INTERNATIONAL JOURNAL OF ENHANCED RESEARCH PUBLICATIONS
Vol.2 Issue X, Month, 2013
ISSN No: XXXX-XXXX
Commutative Rule involving the Laplace Operator
The-Nan Chang , Fu-Lai Chu
Tatung University
Taipei,Taiwan
tnchang@ttu.edu.tw
Abstract: Commutation rules between the Laplace operator and other basic operators (include the divergence, the curl, and the
gradient operator) are established. These rules were seldom noticed in the past. However, it is shown that they lead to an alternative
derivation of the electromagnetic fields due to an arbitrary current distribution in a homogeneous environment without referring to the
vector potential concept. The derived mathematical result can still be linked to show the physical insight of the original problem.
Keywords: vector analysis, operator, potential, commutative, Laplace
Introduction
In the course of electromagnetic waves [1-4], both scalar and vector distributions are frequently encountered. The gradient, the divergence,
and the curl are three basic vector operators. The gradient operates on the scalar function and the other two operate on the vector functions.
The Laplace operator operates on both scalar and vector functions. In vector analysis, many vector analysis formulas are derived. However,
the commutative properties involving the Laplace operator are seldom noticed.
In a scalar system, we know that the impulse response of a system is the derivative of the unit-step response. For a unit-step response of
f(x), we write L (f(x)) =u(x), where u(x) is the unit-step function and L is the operator. Taking the derivative operation on both sides of the
equation, we get L[df(x)/dx] = δ(x), if the L-operator and the derivative operator are commutative to each other. In a vector system, the
curl, the divergence, and the gradient involve a lot of derivative operations. We expect that commutative rules should also be existed in a
vector system. The vector system could be a static E-field, a static H-field or an electromagnetic wave system.
Derivation
⃗ ) is related to J and ∇ ∙ J. The operator L involves the Laplace operator. In
In electromagnetic waves, vector operator L exists such that L(E
this paper, vector commutative properties involving the Laplace operator will be investigated. We found that the commutative property is
⃗ or ⃗H
⃗ without referring to the conventional vector potential approach. Many commutative properties
useful to get an analytic solution for E
can be derived from the following vector identity formulas
∇ × ∇ × ⃗A = ∇(∇ ∙ ⃗A) − ∇2 ⃗A
⃗ = ∇ψ
(A) Let A
∇ × ∇ × (∇ψ) = ∇(∇ ∙ ∇ψ) − ∇2 (∇ψ)
0 = ∇(∇2 ψ) − ∇2 (∇ψ)
∇(∇2 ψ) = ∇2 (∇ψ)
Therefore, the Laplace and the gradient operators are commutative to each other.
(B) Taking divergence operation to both sides of the vector identity formula, we get
0 = ∇ ∙ [∇(∇ ∙ ⃗A)] − ∇ ∙ (∇2 ⃗A)
∇2 (∇ ∙ ⃗A) = ∇ ∙ (∇2 ⃗A)
Therefore, the Laplace and the divergence operators are commutative to each other.
(C) Let ⃗A = ∇ × ⃗F
The left-hand side of the vector identity becomes:
∇ × ∇ × (∇ × ⃗F) = ∇ × [∇(∇ ∙ ⃗F) − ∇2 ⃗F] = −∇ × (∇2 ⃗F)
The right-hand side of the vector identity becomes:
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INTERNATIONAL JOURNAL OF ENHANCED RESEARCH PUBLICATIONS
Vol.2 Issue X, Month, 2013
ISSN No: XXXX-XXXX
⃗ ) − ∇2 (∇ × F
⃗ ) = −∇2 (∇ × F
⃗)
∇(∇ ∙ ∇ × F
⃗ ) = ∇2 (∇ × F
⃗)
∇ × (∇2 F
Therefore, the Laplace and the curl operators are commutative to each other.
