DC Mathematica 2016

Substituting Eq.24 and 25 into Eq.23, we get  𝑧 =  1

(
)
(−
𝑧) 
𝑣∅ (26)

Then substitute Eq. 26 into the wave equation for  𝑧

Eq. (21) yields

2  1

2

 2
2

1

 1

+ ( 2 −

)  1

+

= 0 (27)

This is the well-known differential equations for Bessel functions. Considering the processes of solving the Eq.27 is too long which will affect the understanding of the whole picture in solving Maxwell’s equations, I will put the process of getting the solution of Bessel functions in the last part. The configuration of the step-index fibre is basically a homogeneous core of refractive index n1 and radius a , which is surrounded by an infinite cladding of index n 2 ,The reason for assuming an infinitely thick cladding is that the guided modes have exponentially decaying fields outside the core and the harmonically varying fields inside the core. For the inside region,
< so the solutions to Eq. (27) are Bessel functions of the first kind  𝑣 (
) where  2 =
2 −  1 2 with  1 = 2𝜋 2  , these expressions are  𝑧 =  𝑣 (
)
𝑣∅ 
(−
𝑧) 𝑧 =
 𝑣 (
)
𝑣∅ 
(−
𝑧) While outside the core the solutions to Eq. (27) are Bessel functions of the second kind  𝑣 (
) , where  2 =
2 −  2 2 with  2 = 2𝜋 2  . These expressions are  𝑧 =  𝑣 (
)
𝑣∅ 
(−
𝑧) 𝑧 =  𝑣 (
)
𝑣∅ 
(−
𝑧) The A, B, C, D are all arbitrary constants. These four short and elegant equations describe the propagation mechanism of waveguide in optical fibres.

 The process of solving Bessel equation Now, I would like to show how to solve the Bessel functions. Bessel equation is in this form  2  ′′ +  ′ + ( 2 −  2 ) = 0 , v ≥ 0 The solution of y will be  = ∑   + Differentiate y, we get ′ = ∑(
+
)    +−1  ′′ = ∑(
+
)(
+
− 1)    +−2 Substitute y, y’ and y’’ in to the Bessel equation, we get

18

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