MA2332 Tutorial Sheet: due on 1st April, 2016
1
1. 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.
I will skip over steps that are in the notes. Separating the variables Φ(x, y) =
X(x)Y (y) gives
X 00 Y + XY 00 = 0
(1)
or
X 00
Y 00
=−
(2)
X
Y
Since the left and right hand sides are functions of different independent variables,
they must be equal to a constant, let’s call it E, giving
X 00 = EX
Y 00 = −EY
(3)
which each have solutions in the three classes as in the notes. Now we need to apply
the boundary conditions,
Φ(0, y) = X(0)Y (y) = 0
1
Φ(0, y) = X 0 (0)Y (y) = sin ny
n
(4)
so we need X(0) = 0 and X 0 (0) =constant. If this holds then we have
Y =
A
sin ny
n
(5)
so E = n2 and hence
X = B sinh nx
(6)
X = C1 enx + C2 e−nx
(7)
where I have put 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) =
1
C
sinh nx sin ny
n
Sinéad Ryan, see http://www.maths.tcd.ie/˜ryan/231.html
1
(8)
And to ensure the second (derivative) BC then C = 1/n so
Φ(x, y) =
1
sinh nx sin ny
n2
(9)
2. Using separation of variables to solve the two-dimensional Laplace equation
∇2 V (x, y) = 0,
subject to the boundary conditions: V (0, y) = V (d, y) = 0 and V (x, 0) = V0 sin
2πx
,
d
with d and V0 ∈ R .
As described in the tutorial you can think of this as describing 2 parallel conducting
sheets separated by distance d and the connecting plate held at V0 sin(2πx/d). You
are asked to solve ∇2 V (x, y) = 0 between the plates i.e. solve for the potential
in between the plates. There is then an additional implicit BC, that we want the
potential to go to zero in the “open” spatial direction i.e. V → 0 for y → ∞.
Now solve Laplace equation as in the notes. Applying the zero boundary condition,
V (0, y) = 0 reduces the set of solutions to
X = A cos(kx) + B sin(kx)
Y = Ceky + De−ky
The implicit BC means we should set C = 0 to eliminate the growing exponential.
That leaves
V (x, y) = Q sin(kx)e−ky .
where I have labelled the two coefficients B and D as Q. Now, using V (d, y) = 0
means that sin(kd) = 0 ⇒ kd = nπ, for integer n. Therefore the k that satisfy
the BC are k = nπ/d still an infinite family of values but labelled now by an integer
index n. The potential is then
Vn (x, y) = Qn sin(nπx/d)e−nπy/d ,
where the subscript n indiciates a solution for any integer. The by linearity
V (x, y) =
∞
X
Qn sin
n=1
nπx d
e−
nπy
d
.
Now the last BC V (x, 0) = V0 sin(2πx/d) is satisfied by taking Qn = 0 for all n
except n = 2 giving
2πx − 2πy
e d .
V (x, y) = V0 sin
d
2
3. Use separation of variables to find a solution to the 1-dimensional wave equation
2
∂ 2φ
2∂ φ
=
c
∂t2
∂x2
∂φ
(x, 0) = g(x)
∂t
Note: you may assume that the formula for the coefficients of a Fourier sine series
for a function h(x) defined on 0 ≤ x ≤ L is
with the boundary conditions: φ(x, 0) = f (x), φ(0, t) = 0, φ(L, t) = 0,
2
bn =
L
Z
L
h(x) sin
nπx 0
L
dx, n = 1, 2, 3, . . .
Write φ(x, t) = X(t)T (t) and substitute in the PDE as usual which yields
X 00
1 T 00
= 2 ,
X
c T
and therefore X 00 /X = E and T 00 /T = c2 E for a constant E. Consider the possibilities for E as in notes i.e. E < 0, E > 0, E = 0 and using the zero boundary
conditions eliminate E = 0, E = k 2 , leaving E = −k 2 . Solving the ODEs that result
leads to
X(x) = A sin(kx) + B cos(kx)
T (t) = C sin(ckt) + D cos(ckt)
Now X(0) = 0 ⇒ B = 0 and X(L) = 0 ⇒ k = nπ/L, for integer n. The solution is
then given by
nπx h
nπc nπc i
X
φ(x, t) =
An sin
Cn sin
t + Dn cos
t .
L
L
L
n
Then BC
φ(x, 0) = f (x) =
X
An sin
n
nπx L
Dn =
X
n
Qn sin
nπx L
(10)
and
nπx h nπc
nπc nπc
nπc i
∂φ X
=
An sin
Cn cos
t −
Dn sin
r ,
∂t
L
L
L
L
L
n
so that
nπx nπc X nπc
nπx X
∂φ(x, 0)
= g(x) =
An sin
Cn
=
Pn sin
.
∂t
L
L
L
L
n
n
3
(11)
Eqns(10, 11) are both Fourier series and coefficients Qn and Pn can be found in the
usual way
Z
nπx 2 L
Qn =
f (x) sin
,
L 0
L
Z
nπx nπc
2 L
Pn =
g(x) sin
.
L
L 0
L
4