ECE 6382
The Steepest Descent Method
D. R. Wilton
ECE Dept.
Adapted from original notes of
Prof. David R. Jackson
8/24/10
Steepest Descent Method
Complex Integral:
I      f  z  e  g  z  dz
C
We want to obtain an approximate evaluation of the integral
when the real parameter Ω is very large.
The functions f(z) and g(z) are analytic, so that the path C may
be deformed if necessary. The steepest descent path (SDP) of
interest will pass through a saddle point.
Saddle Point:
dg
g '  z0  
dz
g '  z0   0
g

x
z  z0
g
 i
y
z  z0
0
z  z0
g g

0
y x
Steepest Descent Method (cont.)
Path deformation:
If the path does not go through a SDP, we assume that it can be
deformed to do so.
Any singularities encountered during the path deformation must be
accounted for (e.g., residue of captured poles).
y
C
z0
  
C
C
Cp
 F  z  dz  2 i Res F  z 
p
X
Cp
Cp
x
C
Steepest Descent Method (cont.)
Let
g  z  u  z  i v z
u v v u
(Cauchy Riemann eqs.)
i  i
x x y y
 all first derivatives vanish at saddle point z  z0 !
 g '  z0   0 
Also, by C.R. conditions, wherever g(z) is analytic,
 2u   v    v 
    
2
x
x  y  y  x 
 2u
  2   2u  0
y
Steepest Descent Method (cont.)
u  z0 
 2u  2u
or
 2 0
2
x
y
 if uxx  0, then
Near the saddle point:
u yy  0
“Rubber sheet”
models of u(z).
1
2
u  x, y   u  x0 , y0   u xx  x  x0 
2
1
2
 u yy  y  yo   u xy  x  x0  y  y0 
2
Ignore (rotate coordinates
to eliminate)
Steepest Descent Method (cont.)
The u(x,y) function has a “saddle” shape near the SDP:
u ( x, y )
u  z0 
x
y
z0
Note: the saddle does not necessarily open along one of the principal
axes (only when uxy (x0, y0) = 0).
Steepest Descent Method (cont.)
Along any descending path (where the u function decreases):
I   e
 g  z0 
 f ze
  g  z   g  z0  
dz
contribution is from z0
C
e
 g  z0 

f ze
 is so large most of
i v  z   v  z0  
 u  z  u  z0 
e
0
dz
C
Note in general, both the phase and amplitude of the integrand change.
If we can find a path along which the phase does not change,
 u z u z 
the integrand will look like  f  z  e      with most of the
contribution coming from the neighborhood of z0 .
0
Steepest Descent Method (cont.)
Consider a path of constant phase:
C0 : v  z   v  z0   constant
 v  z   v  z0   0
y
C0
z0
x
This eliminates any variation in the phase term eiv z v z0  inside the integral.
Steepest Descent Method (cont.)
u(z) landscape near z0:
u( z)  u( z0 )
y
hill
valley
v( z )  v( z0 )
z0
C0
(SDP)
valley
hill
x
u( z)  u( z0 )
Steepest Descent Method (cont.)
u(z) and v(z) landscape near z0:
v( z )  v( z0 )
(SAP)
u( z)  u( z0 )
y
z0
C0
v( z )  v( z0 )
(SDP)
x
u( z)  u( z0 )
Constant phase paths through z0 are also steepest descent
paths (SDPs) or steepest ascent paths (SAPs)
Steepest Descent Method (cont.)
Property:
u  x, y  is parallel to C0
Hence C0 is either a “path of steepest descent”
(SDP) or a “path of steepest ascent” (SAP).
Steepest Descent Method (cont.)
proof
u
u
u  x  y
x
y
v
v
v  x  y
x
y
u v u v
u v 

x x y y
u  u  u  u 
    
x  y  y  x 
0
Hence,
Also,
v  u
v  C0 (v is constant on C0 )
Hence
u  C0
y
C0
v
u
x
Steepest Descent Method (cont.)
Local behavior near SP:
1
2
g  z   g  z0   g   z0  z  z0 
2
1
2
so g  z   g  z0   g   z0  z  z0 
2
Let
 g   z0   R ei  R  g   z0  ,   arg g   z0 
 z  z0  rei
1
i   2 
2
g  z   g  z0    Rr  e
2
v
y
z
z0
SDP
C0
Compare variation of real and imag parts of g(z) - g(z0)
above with “landscape” figure near the SP!
x
Steepest Descent Method (cont.)
1
i   2 
2
Rr
e
 
2
0
 g  z   g  z0  
1
 Im  g  z   g  z0     v( z )  v( z0 )    Rr 2  sin   2 
2
 sin   2   0    2  0 or    multiples of 2
1 2
u ( z )  u ( z0 )   Rr
2
v( z)  v( z0 )  0
SAP:   2 SAP  0  2 n
  SAP  

