\partial x_i\over \partial x_i}\,{1\over r} +
v'\, {x_i}\, {\partial \over \partial x_i}\left({1\over r}\right )

$$Three terms. First one:$$
{\partial v'\over \partial x_i}\, {x_i\over r} = {\partial v'\over \partial r} \, {\partial r \over \partial x_i}\, {x_i\over r}= v'' \, \left ( {x_i^2\over r^2} \right )

$$Second one: $$
v'\,{\partial x_i\over \partial x_i}\,{1\over r} = v'\,{1\over r}

$$ Third:$$
v'\, {x_i}\, {\partial \over \partial x_i}\left({1\over r}\right ) = v'\, {x_i}\, \left ( -1\over r^2 \right ) {\partial r \over \partial x_i} = v'\, {x_i}\, \left ( -1\over r^2 \right ) \, {x_i\over r} = - v'\, {x_i^2 \over r^3}

$$Collect the three again: $$
{\partial^2\, u(\vec x)\over \partial x_i^2}=
v'' \, \left ( {x_i^2\over r^2} \right ) +
v'\,{1\over r} -
v'\, {x_i^2 \over r^3}

$$ and now we do some summing up to write down $\Delta u$ (see post #10) : $$
\begin{align*}
\Delta u(\vec x) = \sum {\partial^2 u\over \partial x_i^2} &=
\sum v'' \, \left ( {x_i^2\over r^2} \right ) +
\sum v'\,{1\over r} -
\sum v'\, {x_i^2 \over r^3} \\ \mathstrut \\ &=
v'' + v'\,{n\over r} - v'\,{1\over r} \\ \mathstrut \\ &= v'' + {n-1\over r}\,v'
\end{align*}

$$as dangled in front of us in the exercise.

Now it's me looking cross-eyed from all the cut&paste $\TeX$. It was fun, though.
I promise to look at your post #11 when I recover tomorrow, and point out where things derail (if at all ).
Although it seems your only issue was with the cross terms, and I hope I have shown that they don't pop up at all.

$\$

Although no one asked me to, I want to post a follow-up to show how your explanations are helping me in my PDE course. Thank you.

I just worked out a similar example by mimicking your method.

Problem: find the $\Delta u$ of $$u(x)=\frac{||x||^2(1-||x||^2)^2}{3}+\frac{(1-||x||^2)^3}{2}$$

let $||x||=r$ and compute

Chain rule:

$$\partial_{x_i}u(r)=2r^4x_i-\frac{2}{3}r^2x_i^2-\frac{4}{3}x_i$$

Chain rule again:

$$\partial_{x_i}^2u(r)=8r^2x_i^2-\frac{4}{3}x_i^2-\frac{4}{3}x_i^2-\frac{4}{3}$$

summing up for $n=3$

$$\Delta u = 8r^4-\frac{4}{3}r^2-4$$ $$=8||x||^4-\frac{4}{3}||x||^2-4$$