Harmonic Maps with Potential from $\mathbb{R}^2$ into $S^2$

0.22 MB·English

by  Ruiqi Jiang

Harmonic Maps with Potential from R2 into S2
Ruiqi JIANG∗
3 Abstract
1
0 Westudythe existenceproblemofharmonicmapswithpotentialfromR2 intoS2. Fora
2 specific classofpotentialfunctions onS2,wegivethesufficientandnecessaryconditionsfor

n the existence of equivariant solutions of this problem. As an application, we generalize and
a improvethe resultsonthe Landau-LifshitzequationfromR2 intoS2 in[7]duetoGustafson
J
and Shatah.
6

]
G 1 Introduction
D
. Let (M,g) and (N,h) be two Riemannian manifolds. As is well known, a map u : M → N is
h 0
t called a harmonic map iff it is critical with respect to the energy functional E(u). See [5] for
a
m the precise definitions. The notion of harmonic maps with potential is first suggested by Ratto
[ ([10]). Given a “potential” function H : N → R, a map u : M → N is called a harmonic map
0

1 with potential H iff it is critical with respect to the functional
v
4
F(u) ≡ E(u)+ H(u)dV .
1 g
0 ZM
1
In this paper, we consider the existence problem of harmonic maps with potential in the
.
1
special case where (M,g) is the Euclidean 2-plane R2, and (N,h) is the unit 2-sphere S2 in R3,
0
3 i.e.
1
S2 = {x ∈R3 : |x|2 = 1}.
:
v
i Then, if we set u= (u ,u ,u )∈ R3, the energy is simply
X 1 2 3
r
a E(u) = 1 |∇u|2dx,
2 R2
Z
where
3
|∇u|2 = |∇u |2.
i
i=1
X
Also, we will assume H(u) = G(d(u)) for a function G : [0,π] → R, where d(u) denotes the
geodesic distance from u∈ S2 to the north pole P =(0,0,1). This assumption on H enables us
to seek for solutions which are equivariant with respect to the S1 = O(2) actions on both the
∗Supported byNSFC, Grant No. 10990013
1
domain R2 and the target S2. If we identify the (x ,x )-plane with the complex plane C, and
1 2
consider R3 = C⊕R1, then an m-equivariant map u takes the following form.
u(r,θ) = sinh(r)eimθ +cosh(r)·e (1.1)
3
where h : [0,∞) → R1 with h(0) = 0, m is an non-zero integer, and e = (0,0,1) is the unit
3
vector. For such m-equivariant maps, the energy F reduces to a functional on the function h as
follows. (We omit the factor 2π in the integrals.)
1 ∞ m2sin2h ∞
J(h) = h′2+ rdr+ G(h)rdr. (1.2)
2 r2
Z0 (cid:18) (cid:19) Z0
Moreover, if h is a critical point of J, then the map u defined in (1.1) is a critical point of F,
hence a harmonic map with potential for the chosen potential function H. Thus, the question
of finding harmonic maps with potential is reduced to solving the following O.D.E., which is
the Euler-Lagrange equation of J(u), with suitable boundary conditions at r = 0 and r = ∞.
1 m2sinhcosh
h′′+ h′− = g(h) (1.3)
r r2
where g(h) = G′(h). The boundary conditions we assume will be
h(0) =0, h(∞) = π (1.4)
Such a problem has been considered by some authors in connection with Landau-Lifshitz type
equations. Gustafson and Shatah ([7]), when looking for periodic solutions to certain Landau-
Lifshitz type equation, studied the above problem with g(h) = λsinhcosh + ωsinh. Their
result shows that, if λ > 0, and ω > 0 is small, then the problem has a solution which has
finite energy and increases monotonely from 0 to π. In [8], Hang and Lin considered the same
equation with g(h) = −λsinhcosh, but the condition at infinity is replaced by h(∞) = π/2.
They prove for each λ > 0 there exists a unique solution of infinite energy for their problem.
Inthispaper,wewillconsideraclass offunctionsg in(1.3)forwhichthesolvability question
of problem (1.3)-(1.4) can be completely answered. The functions g in this class satisfy the
following conditions.
(i) There exists ξ ∈(0,π) such that
g(0) = g(ξ) = g(π) = 0,
g(x) > 0, x ∈ (0,ξ),

