# Talk:Laplace operator

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 To-do list for Laplace operator: Give an explanation of what the Laplacian actually is and/or does in the introduction. Quick example of the Laplacian operator on a simple function

## Add a reference explaining the confusing (but universally used) $\nabla^2$notation?

(Disclaimer: I'm neither an experienced Wikipedia editor nor a mathematician). The $\nabla^2$ notation universally used for Laplacian is, it seems to me unfortunate, since it is neither the square of anything nor a function composited with itself. I think a really good aside on this anomalous notation is here: http://physics.ucsd.edu/~emichels/FunkyQuantumConcepts.pdf#page=118 . Would it be appropriate to add a footnote linking to this, or an aside in the article? Ma-Ma-Max Headroom (talk) 16:57, 28 July 2009 (UTC)

• It's not that bad really; no worse than anything else the physicists come up with. is just a vector of operators, and ∇2 really is that squared, the appropriate product (vector dot product). The author of your link gets a bit attached to his idea of thinking of his notation for operators in terms of what they return, not what they are themselves. So, he states that ∇• is "never" written •, although mathematicians tend to use that notation (assuming they are bothering to mark any vectors specially). The rationale is that is a column vector of the partial derivative operators, so should be always emboldened if we are doing that to vectors, even though as an operator ∇• returns a scalar. Either way, this is not confusing, as the symbols are just grungy shorthand notation, and what they mean is obvious and unambiguous. You can add a footnote if you want, but I can't see a particular need.— Kan8eDie (talk) 22:31, 28 July 2009 (UTC)

Wikipedia has any standard on this? We see the triangle up a lot (an instance this very page) but the triangle down squared is also there (wave equation, for example). I like the triangle squared, and have a particular dislike for the box representing dalembertian (notably absent in wave equation) — Preceding unsigned comment added by 201.9.205.23 (talk) 17:00, 13 January 2013 (UTC)

## Section removed

I removed the following from the article. It seems to have no focus whatsoever, and is of only peripheral relevance to the article. The "identity" listed is an obvious one, and the spherical harmonic context is not terribly compelling. Spherical harmonics are already mentioned elsewhere in the article. Here is the offending section. Sławomir Biały (talk) 14:13, 2 September 2009 (UTC)

Identities
• If f and g are functions, then the Laplacian of the product is given by
$\Delta(fg)=(\Delta f)g+2((\nabla f)\cdot(\nabla g))+f(\Delta g).$

Note the special case where f is a radial function $f(r)$ and g is a spherical harmonic, $Y_{lm}(\theta,\phi)$. One encounters this special case in numerous physical models. The gradient of $f(r)$ is a radial vector and the gradient of an angular function is tangent to the radial vector, therefore

$2(\nabla f(r))\cdot(\nabla Y_{lm}(\theta,\phi))=0.$

In addition, the spherical harmonics have the special property of being eigenfunctions of the angular part of the Laplacian in spherical coordinates.

$\Delta Y_{\ell m}(\theta,\phi) = -\frac{\ell(\ell+1)}{r^2} Y_{\ell m}(\theta,\phi).$

Therefore,

$\Delta( f(r)Y_{\ell m}(\theta,\phi) ) = \left(\frac{d^2f(r)}{dr^2} + \frac{2}{r} \frac{df(r)}{dr} - \frac{\ell(\ell+1)}{r^2} f(r)\right)Y_{\ell m}(\theta,\phi).$

## Boundedness

Could someone add information about the boundedness of the Laplacian as a linear operator? As I understand it, it is unbounded when defined on the functions with bounded domain in R^n, but according to bounded operator it is bounded when the domain is R^n itself (and I presume other unbounded domains). Some mention of Sobolev spaces would be nice too. I will try to add what I know about these myself. —Preceding unsigned comment added by Slimeknight (talkcontribs) 21:31, 15 May 2010 (UTC)

This is wrong. As an operator on, e.g., L2(Ω) (of any domain Ω whatsoever—bounded, unbounded, all of Rn), the Laplacian is an unbounded densely-defined operator. However, as an operator
$\Delta : H^2(\Omega)\to L^2(\Omega)$
it is bounded. But this statement has very little content: it is simply how the norm on the Sobolev space H2 is defined. Sławomir Biały (talk) 12:15, 16 May 2010 (UTC)
Ah, okay, I was confused about which norm was was being talked about, I see. That makes a lot more sense to me, as I originally made the comment after being confused as to why it wasn't always unbounded. Thanks very much. slimeknight (talk) 19:04, 16 May 2010 (UTC)

## Azimuth and zenith

Since there is persistent confusion over the meaning of the words "zenith" and "azimuth", allow me to clarify to prevent further erroneous edits to the section on spherical coordinates. The angle φ measured with respect to the north pole is known as the zenith angle; the z axis itself is the zenith axis. The angle θ made between the positive x axis and the the orthogonal projection of the point into the xy plane is known as the azimuth angle, where the x axis itself is the azimuth axis. Now, the article at present says that

$\Delta f = {1 \over r^2} {\partial \over \partial r} \left( r^2 {\partial f \over \partial r} \right) + {1 \over r^2 \sin \varphi} {\partial \over \partial \varphi} \left( \sin \varphi {\partial f \over \partial \varphi} \right) + {1 \over r^2 \sin^2 \varphi} {\partial^2 f \over \partial \theta^2}.$

