First Advances of Thesis

Gerardo Jimenez Trejo

April 17, 2020

First we are going to solve the Paraxial Helmholtz Equation as follows:

i ∂U −1 n0
= 2 ∇⊥ U + Ω2 r 2 U (1)
k0 ∂z 2k0 n0 2
Now with this we will purpose the following ansatz. Let U (r, z) = N (z)eiS(z)r where N (z) is just
2

a normalization factor and S(z) is a function top be determined.

With this let us calculate the solutions
 
dU iS(z)r2 dN (z) iS(z)r2 2 dS(z) dN (z) dS(z)
=e + N (z)e ir =e iS(z)r2 2
+ ir N (z) (2)
dz dz dz dz dz
     
1 d d 1 d d iS(z)r2 N (z) d d iS(z)r2
∇⊥ U = r U = r N (z)e = r e
r dr dr r dr dr r dr dr

N (z) d  2
 N (z)S(z) [2i] d  2
 N (z)S(z) [2i]
irS(z)(2r)eiS(z)r = r2 eiS(z)r = 2r + 2ir3 S(z)


=
r dr r dr r

(3)
2
∇⊥ U = N (z)S(z)[4i]eiS(z)r (1 + ir2 S(z))
Then we replace this results in the Paraxial Helmholtz Equation and then we obtain the follow-
ing.
  
i iS(z)r2 dN (z) 2 dS(z) −1 h 2
i
e + ir N (z) = 2 N (z)S(z)[4i]eiS(z)r (1 + ir2 S(z)) +
k0 dz dz 2k0 n0

n0 2 2  2

+ Ω r N (z)eiS(z)r
2

 
i dN (z) dS(z) −1   n0
+ ir2 N (z) = 2 N (z)S(z)[4i](1 + ir2 S(z)) + Ω2 r2 (N (z))
k0 dz dz 2k0 n0 2

Then if we divide by N (z) the whole expression we will obtain the following terms
i dN (z) dS(z) 2 2iS(z) k0 n0 2 2
− r2 = S(z)2 r2 − + Ω r
N (z) dz dz k0 n0 k0 n0 2
Then we will separate the imaginary and real terms of each other and we will obtain the following
1 dN (z) −2
= S(z)
N (z) dz k0 n0
d −2
⇒
ln(N (z)) = S(z) (4)
dz k0 n0
The next equation do not have any relation with r because this expressions are in the ask Dra.
Sara

1
 dS(z) 2 k 0 n0 2 2
−r2 = S(z)2 r2 + Ω r
dz k0 n0 2
dS(z) 2 k 0 n0 2
= S(z)2 + Ω
dz k0 n0 2
   2
d 2 2
S(z) + S(z) + Ω2 = 0 (5)
dz k0 n0 k0 n0
The last equation 5 can be reduced to other one, that we know the solutions now. Let us purpose
the following change of variable: let α = k02n0 and β = S(z). With this we will make the following
assumptions. β ∈ C ⇒ αβ = αβR + iαβI where αβ 0 = αβR0 + iαβI0 is the derivative of β function.
Then if we replace this solution to the equation.
d
(αβ 0 ) + (α2 β 2 ) + Ω2 = 0
dz
0
⇒ α(βR + iβI0 ) + α(βR
2
+ 2iβR βI − βI2 ) + Ω2 = 0 (6)
Taking the imaginary part of the last equation.
βI0
βI0 + α(2βR βI ) = 0 ⇒ = −2αβR ⇒
βI
d
−2αβR = (ln βI ) (7)
dz
The last expression tell us that the function βR ∝ d
dz ln w ⇒

d
βR = κ ln w (8)
dz
Then replacing the expression of βR in 7 we will obtain the following

−2ακ(ln βR )0 = ln(βI )0 ⇒ −2ακ(ln βR ) + ln(λ) = ln(βI )

ln w−2ακ λ = ln(βI )

λ
⇒ βI = (9)
w−2ακ
 
The equation 9 and 8 will be replaced in the real part of 6 with the term βR0 = κ w00
w − w02
w2
and we will obtain the following.
0
α(βR + α2 βR
2
− αβI2 ) + Ω2 = 0
 00  0 2
w02
  
w w λ
ακ − 2 + α2 κ −α + Ω2 = 0
w w w w2ακ

w00 w02 2 2 λ
+ 2 κ α − ακ − α2 2ακ + Ω2 = 0


ακ
w w w
In order to obtain a new dierential equation, we have to purpose that κ2 α2 − ακ = 0 this

mean that κ = α1 , then if we replace the value of it, and make the term λ in a convenient way, such
that λ = 1 we will obtain the following equation.
d2 w 4
2
+ Ω2 w = 2 2 3 (10)
dx k0 n0 w
The equation 10 is the well known Ermakov equation. Then going back to 5 we can write the
complete solution in terms of our new function w. With αβ = k02n0 S(z) = αβR +iαβI . The equation
9 has the form of βI = w12 and 8 will be βR = ww .
0

