J. of Electromagn. Waves and Appl., Vol. 22, 399-410, 2008

FRACTIONAL DUAL SOLUTIONS USING CALCULUSOF DIFFERENTIAL FORMS

F. Mustafa, M. Faryad, and Q. A. Naqvi

Electronics DepartmentQuaid-i-Azam UniversityIslamabad, Pakistan

Abstract—Fractional dual solutions to Maxwell equations are derivedusing differential forms formalism. Fractional dual solutions may beregarded as intermediate step between the original solutions and dualto the original solutions. Fractional differential operator has been usedto obtain the fractional dual solutions.

1. INTRODUCTION

Classically, Maxwell equations are written and analyzed using Gibbsianvector and dyadic analysis [1, 2]. The calculus of differential formsmay be used as an alternate language to represent and analyzeelectromagnetic problems [3–5]. The calculus of differential forms isthe calculus of quantities that can be integrated. The degree of aform is the dimension of the region over which it is integrated, so thatin R3 there are 0-forms, 1-forms, 2-forms and 3-forms. 1-forms areintegrated over paths, and are represented graphically by surfaces. Forexample, the surfaces of dx are perpendicular to x-axis and infinite iny and z-plane and spaced a unit distance apart [6]. The integral ofa 1-form over a path is the number of surfaces pierced by the path,taking into account the orientation of the surfaces and the directionof integration. Graphically, 2-forms are tubes. A 3-form is a volumeelement, represented by boxes.

Deschamps [7] was the first to present a comprehensive paperon the use of differential forms in engineering electromagnetics.Differential forms have interesting features like: Firstly, differentialforms posses natural algebraic structure to represents the equationsof electromagnetics. Secondly, the operation of exterior derivativegeneralizes the familiar operations of curl, divergence and gradient.Thirdly, the differential forms remain invariant under the change

400 Mustafa, Faryad, and Naqvi

of co-ordinate systems. Fourthly, differential forms posses a cleargeometrical significance, so we have the opportunity to geometrizethe whole electromagnetics by using differential forms [7] etc. Lindelland co-workers have addressed many issues of electromagnetics usingdifferential forms [8–11].

Fractional calculus [13] is branch of mathematics involvingdifferentiation and integration to arbitrary non-integer orders.Engeta applied the tools of fractional calculus in various problemsof electromagnetic fields and waves, and obtained interestingresults [14, 15]. These results highlight certain notable features andpromising potential applications of these operators in electromagnetictheory. He investigated the notion of fractionalization of some otherlinear operators in electromagnetic theory, e.g., curl operator, kernel ofintegral transform [16–19]. Fractionalization of such operators has ledus to novel solutions, interpretable as “fractional solutions”, for certainelectromagnetic problems. An interesting and useful work done byEngheta is fractionalization of curl operator [16]. Mathematical recipeto fractionalize a linear operator is available in [16, 20]. Fractional curloperator has been applied to analyzed various problems [21–32].

In present work, fractional dual solutions to Maxwell equationsobtained by Engheta [16] have been re-derived in differential formformalism In Section 2, we review some of the basics of differentialforms and their use in representing the Maxwell equations. InSection 3, fractional duality in electromagnetics using differential formshas been discussed. Section 4 deals with two examples in this regard.

2. MAXWELL EQUATIONS IN DIFFERENTIAL FORMS

Using differential forms, Ampere-Faraday Maxwell equations in thefrequency domain, can be written as

d ∧ H = jwD (1a)d ∧ E = −jwB (1b)

In the above equations, E represents the electric field and H representsthe magnetic field. Both E and H are the one-form quantities. E hasthe units of Volt and H has the units of Ampere, and d is the exteriorderivative defined as

d =∂

∂xdx +

∂

∂ydy +

∂

∂zdz (2)

and wedge operator ∧ is defined by the anti-commutative relation

dx ∧ dy = −dy ∧ dx (3a)dxdy = −dydx (3b)

Fractional dual solutions using calculus of differential forms 401

It may be noted that Equation (3a) and Equation (3b) describe thesame using two representations used in literature. In representation(3a), wedge operator is present while in (3b) wedge is absent.