 g(x) < 0, x ∈ (ξ,π);
(ii) πg(x)dx > 0; 
0
R
(iii) g′(π) > 0.
We choose the potential function G in (1.2), which is a primitive of g, to be
π
G(x) = − g(t)dt, (1.5)
Zx
2
so that G(π) = 0 and G(0) < 0.
Notice that, the function g in [7] falls into our class. Hence, our result is a generalization
and improvement of theirs.
It is well known that, when g ≡ 0 in (1.3), there is a family of solutions ϕ to (1.3)-(1.4)
λ
which corresponds to a family of harmonic maps from R2 onto S2 of degree m > 0. These
solutions have the following explicit expression.
ϕ (r)= 2arctan[(λr)m]. λ >0
λ
Now we can state our main result.
Theorem 1.1. Assume that the function g ∈ C∞([0, π]) satisfies (i)−(iii), m 6= 0 is an integer
and that G is as in (1.5). The problem (1.3)-(1.4) has solutions with 0< h(r) < π on (0,∞) if
and only if we have
∞
0 < G(ϕ (r))rdr ≤ ∞.
1
Z0
Moreover, the solutions we obtain satisfy h′(r) > 0 on (0,∞) and converge to π exponentially
as r → ∞.
Remark 1. Since
∞ 1 ∞
G(ϕ (r))rdr = G(ϕ (r))rdr,
λ λ2 1
Z0 Z0
we can replace ϕ (r) by ϕ (r) for λ > 0.
1 λ
Remark 2. In fact, in Theorem 1.1 and throughout the paper, we only need to assume that
g ∈ Cα(m)([0,π]), where α(m) =max{1,|m|−2}, if m 6= 0 is fixed.
Ourmethodfortheproofisbasically acombination oftheshootingmethodforO.D.E.’s, the
variational method for obtaining solutions to certain boundary value problems and the blow-up
analysisfordeterminingthebehaviorofsolutionswithlargeinitialdata. Wewillalsorepeatedly
use a Pohozaev type identity in our analysis. (The name “Pohozaev” usually means such an
identity can be obtained by a domain variation along a conformal vector field, and in our case
the vector field is r ∂ .)
∂r
In the next section we would consider an initial value problem of O.D.E. (1.3) with the
singularity at r = 0 and prove its existence, uniqueness and continuous dependence on initial
data. In Section 3, by qualitative analysis, we will establish a series of lemmas to characterize
thebehaviorofsolutionsofO.D.E.(1.3) undersuitableassumptions. InSection 4wediscussthe
existence of the boundary value problems of O.D.E. (1.3) by variational methods. In Section 5,
we give the proof of Theorem (1.1) by shooting method. Finally, we apply our result to certain
Landau-Lifshitz type equations in section 6.
Convention: For convenience, we always assume that m > 0 without further comment.
3
2 The existence of solutions to the initial value problems
Inthis section, we needto consider the initial valueproblem of O.D.E. (1.3) with thesingularity
atr = 0andwanttoproveitsexistence,uniquenessandthecontinuousdependenceontheinitial
data.
