The University of Chicago, Department of Physics
Page 1 of 8
Problem set 8 solutions
[50]
Michael A. Fedderke
Standard disclaimer: while every effort is made to ensure the correctness of these
solutions, if you find anything that you suspect is an error, please email me at
[email protected]
.
All problem references to Arfken and Weber, 7th edition.
I. 10.1.4
[10]
I assume that
t
∈
(0
, π
)
, else the Green’s Function is trivially identically zero on the whole domain.
Consider the boundary value problem

y
00

1
4
y
=
δ
(
x

t
);
y
(0) = 0;
y
(
π
) = 0
.
(1)
In the domains
x
≶
t
, the differential equation is homogeneous and secondorder, so the solution space is
spanned by two linearly independent solutions:

y
00

1
4
y
= 0
⇒
y
(
x
) =
A
cos(
x/
2) +
B
sin(
x/
2)
,
(2)
so if we split the domain
x
∈
[0
, π
]
into subdomains
x
∈
[0
, t
)
and
x
∈
(
t, π
]
, we have the solution
G
(
x
;
t
) =
A
(
t
) cos(
x/
2) +
B
(
t
) sin(
x/
2)
0
≤
x < t
C
(
t
) cos(
x/
2) +
D
(
t
) sin(
x/
2)
t < x
≤
π
,
(3)
where
A, B, C, D
depend in general on
t
but not on
x
. Imposing the BCs
y
(0) = 0
⇒
G
(
x
= 0
, t >
0) = 0
and
y
(
π
) = 0
⇒
G
(
x
=
π, t < π
) = 0
sets
A
=
D
= 0
, while continuity at
x
=
t
demands that
B
(
t
) sin(
t/
2) =
C
(
t
) cos(
t/
2)
⇒
C
(
t
) =
B
(
t
) tan(
t/
2)
.
(4)
The value of
B
(
t
)
can be found by integrating both sides of the ode in the small domain
(
t

, t
+
)
for some
small positive
and then sending
→
0
:
Z
t
+
t

dx

G
00

1
4
G
=
Z
t
+
t

dx δ
(
x

t
)
(5)
⇒ 
G
0

t
+
t

 O
( ) = 1
(6)
⇒
G
0


G
0
+
= 1
(7)
⇒
1
2
B
(
t
) cos(
t/
2) +
1
2
C
(
t
) sin(
t/
2) = 1
.
(8)
Substituting (4) into (8), we have
B
(
t
) = 2 cos(
t/
2)
(9)
C
(
t
) = 2 sin(
t/
2)
(10)
Therefore, we have the Green’s Function
G
(
x
;
t
) =
2 cos(
t/
2) sin(
x/
2)
0
≤
x
≤
t
2 sin(
t/
2) cos(
x/
2)
t < x
≤
π
.
(11)
Mathematical Methods of Physics
v1.0
2 December 2015, 6:00pm
Autumn 2015
The University of Chicago, Department of Physics
Page 2 of 8
II. 10.1.7
[10]
I will initially assume that
k
6
= 0
. Let’s find the Green’s function by solving
¨
ψ
+
k
˙
ψ
=
δ
(
t

t
0
);
ψ
(0) =
˙
ψ
(0) = 0
.
(12)
I will also assume that
t
0
>
0
otherwise the Green’s function is again trivially identically zero on
t
≥
0
. (I am
implicitly assuming that
t
≥
0
is the only domain we are interested in since we have
initial
conditions imposed at
t
= 0
.)
We split the domain up into
t
∈
[0
, t
0
)
and
t
∈
(
t
0
,
∞
]
.
In either subdomain, the ode is homogeneous:
¨
ψ
+
k
˙
ψ
= 0
, and has a solution space spanned by two linearly independent solutions (
they are only linearly
independent for
k
6
= 0
):
ψ
(
t
) =
Ae

kt
+
B
. So the solution is
G
(
t
;
t
0
) =
A
(
t
0
)
e

k
(
t

t
0
)
+
B
(
t
0
)
0
≤
t < t
0
C
(
t
0
)
e

k
(
t

t
0
)
+
D
(
t
0
)
0
< t
0
< t
≤
π
,
(13)
where I have factored out
e
kt
0
from
A
(
t
0
)
and
C
(
t
0
) for reasons to become clear shortly. The IC’s
ψ
(0) =
˙
ψ
(0) = 0
demand that
A
=
B
= 0
, which can be thought of as a statement of causality:
G
(
t
;
t
0
) = 0
for
t < t
0
. We demand
continuity at
t
=
t
0
so
A
+
B
=
C
+
D
⇒
C
=

D
, giving
G
(
t
;
t
0
) =
0
0
≤
t < t
0
D
(
t
0
) 1

e

k
(
t

t
0
)
0
< t
0
< t
≤
π
,
(14)
To find
D
(
t
0
)
we again perform the trick of integrating the ode around
t
0
in an
domain, to find that
˙
ψ

+

+
kψ

+

= 1
⇒
kD
(
t
0
) = 1
⇒
D
= 1
/k,
(15)
so
G
(
t
;
t
0
) =
0
0
≤
t < t
0
1
k
1

e

k
(
t

t
0
)
0
< t
0
< t
≤
π
.
(
k
6
= 0)
(16)
I will not find the Green’s function for
k
= 0
because that case is much more easily handled by direct integration.
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