To see that this is correct for our conventions, consider the Laplacian of $f(x,y,z)=z^2=r^2\cos^2\phi.$ In Cartesian coordinates, the Laplacian is 2. Working this out in spherical coordinates also gives

\begin{align} \Delta f &= \Delta (r^2\cos^2\phi) = 6\cos^2\phi + (-4\cos^2\phi + 2\sin^2\phi) + 0\\ &=2 \end{align}

as expected. However, if instead we were to misunderstsand the meaning of "zenith" and "azimuth" and considered the function $f=r^2\cos^2\theta,$ then

\begin{align} \Delta f &= \Delta (r^2\cos^2\theta) = 6\cos^2\theta + 0 + (2\sin^2\theta-2\cos^2\theta)\\ &=2\cos^2\theta + 2 \end{align}

which is not equal to 2. Sławomir Biały (talk) 11:07, 27 July 2010 (UTC)

Comment: that is inconsistent with the usage in Spherical coordinate system, where φ is used for azimuth. ...ah, but reading on, I see that

However, some authors (including mathematicians) use φ for inclination (or elevation) and θ for azimuth, which "provides a logical extension of the usual polar coordinates notation" Some authors may also list the azimuth before the inclination (or elevation), and/or use ρ instead of r for radial distance.

So, as so often, it's a case of "standards are wonderful, you can't have too many of them". Regards, JohnCD (talk) 16:31, 27 July 2010 (UTC)
Up until quite recently, we used the other convention. But I changed to this one because of this very issue. People were still getting them mixed up even in the convention consistent with our spherical coordinate system article. I think the current conventions are preferable, because we should use the same letter to designate the azimuth in the polar, cylindrical, and spherical coordinates. I believe that this letter is almost always θ for the polar and cylindrical systems, and so we should use the convention for the spherical coordinate system in which it is also θ. It would seem to me to be far too confusing to have θ an azimuth in one paragraph, and then a zenith in the very next part. I thought changing to the "mathematical" convention would resolve the confusion, but apparently not. Sławomir Biały (talk) 17:50, 27 July 2010 (UTC)

### varphi vs phi

Starting from the convention that theta is azmuthal and phi is latitude... I noticed you had two different versions of the equation, and one used \varphi for the latitude angle. I think these must have been intended to be the same angle (else there is no definition for the two), so I changed it. -Theanphibian (talkcontribs) 12:32, 14 May 2013 (UTC)

## 'Spheres centred at'...

Firstly it's not clear if it's talking about the surface or the solid (I know that it's surface technically but I don't know if the editor knew this).

Secondly, shouldn't really be in the intro, followed by zero follow up. Should probably be in the article somewhere, with at least some discussion of importance or derivation; I'm okay at vector calculus but I've never heard this interpretation before. — Preceding unsigned comment added by 86.131.51.193 (talk) 17:00, 12 July 2011 (UTC)

As usual in mathematics, "sphere" means the surface of the ball. This characterization was chosen for the lead paragraph since it seemed to be the simplest way to say what the Laplacian is for a general audience (no assumption of vector calculus). Sławomir Biały (talk) 17:53, 12 July 2011 (UTC)
Let's revisit this. The average of a function f over the sphere of radius r centered at the origin in $\mathbb{R}^n$ is
$\overline{f}(r)=\frac{1}{\alpha_nr^{n-1}} \int_{S^{n-1}_r(0)} f(x)\,d\sigma(x).$
Apply a change of variables to convert to spherical coordinates:
$\overline{f}(r)=\frac{1}{\alpha_n} \int_{S^{n-1}(0)} f(r\xi)\,d\sigma(\xi).$
Differentiating with respect to r gives
$\overline{f}'(r) = \frac{1}{\alpha_n} \int_{S^{n-1}(0)} \nabla f(r\xi)\cdot\xi\,d\sigma(\xi) = \frac{1}{\alpha_nr^{n-1}} \int_{S^{n-1}_r(0)} \nabla f(x)\cdot\frac{x}{r}\,d\sigma(x).$
Applying the divergence theorem gives
$\overline{f}'(r) = \frac{1}{\alpha_nr^{n-1}}\int_{B^n_r(0)}\Delta f(x)\,dx.$
So $\overline{f}'(r)$ is equal to $r/n$ times the average value of $\Delta f$ over the ball of radius r. Sławomir Biały (talk) 12:10, 13 July 2011 (UTC)

So... the introductory statement is incorrect? I can't see how the above is relevant. --81.152.176.83 (talk) 16:04, 13 July 2011 (UTC)
No, the precise statement is: $\overline{f}(r)=f(0)+c_n r^2\Delta f(0) +o(r^2).$ Sławomir Biały (talk) 03:07, 14 July 2011 (UTC)
...okay, for now, I'm removing the statement. The above in no way shows equality to 'the rate at which the average value of ƒ over spheres centered at p, deviates from ƒ(p) as the radius of the sphere grows', and as such is extremely unhelpful. — Preceding unsigned comment added by 86.131.49.172 (talk) 14:43, 19 July 2011 (UTC)
I've rephrased the statement. Sławomir Biały (talk) 16:42, 19 July 2011 (UTC)
Rephrase it again. It is not even roughly equal 'up to a factor'; there is an r^2 term. The true statement is now so lacking in intuitive meaning and consequence that it seems to me totally bizarre having in the beginning as if it's either important or helpful. — Preceding unsigned comment added by 86.131.49.172 (talk) 15:57, 20 July 2011 (UTC)