UNIVERSITY OF DUBLIN
XMA2331/2
TRINITY COLLEGE
Faculty of Science
school of mathematics
SF Mathematics
SF Theoretical Physics
SF Two Subject Mod
Trinity Term 2010
Course 2331/2332: Equations of Mathematical Physics I/II
Dr. S. Ryan
ATTEMPT SIX QUESTIONS:
THREE FROM SECTION A AND THREE FROM SECTION B
All questions carry equal marks.
Log tables are available from the invigilators, if required.
Non-programmable calculators are permitted for this examination,—please indicate the make
and model of your calculator on each answer book used.
2
XMA2331/2
SECTION A
1. Sketch the triangular wave function defined by f (x) = |x| for −1 ≤ x ≤ 1 and
f (x + 2) = f (x), ∀x.
Determine the (real) Fourier series of this function.
Use this result (with Parseval’s theorem) to show that
1+
1
1
π2
1
.
+
+
+
.
.
.
=
32 52 72
8
2
=
2
= 1.
a0
Z
1
dx|x|
−1
1
Z
dx|x| cos(πnx)
an =
−1
2 ∈10
=
2
=
n2 π 2
= −
x cos(πnx)
(cos(nπx) − 1)
4
n2 π 2
, n odd.
Also,
Z
1
|x| sin(nπx) = 0
bn =
−1
since the integrand is odd. So
∞
f (x) =
1 X 4
−
cos(πnx)
2 n=1 π 2 n2
Then, rearranging the above gives
π2
=
8
X
n=1,n
odd
1
1
1
cos(nπx) = 1 + 2 + 2 + . . . .
2
n
3
5
3
XMA2331/2
RR
2. Evaluate the surface integral,
F · dS, where F = xi + yj + z 4 k and the surface
S
S is the upper half of the sphere x2 + y 2 + z 2 = 9 and the disk x2 + y 2 ≤ 9 in the
plane z = 0, with positive orientation.
The sphere is parameterised as 3 sin θ cos φi + 3 sin θ sin φj + 3 cos θ. Then
∂r
∂r
= 3 cos θ cos φi + 3 cos θ sin φj − 3 sin θ
= −3 sin θ sin φi + 3 sin θ cos φj + 0
∂θ
∂φ
so that the cross product is
∂r
∂r
×
= 9(sin2 θ cos φ, sin2 θ sin φ, cos θ sin θ).
∂θ ∂φ
Then the surface integral for the sphere is
Z
π/2
Z
dθ
0
0
2π
∂r
∂r
dφF ·
×
= 2π(27)
∂θ ∂φ
For the disk (at z = 0) S =
sum of the two parts.
Z
π/2
(sin3 θ + cos5 θ sin θ)
0
2 9
= 2π(27)
+
3 2
= 279π.
R 3 R 2π
0
0
r2 = 18π and the total surface integral is the
4
XMA2331/2
3. Prove that the Fourier transform of an even function is even.
Express as a Fourier integral, the function
cos(x) |x| <
f (x) =
0
|x| >
π
2
π
2
.
Proof: Have f (x) = f (−x) so substituting in the fourier transform
Z ∞
1
f (k) =
dxf (−x)e−ikx
2π −∞
Now write u = −x so du = −dx and
Z u=−∞
−duf (u)e−ik(−x)
f˜(k) =
Zu=∞
∞
=
duf (u)e−i(−k)x
−∞
= f˜(−k).
R∞
Writing f as a Fourier integral f (x) = −∞ dk eikx f˜(k). We require the Fourier
transform:
Z ∞
Z π
2
1
1
eix + e−ix
−ikx
f˜(k) =
dk e
f (x) =
dk e−ikx
2π −∞
2π − π2
2
π
i(1−k)x
1
ei(−1−k)x 2
e
=
+
4π i(1 − k) i(−1 − k) − π
2
1 ie−ikπ/2 + ieikπ/2 −ie−ikπ/2 − ieikπ/2
=
+
4π
i(1 −k)
i(−1− k)
1
kπ
1
1
1
kπ
1
=
2 cos
+
= cos
.
4π
2
1−k 1+k
π
2
1 − k2
Therefore
1
f (x) =
π
Z
∞
dk cos
−∞
Remark: f˜(k) is well behaved at k = ±1.
kπ
2
eikx
.
1 − k2
5
XMA2331/2
4. Consider the rectangular box, B defined by 0 ≤ x ≤ a, 0 ≤≤ b, 0 ≤ z ≤ c, for
a, b, c ∈ R. Evaluate
Z Z Z
I=
(xy 2 + z 3 )dV.
B
Z Z Z
I =
=
=
=
=
=
(xy 2 + z 3 )dV
B
Z c Z b Z a
dz
dy
dx(xy 2 + z 3 )
0
0
0
x=a
Z c Z b 2 2
xy
3 dy
dz
+ xz 2
0
0
x=0
y=b
Z c 2 3
ay
dz
+ ayz 3 6
0
y=0
2 3
4 z=c
a b z abz +
6
4 z=0
a2 b3 c abc4
+
.
6
4
Consider the tetrahedron T which has vertices (0, 0, 0), (1, 0, 0), (0, 1, 0), (0, 0, 1).
