# Radial solutions to the Laplace equation

• docnet
In summary: I think the substitution ##v'=a## is supposed to do this.Yes, that is correct! The substitution ##v'=a## is precisely meant to eliminate the cross terms, and leaves you with just ##r^2## in the denominator. This allows you to solve for ##a## and then integrate to find ##v##. Good job!

#### docnet

Gold Member
Homework Statement
Relevant Equations

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 $$v'=a$$
$$0=a'+\frac{n-1}{r}a$$
$$\frac{a'}{a}=\frac{1-n}{r}$$
$$\frac{d}{da} ln(a)=\frac{1-n}{r}$$
$$ln(a)=\int\frac{1-n}{r}dr$$
$$ln(a)=(1-n)ln(r)+C$$
$$a=Cr^{1-n}$$
$$v'=Cr^{1-n}$$
$$v=Cr^{-n}+D$$

Last edited:
##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##

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

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

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

docnet said:
when ##n=2##, ##\ v=C+D##.
I don't think that with your answer the equation ##v^′=Cr^{1−n}## is satisfied .

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

oops.. yes you're so right.

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

With ##n=2## ?

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##

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 ?

##\ ##

BvU said:
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
$$v''(|x|)+v'(|x|)\frac{n-1}{r}$$

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

docnet said:
And now my eyes are numb from typing in ##\LaTeX## .
I believe you. And that much isn't even necessary !

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

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

## What is the Laplace equation?

The Laplace equation is a partial differential equation that describes the behavior of a scalar field in a given region. It is typically used to model physical phenomena such as heat flow, fluid dynamics, and electrostatics.

## What are radial solutions to the Laplace equation?

Radial solutions to the Laplace equation are solutions that depend only on the distance from the center of a given region. This means that the solution is symmetric about the center and does not change with respect to direction.

## Why are radial solutions important?

Radial solutions are important because they simplify the Laplace equation and make it easier to solve. They also have many real-world applications, such as in the study of spherical objects and in the design of radial systems.

## How can radial solutions be found?

Radial solutions to the Laplace equation can be found using the separation of variables method. This involves separating the variables into radial and angular components, solving each component separately, and then combining them to form the final solution.

## What are some examples of radial solutions?

Some examples of radial solutions to the Laplace equation include the potential inside a spherical capacitor, the temperature distribution in a spherical object, and the electric potential around a point charge.