APPM 4360/5360
Homework Assignment #3 Solutions
Solution: If u is harmonic, then ∇2 u = u xx + u y y = 0.
Since
u x = 3x 2 + a y 2 ,
u y = 2ax y,
Problem #1 (16 points): Check the Caucy–Riemann
(C-R) conditions for the following. If they satisfy the
C-R conditions, find f ′ (z).
(a)
(b)
(c)
(d)
Spring 2014
u xx = 6x,
f (x, y) = x − i y + 1.
f (x, y) = y 3 − 3x 2 y + i (x 3 − 3x y 2 + 2).
f (x, y) = e y (cos x + i sin y).
f (r, θ) = log r + i θ.
u y y = 2ax,
we must chose a = −3. Then
u x = 3(x 2 − y 2 ) = v y
=⇒
u y = −6x y = −v x
Solution:
v = 3x 2 y + g (y),
v = 3x 2 y − y 3 + const.
u y = 0 = −v x ,
Problem #3 (16 points): Given the real part, u(x, y), of
the analytic function, f (z) = u(x, y) + i v(x, y), find the
imaginary part, v(x, y).
f ′ (z) does not exist.
(b) f (x, y) = y 3 − 3x 2 y + i (x 3 − 3x y 2 + 2) = u + i v, so
u = y 3 − 3x 2 y, v = x 3 − 3x y 2 + 2. Since
(a)
(b)
(c)
(d)
u y = 3y 2 − 3x 2 = −v x ,
u x = −6x y = v y ,
=⇒
and so
(a) f (x, y) = x − i y + 1 = u + i v, so u = x + 1, v = −y.
Since
u x = 1 ̸= v y = −1,
v = 3x 2 y − y 3 + f (x),
u(x, y) = 3x 2 y − y 3 .
u(x, y) = 2x(c − y), where c is a constant.
y
u(x, y) = x 2 +y 2 .
u(x, y) = cos x cosh y.
we have that
′
2
f (z) = u x + i v x = 3i (x + i y) = 3i z
Solution: For all of these, we use the C-R condition to
find the harmonic conjugate:
2
(a) u x = 6x y = v y =⇒ v = 3x y 2 + f (x) and
u y = 3x 2 − 3y 2 = −v x =⇒ v = −x 3 + 3x y 2 + g (y),
so
v = 3x y 2 − x 3 + const.
(as f (x, y) = f (z) = i z 3 + 2i ).
(c) f (x, y) = e y (cos x + i sin y) = u + i v, so
u = e y cos x, v = e y sin y. Since
(b) u x = 2(c − y) = v y =⇒ v = 2c y − y 2 + f (x) and
u y = −2x = −v x =⇒ v = −x 2 + g (y), so
u x = −e y sin x ̸= v y = e y sin y + e y cos y,
u y = e y cos x ̸= −v x = 0,
v = −x 2 + 2c y − y 2 + const.
f ′ (z) does not exist.
(d) f (r, θ) = log r + i θ = u + i v, u = log r , v = θ. Since
1 vθ
ur = =
,
r
r
(c)
ux = −
uθ
vr = 0 = − ,
r
2x y
(x 2 + y 2 )2
= v y =⇒ v =
x
x2 + y 2
+ f (x)
and
we have that
f ′ (z) = e −i θ (u r + i v r ) =
uy =
e −i θ 1
= .
r
z
x2 − y 2
x
= −v x =⇒ v = 2
+ g (y),
2
2
(x + y )
x + y2
so
v=
Problem #2 (6 points): Choose the constant a so that
the function
3
u(x, y) = x + ax y
x
x2 + y 2
+ const.
(d) u x = − sin x cosh y = v y =⇒ v =
− sin x sinh y + f (x) and u y = cos x sinh y =
−v x =⇒ v = − sin x sinh y + g (y), so
2
is harmonic. Find a conjugate function v in that
case.
v = − sin x sinh y + const.
1
Solution:
Problem #4 (24 points): Determine where, if
anywhere, the following functions are analytic.
Discuss if they have any singular points or if they’re
entire.
(a)
(b)
(c)
(d)
(e)
(f)
Ω(z) = −
k
k(x − i y)
= ϕ+iψ = −
,
2πz
2π(x 2 + y 2 )
so the velocity potential is
ϕ(x, y) = −
tan z.
exp(sin z).
exp{1/(z − 1)}.
exp(z).
z
.
z 4 +1
cos x cosh y − i sin x sinh y.
kx
,
2π(x 2 + y 2 )
the stream function is
ψ(x, y) =
ky
,
2π(x 2 + y 2 )
and the velocity field is (v 1 , v 2 ) where
Solution:
v 1 = ϕx = ψ y =
(a) Since
tan z =
(b)
(c)
(d)
(e)
sin z
cos z
k(x 2 − y 2 )
2π(x 2 + y 2 )2
and
v 2 = ϕ y = −ψx =
kx y
.
