Please read the instructions below before you start the exam.
April/May 2020 MA3PD2 2019/0 A 800
UNIVERSITY OF READING
PARTIAL DIFFERENTIAL EQUATIONS II (MA3PD2)
Two hours
Answer ALL questions in section A and at least ONE question from section
B. (If more than one question from section B is attempted then marks from the
BETTER section B question will be used. If the exam mark calculated in this
way is less than 40%, then marks from the other section B question which has
been attempted will be added to the exam mark until 40% is reached).
vc2
Page 2
SECTION A
1. Let (r, ✓) denote polar coordinates, and suppose that u(r, ✓) satisfies the
Neumann Laplace problem⇢
urr + r1ur + r2u✓✓ = 0 in D = {0 < r < 1, 0  ✓ < 2⇡},
@u/@r = F (✓) on C = {r = 1, 0  ✓ < 2⇡},
in which the boundary forcing function F (✓) satisfies
R 2⇡
0 F (✓) d✓ = 0.
The solution u must be bounded in D and 2⇡-periodic in ✓.
(a) Use separation of variables to show that the most general solution
u(r, ✓) which is bounded and satisfies Laplace’s equation in D, and is
2⇡-periodic in ✓, can be written as
u(r, ✓) =
1
2
a0 +
1X
n=1
rn[an cos(n✓) + bn sin(n✓)]
for constants an, bn.
[27 marks]
(b) Now apply the boundary condition on C, determine the constants an,
and hence show that the solution can be written as
u(r, ✓) =
1
2
a0 +
1
⇡
Z 2⇡
0
F (✓0)
1X
n=1
rn
n
cos[n(✓ ✓0)] d✓0. (1)
Explain why the constant a0 is indeterminate.
[10 marks]
(c) Use the identity
P1
n=1 z
n/n = ln(1 z) (which is valid for |z| < 1)
to explicitly evaluate the summation in (1), and hence show that
u(r, ✓) =
1
2
a0 1
2⇡
Z 2⇡
0
F (✓0) ln[1 2r cos(✓ ✓0) + r2] d✓0,
for r < 1. [10 marks]
(d) Write down the solution u(r, ✓) in the case F (✓) = cos(4✓) + sin ✓.
[13 marks]
MA3PD2 2019/0 A 800 vc2
Page 3
SECTION B
2. (a) Let D = {0 < x < 1, 0 < y < ⇡}, and consider the Dirichlet Green’s
function problem⇢ r2GD = (x x0)(y y0) in D, with (x0, y0) 2 D,
GD = 0 on C, the boundary of D,
By expanding GD as a Fourier sine series in y, derive the
representation
GD(x, y|x0, y0) =
1X
n=1
sin(ny) sin(ny0)
n⇡ sinhn
(cosh[n(1 x x0)] cosh[n(1 |x x0|)])
for distinct points (x, y), (x0, y0) in D.
[27 marks]
[The identity cosh(A+B) cosh(A B) = 2 sinhA sinhB may
prove useful. You may also use the symmetry property
GD(x, y|x0, y0) = GD(x0, y0|x, y) without proof, if required..]
(b) Now consider the general Neumann Helmholtz Green’s function
problem⇢ r2GN + cGN = (x x0)(y y0) in D, with (x0, y0) 2 D,
@GN/@n = 0 on C,
(2)
in which C is the boundary of D, D is a bounded domain, and c < 0.
Show that the solution to (2) satisfies the symmetry property
GN(x, y|x0, y0) = GN(x0, y0|x, y) for distinct (x, y), (x0, y0) in D.
[13 marks]
[Green’s first identity may be assumed in the formRR
D

r2 +r ·r dxdy = RC @ @n dc.]
MA3PD2 2019/0 A 800 vc2 Turn over
Page 4
3. Let D = {x > 0, 0 < y < ⇡}, and C be its boundary. Suppose that
u = u(x, y) satisfies the Neumann Helmholtz problem8>><>>:
r2u+ k2u = 0 in D,
@u/@n = F on C1 = {0 < x < 1, y = 0},
@u/@n = 0 on remainder of boundary C2 = C\C1,
u outgoing and bounded as x!1 in D.
It is given that the corresponding Green’s function GN(x, y|x0, y0) satisfies8<:
r2GN + k2GN = (x x0)(y y0) in D, with (x0, y0) 2 D,
@GN/@n = 0 on C,
GN outgoing and bounded as x!1 in D,
and can be written as
GN(x, y|x0, y0) =
1X
n=0
✏n cos(ny) cos(ny0)
2↵n⇡
n
e↵n|xx0| + e↵n(x+x0)
o
,
(3)
for (x, y) 6= (x0, y0), where ✏0 = 1, ✏n = 2 for n 1, and
↵n =
⇢ p
n2 k2 if n > k
ipk2 n2 ⌘ in if n < k. (4)
(a) Write DL = {0 < x < L, 0 < y < ⇡}, for L > 1. Use a suitable
Green’s identity to show that
u(x0, y0) =
Z 1
0
F (x)GN(x, 0|x0, y0) dx+
Z ⇡
0
✓
u
@GN
@x
GN @u
@x
◆
x=L
dy
(5)
for (x0, y0) 2 DL. [Green’s first identity may be assumed in the formRR
D

r2 +r ·r dxdy = RC @ @n dc.]
[12 marks]
MA3PD2 2019/0 A 800 vc2
Page 5
(b) Starting from (3), explain why
GN(x, y|x0, y0) ⇠
NX
n=0
✏n cos(ny) cos(ny0)
in⇡
einx cos(nx0) (6)
as x!1, where n is defined in (4). How is the integer N
determined? Explain why a suitable expansion of u(x, y) as x!1 is
u(x, y) ⇠
NX
n=0
✏nB
+
n e
inx cos(ny), (7)
for some constants B+n .
[11 marks]
(c) Finally, use the expansions (6) and (7) to show that the second integral
on the right-hand side of (5) vanishes in the limit as L!1.
[17 marks]
[End of Question Paper]
MA3PD2 2019/0 A 800 vc2