A. Appendix

A.1 Analytical results of two-dimensional slab waveguide

The analytic solution of an electromagnetic wave in a slab waveguide shown in Figure 3 is provided in various textbooks regarding optical waveguides and related fields [5–8]. In this appendix, the analytical results of Eqs. (49)–(54) are derived in accordance with these references.

A propagating electromagnetic wave with no y-dependence in the z-direction with angular frequency ω and propagation constant β, the wave number in the z-direction, is written as

$$E(t, \varkappa, y, z) = e(\varkappa)e^{i(\alpha t - \beta x)},\tag{68}$$

$$\mathbf{H}(t, \mathbf{x}, \mathbf{y}, \mathbf{z}) = \mathbf{h}(\mathbf{x}) e^{i(\alpha t - \beta \mathbf{z})}.\tag{69}$$

Maxwell's equations in dielectrics using Eqs. (13) and (22) become

$$i\mu\_0 a h\_{\mathbf{x}}(\mathbf{x}) = -i\beta e\_{\mathcal{Y}}(\mathbf{x})\tag{70}$$

$$i\mu\_0 a h\_\gamma(\varkappa) = i\beta \varepsilon\_\mathbf{x}(\varkappa) + \partial\_\mathbf{x} \varepsilon\_\mathbf{z}(\varkappa),\tag{71}$$

$$i\mu\_0 a h\_x(\mathbf{x}) = -\partial\_\mathbf{x} e\_\mathbf{y}(\mathbf{x}),\tag{72}$$

ezð Þ¼ x

Electro-magnetic Simulation Based on the Integral Form of Maxwell's Equations

DOI: http://dx.doi.org/10.5772/intechopen.81338

quently, Hyð Þ <sup>x</sup> and <sup>∂</sup>xhyð Þ <sup>x</sup> in Eqs. (76)–(78) are continuous.

In case 1, the solutions of Eq. (76) are

<sup>Q</sup> cos Pd 2

2

� <sup>P</sup> sin Pd

� � cos Qd

� � cos Qd

2 � �

2 � �

1. ∣β∣ < ncl

Figure 16.

Index distribution of the slab waveguide.

3. nco ≤ ∣β∣

hyð Þ¼ x

and

83

2. ncl ≤ ∣β∣ < nco

8 >>>>>>><

>>>>>>>:

Solving Eq. (76) requires considering the following three cases:

1 in xð Þ<sup>2</sup>

that the boundary condition at x ¼ �d=2 be considered. Because of the integral form of Maxwell's equations, the components of electric and magnetic fields parallel to the boundary surface of the indices are continuous, and the components of the electric and magnetic flux densities normal to the surface are continuous. Conse-

The discontinuity of n xð Þ at x ¼ �d=2 and x ¼ d=2, shown in Figure 16, requires

Q cosð Þ Px ð Þ jxj≤d=2

2 � � sin Qd

2 � � sin Qd

2

ð Þ jxj≥d=2

(79)

,

2

<sup>þ</sup> <sup>P</sup> sin Pd

� � � � cosð Þ Qx

� <sup>Q</sup> cos Pd

� � � � sin ð Þ Qx

<sup>ε</sup>0ω∂xhyð Þ <sup>x</sup> : (78)

$$i\boldsymbol{n}(\boldsymbol{\kappa})^2 \boldsymbol{e}\_0 \boldsymbol{\alpha} \mathbf{e}\_{\boldsymbol{\kappa}}(\boldsymbol{\kappa}) = i\beta \boldsymbol{h}\_{\boldsymbol{\gamma}}(\boldsymbol{\kappa}),\tag{73}$$

$$i\boldsymbol{n}(\boldsymbol{\kappa})^2 \boldsymbol{e}\_0 a \boldsymbol{e}\_\mathcal{V}(\boldsymbol{\kappa}) = -i\beta \boldsymbol{h}\_\mathbf{x}(\boldsymbol{\kappa}) - \partial\_\mathbf{x} \boldsymbol{h}\_\mathbf{z}(\boldsymbol{\kappa}),\tag{74}$$

$$
\dot{m}(\mathbf{x})^2 \varepsilon\_0 a \varkappa\_x(\mathbf{x}) = \partial\_\mathbf{x} h\_\mathcal{\mathcal{Y}}(\mathbf{x}),\tag{75}
$$

where n xð Þis the index distribution shown in Figure 16. As shown in Eqs. (70)–(75), there are two closed equation classes. The first class contains Eqs. (70), (72), and (74), which have components of electric and magnetic fields transverse to the propagation direction and a longitudinal component of the magnetic field in that direction. The second class contains Eqs. (71), (73), and (75), which have components of electric and magnetic fields transversed to the propagation direction and a longitudinal component of the electric field in that direction. Solution to the first class comprise the transverse electric (TE) mode because the electric field has only a component transversed to the propagation direction, and solution to the second class comprises the transverse magnetic (TM) mode because the magnetic field has only a component transversed to that direction. In Section 4, numerical and analytical results of Hy are shown, and they are TM modes.

Hereafter, our discussion is limited to the TM mode. Then, Eqs. (71), (73), and (75) are rewritten as

$$\partial\_{\mathbf{x}}^2 h\_{\mathcal{V}}(\mathbf{x}) = -\left[ \left( \frac{2\pi m(\mathbf{x})}{\lambda} \right)^2 - \beta^2 \right] h\_{\mathcal{V}}(\mathbf{x}),\tag{76}$$

$$
\epsilon\_{\mathbf{x}}(\mathbf{x}) = \frac{\beta}{n(\mathbf{x})^2 \epsilon\_0 a} h\_{\mathbf{y}}(\mathbf{x}),
\tag{77}
$$

Electro-magnetic Simulation Based on the Integral Form of Maxwell's Equations DOI: http://dx.doi.org/10.5772/intechopen.81338

Figure 16.

Conflict of interest

Recent Advances in Integral Equations

accordance with these references.

z-direction, is written as

TM modes.

82

(75) are rewritten as

A. Appendix

The author declares no conflicts of interest associated with this manuscript.

The analytic solution of an electromagnetic wave in a slab waveguide shown in Figure 3 is provided in various textbooks regarding optical waveguides and related fields [5–8]. In this appendix, the analytical results of Eqs. (49)–(54) are derived in

A propagating electromagnetic wave with no y-dependence in the z-direction with angular frequency ω and propagation constant β, the wave number in the

where n xð Þis the index distribution shown in Figure 16. As shown in Eqs. (70)–(75), there are two closed equation classes. The first class contains Eqs. (70), (72), and (74), which have components of electric and magnetic fields transverse to the propagation direction and a longitudinal component of the magnetic field in that direction. The second class contains Eqs. (71), (73), and (75), which have components of electric and magnetic fields transversed to the propagation direction and a longitudinal component of the electric field in that direction. Solution to the first class comprise the transverse electric (TE) mode because the electric field has only a component transversed to the propagation direction, and solution to the second class comprises the transverse magnetic (TM) mode because the magnetic field has only a component transversed to that direction. In Section 4, numerical and analytical results of Hy are shown, and they are

Hereafter, our discussion is limited to the TM mode. Then, Eqs. (71), (73), and

2πn xð Þ λ � �<sup>2</sup>

> β n xð Þ<sup>2</sup>

" #

� <sup>β</sup><sup>2</sup>

hyð Þ x , (76)

<sup>ε</sup>0<sup>ω</sup> hyð Þ <sup>x</sup> , (77)

<sup>i</sup>ð Þ <sup>ω</sup>t�β<sup>z</sup> , (68)

<sup>i</sup>ð Þ <sup>ω</sup>t�β<sup>z</sup> : (69)

iμ0ωhxð Þ¼� x iβeyð Þ x (70)

<sup>i</sup>μ0ωhzð Þ¼� <sup>x</sup> <sup>∂</sup>xeyð Þ <sup>x</sup> , (72)

<sup>ε</sup>0ωeyð Þ¼� <sup>x</sup> <sup>i</sup>βhxð Þ� <sup>x</sup> <sup>∂</sup>xhzð Þ <sup>x</sup> , (74)

ε0ωexð Þ¼ x iβhyð Þ x , (73)

<sup>ε</sup>0ωezð Þ¼ <sup>x</sup> <sup>∂</sup>xhyð Þ <sup>x</sup> , (75)

<sup>i</sup>μ0ωhyð Þ¼ <sup>x</sup> <sup>i</sup>βexð Þþ <sup>x</sup> <sup>∂</sup>xezð Þ <sup>x</sup> , (71)

Eð Þ¼ t; x; y; z eð Þ x e

Hð Þ¼ t; x; y; z hð Þ x e

Maxwell's equations in dielectrics using Eqs. (13) and (22) become

in xð Þ<sup>2</sup>

in xð Þ<sup>2</sup>

in xð Þ<sup>2</sup>

∂2

xhyð Þ¼� x

exð Þ¼ x

A.1 Analytical results of two-dimensional slab waveguide

Index distribution of the slab waveguide.

$$
\varepsilon\_{\mathbf{z}}(\mathbf{x}) = \frac{1}{\dot{m}(\mathbf{x})^2 \varepsilon\_0 \alpha} \partial\_{\mathbf{x}} h\_{\mathbf{y}}(\mathbf{x}). \tag{78}
$$

The discontinuity of n xð Þ at x ¼ �d=2 and x ¼ d=2, shown in Figure 16, requires that the boundary condition at x ¼ �d=2 be considered. Because of the integral form of Maxwell's equations, the components of electric and magnetic fields parallel to the boundary surface of the indices are continuous, and the components of the electric and magnetic flux densities normal to the surface are continuous. Consequently, Hyð Þ <sup>x</sup> and <sup>∂</sup>xhyð Þ <sup>x</sup> in Eqs. (76)–(78) are continuous.

Solving Eq. (76) requires considering the following three cases:


In case 1, the solutions of Eq. (76) are

$$h\_{\mathbf{y}}(\mathbf{x}) = \begin{cases} Q\cos(P\mathbf{x}) & (|\mathbf{x}| \le d/2) \\\\ \begin{bmatrix} Q\cos\left(\frac{Pd}{2}\right)\cos\left(\frac{Qd}{2}\right) + P\sin\left(\frac{Pd}{2}\right)\sin\left(\frac{Qd}{2}\right) \\\\ -\left[P\sin\left(\frac{Pd}{2}\right)\cos\left(\frac{Qd}{2}\right) - Q\cos\left(\frac{Pd}{2}\right)\sin\left(\frac{Qd}{2}\right)\right]\sin\left(Q\mathbf{x}\right) \end{cases} & (|\mathbf{x}| \ge d/2) \end{cases} \tag{79}$$

and

$$h\_{\mathcal{V}}(\mathbf{x}) = \begin{cases} Q \sin(P\mathbf{x}) & (|\mathbf{x}| \le d/2) \\\\ \begin{bmatrix} \text{sign}(\mathbf{x}) \left\{ \left[ Q \sin\left(\frac{Pd}{2}\right) \cos\left(\frac{Qd}{2}\right) - P \cos\left(\frac{Pd}{2}\right) \sin\left(\frac{Qd}{2}\right) \right] \cos(Q\mathbf{x}) \\\\ \end{bmatrix} & \begin{aligned} \text{if } \mathbf{q} = \mathbf{p} \text{ (} \mathbf{x} \text{)} \\\\ \mathbf{-} \left[ P \cos\left(\frac{Pd}{2}\right) \cos\left(\frac{Qd}{2}\right) + Q \sin\left(\frac{Pd}{2}\right) \sin\left(\frac{Qd}{2}\right) \right] \sin(Q\mathbf{x}) \end{cases} \end{cases} \tag{80}$$

where

$$P = \sqrt{\left(\frac{2\pi m\_{co}}{\lambda}\right)^2 - \beta^2},\tag{81}$$

and

where

Defining

where

and

where

85

hyð Þ¼ x

DOI: http://dx.doi.org/10.5772/intechopen.81338

In case 3, the solutions of Eq. (76) are

hyð Þ¼ x

hyð Þ¼ x

As a result of Eqs. (94) and (95),

8 < 8 <

8 <

Qd

<sup>2</sup> ¼ � ncl

sin ð Þ Px ∣x∣ ≤d=2

<sup>2</sup> cot

cosh ð Þ Px ∣x∣ ≤d=2

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi <sup>β</sup><sup>2</sup> � <sup>2</sup>πnco λ � �<sup>2</sup>

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi <sup>β</sup><sup>2</sup> � <sup>2</sup>πncl λ � �<sup>2</sup>

2

sinh ð Þ Px ∣x∣ ≤ d=2

2

� <sup>2</sup>πncl λ � �<sup>2</sup>

� �e�Qð Þ <sup>j</sup>xj�d=<sup>2</sup> <sup>∣</sup>x<sup>∣</sup> <sup>&</sup>gt; <sup>d</sup>=<sup>2</sup>

: , (97)

� �e�Qð Þ <sup>j</sup>xj�d=<sup>2</sup> <sup>∣</sup>x<sup>∣</sup> <sup>&</sup>gt; <sup>d</sup>=<sup>2</sup>

: , (93)

<sup>d</sup> ∣x∣ > d=2

� �: (90)

<sup>2</sup> , (91)

<sup>2</sup> , (92)

, (94)

, (95)

� �, (96)

� �: (98)

, (99)

: , (89)

Pd 2

2 � �e�Q2∣x∣�<sup>d</sup>

signð Þ <sup>x</sup> sin Pd

Electro-magnetic Simulation Based on the Integral Form of Maxwell's Equations

nco � �<sup>2</sup> Pd

<sup>u</sup> <sup>¼</sup> Pd

<sup>w</sup> <sup>¼</sup> Qd

yields Eqs. (53), (55), (56), and (59). This is the solution we want.

cosh Pd 2

s

s

<sup>Q</sup> <sup>¼</sup> <sup>P</sup> tanh Pd

2

<sup>Q</sup> ¼ �<sup>P</sup> coth Pd

λ � �<sup>2</sup>

P ¼

Q ¼

signð Þ <sup>x</sup> sinh Pd

�P<sup>2</sup> <sup>þ</sup> <sup>Q</sup><sup>2</sup> <sup>¼</sup> <sup>2</sup>πnco

$$Q = \sqrt{\left(\frac{2\pi n\_{cl}}{\lambda}\right)^2 - \rho^2}.\tag{82}$$

However, this solution does not show electromagnetic wave propagation in the z-direction but shows a reflection and transmission problem of the film when the incident angle is less than the critical angle. Let us discuss this in more detail. When ∣x∣ ≤d=2, Hyð Þ t; x; y; z is a linear combination of

$$\mathbf{e}^{i(\alpha t - P\mathbf{x} - \beta \mathbf{x})}, \quad \text{and} \quad \mathbf{e}^{i(\alpha t + P\mathbf{x} - \beta \mathbf{x})},\tag{83}$$

a plane wave whose wavenumber is ð Þ P; 0; β and ð Þ �P; 0; β , respectively. When ∣x∣ > d=2, Hyð Þ t; x; y:z is a linear combination of

$$e^{i(\alpha t - Q\mathbf{x} - \beta \mathbf{z})}, \quad \text{and} \quad e^{i(\alpha t + Q\mathbf{x} - \beta \mathbf{z})},\tag{84}$$

a plane wave whose wavenumbers are ð Þ Q; 0; β and ð Þ �Q; 0; β , respectively. Therefore, with a suitable linear combination of Eqs. (79) and (80), the solution becomes that a plane wave with wavenumber ð Þ P; 0; β is incident from the x < � d=2 region to be reflected and transmitted by a film of the ∣x∣ ≤d=2 region with a reflected wave propagated in the x < � d=2 region and a transmitted wave propagated in the x > d=2 region. Therefore, this solution is not what we want.

In case 2, the solutions of Eq. (76) are

$$h\_{\mathcal{Y}}(\mathbf{x}) = \begin{cases} \cos\left(P\mathbf{x}\right) & |\mathbf{x}| \le d/2\\ \cos\left(\frac{P d}{2}\right) e^{-Q\frac{2|\mathbf{x}| - d}{d}} & \text{,} \end{cases},\tag{85}$$

where

$$P = \sqrt{\left(\frac{2\pi m\_{co}}{\lambda}\right)^2 - \beta^2},\tag{86}$$

$$Q = \sqrt{\beta^2 - \left(\frac{2\pi m\_{cl}}{\lambda}\right)^2},\tag{87}$$

$$\frac{Qd}{2} = \left(\frac{n\_{cl}}{n\_{cl}o}\right)^2 \frac{Pd}{2} \tan\left(\frac{Pd}{2}\right),\tag{88}$$

Electro-magnetic Simulation Based on the Integral Form of Maxwell's Equations DOI: http://dx.doi.org/10.5772/intechopen.81338

and

hyð Þ¼ x

8 >>>>>>>><

>>>>>>>>:

where

where

84

signð Þ <sup>x</sup> <sup>Q</sup> sin Pd

<sup>þ</sup> <sup>P</sup> cos Pd 2 � �

∣x∣ ≤d=2, Hyð Þ t; x; y; z is a linear combination of

∣x∣ > d=2, Hyð Þ t; x; y:z is a linear combination of

In case 2, the solutions of Eq. (76) are

hyð Þ¼ x

e

e

�

Recent Advances in Integral Equations

2 � � cos Qd 2 � �

cos Qd 2 � �

P ¼

Q ¼

Q sin ð Þ Px ð Þ jxj≤ d=2

sin Qd 2

sin Qd 2

cosð Þ Qx

�

, (81)

: (82)

<sup>i</sup>ð Þ <sup>ω</sup>tþPx�β<sup>z</sup> , (83)

<sup>i</sup>ð Þ <sup>ω</sup>tþQx�β<sup>z</sup> , (84)

