Course 2 - Homework Assignment 1 Solution
1. (a) Using the equation
w00 = w − wp
and its first integral
(w0 )2 = w2 −
we have
φ001
2
wp+1 ,
p+1
µ
¶
p−1
p+1
p − 1 p−3 0 2
00
=
w 2 w +
w 2 (w )
2
2
µ
¶
p+1
3p−1
p+1
p − 1 p+1 p − 1 3p−1
=
w 2 −w 2 +
w 2 −
w 2
2
2
p+1
¶2
µ
p+1
3p−1
p+1
=
w 2 − pw 2 .
2
Hence,
õ
L0 (φ1 ) =
p+1
2
!
¶2
−1 w
p+1
2
= λ1 φ1 .
Since the first eigenfunction is necessarily positive and other eigenfunctions are orthogonal
to it, φ1 is the principal eigenfunction and λ1 is the principal eigenvalue.
(b) From (a),
L0 (wr−1 ) = λ1 wr−1 ,
so
r−1
r−1
L−1
) = λ−1
.
0 (w
1 w
(c) It is clear that
Z
w
r−1
r−1
L−1
)
0 (w
1
=
λ1
Z
wp+1 .
2. First we show that w decays exponentially. In fact since w(∞) = 0 we see that for r > r1 wp−1 <
and hence for r > r1 w satisfies
1
∆w = (1 − wp−1 )w ≥ w
4
1
Next we note that the function e− 2 r satisfies
1
1 1
∆(e− 2 r ) ≤ e− 2 r
4
By comparison principle for r > r1
1
w(r) ≤ w(r1 )e− 2 (r−r1 )
Similarly since
∆wr − wr + pwp−1 wr =
1
N −1
wr
r2
3
4
and pwp−1 is small near ∞, wr decays exponentially at ∞. This fact implies that when we integrate
by parts below, the boundary terms all vanish.
Following the hint and integrating on (0, ∞), we have
Z
Z
N −1
0 = rwr (r
wr )r + rN wr (−w + wp )
µ
¶
Z
Z
w2
wp+1
N −1
N
=− r
wr (rwrr + wr ) + r
−
+
2
p+1 r
µ 2
¶
Z
Z
Z
³w ´
w
wp+1
r
N −1 2
N −1
N
wr + N r
=− r
− r
−
2 r
2
p+1
µ
¶Z
Z
Z
N
N
N
rN −1 w2 −
rN −1 wp+1 ,
=
−1
rN −1 wr2 +
2
2
p+1
and
Z
Z
w(r
wr )r + rN −1 w(−w + wp )
Z
Z
Z
N −1 2
N −1 2
=− r
wr − r
w + rN −1 wp+1 .
0=
N −1
Multiplying the second by (N/2 − 1) and adding it to the first, we have
µ
¶Z
Z
N
N
N −1 2
rN −1 wp+1
0= r
w +
−1−
2
p+1
Z
Z
(N − 2)p − (N + 2)
N −1 2
= r
w +
rN −1 wp+1 .
2(p + 1)
Therefore, if p ≥
N +2
,
N −2
then the right hand side is strictly positive, unless w ≡ 0.
3. Let
Z
f (t) =
and
(|∇u + tφ|2 + (u + tφ)2 )
µZ
g(t) =
|u + tφ|
2
p+1
2
¶− p+1
.
We compute
Z
Z
(∇u · ∇φ + uφ) + 2t (|∇φ|2 + φ2 ),
R
|u + tφ|p−1 (u + tφ)φ
0
g (t) = −2 ¡R
¢ 2 +1 ,
|u + tφ|p+1 p+1
Z
f (0) = (|∇u|2 + u2 ),
Z
0
f (0) = 2 (∇u · ∇φ + uφ),
Z
00
f (0) = 2 (|∇φ|2 + φ2 ),
0
f (t) = 2
µZ
g(0) =
|u|
R
p+1
2
¶− p+1
,
|u|p−1 uφ
,
2
p+1 ¢ p+1 +1
|u|
¡R p−1 ¢2
R p−1 2
|u| uφ
|u| φ
00
g (0) = 2(p + 3) ¡R
−
2p
.
2
2
¢
¡R
p+1 p+1 +2
p+1 ¢ p+1 +1
|u|
|u|
g 0 (0) = −2 ¡R
The first derivative is therefore
2 µµZ
¶− p+1
¶
¶¶
µZ
µZ
p+1
p−1
0
|u|
(∇u · ∇φ + uφ) − c1
|u| uφ
ρ (0) = 2
where
R
c1 =
(|∇u|2 + u2 )
R p+1 .
|u|
Since u is the minimizer of E[u], we have ρ0 (0) and so
Z
(−∆u + u − c1 |u|p−1 u)φ = 0
for any test function φ, from which we get the Euler-Lagrange equation
∆u − u + c1 |u|p−1 u = 0.