Flux densities D and B are of two-form quantities and are relatedwith electric and magnetic field (one forms) as

D = �εE (4a)B = �µH (4b)

where Hodge operator � relates one-form quantities with two-formquantities as

�dx = dy ∧ dz = dydz (5a)�dy = dz ∧ dx = dzdx (5b)�dz = dx ∧ dy = dxdy (5c)�(dxdy) = dz (5d)�(dydz) = dx (5e)�(dzdx) = dy (5f)

It is assumed that medium is homogeneous, isotropic and non-dispersive. From (5) it may be noted that

� � f = f (6)

where f is differential form of any grade.For homogeneous and isotropic medium, Equation (1) may be

written as

ηd ∧ H = ηjw�εE (7a)

d ∧ E = −jw�µ

ηηH (7b)

where η =√

µε . Above equation may be written as

d ∧ ηH = jw√

εµ�E (8a)d ∧ E = −jw

√µε�ηH (8b)

It may be noted that ηH is one form quantity.

3. FRACTIONAL DUALITY USING DIFFERENTIALFORM FORMALISM

Duality in electromagnetics states that if (E, ηH) is one set of solutionsto the Maxwell equations then (ηH,−E) is another set of solutions

402 Mustafa, Faryad, and Naqvi

to Maxwell’s equations. Solution set (E, ηH) may be termed as theoriginal solutions while solution set (ηH,−E) may be termed as thedual to the original solutions. The solutions, which may be regardedas the intermediate steps between original solutions and dual to theoriginal solutions may be termed as fractional dual solutions. To findthe dual solutions, we apply the Hodge star operator on both sides ofthe (8) and get get

�d ∧ ηH = jw√

µε��E (9a)�d ∧ E = −jw

√µε��ηH (9b)

or

�d ∧ ηH = jw√

µεE (10a)�d ∧ E = −jw

√µεH (10b)

or

E =1jk

[�d ∧ ηH] (11a)

−ηH =1jk

[�d ∧ E] (11b)

where w√

µε = k. It may be noted that operator (d∧) maps one formquantities onto two form quantities, while operator

(1jk�d∧

)maps

one form quantities onto one form quantities. Equation (11) givesthe dual to the original solutions to Maxwell equations, so operator(

1jk�d∧

)may be termed as duality operator in differential forms

electroamgnetics.The solutions which may be regarded as intermediate solutions

between original solution and dual to the original solution can be foundby fractionalizing the duality operator. The fractionalized dualityoperator is termed as fractional duality operator. What is meant byfractionalization of a linear operator and the recipe to fractionalize alinear operator is given in [16]. According to [16], a linear operatormay be fractionalize using three step recipe. Firstly, the eigenvaluesand eigenforms of the operator are required. Secondly, original solutionhave to expand in terms of eigenform of the operator. Thirdly,fractionalization of operator is obtained by fractionalization of theeigenvalues of the operator. It is important to note that fractionalduality operator has been fractionalized in spectral domain. In spectraldomain, it becomes equivalent to a simple operator.

Fractional dual solutions using calculus of differential forms 403

To find the fractional dual solutions to the Maxwell equations, wehave to fractionalize the duality operator 1

jk�d∧ as follow

Efd =[

1(jk)α

(�d∧)αE]

(12a)

ηHfd =[

1(jk)α

(�d∧)αηH]

(12b)

where α is the fractional parameter. Fractional parameter α may bereal/complex. Equation (12) provides new solutions to the source-freeMaxwell equations. For α = 0, we have Efd = E and ηHfd = ηH,which are the original set of the solutions satisfying (5). When α = 1,we obtain

Efd =1

(jk)(�d∧)E = ηH (13a)

ηHfd =1

(jk)(�d∧)ηH = −E (13b)

which are dual to the original solutions of Maxwell equations.Therefore, for all values of α between zero and unity, (12) providesus new set of solutions which can effectively be regarded as theintermediate solutions between original and the dual solutions.