, (86)

, (87)

, (88)

ð Þ jxj≥ d=2 ,

(80)

sin ð Þ Qx

� <sup>P</sup> cos Pd 2 � �

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2πnco λ � �<sup>2</sup>

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2πncl λ � �<sup>2</sup>

However, this solution does not show electromagnetic wave propagation in the z-direction but shows a reflection and transmission problem of the film when the incident angle is less than the critical angle. Let us discuss this in more detail. When

a plane wave whose wavenumber is ð Þ P; 0; β and ð Þ �P; 0; β , respectively. When

a plane wave whose wavenumbers are ð Þ Q; 0; β and ð Þ �Q; 0; β , respectively. Therefore, with a suitable linear combination of Eqs. (79) and (80), the solution becomes that a plane wave with wavenumber ð Þ P; 0; β is incident from the x < � d=2 region to be reflected and transmitted by a film of the ∣x∣ ≤d=2 region with a reflected wave propagated in the x < � d=2 region and a transmitted wave propagated in the x > d=2 region. Therefore, this solution is not what we want.

cosð Þ Px ∣x∣ ≤d=2

e�Q2∣x∣�<sup>d</sup> d

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2πnco λ � �<sup>2</sup>

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi <sup>β</sup><sup>2</sup> � <sup>2</sup>πncl λ � �<sup>2</sup>

<sup>2</sup> tan

� <sup>β</sup><sup>2</sup>

Pd 2 � �

: , (85)

� <sup>β</sup><sup>2</sup>

� <sup>β</sup><sup>2</sup>

� � � �

� � � �

s

s

<sup>i</sup>ð Þ <sup>ω</sup>t�Px�β<sup>z</sup> , and e

<sup>i</sup>ð Þ <sup>ω</sup>t�Qx�β<sup>z</sup> , and e

cos

s

s

P ¼

Q ¼

Qd <sup>2</sup> <sup>¼</sup> ncl nco � �<sup>2</sup> Pd

8 <

Pd 2 � �

<sup>þ</sup> <sup>Q</sup> sin Pd 2 � �

$$h\_{\mathcal{Y}}(\mathbf{x}) = \begin{cases} \sin\left(P\mathbf{x}\right) & |\mathbf{x}| \le d/2\\ \text{sign}(\mathbf{x})\sin\left(\frac{Pd}{2}\right)e^{-Q\frac{2|\mathbf{x}|-d}{d}} & |\mathbf{x}| > d/2 \end{cases} \tag{89}$$

where

$$\frac{Qd}{2} = -\left(\frac{n\_{cl}}{n\_{cl}o}\right)^2 \frac{Pd}{2} \cot\left(\frac{Pd}{2}\right). \tag{90}$$

Defining

$$
u = \frac{Pd}{2},
\tag{91}$$

$$w = \frac{Qd}{2},\tag{92}$$

yields Eqs. (53), (55), (56), and (59). This is the solution we want. In case 3, the solutions of Eq. (76) are

$$h\_{\mathcal{V}}(\mathbf{x}) = \begin{cases} \cosh\left(P\mathbf{x}\right) & |\mathbf{x}| \le d/2\\ \cosh\left(\frac{Pd}{2}\right) e^{-Q\left(|\mathbf{x}| - d/2\right)} & |\mathbf{x}| > d/2 \end{cases} \tag{93}$$

where

$$P = \sqrt{\beta^2 - \left(\frac{2\pi m\_{co}}{\lambda}\right)^2},\tag{94}$$

$$Q = \sqrt{\beta^2 - \left(\frac{2\pi n\_{cl}}{\lambda}\right)^2},\tag{95}$$

$$Q = P \tanh\left(\frac{Pd}{2}\right),\tag{96}$$

and

$$h\_{\mathcal{Y}}(\mathbf{x}) = \begin{cases} \sinh\left(P\mathbf{x}\right) & |\mathbf{x}| \le d/2\\ \operatorname{sign}(\mathbf{x})\sinh\left(\frac{Pd}{2}\right)e^{-Q\left(|\mathbf{x}|-d/2\right)} & |\mathbf{x}| > d/2 \end{cases} \tag{97}$$

where

$$Q = -P \coth\left(\frac{Pd}{2}\right). \tag{98}$$

As a result of Eqs. (94) and (95),

$$-P^2 + Q^2 = \left(\frac{2\pi m\_{co}}{\lambda}\right)^2 - \left(\frac{2\pi m\_{cl}}{\lambda}\right)^2,\tag{99}$$
