- #1

Does anybody out there know what the Laplacian is for two dimensions?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter MalleusScientiarum
- Start date

- #1

Does anybody out there know what the Laplacian is for two dimensions?

- #2

- 13,242

- 1,017

[tex] \Delta =g^{ij}\nabla_{i}\nabla_{j} [/tex]

,where [itex] i,j=1,2 [/itex]. Incidentally, for a manifold on which Christoffel's symbols are 0, it reduces to the standard form

[tex] \Delta=g^{ij}\partial_{i}\partial_{j} [/tex]

Daniel.

- #3

That is probably one of the most complicated answers to a simple question I've ever seen.

- #4

HallsofIvy

Science Advisor

Homework Helper

- 43,021

- 970

It depends upon the coordinate system. Dextercioby gave the general formula for any coordinate system, thereby making it, as MalleusScientarum said, "one of the most complicated answers to a simple question I've ever seen"!

In Cartesian coordinates it is [tex]\frac{\partial^2}{\partial x^2}+ \frac{\partial^2}{\partial y^2}[/tex].

By the way, the Laplacian is the simplest differential operator that is "invarient under rigid motions".

In Cartesian coordinates it is [tex]\frac{\partial^2}{\partial x^2}+ \frac{\partial^2}{\partial y^2}[/tex].

By the way, the Laplacian is the simplest differential operator that is "invarient under rigid motions".

Last edited by a moderator:

- #5

mathwonk

Science Advisor

Homework Helper

- 11,389

- 1,613

lets see now, isn't the laplacian the differential operator that vanishes on harmonic functions? if so, wouldn't that say it is the one that governs steady state heat flow? and characterizes the real and imaginary parts of holomorphic functions?

harmonic forms on the other hand are very sueful in discussing cohomology. the de rham theorem says that every cohomology class on a compact oriented? diff manifold can be represented by a smooth differential form, but there is no uniqueness.

by imposing a metric and hence defining a laplacian, one can define harmonic foirms and thewn there is a unique harminic representative for each cohomology class.

e.g. on an elliptic curve, formed as a quotient of the complex numbers by a lattice, one has the natural harmonic basis dz and "dzbar".

harmonic forms on the other hand are very sueful in discussing cohomology. the de rham theorem says that every cohomology class on a compact oriented? diff manifold can be represented by a smooth differential form, but there is no uniqueness.

by imposing a metric and hence defining a laplacian, one can define harmonic foirms and thewn there is a unique harminic representative for each cohomology class.

e.g. on an elliptic curve, formed as a quotient of the complex numbers by a lattice, one has the natural harmonic basis dz and "dzbar".

Last edited:

- #6

fourier jr

- 757

- 13

- #7

quetzalcoatl9

- 537

- 1

[tex]T_p = k\Delta T[/tex]

where in cartesian coordinates [itex]T_p[/itex] is the change in temperature at the point [itex]p[/itex]

[tex]T(t,x,y,z)[/tex] and [tex]p: (x,y,z)[/tex]

[tex]T_p = k(\frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2} + \frac{\partial ^2 T}{\partial z^2})[/tex]

- #8

Apologies, I guess I should specify for polar coordinates.

- #9

inha

- 576

- 1

Just ignore the z-component. Or I suppose you could plug in [tex]x=r\cos\theta[/tex] and [tex]y=r\sin\theta[/tex] to the cartesian version, apply the chain rule and calculate .

- #10

Share:

- Replies
- 7

- Views
- 173

- Last Post

- Replies
- 0

- Views
- 650

- Last Post

- Replies
- 3

- Views
- 371

- Replies
- 14

- Views
- 880

- Last Post

- Replies
- 12

- Views
- 2K

- Last Post

- Replies
- 8

- Views
- 1K

- Last Post

- Replies
- 13

- Views
- 1K

- Replies
- 3

- Views
- 926

- Last Post

- Replies
- 1

- Views
- 862

- Last Post

- Replies
- 6

- Views
- 3K