⃗ , then use the result of (B),
(D) Replacing ψ in formula (A) by ∇ ∙ F
⃗ )] = ∇2 [∇(∇ ∙ F
⃗ )]
∇[∇2 (∇ ∙ F
⃗ ) = ∇2 (∇∇ ∙ F
⃗)
∇∇ ∙ (∇2 F
(E) Let ⃗F = ∇ × ⃗G in formula (C),
∇ × [∇2 (∇ × ⃗G)] = ∇2 (∇ × ∇ × ⃗G)
⃗ )= ∇2 (∇ × ∇ × G
⃗)
∇ × ∇ × (∇2 G
The last two formulas state that the Laplace operator is also commutative with respect to the operator ∇(∇ ∙) and the operator ∇ × (∇ ×).If
we denote ∇ × (∇ ×) × ⋯ ⋯ × (∇ ×) = (∇ ×)n for n-times of curl operation, it is easy to show that ∇2 (∇ ×)n = (∇ ×)n (∇2 ). We can also
show that ∇2 [∇(∇ ∙)]n = [∇(∇ ∙)]n (∇2 ) for n-times operation of ∇(∇ ∙).
⃗ and ⃗H
⃗ according to the Maxwell’s
In a simple media with the constitutive parameters of ε0 , 𝜇, the impressed source Ji generates fields E
equation:
⃗
∇ × ⃗E = −ẑ⃗H
⃗ = Ji + ŷE
⃗
∇ × ⃗H
, where ẑ = jω𝜇 , and ŷ = jω𝜀0 .The 𝐸⃗ -field satisfies:
𝛻 × 𝛻 × 𝐸⃗ + 𝑧̂𝑦̂𝐸⃗ = −𝑧̂ 𝐽𝑖
⃗ = (∇ ∙ D
⃗⃗ )⁄𝜀0 = ρ/𝜀0 , wher ρ = −∇ ∙ Ji /(𝑗𝜔), and k 2 = −ẑŷ.
Since, ∇ ∙ E
1
⃗ + k2E
⃗ = ẑJi − [∇(∇ ∙ Ji )]
We get, ∇2 E
̂
𝑦
With the property of superposition, ⃗E can be expressed as ⃗E = ⃗E1 + ⃗E2 , where,
⃗ 1 + k2 E
⃗ 1 = ẑJi , and
∇2 E
1
⃗ 2 + k2 E
⃗ 2 = − [∇(∇ ∙ Ji )]
∇2 E
𝑦̂
⃗ 1 is:
The analytic form for E
⃗⃗⃗′
⃗ 1 = −𝑎 [
E
−𝑗𝑘|𝑟 −𝑟 |
ẑ
Ji (𝑟⃗⃗⃗′ )𝑒
∭
𝑑𝜏 ′ ]
4𝜋
|𝑟 − ⃗⃗⃗
𝑟 ′|
, where 𝑎 is a constant vector. The integral takes on the volume where the current exists. For arbitrarily located current distribution, it can
be decomposed into three components in a rectangular coordinate system. The derivation of this equation will be discussed in the appedix.
⃗ 2 can be expressed in terms of E
⃗ 1 , a simple proof is given below:
E
⃗ 1 + k2 E
⃗ 1 = ẑJi
Since, ∇2 E
We get : ∇∇ ∙ [∇2 ⃗E1 + k 2 ⃗E1 ] = ∇∇ ∙ (ẑJi )
By commutative property, the above equation can be written as :
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INTERNATIONAL JOURNAL OF ENHANCED RESEARCH PUBLICATIONS
Vol.2 Issue X, Month, 2013
ISSN No: XXXX-XXXX
𝛻 2 [𝛻𝛻 ∙ 𝐸⃗1 ] + 𝑘 2 [𝛻𝛻 ∙ (𝐸⃗1 )]= 𝛻𝛻 ∙ (𝑧̂ 𝐽𝑖 )
Therefore, ⃗E2 =
⃗ 1)
−∇∇∙(E
ẑ𝑦̂
Conventionally, the 𝐸⃗ -field is derived with the potential approach. The potential approach gives its physical appeal. In our expression, one
can easily show that 𝐸⃗1 is the expression for the conventional vector potential multiplied by –𝑗𝜔, and 𝐸⃗2 is the negative gradient of the
conventional scalar potential. Therefore, the conventional vector and scalar potentials can easily be extracted out.
CONCLUSION
In general, αβ ≠ βα for two linear operators α and β . But, if they are equal, α and β are said to commute [5]. To the authors’ best
knowledge, the commutative properties between several useful operators are seldom noticed in the EM community. In this short note, we
revisit those basic vector operators. Starting from the definition of the vector Laplace operator, we found that many commutative rules
exist between the vector Laplace operator and the basic operators. The vector Laplace operator is also commutative with the (∇∇ ∙ )
⃗ -field in free space without explicitly referring to the vector potential approach. The
operator. It leads to derive the analytic solution for E
Lorentz-Gauge condition is also not required in our derivation. We can also show that the scalar Laplace operator is also found
commutative with the gradient operator. Though no new expressions are derived, the mathematical skill paves a different and straightforward way to solve the problem.