For the convenience, we rewrite (1.3) as following form:
1 m2
h′′+ h′− sinhcosh−g(h) = 0.
r r2
We consider the following initial value problem:
1 m2
h′′+ h′− sinhcosh−g(h) = 0, r ∈ (0,+∞) (2.1)
r r2
h(0) = 0, h(m)(0) = m!a, (2.2)
where g(x) ∈ C∞(R), kgk ≤ C < ∞, a ∈ R and h(m) denotes the m-order derivative of h.
C1
Definition 2.1. If h(r) ∈ Cm[0,+∞)∩C∞(0,+∞) satisfies (2.1)-(2.2), then h(r) is called a
solution to (2.1)-(2.2).
Remark 3. If h(r) is a solution of (2.1)-(2.2), by substituting the asymptotic expansion of h(r)
at r = 0 to the equation (2.1), we see that k-th derivative of h(r) evaluated at the point 0 with
k 6 m−1 is zero, i.e.
h(k)(0) = 0, 0 6 k 6 m−1.
So the initial value (2.2) is given reasonably for (2.1).
We shall adopt the contraction map principle, whic is different from that in [6, 8], to address
the problem of existence and uniqueness of local solutions to (2.1)-(2.2). Then, by means of the
standard existence and uniqueness theory on ordinary differential equation, we can extend the
local solution to the whole interval [0,+∞).
It is easy to see that (2.1) can be expressed in the following form:
m2
(rh′)′ = sinhcosh+g(h)r (2.3)
r
Hence, (2.3) is equivalent to the integral equation:
r 1 s m2
h(r) = ( sin2h+g(h)t)dtds. (2.4)
s 2t
Z0 Z0
Let
h(r) = arm+rm+2φ (2.5)
and substitute it into (2.4), then we get
1 r 1 s m2
φ = sin2(atm+tm+2φ)+g(atm +tm+2φ)t dtds−arm .
rm+2 s 2t
(cid:26)Z0 Z0 (cid:27)
(cid:2) (cid:3)
4
Define a map
T :C[0, δ] → C[0, δ]
by
1 r 1 s m2
T(φ) = sin2(atm+tm+2φ)+g(atm+tm+2φ)t dtds−arm , (2.6)
rm+2 s 2t
(cid:26)Z0 Z0 (cid:27)
(cid:2) (cid:3)
where δ would be determined later.
First, we need to verify that T is well defined. Since, for any fixed continuous function φ,
there holds true
r 1 s m2
sin2(atm +tm+2φ)+g(atm+tm+2φ)t dtds
s 2t
(cid:12)Z0 Z0 (cid:12)
(cid:12) r 1 s (cid:2)m2 (cid:3) (cid:12)
≤ (cid:12)(cid:12) 2(|a|tm+tm+2|φ|)+C(|a|tm+tm+2|φ|)t dt(cid:12)(cid:12)ds
s 2t
Z0 Z0
(cid:2) r 1 s (cid:3)
≤ C(|a|+kφk ) (tm−1+tm+3)dtds
C[0,δ] s
Z0 Z0
≤ C(|a|+kφk )(rm+rm+4)
C[0,δ]
≤ C(|a|+kφk )(δm +δm+4)< +∞, (2.7)
C[0,δ]
where C is independent of φ, therefore we know that T(φ)(·) is continuous on (0,δ]. The
remaining is to verify that T(φ)(·) is also continuous at r = 0. Indeed,
r 1 s[m2 sin2(atm +tm+2φ)+g(atm+tm+2φ)t]dtds−arm
limT(φ) = lim 0 s 0 2t
r→0 r→0 rm+2
R R
1 r[m2 sin2(atm+tm+2φ)+g(atm +tm+2φ)t]dt−marm−1
= lim r 0 2t
r→0 (m+2)rm+1
R
r[m2 sin2(atm +tm+2φ)+g(atm+tm+2φ)t]dt−marm
= lim 0 2t
r→0 (m+2)rm+2
R
m2 sin2(arm+rm+2φ)+g(arm +rm+2φ)r−m2arm−1
= lim 2r
r→0 (m+2)2rm+1
m2 sin2(arm+rm+2φ)+g(arm +rm+2φ)r2−m2arm
= lim 2
r→0 (m+2)2rm+2
1(φ(0)− 2a2+g′(0)a), m =1,
9 3
= (2.8)

