Course 2 - Homework Assignment 3 Solution
1. (a)
u00 (x) = f (x)
Z ∞
0
u (x) = −
f (t) dt
x
Z ∞Z ∞
u(x) =
f (s) dsdt
x
t
As x → ∞, the dummy variables t, s → ∞ so
Z ∞Z ∞
|u(x)| ≤
Ce−s dsdt ≤ Ce−x .
x
t
(b) Similarly,
Z
Z
x
t
u(x) =
f (s) dsdt.
−∞
(c) The condition is
Z
∞
∞
f (x) dx = 0.
−∞
Indeed, using the formula in (b),
¯Z t
¯ ¯ Z
¯
¯ ¯
¯
f (s) ds¯¯ = ¯¯−
¯
−∞
for large t, hence for all t. Then
Z x
|u(x)| ≤
−∞
(d) The condition is
Z
¯Z
¯
¯
¯
t
−∞
∞
t
¯ Z
¯
f (s) ds¯¯ ≤
∞
t
¯
Z
¯
¯
f (s) ds¯ dt ≤ C
Z
∞
Ce−s ds = Ce−t
∞
e−|t| dt ≤ C.
−∞
∞
f (x) dx =
xf (x) dx = 0,
−∞
−∞
d2
. Note also that the second also reads
dx2
that is, f is orthogonal to the kernels of
Z
∞
Z
t
f (s) dsdt = 0.
−∞
−∞
These allow us to write, for x → ∞,
Z x Z t
Z
u(x) =
f (s) dsdt = −
−∞
∞
∞
Z
Z
t
∞
Z
∞
f (s) dsdt =
x
∞
1
x
t
f (s) dsdt ≤ Ce−x .
2. (a)
u0 (x) = −2 log cosh(x) + log 2
2 sinh(x)
u00 (x) = −
cosh(x)
2
u000 (x) = −
= −eu0 (x) .
cosh2 (x)
(b) The translation invariance is clear, since the equation involves no x. For the scaling invariance,
we see that
(u0 (λx) + 2 log λ)00 + eu0 (λx)+2 log λ = λ2 u000 (λx) + λ2 eu0 (λx) = 0.
(c) Applying −
∂
to (u0 (x − a))00 + eu0 (x−a) = 0, we have
∂a
(u00 (x − a))00 + eu0 (x−a) u00 (x − a) = 0,
that is,
(u00 )00 + eu0 u00 = 0.
The first kernel is φ1 = u00 (x) = −
2 sinh(x)
= −2 tanh(x).
cosh(x)
The second one is given by
¯
∂ ¯¯
φ2 (x) =
(u00 (λx) + 2 log λ) = xu00 (x) + 2 = −2x tanh(x) + 2.
¯
∂λ λ=1
(d) First we compute the Wronskian
W = φ1 φ02 − φ01 φ2 = (u00 )2 − 2u000 =
4 sinh2 (x) + 4
= 4.
cosh2 (x)
By the variation of parameters formula, we have for any given a, b,
Z x
Z x
φ2 (t)f (t)
φ1 (t)f (t)
φ(x) = −φ1 (x)
dt + φ2 (x)
dt
W (t)
W (t)
a
b
Z x
Z x
= − tanh(x) (t tanh(t) + 1)f (t) dt + (x tanh(x) + 1)
tanh(t)f (t) dt
a
b
For decay at the positive end, we take a = b = ∞ to obtain
Z ∞
Z
φ(x) = tanh(x)
(t tanh(t) + 1)f (t) dt − (x tanh(x) + 1)
x
∞
tanh(t)f (t) dt.
x
(e) We require that f be orthogonal to the kernels φ1 and φ2 , that is,
Z ∞
Z ∞
tanh(x)f (x) dx =
(x tanh(x) + 1)f (x) dx = 0.
−∞
−∞
In fact since φ2 (x) grows like x, the decay is a bit weaker:
|φ(x)| ≤C(1 + |x|)e−|x| for all x.
2
3. If we define ã(r) = rN −1 a(r), then this is what has been discussed in the lectures. The necessary
condition is (log ã)0 (r0 ) = 0 and a sufficient condition is the non-degeneracy (log ã)00 (r0 ) 6= 0. By
direct computations,
ã0
ã
µ 0 ¶2
ã
ã00
00
−
(log ã) =
ã
ã
0
0
ã
a
N −1
= +
ã
a
r
00
00
ã
a
2(N − 1) a0 (N − 1)(N − 2)
.
