so that if φrepresents either of the components F 1 or F 2 one can write
(h·∇ )φ=h (^1) ∂x∂φ
1
+h (^2) ∂x∂φ
2
Observe that the operator
(h·∇ )^2 φ=(h·∇ )(h·∇ )φ=h 1 ∂
∂x 1(
h 1 ∂φ
∂x 1+h 2 ∂φ
∂x 2)
+h 2 ∂
∂y(
h 1 ∂φ
∂x 1+h 2 ∂φ
∂x 2)=h^21 ∂(^2) φ
∂x 12
- 2h 1 h 2 ∂
(^2) φ
∂x 1 ∂x 2
+h^22 ∂
(^2) φ
∂x 22
In a similar fashion one can show
(h·∇ )^3 φ=(h·∇ )(h·∇ )^2 φ=h^31∂^3 φ
∂x^31 + 3 h2
1 h^2∂^3 φ
∂x^21 ∂x 2 + 3h^1 h2
2∂^3 φ
∂x 1 ∂x^22 +h3
2∂^3 φ
∂x^32and in general for any positive integer none can use the binomial expansion to
calculate the operator
(h·∇ )nφ=(
h 1∂
∂x 1 +h^2∂
∂x 2)n
φThis operator can be used to represent the Taylor series expansion of a function
F =F(x )where x= (x 1 ,x 2 ). If x 0 = (x^01 ,x^02 )is a constant and h= (h 1 ,h 2 )denotes a
small vector displacement from the point x 0 , then the Taylor series expansion can
be written
F(x 0 +h) =∑nm=11
m!(h·∇ )mF(x )
x=x 0+1
(n+ 1)!(h·∇ )n+1F(x )
x=x 0(6 .84)where all derivatives are to be evaluated at the point x 0.
In three dimensions vectors of the form
F=F(x ) = F(x 1 ,x 2 ,x 3 ) = F 1 (x 1 ,x 2 ,x 3 )ˆe 1 +F 2 (x 1 ,x 2 ,x 3 )ˆe 2 +F 3 (x 1 ,x 2 ,x 3 )ˆe 3which have (n+1) partial derivatives can be expanded in a Taylor series by expanding
each of the components in a Taylor series. Associated with the vector displacement
h= (h 1 ,h 2 ,h 3 )one can define the operator
(h·∇ ) = h 1 ∂
∂x 1+h 2 ∂
∂x 2+h 3 ∂
∂x 3