Source : ct|01.06.13

< Mathematics and physics

Laplacian of a scalar field

The Laplacian does not appear in the words commonly used in the dictionary. In mathematics and physics, it represents a differential operator used in vector analysis. Its name pays homage to the work done by the mathematician, astronomer and French physicist Pierre-Simon Laplace (1749-1827).
What is the precise definition of this operator and what is it exactly ?…

1) Definition

Consider a scalar field (or scalar potential) U(x,y,z). We call Laplacian of this field U(x,y,z) the divergence of the gradient of U(x,y,z)

We write the Laplacian of a scalar field U

with the Laplacian operator equal to

and i =(1,0,0), j=(0,1,0), k=(0,0,1)

2) Meaning

To illustrate what represents the Laplacian of a scalar field U (x,y,z) at a point M (x,y,z), we use the simple case of a scalar field U (x) with one dimension at a point M (x) and the case of a scalar field U (x,y) with two dimensions at a point M (x,y).

• With one dimension, the Laplacian of a scalar field U(x) at a point M(x) is equal to the second derivative of the scalar field U(x) with respect to the variable x.

dU/dx, derivative of U(x) at the point M(x) is the slope of the tangent to the curve U(x) in this point. It represents the infinitesimal variation of U(x) relative to an infinitesimal change in x at this point.
d²U/dx², second derivative of U(x) at the point M(x) is the slope of the derivative U'(x)=dU/dx at this point. It represents the infinitesimal variation of U'(x) relative to an infinitesimal change in x at this point.

If one is interested in the local form of the function y =U(x) over an interval of width d we distinguish the following three cases :

• Locally concave function
  In this case, the derivative dU/dx of the function U(x) decreases on [x-d, x + d] U and the Laplacian of U (second derivative of U) is negative.


• Locally convex function
  In this case, the derivative dU/dx of the function U(x) is increases on [x-d, x + d] U and the Laplacian of U (second derivative of U) is positive.


• Locally "flat" function
  In this case, the derivative dU/dx of the function U(x) is constant on [x-d, x + d] U and the Laplacian of U (second derivative of U) is equal to 0.

The Laplacian of a scalar field U(x) at a point M(x) indicates the concavity of this field at this point.

We can extend this reasoning to a two-dimesnional scalar field U(x, y).

• If ΔU < 0 at point M(x,y)
  That means that (globally, with respect to x and y) the function U (x,y) is concave at this point. In that case, the value taken by U(x,y) at point M (x,y) is greater than the average of the values of a set of points situated at a distance d of point M.
• If ΔU = 0 at point M(x,y)
  That means that (globally, with respect to x and y) the function U (x,y) is "flat" at this point. In that case, the value taken by U(x,y) at point M (x,y) is equal to the average of the values of a set of points situated at a distance d of point M.
• If ΔU > 0 at point M(x,y)
  That means that (globally, with respect to x and y) the function U (x,y) is convex at this point. In that case, the value taken by U(x,y) at point M (x,y) is less than the average of the values of a set of points situated at a distance d of point M.

In conclusion
The Laplacian of a scalar field U(x,y,z) at a point M(x,y,z) indicates the concavity of this field at this point.

Generalization

Considering a vector field V(x,y,z) = Vx(x,y,z) i + Vy(x,y,z) j + Vz(x,y,z) k

The Laplacian ΔV(x,y,z) of this vector field is a vector whose components are equal to the Laplacians of the components of the vector V(x,y,z).

Example

Consider the scalar field U(x,y) = x³ defined on RxR.
Calculate the Laplacian of this scalar field. Deduce the concavity of this field on RxR.

Answer

We havea
ΔU ‹ 0 for x є [-∞,0[ x [-∞,+∞]
ΔU=0 for x = (0,0)
ΔU › 0 for x є ]0,+ ∞] x [-∞,+∞]

Laplacian in cylindrical coordinates

Consider, in cylindrical coordinates, a scalar field U(r,θ,z) and a vector
E (r,θ,z) = Er (r,θ,z) u + Eθ(r,θ,z) v + Ez(r,θ,z) k = grad U
ΔU = div (grad U) = div E

hence

the Laplacian of U is equal to the divergence of E, hence

Laplacian in spherical coordinates

Consider, in spherical coordinates, a scalar field U(r,θ,φ) and a vector
E (r,θ,φ) = Er (r,θ,φ) u + Eθ(r,θ,φ) v + Eφ(r,θ,φ) w = grad U
ΔU = div (grad U) = div E

hence

the Laplacian of U is equal to the divergence of E, hence