Before computing ρ00 (0), we wish to simply the calculations by choose c1 = 1 by a scaling argument.
In fact, E[u] is scaling invariant, i.e. E[u] = E[λu] for any λ > 0. By going through the whole
argument with λu instead, the Euler-Lagrange equation is
λ∆u − λu + c1 λp |u|p−1 u = 0.
If we choose λ such that c1 λp−1 = 1, then we could have assumed, without loss of generality, that
c1 = 1. Note also the consequence
Z
Z
(∇u · ∇φ + uφ) = |u|p−1 uφ.
3
Finally, we compute the second derivative by
R
(|∇φ|2 + φ2 )
00
f (0)g(0) = 2 ¡R
¢ 2 ,
|u|p+1 p+1
¡R p−1 ¢2
|u| uφ
0
0
2f (0)g (0) = −8 ¡R
¢ 2 +1 ,
|u|p+1 p+1
¡R p−1 ¢2
R p−1 2
|u| uφ
|u| φ
00
f (0)g (0) = 2(p + 3) ¡R
−
2p
2
2 ,
¢ +1
¡R p+1 ¢ p+1
|u|p+1 p+1
|u|
ρ00 (0) = f 00 (0)g(0) + 2f 0 (0)g 0 (0) + f (0)g 00 (0)

¡R p−1 ¢2 
2
µZ
¶− p+1
Z
Z
|u| uφ
 (|∇φ|2 + φ2 ) − p |u|p−1 φ2 + (p − 1)
.
=2
|u|p+1
2
¡R p+1 ¢ p+1
+1
|u|
4. (a)
µ
¶
t
H(t) = tanh √
2
µ
¶
1
t
0
2
√
H (t) = √ sech
2
2
µ
¶
µ
¶
t
t
00
2
√ tanh √
H (t) = sech
2
2
µ
µ
¶¶
µ
¶
t
t
2
= 1 − tanh √
tanh √
2
2
3
= H(t) − H (t)
(b) Write ρ(t) = E[u + tφ].
¶
Z µ
1
2
ρ (0) =
∇u · ∇φ + (1 − u )(−2uφ)
2
Z
= (−∆u − u + u3 )φ.
0
Hence the result follows.
(c) For steady states H 00 = 0, so
H2
=0
1 + aH 2
−H(1 + aH 2 − H) = 0
−H +
h± =
4
1±
√
1 − 4a
.
2a
Now we compute the integral
¶
H2
F (h+ ) =
−H +
dH
1 + aH 2
0
¶
Z h+ µ
1
1
=
−H + −
dH
a a(1 + aH 2 )
0
√
h2
h+
1
=− + +
− √ tan−1 ( ah+ ).
2
a
a a
Z
h+ µ
Since ah2+ = h+ − 1, the equation for which the integral is zero is
−
√
h+ − 1
1
+ h+ − √ tan−1 ( ah+ ) = 0,
2
a
or
√
1
h +1
√ tan−1 ( ah+ ) = +
.
2
a
For a = a0 , we integrate the equation as follows.
(H 0 )2
+ F (H) = 0
2
p
H 0 = ± −2F (H)
An implicit solution is given by
Z
t
H0
p
t=
Z
−2F (H)
0
H(t)
p
=
H(0)
dt
ds
−2F (s)
.
5. (a) If u(x) = H(a · x + b), then ∆u = |a|2 H 00 (a · x + b). Therefore the condition is |a| = 1.
(b) This is the same as 4(b).
∂u
= aj H 0 (a · x + b) 6= 0, which satisfies
∂xj
the equation ∆ψ = (3u2 − 1)ψ. Let φ be any smooth function with compact support. Testing
the linearized equation with φ2 /ψ (tricky!), we have
Z
Z 2
φ ∆ψ
2
2
(3u − 1)φ =
ψ
µ 2¶
Z
φ
= − ∇ψ · ∇
ψ
µ
¶
Z
2ψφ∇φ − φ2 ∇ψ
= − ∇ψ ·
ψ2
(c) Since |a| = 1, there is a j such that aj 6= 0. Let ψ =
5
Notice that the integration by parts is justified by the (exponential) decay of ψ near ∞. Now
we complete the trick by
!
Z
Z Ã
2
2
φ∇ψ
φ
|∇ψ|
2
2
(|∇φ| + (3u2 − 1)φ2 ) =
|∇φ| − 2∇φ ·
+
ψ
ψ2
¯2
Z ¯
¯
¯
φ∇ψ
¯
= ¯¯∇φ −
ψ ¯
≥0
If equality holds, then
∇φ −
This implies
φ∇ψ
≡ 0.
ψ
µ ¶
ψ∇φ − φ∇ψ
φ
=
∇
≡ 0,
ψ
ψ2
which is impossible unless φ ≡ 0 since ψ does not have a compact support.
6