Question

Find \(\dfrac{\partial^2}{\partial x_1^2}(log|x|\)
Where |\(|x|=\sqrt{x_1^2+x_2^2+x_3^2}\)
Please, help

Answer 1

try calculating just the first derivative, and use the chain rule. Look at \(x_2\) and \(x_3\) as if they're just constants.

Answer 2

\(\dfrac{\partial}{\partial x_1}(log|x|) = \dfrac{1}{|x|}*|x|_{x_1}' \)
\(= \dfrac{1}{|x|}\dfrac{1}{2|x|}*2x_1=\dfrac{x_1}{|x|^2}\)

Answer 3

Am I ok so far?

Answer 4

Yeah, this looks good so far, you can cancel the square root with the square now since the thing under the square root is all squares which are positive.

Answer 5

Now, second derivative w.r.t x_1
\((x_1|x|^{-2})'= |x|^{-2}+x_1(-2)|x^{-3}*\dfrac{x_1}{|x|}=\dfrac{1}{|x|^2}-\dfrac{2x_1^2}{|x|^4}\)
\(=\dfrac{|x|^2-2x_1^2}{|x|^4}\)
Am I right?

Answer 6

Yeah that looks correct to me!

Answer 7

Since |x| is symmetric, they will be the same for x2,x3, right?

Answer 8

Yeah, since trading out the labels on the variables doesn't change the equation, you're right, you derived it for any of the three variables

Answer 9

Note: I am working on Laplace equation and its fundamental solution in \(\mathbb R^n\)
I am starting at the basic case n =2, 3.

Answer 10

In \(\mathbb R^2\)
\[\triangle x = \dfrac{|x|^2-2x_1^2+|x|^2-2x_2^2}{|x|^4}=\dfrac{2|x|^2-2(x_1^2+x_2^2)}{|x|^4}=0\]
because \(|x|=\sqrt{x_1^2+x_2^2}\rightarrow |x|^2=x_1^2+x_2^2\)
As above, it does not hold in \(\mathbb R^3\) , why?

Answer 11

I don't see why you think that it does not hold in \(\mathbb{R}^3\).

Answer 12

Show me it works, please.

Answer 13

\[=\dfrac{|x|^2-2x_1^2}{|x|^4}+\dfrac{|x|^2-2x_2^2}{|x|^4}+\dfrac{|x|^2-2x_3^2}{|x|^4}\]
\[\triangle x = \dfrac{|x|^2-2x_1^2+|x|^2-2x_2^2+|x|^2-2x_3^2}{|x|^4}=\dfrac{3|x|^2-2(x_1^2+x_2^2+x_3^2)}{|x|^4}\neq0\]

Answer 14

Wait, I'm confused as to what you're doing here. Are you saying that
\[\log(|x|)\] is supposed to be a solution to the differential equation \[\Delta x = 0\]

For example, there's no problem with saying a+b+c=0 since this means a=-b-c and the variables are related to each other in some way.

Answer 15

actually, it is \(-\dfrac{1}{2\pi} log|x|\) , not just log|x|, but to me \(-1/2pi\) is just a constant. It doesn't effect the derivative, so, I let it aside.

Answer 16

Yeah totally I agree with that.

Answer 17

Let I show you the paper I am trying to understand. :)

Answer 18

Yeah sure, sounds good :D

Answer 19

page 33 , 2.6

Answer 20

Looks like they are saying the solution for \(n \ge 3\) is:
\[\frac{1}{n(n-2)\alpha_n} \frac{1}{|x|^{n-2}}\]
and the solution for \(n=2\) is
\[\frac{-1}{2\pi}\log|x|\]

so maybe this answers your question? Or are you asking where do any of these solutions come from?

Answer 21

for n=2, I want to check whether it satisfies laplace equation or not before going further. But as shown, it doesn't make sense to me.
for n=3, I got until second derivative but stuck at " It follows that \(\triangle \Gamma =0\) if \(x\neq 0\) "
I don't understand why.

Answer 22

You showed that the solution in \(\mathbb R^2\) is satisfied:
\[\triangle x = \dfrac{|x|^2-2x_1^2+|x|^2-2x_2^2}{|x|^4}=\dfrac{2|x|^2-2(x_1^2+x_2^2)}{|x|^4}=0\]

But if you want to look at Laplace's equation in \(\mathbb R^3\) then you have to use an entirely different equation since the 2D solution is not a solution in 3D.

Answer 23

Yes, I understood. For this particular problem, I worked on log |x|, but the result looks not good.

Answer 24

OOOOOOOOOOOOOOOH, I know where I am stupid. It works for R^2 only. Got you. thanks a ton

Answer 25

Please, help me on R^3, why \(\triangle \Gamma =0\) if \(x\neq 0\)?

Answer 26

Haha ok glad I could help, let me see about this next thing one sec

Answer 27

I think it's just cause you can't divide by 0, since Gamma has this \(\frac{1}{|x|^{n-2}}\) term

Answer 28

One more question : what is D in \(D\Gamma\)? it is derivative / gradient, right?
And why \(D\Gamma \bullet normal~vector =\) the RHS?

Answer 29

For the first question, yeah, D is the gradient, they say this on the first page right after they define the Laplacian,

\[\Delta = \mathrm {div} D\]
Which says the Laplacian is the Divergence of the Gradient.

The other thing comes from the Divergence theorem, and apparently this is where the normalization constant \(\alpha_n\) comes from. Otherwise it would be on the right hand side and look like Gauss's law, the \(\alpha_n\) normalization constant is essentially like the "charge contained" in the volume \(B_r(0)\) if that helps.