Solutions to Problem Session 5

Solutions to Problem Session 5
Preben Alsholm
October 4, 2007
1 Problem session
Solution 1 (A) Consider the real functions u j : R 2 ! R; j = 1; 2; given by
u 1 (x; y) = 2x(1 y) and u 2 (x; y) = x 3 3xy 2
1. We show that these functions are harmonic by applying the de…nition:
@ 2 u 1
@x 2 + @2 u 1
@y 2 = @2
@2
(2x(1 y)) + (2x(1 y)) = 0 + 0 = 0
@x2 @y2 @ 2 u 2
@x 2 + @2 u 2
@y 2 = @2
@x 2 x3 3xy 2 + @2
@y 2 x3 3xy 2 = 6x 6x = 0
2. We determine for each of the functions u j all harmonic conjugates v j so that f j = u j + iv j
is an analytic function. We use the Cauchy-Riemann equations:
For u = u 1 = 2x(1
y) we …nd
v y = u x
v x = u y
v y = u x = 2
from which follows v = 2y y 2 + h (x). This implies v x = h 0 (x). However, we also have
v x = u y = 2x, so h 0 (x) = 2x and consequently h (x) = x 2 + C. Therefore the harmonic
conjugates of u 1 are given by
2y
v 1 (x; y) = x 2
y 2 + 2y + C
where C 2 R. We have
f 1 (z) = f 1 (x + iy) = u 1 (x; y) + iv 1 (x; y) = 2x(1 y) + i x 2 y 2 + 2y + C
= i (x + iy) 2 + 2 (x + iy) + iC = iz 2 + 2z + iC
For u = u 2 = x 3
3xy 2 we …nd
v y = u x = 3x 2 3y 2
from which follows v = 3x 2 y y 3 + h (x). This implies v x = 6xy + h 0 (x). However, we also
have v x = u y = 6xy, so h 0 (x) = 0 and consequently h (x) = C. Therefore the harmonic
conjugates of u 2 are given by
v 2 (x; y) = 3x 2 y
y 3 + C
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where C 2 R. We have
f 2 (z) = f 2 (x + iy) = u 2 (x; y) + iv 2 (x; y) = x 3 3xy 2 + i 3x 2 y y 3 + C
= (x + iy) 3 + iC = z 3 + iC
Solution 2 (B) Consider any branch of the logarithm of the form
L (z) = ln jzj + i arg (z);
where arg (z) 2]; + 2]:
v(z) = arg (z) is a harmonic function in the slit plane D = C n fre i j r 0g; because L (z)
is analytic in D . It is not harmonic in C n f0g for the simple reason that it is not continuous
there.
Let D be a domain given as the union of two domains D = D 1 [ D 2 . If u : D ! R is harmonic
in D 1 and in D 2 , then u is harmonic in D, since to be harmonic in some set means to satisfy
Laplace’s di¤erential equation u = 0 in each point of the set. This simple fact implies that
u(z) = ln jzj is a harmonic function in C n f0g it is the real part of both Log (z) and L 0 (z).
Solution 3 (C) We shall …nd a function u which is harmonic in the region sketched in …g.
3.12. We use
(x; y) = A arg (z 1 i) + B
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since arg (z 1 i) is the imaginary part of L (z 1 i) which is analytic in the region.
4 4
Imposing the boundary condtions gives
10 = (x; 1) = A 0 + B; x > 1
3
0 = (1; y) = A + B; y < 1
2
Thus B = 10 and A = 20
3 so (x; y) = 20
3 arg (z 1 i) + 10
4
The value at (0; 0) is (0; 0) = 20
3 arg 20 5
( 1 i) + 10 =
4 3 4 + 10 = 5 3 .
Solution 4 (D)
1. Consider the …rst Example on p. A-22:
w = f(z) = z 2
and the domain, the half strip,
D = f z = x + iy j 0 < x < 1 ^ y > 0 g
The domain D is bounded by the three curves
1 : z = iy; y 0 ; 2 : z = x; 0 x 1 ; 3 : z = 1 + iy; y 0
We …nd
f (iy) = y 2 so f ( 1 ) = R [ f0g
f (x) = x 2 so f ( 2 ) = [0; 1]
f (1 + iy) = (1 + iy) 2 = 1 y 2 + 2iy
2

so f ( 3 ) is the upper branch of the parabola u = 1
1
4 v2 in the uv-plane.
That f is conformal in D follows from f 0 (z) = 2z 6= 0 in D. That f maps D injectively,
hence bijectively onto E = f(D) follows from
z 2 1 = z 2 2 , z 1 = z 2
and from the fact that not both of z 2 can belong to D.
The inverse function f 1 : E ! D is a square root, and it is the principal branch of the
square root, since D belongs to the right half plane.
2. Consider the last Example on p. A-22:
and the domain, the strip,
w = f(z) = e z
D = f z j 0 < Im z < g
f is conformal in D since the derivative is clearly di¤erent from zero. f maps D injectively,
hence bijectively onto the upper half plane E = f(D), since it is injective in the larger strip
f z j < Im z g. The inverse function f 1 : E ! D is the well-known function
Log.
Solution 5 (E) We consider the sine function in the …rst Example on p. A-24.
1. The vertical half-lines which bound the half strip S + given as
S + = f z = x + iy j jxj < 2
are given by z = 2
+ iy; y > 0. We …nd
sin
2 + iy = 1 2i
exp
i 2
y
^ y > 0 g
exp
i 2 + y
= 1 2i
ie y + ie y = cosh y
and also
sin
2 + iy
= 1 2i
exp
i 2
y
exp
i
2 + y
= 1 2i
ie y ie y = cosh y
Thus the vertical half-line to the right is mapped onto ]1; 1[ and the vertical half-line to
the left is mapped onto ] 1; 1[.
The sine function is conformal in S + since its derivative cos is di¤erent from zero in S + .
We have
sin (x + iy) = sin x cosh y + i cos x sinh y
Since the imaginary part is positive sin(S + ) is at least a part of the upper half-plane.
Keeping y > 0 …xed and letting x 2
2 ;
2 vary we see that u + iv = sin (x + iy) takes on
all values on the upper half of the ellipse
u 2
cosh 2 y +
v2
sinh 2 y = 1
When y increases from 0 and up the ellipse expands from being just the real line segment
] 1; 1[ to being larger and larger having semi-axes cosh y and sinh y. Thus we see that
sin(S + ) is the upper half-plane.
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2. By using that exp z = exp (z) we get
sin z = 1 2i (eiz e iz ) = 1 2i
e iz
e iz = sin z
3. By using sin z = sin z we see that the lower strip S = f z = x + iy j jxj < 2 ^ y < 0 g is
mapped bijectively onto the lower half-plane.
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