Without loss of generality, we can consider the the electromagneticfield to be in terms of plane waves, in which case the fractional dualityoperator 1

(jk)α (�d∧)α may take a simpler form. In differential forms,the plane waves are represented as [8]

E(r) = E0 exp(κ|r) (14a)ηH(r) = ηH0 exp(κ|r) (14b)

where the duality product | of a vector r = a1ux + a2uy + a3uz and aone form κ = α1dx + α2dy + α3dz is defined as

κ|r = r|κ = a1α1 + a2α2 + a3α3

For plane waves, (11) and (12) takes the form, respectively

E =1jk

[�κ ∧ ηH] (15a)

−ηH =1jk

[�κ ∧ E] (15b)

404 Mustafa, Faryad, and Naqvi

Efd =[

1(jk)α

(�κ∧)αE]

(16a)

ηHfd =[

1(jk)α

(�κ∧)αηH]

(16b)

So, for plane wave propagation 1(jk)α (�d∧)α is equivalent to

1(jk)α (�κ∧)α. Following [8], we can fractionalize this operator byfractionalizing its eigen values. The eigenvalues and eigen forms ofoperator 1

jk�κ∧ are

A1 =(dx − jdy)√

2, a1 = +j (17a)

A2 =(dx + jdy)√

2, a2 = −j (17b)

A3 = dz, a3 = 0 (17c)

Now fractional dual solutions can be found by first expressing the fieldin terms of eigen forms and then fractionalizing the eigen values

E(r) = (PA1 + QA2 + RA3) exp(κ|r) (18a)ηH(r) = (SA1 + TA2 + UA3) exp(κ|r) (18b)

where P , Q, R, S, T and U are expansion coefficients. The fractionaldual fields are found as

Efd(r) = (aα1 PA1 + aα

2 QA2 + aα3 RA3) exp(κ|r) (19a)

ηHfd(r) = (aα1 SA1 + aα

2 TA2 + aα3 UA3) exp(κ|r) (19b)

4. EXAMPLES

4.1. TEM Plane Wave

Consider linearly polarized TEM uniform plane wave propagatingalong the negative z-direction. The electric and magnetic field canbe written as

E = dxE0 exp(κ|r) (20a)ηH = dyE0 exp(κ|r) (20b)

where κ = jkdz, r = zuz and Eo is the density of the electric field one-form. Equation (20a) represents surfaces perpendicular to x-axis while

Fractional dual solutions using calculus of differential forms 405

(20b) represents surfaces perpendicular to y-axis. The above equationsmay be written as

E = E0

{1√2A1 +

1√2A2

}exp(κ|r) (21a)

ηH = jE0

{1√2A1 −

1√2A2

}exp(κ|r) (21b)

and the fractional dual fields are

Efd = E0

{jα

√2A1 +

(−j)α

√2

A2

}exp(κ|r) (22a)

ηHfd = jE0

{jα

√2A1 −

(−j)α

√2

A2

}exp(κ|r) (22b)

or

Efd = E0

{cos

(απ

2

)dx + sin

(απ

2

)dy

}exp(κ|r) (23a)

ηHfd = E0

{− sin

(απ

2

)dx + cos

(απ

2

)dy

}exp(κ|r) (23b)

This set of solutions, which also satisfy the Maxwell equations,represents another uniform TEM plane wave propagating in the +z-direction. However surfaces have been rotated in the x-y plane by anangle απ

2 . When α = 0, (23) results in the original solutions and forα = 1, we get dual fields

Efd = dyE0 exp(κ|r)Hfd = −dxE0 exp(κ|r)

