Laplacian of 1/r is zero at orign

Hey All,

In my vector calculus class my lecturer was showing that the laplacian of 1/r is zero. He further said that since 1/r and its derivatives are not defined at the origin we state that the Laplacian of 1/r is zero for all values of r not equal to zero. He then says that this caveat is extremely important in advanced courses.

Can someone please tell me in what sort of courses this is important, and if possible why ?

Cheers,
Thrillhouse

HallsofIvy
Homework Helper
?? In any course where you want to be right! 1/r is defined and smooth for any non-zero value of r but is not defined and so not continuous or differentiable at r=0. The Laplacian of 1/r does not exist at r= 0. There might be some cases in which you can get away with smoothing the Laplacian by assuming taking the value of the limit as r goes to 0 to be the value at r= 0, but you should always show that that is possible in that particular case, not just assume it.

D H
Staff Emeritus
This is a physics course, so of course the math is a bit looser there than Halls might like.

The Laplacian of 1/r is *not* zero at the origin. The Laplacian of 1/r is

$$\nabla^2\frac 1 r = -\,\frac{\delta(r)}{r^2}$$

For example, see http://www.iop.org/EJ/article/0143-...quest-id=31cf705f-8110-49fa-930d-e5ba7a4b95a5. This becomes quite important in advanced (upper undergrad and above) physics classes.

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If a function has laplacian zero everywhere in a domain, then the norm of the function inside the domain is always <= the maximum norm on the boundary. The Maximum Principle. But that fails for 1/r. For example, the unit ball, we have norm 1 on the boundary, but norm 2 at a point with r=1/2 .

arildno
Homework Helper
Gold Member
Dearly Missed
This is a physics course, so of course the math is a bit looser there than Halls might like.

The Laplacian of 1/r is *not* zero at the origin. The Laplacian of 1/r is

$$\nabla^2\frac 1 r = -\,\frac{\delta(r)}{r^2}$$

For example, see http://www.iop.org/EJ/article/0143-...quest-id=31cf705f-8110-49fa-930d-e5ba7a4b95a5. This becomes quite important in advanced (upper undergrad and above) physics classes.

No, it is not, DH.

It is a completely wrong statement, the Laplacian of 1/r is 0 wherever it is defined.

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D H
Staff Emeritus
No, it is not, DH.

It is a completely wrong statement, the Laplacian of 1/r is 0 wherever it is defined.

Take your beef up with the author of the reference I gave in my previous post. Or with Eric Weisstein in either the CRC Concise Encyclopedia of Mathematics (http://books.google.com/books?id=_8TyhSqHUiEC&pg=PA1702#v=onepage&q=&f=false) or at MathWorld (http://mathworld.wolfram.com/Laplacian.html).

And a few more: