Teorema de transformación de
funciones armónicas
Sea f una función analítica que transforma un dominio
D en un dominio D. Si U es armonica en D, entonces
la función real u(x, y) = U(f(z)) es armonica en D.
Proof
We will give a special proof for the special case in which D
is simply connected. If U has a harmonic conjugate V in D,
then H = U + iV is analytic in D, and so the composite
function H(f(z)) = U(f(z)) + iV(f(z)) is analytic in D. It follow
that the real part U(f(z)) is harmonic in D.
Solving Dirichlet Problems Using Conformal Mapping
1. Find a conformal mapping w = f(z) that transform s the
original region R onto the image R. The region R may be
a region for which many explicit solutions to Dirichlet
problems are known.
2. Transfer the boundary conditions from the R to the
boundary conditions of R. The value of u at a boundary
point  of R is assigned as the value of U at the
corresponding boundary point f().
3. Solve the Dirichlet problem in R. The solution may
be apparent from the simplicity of the problem in R
or may be found using Fourier or integral transform
methods.
4. The solution to the original Dirichlet problems is u(x,
y) = U(f(z)).
Ejemplo: The function U(u, v) = (1/)
Arg w is harmonic in the upper halfplane v > 0 since it is the imaginary
part of the analytic function g(w) =
(1/) Ln w. Use this function to solve
the Dirichlet problem en la figura
superior (a).
Solution
The analytic function f(z) = sin z
maps the original region to the upper
half-plane v  0 and maps the
boundary segments to the segments
shown in Fig 20.14(b). The harmonic
function U(u, v) = (1/) Arg w satisfies
the transferred boundary conditions
1
1  cos x sinh y 
U(u, 0) = 0 for u > 0 and U(u, 0) = 1 u ( x, y )  tan 


 sin x cosh y 
for u < 0.
Example 7
From C-1 in Appendix IV, the analytic
za
72 6
function f ( z ) 
, a
az  1
where r0  5  2 6
5
maps the region outside the two open disks
|z| < 1 and |z – 5/2| < ½ onto the circular
region r0  |w|  1,
. See Fig 20.15(a) and (b).
Fig 20.15
Example 7 (2)
In problem 10 of Exercise 14.1, we found
U (r ,  )  (log e r ) /(log e r0 ) z )
is the solution to the new problem. From
Theorem 20.2 we conclude that the solution
to the original problem is
1
z  (7  2 6 ) / 5
u ( x, y ) 
log e
log e (5  2 6 )
(7  2 6 ) z / 5  1
• A favorite image region R for a simply
connected region R is the upper half-plane
y  0. For any real number a, the complex
function
Ln(z – a) = loge|z – a| + i Arg (z – a)
is analytic in R and is a solution to the
Dirichlet problem shown in Fig 20.16.
Fig 20.16
• It follows that the solution in R to the
Dirichlet problem cwith
, a xb
 0
U ( x, 0)  
 0, otherwise
is the harmonic function
U(x, y) = (c0/)(Arg(z – b) – Arg(z – a))
Fig 20.35(a)
Example 6
Solve the Dirichlet problem in Fig 20.35(a)
using conformal mapping by constructing a
linear fractional transformation that maps the
given region into the upper half-plane.
Example 6 (2)
Solution
The boundary circles |z| = 1 and |z – ½| = ½
each pass through z = 1. We can map each
boundary circle to a line by selecting a linear
fractional transformation
that
a pole at z
z  i 11
z has
i
T
(
z
)


(
1

i
)
= 1. If we require
z  1  1T(i)
 i = 0 and
z  1T(-1) = 1, then
T (0)  1  i, T ( 12  12 i )  1  i
Since
, T maps
the interior of |z| = 1 onto the upper half-
Example 6 (3)
The harmonic function U(u, v) = v is the
solution to the simplified Dirichlet problem in
the w-plane, and so u(x, y) = U(T(z)) is the
solution to the original Dirichlet problem
in
z i
Since the imaginary part of T ( z )  (1  i )
is
the z-plane.
z 1
1  x2  y 2
1  x2  y 2
, the solution is u ( x, y ) 
2
2
( x  1)  y
( x  1) 2  y 2
Example 6 (4)
The level curves u(x,
2 y) = c can2 be written as
 x  c   y2   1 




 1 c 
1  c 
and are therefore circles that pass through z
= 1. See Fig 20.36.
Fig 20.36
20.4 Schwarz-Christoffel
Transformations
• Special Cases
First we examine the effect of f(z) = (z –
x1)/, 0 <  < 2, on the upper half-plane y
 0 shown in Fig 20.40(a).
• The mapping is the composite of  = z – x1
and w = /. Since w = / changes the
angle in a wedge by a factor of /, the
interior angle in the image is (/) = .
See Fig 20.40(b).
•
Note that f'(z)  A( z  x1 )( /  )1, A   / 
Next assume that f(z) is a function that is
(1 /  )1
2

f
(
z
)

A
(
z

x
)
(
z

x
)(

/ and
)  1that
analytic in the upper
half-plane
1
2
jas the derivative
(1)
where x1 < x2. We use the fact that a curve
(t ) is a line segment when the
warg= fw(t)
argument of1 its tangent vector
is
 2 w(t)


 Arg A    1 Arg (t  x1 )    1 Arg (t  x2 )
constant. From

 (1) we get  

• Since Arg(t – x) =  for t < x, we can find
the variation of f (t) along the x-axis. They
are
shown in theargfollowing
table.
Interval
f ' (t )
Change in arg.
(, x1 )
Arg A  (α1  π )  (α2  π )
0
( x1 , x2 )
Arg A  (α2  π )
π  α1
( x2 , )
Arg A
π  α2
See Fig 20.41.
Fig 20.41
THEOREM 20.4
Schwarz-Christoffel Formula
Let f(z) be a function that is analytic in the upper
half-plane y > 0 and that has the derivative
f ( z )
 A( z  x1 )
(1 /  ) 1
( z  x2 )
( 2 /  ) 1
 ( z  xn )
( n /  ) 1
(3)
where x1 < x2 < … < xn and each i satisfies
0 < i < 2. Then f(z) maps the upper half-plane y  0
to a polygonal region with interior angles 1, 2, …,
n.
Comments
(i) One can select the location of three of the
points xk on the x-axis.
(ii) A general form for f(z) is

f ( z )  A  ( z  x1 )
(1 /  ) 1
...( z  xn )
( n /  ) 1

dz  B
and can be considered as the compisite of
g ( z )   ( z  x1 )(1 /  )1...( z  xn )( n /  )1 dz
and w  Az  B.
(iii) If the polygonal region is bounded only
n – 1 of the n interior angles should be
Example 1
Use the Schwarz-Christoffel formula to
construct a conformal mapping from the
upper half-plane to the strip |v| 1, u  0.
See Fig 20.42.
Example 1 (2)
Solution
We may select x1 = −1, x2 = 1 on the x-axis,
and we will construct a mapping f with f(−1)
1 /
2 1 = 21=
/ 2 /2, (3) gives
= −i,ff(1)
=
i.
Since
' ( z )  A( z  1) ( z  1)
1
A
1
A 2

.
1/ 2
2 1/ 2
i (1  z )
( z  1)
Example 1 (3)
Thus f(z)   Ai sin 1 z  B.
Since f (1)  i, f (1)  i, then
 i  Ai

2
 B, i   Ai
then f ( z ) 
2


2
i sin 1 z.
 B imply B  0, A  2 / 
Example 2
• Use the Schwarz-Christoffel formula to
construct a conformal mapping from the
upper half-plane to the region shown in Fig
20.43(b).
Example 2 (2)
Solution
We may select x1 = −1, x2 = 1 on the x-axis,
and we will require f(−1) = ai, f(1) = 0. Since
1 = 3/2
f ' ( z )  A( z  1)1/ 2 ( z  1) 1/ 2 . If we write
2 = /2, (3) gives


z
1
f ' ( z )  A 2
 2
, then
1/ 2
1/ 2 
( z  1) 
 ( z  1)


f ( z )  A ( z 2  1)1/ 2  cosh 1 z  B
Example 2 (3)
Notice that cosh 1 (1)  i and cosh 1 1  0,
and so ai  f (1)  A(i )  B,0  f (1)  B.
Therefore f ( z ) 

(z

a
2
 1)1/ 2  cosh 1 z

Example 3
• Use the Schwarz-Christoffel formula to
construct a conformal mapping from the
upper half-plane to the region shown in Fig
20.44(b).
Example 3 (2)
Solution
Since the region is bounded, only two of the
60
f ' ( z )interior
 Az 2 / 3 (angles
z  1) 2 / 3should be included. If x1 =
0,
1, Theorem
we obtain
Wexcan
18.8 to get the
2 =use
antideriva tive
1
f ( z )  A 2 / 3
ds  B.
2/3
0s
( s  1)
z
Example 3 (3)
We require f (0)  0, f (1)  1, then B  0 and
1
1  A 2 / 3
dx  (1 / 3)
2/3
0x
( x  1)
z
1
1
Thus f ( z ) 
ds

2
/
3
2
/
3
(1 / 3) 0 s ( s  1)
1
Example 4
Use the Schwarz-Christoffel formula to
construct a conformal mapping from the
upper half-plane to the upper half-plane with
the horizontal line v = , u  0, deleted. See
Fig 20.45.
Example 4 (2)
Solution
The nonpolygonal target region can be
approximated by a polygonal region by
adjoining a line segment from w = i to a pint
)1
u0 on the fnegative
20.45(b).
' ( z )  A( z u-axis.
1)(1 /  )1See
z (2 / Fig
If we require f(−1) = i, f(0) = u0, then
Note that as u0 approaches −, 1 and 2
approach 2 and 0, respectively.
Example 4 (3)
This suggests we examine the mappings
that satisfy w = A(z + 1)1z-1 = A(1 + 1/z) or
w = A(z + Ln z) + B.
First we determine the image of the upper
half-plane under g(z) = z + Ln z and then
translate the image region if needed. For t
real
g(t) = t + loge |t| + i Arg t
If t < 0, Arg t =  and u(t) = t + loge |t| varies
from − to −1. It follows that w = g(t) moves
Example 4 (4)
When t > 0, Arg t = 0 and u(t) varies from −
to . Therefore g maps the positive x-axis
onto the u-axis. We can conclude that g(z) =
z + Ln z maps the upper half-plane onto the
upper half-plane with the horizontal line v =
, u  −1, deleted. Therefore w = z + Ln z +
1 maps the upper half-plane onto the
original target region.
20.5 Poisson Integral Formulas
• Introduction
It would be helpful if a general solution
could be found for Dirichlet problem in
either the upper half-plane y  0 or the unit
disk |z| = 1. The Poisson formula fro the
upper half-plane provides such a solution
expressing the value of a harmonic
function u(x, y) at a point in the interior of
the upper half-plane in terms of its values
on the boundary y = 0.
Formula for the Upper HalfPlane
• Assume that the boundary function is
given by u(x, 0) = f(x), where f(x) is the
step function indicated in Fig 20.55.
• The solution of the corresponding Dirichlet
problem in the
upper
half-plane
is
u
u ( x, y ) 
i

[Arg ( z  b)  Arg ( z  a)]
(1)
Since Arg(z – b) is an exterior angle
formed by z, a and b, Arg(z – b) = (z) +
ui
ui
z b

Arg(z –u (a),
< , and we can
x, ywhere
)   (0
z) <
  Arg


 z  a
write
• The superposition principle can be used to
solve the more general Dirichlet problem
in Fig 20.56.
• If u(x, 0) = ui for xi-1  x  xi, and u(x, 0) =
0 outside nthe interval [a, b], then from (1)
u ( x, y )  
i 1

1
ui

[Arg ( z  xi )  Arg ( z  xi 1 )]
n
uii ( z )


i 1
(3)
Note that Arg(z – t) = tan-1(y/(x – t)), where
tan-1 is selected between 0 and , and
From (3),
1
n
dt
u ( x, y )    ui Arg ( z  t )dt
 i 1 xi 1 d
1
n
xi
ui y
 
dt
2
2
 i 1 xi 1 ( x  t )  y
Since u ( x, 0)  0 outside the interval [a,b],
xi
y

u (t ,0)
we have u ( x, y )  
dt
2
2
  ( x  t )  y
(4)
THEOREM 20.5
Poisson Integral Formula for the
Upper Half-Plane
Let u(x, 0) be a piecewise-continuous function on every
finite interval and bounded on － < x < . Then the
function defined by
y 
u (t , 0)
u ( x, y )  
dt
2
2
  ( x  t )  y
is the solution of the corresponding Dirichlet problem
on the upper half-plane y  0.
Example 1
Find the solution of the Dirichlet problem in
the upper half-plane that satisfies the
boundary condition u(x, 0) = x, where |x| < 1,
and u(x, 0) = 0 otherwise.
t
Solution u ( x, y )  y 1
dt.

2
2

1
 ( x formula
t)  y
By the Poisson integral
Example 1 (2)
Using s  x  t , then
1 y
2
2
1  x  t   t  1
u ( x, y )   log e (( x  t )  y )  x tan 

 2
 y  t  1
 ( x  1) 2  y 2  x  1  x  1 
y
1  x  1  

log e 
  tan 
  tan 

2
2
2
 y 
 y 
 ( x  1)  y   
Example 2
The conformal mapping f(z) = z + 1/z maps
the region in the upper half-plane and
outside the circle |z| = 1 onto the upper halfplane v  0. Use the mapping and the
Poisson integral formula to solve the
Dirichlet problem shown in Fig 20.57(a).
Fig 20.57
Example 2 (2)
Solution
Using the result of Example 4 of Sec 20.2,
we can transfer the boundary conditions to
the w-plane. See Fig 20.57(b). Since U(u, 0)
is a step function, we will use the integrated
1rather than 1the Poisson integral.
solution
(3)
U (u , v)  Arg ( w  2)  [  Arg ( w  2)]
 to the newDirichlet problem is
The solution
w2
 1  Arg (
)

w2
1
Example 2 (3)
and therefore
1
1
z 1 z  2
u ( x, y )  U ( z  )  1  Arg (
)
z

z 1 z  2
z  1

u ( x, y )  1  Arg 


 z  1
1
2
THEOREM 20.6
Poisson Integral Formula for the
Unit Disk
Let u(ei) be a bounded and piecewise continuous for
－    . Then the solution to the corresponding
Dirichlet Problem on the open units disk |z| < 1 is given
by
2
1 
1

|
z
|
u ( x, y )   u (eit ) it
dt
2
2 
|e z|
(5)
Geometric Interpretation
• Fig 20.58 shows a thin membrane (as a
soap film) that has been stretched across
a frame defined by u = u(ei).
Fig 20.58
• The displacement u in the direction
perpendicular to the z-plane satisfies the
2
2
2
two-dimensional
wave
equation



u

u

u
2
a  2  2   2
 x y  t
and so at equilibrium, the displacement
function u = u(x, y) is harmonic. Formula
(5) provides an explicit solution for u and
has the advantage that the integral is over
Example 3
A frame for a membrane is defined by u(ei)
= | | for −    . Estimate the equilibrium
displacement of the membrane at (−0.5, 0),
(0, 0) and (0.5, 0).
2
1  1 z
Solutionu ( x, y )   t
dt
2


it
2
e
z
Using (5),
Example 3 (2)
• When (x, y) = (0, 0), we get
1 

u ( x, y ) 
t dt 

2 
2
For the other two values of (x, y), the
integral is not elementary and must be
estimated using a numerical solver. We
have u(−0.5, 0) = 2.2269, u(0.5, 0) =
0.9147.
Fourier Series Form
• Note that un(r,) = rn cos n and vn(r,) =
rn sin n are each harmonic, since these
functions are the real and imaginary parts
of zn. If a0, an, bn are chosen to be the

a
n u(ei) for −
n  <  < ,
0
Fourier
coefficients
of
u (r ,  )    an r cos n  bn r sin n  (6)
2 n1
then
We find (6) is harmonic and
a0 
u (1, )    an cos n  bn sin n   u (ei )
2 n1
From (5) we have
1
u ( r , ) 
2


u (eit )
1 r2
e  re
it
i 2
dt
a0 
   an r n cos n  bn r n sin n
2 n1


Example 4
Find the solution of the Dirichlet problem in
the unit disk satisfying the boundary
condition u(ei) = sin 4 . Sketch the level
curve u = 0.
Solution
Rather than using (5), we use (6) which
reduces to u(r, ) = r4 sin 4 . Therefore u =
0 if and only if
sin 4 = 0. This implies u = 0 on the lines x =
0, y = 0 and y = x.
Example 4 (2)
If we switch to rectangular coordinates, u(x,
y) = 4xy(x2 – y2). The surface of u(x, y) =
4xy(x2 – y2), the frame u(ei) = sin 4, and
the system of level curves were sketched in
Fig 20.59.
Fig 20.59
20.6 Applications
• Vector Fields
A vector field F(x, y) = P(x, y)i + Q(x, y)j in
a domain D can also be expressed in the
complex form
F(x, y) = P(x, y) + iQ(x, y)
Pthatdiv
Q F = P/
Px+
QQ/y and curl F
Recall