π(x 2 + y 2 )2
is a ratio of entire functions, so is it except for
the points where cos z = 0 (that is, where
A plot of the streamlines is below
z = π2 + πk, k ∈ Z).
exp(sin z) is entire because it’s the composition
of entire functions.
exp{1/(z − 1)} is analytic everywhere except z = 1
because it’s the composition of analytic
functions.
exp(z) is not analytic anywhere, since z isn’t.
z
is the ratio of two polynomials and
z 4 +1
polynomials are entire functions. So it is analytic
everywhere except for the roots of z 4 + 1 = 0,
z 4 = −1 = e i π =⇒ z = e i (π/4+nπ/2) ,
or
n = 0, 1, 2, 3
p
z =±
2
(1 ± i ).
2
(f) cos x cosh y − i sin x sinh y =
cos x cos i y − sin x sin i y = cos(x + i y) = cos z,
which is an entire function.
Problem #6 (20 points): (Flow at a corner) Consider
the complex potential
Problem #5 (18 points): Consider the following
complex potential
Ω(z) = −
k
,
2πz
Ω(z) = z c ,
where c > 1/2 is a constant. Writing z = r e i θ , show that
the rays θ = 0 and θ = π/c are streamlines, and so can
be thought of barriers. Calculate the corresponding
velocity potential, stream function, and velocity field.
Sketch or use a computer to graph the streamlines in
the sector 0 ≤ θ ≤ π/c for c = 4, c = 1, and
c = 2/3.
k ∈ R,
which is referred to as a doublet. Calculate the
corresponding velocity potential, stream function, and
velocity field. Sketch or use a computer to graph the
streamlines.
2
For c = 2/3:
Solution: Since
Ω(z) = z c = r c e i cθ = r c cos(cθ) + i r c sin(cθ),
the velocity potential and the stream function are,
respectively,
ϕ = r c cos(cθ),
ψ = r c sin(cθ).
For the velocity field, (v 1 , v 2 ), we use that
Ω′ (z) = v 1 + i v 2 and
Ω′ (z) = c z c−1 = cr c−1 {cos[(c − 1)θ] + i sin[(c − 1)θ]} ;
so
v 1 = cr c−1 cos[(c − 1)θ]
and
v 2 = −cr c−1 sin[(c − 1)θ].
Streamlines are the lines where ψ is constant. Since
ψ = 0 for θ = 0 and θ = π/c, they are both streamlines
of the flow. Plots of streamlines are below: For c = 1:
Extra-Credit Problem #7 (4 points): Discuss the flow
represented by Ω(z) = log z. Setting z = r e i θ
gives
Ω(z) = log r + i θ + 2πi k,
k ∈Z
(we’ll discuss the properties of log z, z ∈ C, next week).
Calculate the corresponding velocity potential, stream
function, and velocity field. Sketch or use a computer
to graph the stream function. Discuss the
results.
For c = 4:
Solution: The velocity potential is ϕ = log r and the
stream function is ψ = θ + 2πk, k ∈ Z. We again use
that
Ω′ (z) = v 1 − i v 2 =
1
x −i y
= 2
z x + y2
to get the velocity field (v 1 , v 2 ) where
v1 =
x
x2 + y 2
and v 2 =
y
.
x2 + y 2
So despite the multivaluedness of Ω and ψ, the flow is
well defined and unambiguous.
A plot of the streamlines follow:
3
Thus, the velocity potential is
ϕ = c log(r 1 r 2 ) =
{
}
c
log [(x − a)2 + y 2 ][(x + a)2 + y 2 ]
2
and the stream function is
ψ = c(θ1 + θ2 ) + 2πck
(
= c arctan
y
y )
+ arctan
+ 2πck,
x −a
x +a
where k ∈ Z. The velocity field, (v 1 , v 2 ), is given
by
c(x + a)
c(x − a)
+
v1 =
2
2
(x − a) + y
(x + a)2 + y 2
and
v2 =
Extra-Credit Problem #8 (6 points): (Flow at a wall)
Consider the complex potential
Ω(z) = c log(z − a) + c log(z + a)
= c log(z 2 − a 2 ) + 2kπi c,
k ∈ Z,
where a > 0 and c real. Calculate the corresponding
velocity potential, stream function, and velocity field.
Sketch or use a computer to graph the stream
function. Discuss the results.
Solution: If we let
z − a = r 1 e i θ1 ,
z + a = r 2 e i θ2 ,
then
Ω(z) = c(log r 1 + log r 2 ) + i c(θ1 + θ2 ) + 2πi ck,
k ∈ Z.
4
cy
cy
+
.
(x − a)2 + y 2 (x + a)2 + y 2