It is obvious that fractional dual fields represents planes perpendicularto a line making an angle απ

2 , in clockwise direction. Expression forthe corresponding fractional dual flux densities (two-form quantities)are represented as

Dfd = �εE0

{cos

(απ

2

)dx + sin

(απ

2

)dy

}exp(κ|r) (24a)

ηBfd = �µE0

{− sin

(απ

2

)dx + cos

(απ

2

)dy

}exp(κ|r) (24b)

or

Dfd = εE0

{cos

(απ

2

)dydz + sin

(απ

2

)dzdx

}exp(κ|r) (25a)

ηBfd = µE0

{− sin

(απ

2

)dydz + cos

(απ

2

)dzdx

}exp(κ|r) (25b)

406 Mustafa, Faryad, and Naqvi

Dfd and ηBfd represents tubes of same size as D and ηB but facesare rotated by an angle απ

2 clockwise.

4.2. Reflection from PEC Plane

Consider x-polarized plane wave propagating along z-axis hits theperfect electric conducting plane (PEC) placed at z = 0. The fieldin region z ≥ 0 is

E = Ei + Er (26a)ηH = ηHi + ηHr (26b)

where

Ei = dxE0 exp(κ|r) (27a)Er = −dxE0 exp(−κ|r) (27b)

ηHi =1jk

� κ ∧ Ei = dyE0 exp(κ|r) (27c)

ηHr = − 1jk

� κ ∧ Er = dyE0 exp(−κ|r) (27d)

The fractional solution may be obtained by linear combination

Efd = Eifd + Er

fd (28a)

ηHfd = ηHifd + ηHr

fd (28b)

where

Eifd =

1(jk)α

(�κ∧)αEi

= E0

{jα

√2A1 +

(−j)α

√2

A2

}exp(κ|r)

= E0

{cos

(απ

2

)dx + sin

(απ

2

)dy

}exp(κ|r) (29a)

Erfd =

1(jk)α

(�κ∧)αEr

= −E0

{(−j)α

√2

A1 +jα

√2A2

}exp(−κ|r)

= E0

{− cos

(απ

2

)dx + sin

(απ

2

)dy

}exp(−κ|r) (29b)

Fractional dual solutions using calculus of differential forms 407

ηHifd =

1(jk)α

(�κ∧)αηHi

= jE0

{jα

√2A1 −

(−j)α

√2

A2

}exp(κ|r)

= E0

{− sin

(απ

2

)dx + cos

(απ

2

)dy

}exp(κ|r) (29c)

ηHrfd =

1(jk)α

(�κ∧)αηHr

= jE0

{(−j)α

√2

A1 −jα

√2A2

}exp(−κ|r)

= E0

{sin

(απ

2

)dx + cos

(απ

2

)dy

}exp(−κ|r) (29d)

Using (29) in (28), the (Efd, ηHfd) is given as

Efd = 2j exp(−jαπ

2)E0 sin

(κ|r +

απ

2

)

×[cos

(απ

2

)dx + sin

(απ

2

)dy

](30a)

ηHfd = 2 exp(−jαπ

2)E0 cos

(κ|r +

απ

2

)

×[− sin

(απ

2

)dx + cos

(απ

2

)dy

](30b)

The transverse impedance of wave is

Zfd =Efdx

Hfdy= −Efdy

Hfdx= jη tan

(κ|r +

απ

2

)(31)

where Zfd is a zero form. At z = 0, for α = 0 and z = 0 we havethe impedance of PEC plane and for α = 1 we have the impedance ofPMC (perfect magnetic conductor) plane.

5. CONCLUSION

Using fractional differential operator, we have obtained the fractionaldual solutions. We have obtained the fractional dual solutions for atraveling plane wave and standing wave. It is observed that fractionaldual solutions represents the rotation of surfaces, tubes etc.

408 Mustafa, Faryad, and Naqvi

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