APPENDIX:
Since the operator approach is highlighted in this paper, derivation of ⃗⃗⃗⃗
E1 is shown below using only the basic vector
opertions.
Given 𝑟 2 = 𝑥 2 +𝑦 2 + 𝑧 2 or r=√𝑥 2 + 𝑦 2 + 𝑧 2 ,
1
1
1
2
, it is easy to show that 𝛻𝑟 2 =2𝑟𝑎
⃗⃗⃗⃗𝑟 , 𝛻𝑟=𝑎
⃗⃗⃗⃗𝑟 , 𝛻 𝑟 = − 𝑟 2 ⃗⃗⃗⃗
𝑎𝑟 , 𝛻 𝑟 2 = − 𝑟 3 ⃗⃗⃗⃗
𝑎𝑟 , 𝛻 ∙ 𝑟⃑ = 3, 𝑎𝑛𝑑 𝛻𝑒 −𝑗𝑘𝑟 =
−𝑗𝑘𝑒 −𝑗𝑘𝑟 ⃗⃗⃗⃗
𝑎𝑟 .Therefore, 𝛻 2 𝑒 −𝑗𝑘𝑟 = 𝛻 ∙ (−𝑗𝑘𝑒 −𝑗𝑘𝑟 𝛻𝑟) = 𝛻(−𝑗𝑘𝑒 −𝑗𝑘𝑟 ) ∙ 𝛻𝑟 + (−𝑗𝑘𝑒 −𝑗𝑘𝑟 )𝛻 ∙ 𝛻𝑟 = −𝑘 2 𝑒 −𝑗𝑘𝑟 𝛻𝑟 ∙ 𝛻𝑟 +
2
2
𝑟
𝑟
(−𝑗𝑘𝑒 −𝑗𝑘𝑟 ) = −𝑘 2 𝑒 −𝑗𝑘𝑟 + (−𝑗𝑘𝑒 −𝑗𝑘𝑟 ) , and
𝛻2
𝑒 −𝑗𝑘𝑟
1 1
1
1 1
1
= 𝛻 ∙ (𝑒 −𝑗𝑘𝑟 𝛻 + 𝛻𝑒 −𝑗𝑘𝑟 ) = 𝑒 −𝑗𝑘𝑟 𝛻 2 + 2(𝛻𝑒 −𝑗𝑘𝑟 ) ∙ 𝛻 + 𝛻 2 𝑒 −𝑗𝑘𝑟 = −4𝜋𝛿(𝑟⃑) − 𝑘 2 𝑒 −𝑗𝑘𝑟
𝑟
𝑟 𝑟
𝑟
𝑟 𝑟
𝑟
Eq.(1) can be rewritten as :(𝛻 2 + 𝑘 2 )
𝑒 −𝑗𝑘𝑟
4𝜋𝑟
(1)
= −𝛿(𝑟), or (𝛻 2 + 𝑘 2 )𝜓 (|𝑟 − ⃗⃗𝑟 ′ |) = −𝛿 (𝑟 − ⃗⃗𝑟 ′ )
Mutiplying both sides of the above equation by a constant vector 𝑎
⃗⃗⃗ , it yields :
(𝛻 2 + 𝑘 2 )𝑎
⃗⃗⃗ 𝜓 (|𝑟 − ⃗⃗⃗
𝑟 ′ |) = −𝑎
⃗⃗⃗ 𝛿(𝑟 − ⃗⃗⃗
𝑟′)
REFERENCES
[1]
[2]
[3]
[4]
[5]
David K. Cheng, Field and wave electromagnetics, 2nd edition, , Addison-Weslely Inc. 1989
Roger F. Harrington, Time-Harmonic electromagnetic fields, Wiley-IEEE Press., Sep. 2001
Jin Au Kong, Electromagnetic wave theory, EMW Publishing, Cambridge, Massachusetts, U.S.A., 2005
Robert S. Elliott, Antenna theory and design, revised edition, John Wiley &Sons, Inc., 2003
Jon Mathews, and R.L. Walker, Mathematical methods of Physics, 2nd edition, California Institute of Tech
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