and

xx −P/
y y)k. Ifywe require
x
= (Q/
both of
them are zeros, then
THEOREM 20.7
Vector Fields and Analyticity
(i) Suppose that F(x, y) = P(x, y) + Q(x, y) is a vector
field in a domain D and P(x, y) and Q(x, y) are
continuous and have continuous first partial
derivatives in D. If div F = 0 and curl F = 0, then
complex function
g ( z )  P( x, y )  iQ ( x, y )
is analytic in D.
(ii) Conversely, if g(z) is analytic in D, then F(x, y) =
g (z ) defined a vector field in D for which div F = 0
and curl F = 0.
THEOREM 20.7
Proof
If u(x, y) and v(x, y) denote the real and
u parts
 (v)ofug(z),
 (then
v) u = P and v =
imaginary

, 
; that is,
x
y y
x
−Q. Then
u v u
v
 , 
x y y
x
Equations in (2) are the Cauchy-Riemann
(2)
Example 1
The vector field F(x, y) = (−kq/|z − z0|2)(z −
z0) may be interpreted as the electric field by
a wire that is perpendicular to the z-plane at
z = z0 and carries a charge of q coulombs
length.
kq
 kq
per
unit
The
corresponding
complex
g ( z) 
( z  z0 ) 
2
z  z0
function
z isz0
Since g ( z ) is analytic for z  z0 , div F  0, curl F  0.
Example 2
The complex function g(z) = Az, A > 0, is
analytic in the first quadrant and therefore
V ( x, y )  g ( z )  Ax  iAy
gives rise to the vector field
which satisfies div V  0, curl V  0.
Potential Functions
• Suppose that F(x, y) is a vector field in a
simply connected domain D with div F = 0
and curl F = 0. By Theorem 18.8, the
analytic function g(z) = P(x, y) − iQ(x, y)
G ( z )   ( x , y )  i ( x , y )
has an antiderivative
(4)
in D, which is called a complex potential
for the vector filed F.


Note that g ( z )  G ' ( z )  ( x, y )  i
( x, y )
•
x
x


 ( x, y )  i ( x, y )
x
y


and so
 P,  Q
(5)
x
y
Therefore F =  , and the harmonic
function  is called a (real) potential
function of F.
Example 3
• The potential  in the half-plane x  0
satisfies the boundary conditions (0, y) =
0 and (x, 0) = 1 for x  1. See Fig.
20.68(a). Determine a complex potential,
the equipotential lines, and the field F.
Fig. 20.68
Example 3 (2)
Solution
We knew the analytic function z = sin w
maps the strip 0  u  /2 in the w-plane to
the region R in question. Therefore f(z) = sin1z maps R onto the strip, and Fig 20.68(b)
shows the transferred boundary conditions.
The simplified Dirichlet problem has the
solution U(u, v) = (2/)u, and so (x, y) =
U(sin-1z) = Re((2/) sin-1z) is the potential
function on D, and G(z) = (2/)u sin-1z is a
Example 3 (3)
Note that the equipotential lines  = c are
the images of the equipotential lines U = c in
the w-plane under the inverse mapping z =
sin w. We found2that the
vertical
lines
u
=
a
2
y of the hyperbola
is mapped ontoxa branch

1
sin 2 a
cos 2 a
Example 3 (4)
Since the equipotential lines U = c, 0 < c < 1
is the vertical line u = /2c, it follows that the
2
equipotential lines
right branch of
x 2  = c isythe

1
the hyperbola
2
2
sin (c / 2) cos (c / 2)
Since F  G' ( z ) and (d / dz) sin 1 z  1 /(1  z 2 )1/ 2 ,
2
1
2
1
then F 