Evaluate
Z Z Z
ydV.
I=
T
Z
Z Z Z
ydV =
I =
Z
1−x
1−x
dyy(1 − x − y)
1−x
Z
y2 y3
=
dx (1 − x) −
2
3 0
Z 1
1
1
=
dx (1 − x3 ) = .
6
24
0
=
dx
0
Z
1−x−y
dzy
dy
dx
0
T
Z
Z
0
6
XMA2331/2
SECTION B
5. State Stokes’ theorem.
from notes
R
Use Stokes’ theorem to evaluate S (∇ × F) · dS where the vector field F = z 2 i −
3xyj + x3 y 3 k and S is the surface described by z = 5 − x2 − y 2 above the z = 1
plane and oriented upwards.
Explain why Stokes theorem cannot by used to transform the surface (flux) integral
Z
(xi + yj + zk) · dS
S
into a line integral on the boundary of S.
H
Calculate F · dl where the closed path is the circle x2 + y 2 = 4, a boundary of
the surface S at z = 1. Parameterise the path with x = cos u, y = sin u, z = 1 and
0 ≤ u ≤ 2π. Then,
dr
= (− sin u, cos u, 0)
du
and F · dr/du = − sin u(1 + 3 cos2 u). The line integral is then
I
Z 2π
F · dl =
du(− sin u − 3 cos2 u sin u) = 0.
0
For the second part note that if F = (x, y, z) then divF = 3 6= 0 and since
div(curl)F = 0 by definition then F = (x, y, z) cannot be a curl, so can’t use
Stokes.
6. Using the Frobenius method, find the general solution fo the ODE
4xy 00 (x) + 2y 0 (x) + y(x) = 0
where, as usual, the prime denotes differentiation with respect to x.
Comment on why the “naive” series solution approach will not work.
P∞
P∞
P∞
n+s 0
Write y =
y =
+ s) xn+s ,
y 00 =
n=0 an x
n=0 an (n P
n=0 an (n +
∞
n+s
0
s−1
n+s
s)(n + s − 1) x P. Now: y (x) = a0 sx
+ n=0 an+1 (n + 1 + s)x
xy 00 (x) =
∞
s−1
m+s
a0 s(s − 1)x + m=0 am+1 (m + 1 + s)(m + s)x
.
4xy 00 + 2y 0 + y = a0 [4s(s − 1) + 2s]
∞
X
+
[4(m + 1 + s)(m + s)am+1 + 2(m + 1 + s)am+1 + am ] xm+s
m=0
∞ 1 s−1 X
1
4a0 s(s − )x +
4(m + 1 + s)(m + s + )am+1 + am .
2
2
m=0
Set a0 = 1 Indicial equation: s(s − 21 ) = 0 with roots s = 0 and s = 12 .
7
XMA2331/2
RR
7. Use the Gauss (divergence) theorem to calculate
F · dS, where S is the surface
S
of the box B with vertices (±1, ±2, ±3) with outward pointing normals and F =
(x2 z 3 , 2xyz 3 , xz 4 ).
Verify, by direct calculation, the divergence theorem for the case where F = (x, y, z)
on the surface of the solid sphere of radius R centred at the origin.
8. Use separation of variables to solve the two-dimensional Laplace equation for Φ(x, y)
with boundary conditions given by
Φ(0, y) = 0,
1
∂
Φ(0, y) =
sin(ny), (n ∈ Z).
∂x
n
called the Hadamard conditions.
Φ(x, y) = X(x)Y (y) gives X 00 Y + XY 00 = 0 or
X 00
X
00
= − YY . Then
X 00 = EX
Y 00 = −EY
(1)
which each have solutions in the three classes as in the notes.
Now apply the boundary conditions,
Φ(0, y) = X(0)Y (y) = 0
1
Φ(0, y) = X 0 (0)Y (y) = sin ny
n
(2)
so we need X(0) = 0 and X 0 (0) = constant. If this holds then we have
Y =
A
sin ny
n
(3)
so E = n2 and hence
X = B sinh nx
(4)
X = C1 enx + C2 e−nx
(5)
where I have substituted
into the conditions on X and put the two exponentials together to get the sinh nx.
Now, putting this back together we get
Φ(x, y) =
C
sinh nx sin nx
n
(6)
8
1
XMA2331/2
Some useful formulae
1. A function with period l has a Fourier series expansion
∞
a0 X
f (x) =
+
an cos
2
n=1
2πnx
l
+
∞
X
bn sin
n=1
2πnx
l
where
a0
an
bn
l/2
Z
2
=
l
dxf (x),
−l/2
Z l/2
2πnx
dxf (x) cos
,
l
−l/2
Z
2 l/2
2πnx
=
dxf (x) sin
.
l −l/2
l
2
=
l
2. A function with period l has a Fourier series expansion
f (x) =
∞
X
cn exp
n=−∞
where
1
cn =
l
Z
l/2
dxf (x) exp
−l/2
2iπnx
l
,
−2iπnx
l
.
3. The Fourier integral representation (or Fourier transform) is
Z ∞
f (x) =
dk fg
(k)eikx ,
−∞
Z ∞
1
g
f (k) =
dxf (x)e−ikx .
2π −∞
c UNIVERSITY OF DUBLIN 2011
,