=
+
+
ã
a
r
a
r2
(log ã)0 =
Therefore, the necessary condition is
(log a)0 (r0 ) = −
and a sufficient condition is
(log a)00 (r0 ) 6=
N −1
r0
N −1
.
r02
4. (a)
8
= log 8 − 2 log(1 + r2 )
(1 + r2 )2
4r
u00 (r) = −
1 + r2
4(r2 − 1)
00
u0 (r) =
(1 + r2 )2
1
1
u000 + u00 + eu0 =
(4r2 − 4 − 4 − 4r2 + 8) = 0.
r
(1 + r2 )2
u0 (r) = log
(b) First note that the equation holds at λr:
u000 (λr) +
1 0
u (λr) + eu(λr) = 0.
λr 0
Then
¶
µ
1 0
1
u(λr)
0
u0 (λr)+2 log λ
2
00
= 0.
= λ u0 (λr) + u0 (λr) + e
(u0 (λr)+2 log λ) + (u0 (λr)+2 log λ) +e
r
λr
00
(c) A kernel is
¯
2(1 − r2 )
∂ ¯¯
0
(u
(λr)
+
2
log
λ)
=
ru
(r)
+
2
=
.
0
0
∂λ ¯λ=1
1 + r2
Let us take
Z0 (r) =
3
1 − r2
.
1 + r2
(d) Let Z1 be the other kernel. By Abel’s formula, the Wronskian satisfies, up to a constant,
W (r) = Z0 (r)Z10 (r) − Z00 (r)Z1 (r) = e−
R
1
r
1
= .
r
Solving for Z1 (r) (reduction of order), we have
Z r
W (s)
Z1 (r) = Z0 (r)
ds
Z02 (s)
Z
1 − r2 r (1 + s2 )2
=
ds
1 + r2
s(1 − s2 )2
µ
¶
2
1 − r2
log r − 2
=
1 + r2
r −1
2
1−r
2
=
log r +
2
1+r
1 + r2
By the variation of parameters formula, the solution satisfying
0
φ (0) = 0
is given by
Z
Z r
Z1 (s)f (s)
Z0 (s)f (s)
φ(r) = AZ0 (r) − Z0 (r)
ds + Z1 (r)
ds
W (s)
W (s)
0
0
Z
Z
1 − r2 r s((1 − s2 ) log s + 2)
(1 − r2 ) log r + 2 r s(1 − s2 )
=−
f (s) ds +
f (s) ds.
1 + r2 0
1 + s2
1 + r2
1 + s2
0
r
where A is an arbitrary constant.
5. Note that as r → ∞,
Z0 (r) = −1 + O(
,
1
)
r2
Z1 (r) = − log r + O(
and
Z
r
0
log r
)
r2
Z ∞
s((1 − s2 ) log s + 2)
s((1 − s2 ) log s + 2)
log r
f
(s)
=
f (s) + O( 2 )
2
2
1+s
1+s
r
0
Z
Z r
∞
s(1 − s2 )
1
f (s) =
Z0 (s)f (s)sds + O( 2 )
2
1+s
r
0
0
Hence we have as r → +∞
Z
Z
∞
φ(r) = A −
∞
Z1 (s)f (s)sds − log r
0
Z0 (s)f (s)sds + O(
0
Thus a necessary condition for φ to be bounded is that
Z ∞
Z0 (s)f (s)sds = 0
0
4
log r
)
r2
Under the above condition we can choose
Z
∞
A=
Z1 (s)f (s)sds
0
so that φ decays at +∞:
φ = O(
log r
)
r2
Alternatively, we can define
Z ∞
Z ∞
Z1 (s)f (s)
Z0 (s)f (s)
φ(r) = Z0 (r)
ds − Z1 (r)
ds
W (s)
W (s)
r
r
Z
Z
1 − r2 ∞ s((1 − s2 ) log s + 2)
(1 − r2 ) log r + 2 ∞ s(1 − s2 )
=
f (s) ds −
f (s) ds,
1 + r2 r
1 + s2
1 + r2
1 + s2
r
so for r large
Z
Z
∞
|φ(r)| ≤
∞
s log s |f (s)| ds + log r
r
Z
∞
≤2
s |f (s)| ds
r
s log s |f (s)| ds
r
Z
∞
s log s
ds
1 + s4
Zr ∞
C
log s
≤
ds
r r s2
µ
¶
log r
=O
r2
≤C
However this definition (1) does not satisfy the initial condition
0
φ (0) = 0
unless
R∞
0
Z0 (s)f (s)sds = 0. In fact we calculate
Z
1 ∞
0
Z0 (s)f (s)sds
φ (0) = lim
r→0 r 0
5
(1)