2 1/ 2
 (1  z )
 (1  z 2 )1/ 2
Steady-State Fluid Flow
• The vector V(x, y) = P(x, y) + iQ(x, y) may
also be expressed as the velocity vector of
a two-dimensional steady-state fluid flow
at a point (x, y) in a domain D.
If div V = 0 and curl V = 0, V has a
complex velocity potential
G' (z )  V
G(z) = (x, y) + (x, y)
that satisfies
• Here special importance is placed on the
level curves (x, y) = c. If z(t) = x(t) + iy(t)
dx of a particle,
dy then
is the path
 P ( x, y ),  Q ( x, y )
(6)
Hence
dt
dt
dy / dx  Q ( x, y ) / P ( x, y ) or
 Q( x,y )dx  P ( x, y )dy  0.
 (Q) P
Since div V  0 implies

y
x
•
and by the Cauchy - Riemann equations


 
   Q and

P
x
y
y x
and all solutions of (6) satisfy ψ ( x, y )  c.
The function (x, y) is called a stream
function and the level curves (x, y) = c
are streamlines for the flow.
Example 4
• The uniform flow in the upper half-plane is
defined by V(x, y) = A(1, 0), where A is a
fixed positive constant. Note that |V| = A,
and so a particle in the fluid moves at a
constant speed. A complex potential for
the vector field is G(z) = Az = Ax + iAy,
and so the streamlines are the horizontal
lines Ay = c. See Fig 20.69(a). Note that
the boundary y = 0 of the region is itself
streamline.
Fig 20.69(a)
Example 5
2 gives rise
• The analytic
function
G(z)
=
z
V( x, y)  G' ( z )  (2 x,2 y)
to the vector field
in
the first quadrant.
Since z2 = x2 − y2 + i(2xy), the stream
function is (x, y) = 2xy and the
streamlines are the hyperbolas 2xy = c.
See Fig 20.69(b).
Fig 20.69(b)
THEOREM 20.8
Streamline
Suppose that G(x) = (x, y) + i(x, y) is analytic in a
region R and (x, y) is continuous on the boundary of
R. Then V(x, y) = G(z ) defined an irrotational and
incompressible fluid flow in R. Moreover, if a particle
is placed is placed inside R, its path z = z(t) remains in
R.
Example 6
The analytic function G(z) = z + 1/z maps
the region R in the upper half-plane and
outside the circle |z| = 1onto the upper halfplane v  0. The boundary of R is mapped
onto the u-axis, and so v = (x, y) = y – y/(x2
+ y2) is zero on
G' ( zthe
)  1boundary
 1 / z 2 , andof
so R. Fig 20.70
shows the streamlines. The
velocity field is
1
i
2 i
G' (re )  1  2 e
given by
r
Example 6 (2)
It follows that V  (1, 0) for large values of r,
and so the flow is approximately uniform at
large distance from the circle |z| = 1. The
resulting flow in R is called flow around a
cylinder.
Fig 20.70
Example 7
The analytic function f(w) = w + Ln w + 1
maps the upper half-plane v  0 to the upper
half-plane y  0, with the horizontal line y = ,
x  0, deleted. See Example 4 in Sec 20.4. If
G(z) = f -1(z) = (x, y) + i(x, y), then G(z)
maps R onto the upper half-plane and maps
the boundary of R onto the u-axis. Therefore
(x, y) = 0 on the boundary of R.
Example 7 (2)
It is not easy to find an explicit formula for
(x, y). The streamlines are the images of
the horizontal lines v = c under z = f(w). If
we write
0,t then
z  f (tw =
ic )t +t ic,
 icc >
Ln(
ic )  1the
, that is,
streamlines 1can be2 2
x  t  1  log e (t  c ), y  c  Arg (t  ic )
2
Fig 20.71
Example 8
The analytic function f(w) = w + ew + 1 maps
the strip 0  v   onto the region R shown
in Fig 20.71. Therefore G(z) = f -1(z) = (x, y)
+ i(x, y) maps R back to the strip and from
M-1 in the Appendix IV, maps the boundary
line y = 0 onto the u-axis and maps the
boundary line y =  onto the horizontal line v
= . Therefore (x, y) is constant on the
boundary of R.
Example 8 (2)
The streamlines are the images of the
horizontal lines v = c, 0 < c < , under z =
f(w). If we write w = t + (ic,
t ic )then the
z  f (t  ic )  t  ic  e
 1, that is,
streamlines can be
x  t  1  et cos c, y  c  et sin c
See Fig 20.72.
Fig 20.72