2
 2 1 2i 1 d
S(z) = + where = ln(w(z)) (11)
k0 n0 R(z) k0 n0 w2 R(z) dz
As we can see the new function w is involved in all our unknown functions. In this way we will
replace the equation 11 in 5 and we will obtain the following.
−d d 2i
ln (N (z)) = ln(w(z)) +
dz dz k0 n0 w2
Z z
d d d 2dz̃
− ln (N (z)) = ln(w(z)) + i χ(z) where χ(z) =
dz dz dz k0 n0 w2 (z̃)
N (z) 1
= e−iχ(z)
N0 w
N0 −iχ(z)
N (z) = e (12)
w
Now, let us take the Ermakov equation (10). We know from the literature, that while we know
a particular solution of this equation we are able to nd the general solution in terms of this. Let
µ(z) solution and wp (z) a particular solution of the Ermakov equation, the general solution can be
written as follows.
1
w(z) = ± aµ2 (z) + bµ(z)wp (z) + cwp2 (z) 2 (13)


Where the coecients had to fullll this relation. 4ac − b2 = W2α

0
and W0 is the wronskian
of the equation. In order to nd a linearly independent function, the wronskian of the function
has to be zero. When we nd the solution to the homogeneous equation we are able to nd the
d2 w
complete solution in the form of equation 13. Then solving the following relation 2 + Ω2 w = 0
dz
the particular solution is wp = cos Ωz and its orthogonal complement is µ = sin Ωz .
Then let us calculate the wornskian of the solutions in order to guarantee the orthogonality of
the system.
   
 wp µ   cos Ωz sin Ωz 
W0 =  0 = =Ω
wp µ0  −Ω sin Ωz Ω cos Ωz 
  

The replacing the particular solution of the homogeneous in the equation 13 as follows.
1  1
w(z) = ± a sin2 Ωz + b sin Ωz cos Ωz + c cos2 Ωz 2 = k12 cos Ω(z − z0 ) + k22 sin Ω(z − z0 ) 2 (14)


We can represent, the term in the coecient b as a shift in z. With this we can nd two
possibilities b = 0 ⇒ z0 6= 0 or b 6= 0 ⇒ z0 = 0, we choose the second option and then we expand
the solution w(z) by the trigonometrical options.

w(z) = [k12 (cos2 Ωz cos2 Ωz0 + 2 cos Ωz cos Ωz0 sin Ωz sin Ωz0 + sin2 Ωz sin2 Ωz0 )+
1
+k22 (sin2 Ωz cos2 Ωz0 − 2 sin Ωz cos Ωz0 cos Ωz sin Ωz0 + cos2 Ωz sin2 Ωz0 )] 2

= [sin2 Ωz k12 sin2 Ωz0 + k22 cos2 Ωz0 + cos Ωz sin Ωz 2 cos Ωz0 sin Ωz0 (k12 − k22 )



| {z } | {z }
a b
1
2
k12 cos2 Ωz0 k22 sin2 Ωz0


+ cos Ωz + ] 2
| {z }
c

Then the coecients will be the following

3
 a = k12 sin2 Ωz0 + k22 cos2 Ωz0
b = 2 cos Ωz0 sin Ωz0 (k12 − k22 )
c = k12 cos2 Ωz0 + k22 sin2 Ωz0

And these fulllls the relation 4ac − b2 = 2α and we obtain form the literature k1 = w0


Ω
this mean an initial condition and from the relation between the coecients k12 k22 = Ωα 2 . Then

replacing in the equation 14 we obtain the following.

 21
α2

w(z) = ± w0 cos Ω(z − z0 ) + sin Ω(z − z0 )
w0 Ω 2
 1
1 2
= ±w0 cos Ω(z − z0 ) + sin Ω(z − z0 ) (15)
zR Ω

Where zR = wα0 and z0 can be understood as the place where the beam starts. Now to recover the
homogeneous case, we just can