2
,

2

SDP:   2 SDP    2 n
  SDP  

 SDP  
2

2



2

2
,

2


2

1 2
u ( z )  u ( z0 )   Rr
2
v( z)  v( z0 )  0
where   arg g  z0 
Steepest Descent Method (cont.)
Hence on the SDP,
 
1
1
j   2 SDP 
2
 g  z   g  z0    Rr  e
   Rr 2 
2
2
 z  z0  reiSDP
Steepest Descent Method (cont.)
y
u decreases
SP
90o
 SDP
SAP
u increases
SDP
x
Steepest Descent Method (cont.)
u ( x, y )
y
x
z0
SDP
SAP
Steepest Descent Method (cont.)
Summary:
I      f  z  e  g  z  dz

C
 Saddle point (SP) occurs at g ( z0 )  0
 Steepest descent path (SDP) is defined by v( z )  v( z0 )
where v( z )  Im g ( z ) and passes through the SP at an angle
of
 SDP  

2


2
with respect to the x - axis where
  arg g   z0 
 Near SP, SDP is approximately z  z0  rei SDP ,  dz  ei SDP dr
 As  increases, contributions to the integral become
increasingly concentrated about the SP z  z0 :
I    ~ f  z0  e
g  z0 

SDP
e
1

  R r 2 
2

dz
Steepest Descent Method (cont.)
I    ~ f  z0  e
To evaluate
g  z0 

e
1

 Rr 2 
2

dz ,
SDP
we note that

e
1

 Rr 2 
2

dz ~ 2e
SDP
i SDP

e
1

 Rr 2 
2

dr  e
0
u  x, y 
y
x
r0
i SDP
2
R
Steepest Descent Method (cont.)
y
SDP
 SDP  
 SDP
x
SAP

2


2
Note: The direction of
integration determines
the sign.
The “user” must
determine this.
Steepest Descent Method (cont.)
Finally,
I    ~ f  z0  e
g  z0 
 I    ~ f  z0  e
Recall  SDP  

2


2
2 iSDP
e
,
R
 g  z0 
2
eiSDP
 g "  z0 
where   arg g   z0 
Example
Bessel function:
J0  

 /2
1
 
 /2
1

cos   cos z  dz
Re I   
where
I    
 /2
 / 2
e
 i cos z 
dz
Hence, we identify:
f z 1
g  z   i cos z
g '  z0   0
z0  0  n
Example (cont.)
y
z0  0


x

2
2
g  z   i cos z
g "  z0   i cos z0  i
  arg g "  z0   
 SDP  

2


2


4


2

2
  SDP

3
  or
4
4
Example (cont.)
Identify the SDP and SAP:
g  z   i cos  x  iy 
 i  cos x cosh y  i sin x sinh y 
u  x, y   sin x sinh y
v  x, y   cos x cosh y
SDP and SAP:
v  z   v  z0   constant
cos x cosh y  constant  cos  0  cosh  0 
1
Example (cont.)
SDP and SAP:
cos x cosh y  1
y
SAP

Examination of the u
function reveals which of
the two paths is the SDP.
u  x, y   sin x sinh y
C

2
SDP

2
x
Example (cont.)
The vertical paths are added
so that the original path has
limits at infinity.
y
Cv1
C
SDP = C + Cv1 + Cv2
It is now clear which choice is
correct for the departure angle:
 SDP  

4


SDP
2
SDP

2
Cv 2
x
Example (cont.)
I    ~ f  z0  e

 g  z0 

2
eiSDP
g "  z0 
(ignoring the contributions of the vertical paths)
Hence,
I    ~ 1 e
2
e

I  ~
so
i cos  0 


2  i 4
e
j


i   
4




i



2


4

J 0    ~ Re 
e




 

1
Example (cont.)
Hence
2


J0  ~
cos    

4

Example (cont.)
Examine the path Cv1 (the path Cv2 is similar).
y
Cv1


2

2
 / 2
Iv1   / 2i ei cos z dz
Let
z 

2
 iy
x
Example (cont.)
0
I v1  i


 i
 

i cos   jy 
 2
 dy
e


0

0


 i e  sinh y dy   i e  sinh y dy

0
 1  d  sinh y
e
dy


  cosh y  dy

Using repeated integration by parts:

 

Note : cos    iy   sin(iy)  i sinh y
 2



 1 
d  1   sinh y
 e sinh y
I v1   i 

i
dy

e

0
dy   cosh y 
  cosh y 
0

1
d  1  
1  d  sinh y
 i
i
e
dy




dy   cosh y    cosh y  dy
0
1
1 d  1  d  1   

 
 i
i 2




 dy  cosh y  dy  cosh y   0
1
1
1
I v1  i  
i
 O( 2 )









Example (cont.)
Hence,
I v1
1
 O 

If we want an asymptotic expansion that is accurate to order 1/ ,
then the vertical paths must be considered. Otherwise, we have
2


J0  ~
cos    

4
