# Action of Laplacian Operator

1. Feb 8, 2005

### makris

Consider a function U(x,y) where x, and y are spatial variables (have units of length)
Assume that the symbol V^2 corresponds to the Laplacian operator.

Then V^2U= Uxx + Uyy where the subscript indicates partial differentiation.

Consider now a function F(x,t) where x is spatial variable (has units of length) and t is a temporal variable (has units of time)

I found quite surprising that the action of the Laplacian on this new function is a little bit different than previously.

V^2F= Fxx

Could you please give me a hint as to why does this happen? Why
V^2F= Fxx + Ftt is incorrect?

What springs to mind is that Ftt has units of acceleration and Fxx represents the concavity of F parallel to XX’ axis. So we cannot really add these two different quantities. Apart from this (which I am not if it is mathematically correct) do you have any other explanation?

2. Feb 8, 2005

### dextercioby

If "x","y" and "t" are independent variables (precisely if "t" isn't a function of "x" and/or "y"),then it comes naturally as an application of the definition of the Laplace diff.operator in 2 dimensions in rectangular coordinates.

Daniel.

3. Feb 8, 2005

### Crosson

The laplacian operator is defined as V*V, the dot product of the del operator with itself. In physics, the del operator is a 3 dimensional vector operator. Thus, for a function such as F(x,y,z,t) , The vector V (F) is the gradient at the point x,y,z at various times t.

Part of your problem is that you think the laplacian is "the sum of the second partial derivatives with respect to each variable" when it only resembles this description in cartesian coordinates for a time independent function. The laplacian is defined as V*V, in general.

4. Feb 8, 2005

### makris

The Laplacian operator can be applied to a function of two variables. I agree that we should not see the action of the Laplacian as taking a function find the partial derivatives and add them…

In my question I assume that x, t are independent variables.
To see the Laplacian as V*V (which is absolutely correct) does not help me understand the problem with the V^2F , F=F(x,t). I imagine the time axis to lie on the same plane and at right angle with the x axis. The z axis is perpendicular to the previously mentioned plane. So F(x,t) represents a surface.

5. Feb 8, 2005

### dextercioby

Well,this functional dependence (coordinates and time) is not typical to mathematics,but to physics.In (nonrelativistic) physics,we always think of nabla as an LPD operator ONLY WRT COORDINATES.Similar for the trace of the hessian,the laplaceian.So leave geometric interperation of functional dependence aside and think in the spirit of physics.

Daniel.

6. Feb 8, 2005

### mathwonk

what they are telling you is this is a question of semantics, i.e. of definitions. it only depends on what you mean by the words "laplace operator".