 (m+12)2[m2φ(0)+g′(0)a], m ≥ 2.

Next, we turn to discussing when T is a contraction mapping. We compute
1 r 1 s m2
|T(φ )−T(φ )| ≤ { sin2(atm +tm+2φ )−sin2(atm+tm+2φ )
1 2 rm+2 s 2t 1 2
Z0 Z0
+ g(atm+tm+2φ )(cid:12)−g(atm +tm+2φ ) t}dtds (cid:12)
1 (cid:12) 2 (cid:12)
1 r 1 s m2
≤ rm(cid:12)(cid:12)+2 s { 2t 2tm+2|φ1−φ2|+Ct(cid:12)(cid:12)m+3|φ1−φ2|}dtds
Z0 Z0
5
1 r 1 s
≤ kφ −φ k {m2tm+1+Ctm+3}dtds
rm+2 1 2 C[0,δ] s
Z0 Z0
1 m2 C
≤ kφ −φ k rm+2+ rm+4
rm+2 1 2 C[0,δ] (m+2)2 (m+4)2
(cid:18) (cid:19)
m2 C
≤ + δ2 kφ −φ k , (2.9)
(m+2)2 (m+4)2 1 2 C[0,δ]
(cid:18) (cid:19)
whereC isapositiveconstant independentofφ. It’s easy toseethatT isacontraction mapping
if δ is small enough such that
m2 C
+ δ2 < 1.
(m+2)2 (m+4)2
Thus there exists unique fix point φ∗ ∈ C[0,δ] such that T(φ∗) = φ∗. So,
h(r) = arm+rm+2φ∗
satisfies (2.1). Moreover, by a simple calculation, we can verify that h(r) also satisfies (2.2).
This means that h(r) is a local solution to (2.1)-(2.2). Hence, by a standard argument we can
extend the local solution to a global solution. Thus, we have shown that the following theorem
holds true.
Theorem 2.1. Assume that g(x) ∈ C∞(R) satisfies kgkC1(R) < ∞. Then, (2.1)-(2.2) always
admits a unique global solution.
For simplicity, we denote the solution of (2.1)-(2.2) by h (r) to emphasize the dependence
a
on the initial value a. Now, we discuss the continuous dependence on the initial data of these
obtained solutions. We need to establish the following theorem.
Theorem 2.2. Assume that g(x) ∈ C∞(R) satisfies kgkC1(R) < ∞. If ha0(r) is the solution of
(2.1)-(2.2), then, ∀R > 0, ∀ε > 0, there exists η = η(a ,ε,R) > 0 such that, if |a−a | < η,
0 0
there holds
kh (r)−h (r)k ≤ ε. (2.10)
a a0 C1[0,R]
Proof. By the standard O.D.E. theory we know that the solutions to (2.1) depend continuously
on the initial data. We need only to prove the theorem in the case R = δ, which has been
determined in (2.9).
From the proof of Theorem 2.1, we know that h is of the following form:
a
h (r)= arm+rm+2φ . (2.11)
a a
In order to get (2.10), we need to estimate size of kφ (r)−φ (r)k . Indeed,
a a0 C[0,R]
|φ −φ | = |T(φ )−T(φ )|
a a0 a a0
1 r 1 s m2
≤ { (sin2(atm+tm+2φ )−sin2(a tm+tm+2φ )−2(a−a )tm
rm+2 s 2t a 0 a0 0
Z0 Z0
+ g(atm +tm+2φ )−(cid:12) g(a tm+tm+2φ ) t}dtds (cid:12)
a (cid:12) 0 a0 (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
6
1 r 1 s m2
≤ { [2tm+2|φ −φ |+(C +(a−a )2)t2m
rm+2 s 2t a a0 0
Z0 Z0
+(C +|φ −φ |2)t2m+4]+C(|a−a |tm+1+tm+3|φ −φ |)}dtds
a a0 0 a a0
1 m2
≤ kφ −φ k +Cδ2 rm+2
rm+2 a a0 C[0,δ] (m+2)2
(cid:26) (cid:18) (cid:19)
1
+C|a−a | rm+2 , (2.12)
0 (m+2)2
(cid:27)
where C is a positive constant depending only on m and a . By choosing δ small enough such
0
that
m2
+Cδ2 < 1,
(m+2)2
we can derive
kφ −φ k ≤ C|a−a |, (2.13)
a a0 C[0,δ] 0
where C is a positive constant depending on m and a . It follows that
0
|h (r)−h (r)| ≤ |a−a |rm+rm+2|φ −φ |
a a0 0 a a0
≤ |a−a |δm+Cδm+2|φ −φ |
0 a a0
≤ C|a−a |, (2.14)
0
and
1 r m2
h′(r)−h′ (r) ≤ { (sin2(atm+tm+2φ )−sin2(a tm+tm+2φ )
a a0 r 2t a 0 a0
Z0
(cid:12) (cid:12) + g(atm+(cid:12) tm+2φ )−g(a tm+tm+2φ ) t}dt (cid:12)
(cid:12) (cid:12) (cid:12) a 0 a0 (cid:12)
1 r m2
≤ r (cid:12)(cid:12) { 2t [2|a−a0|tm+2tm+2|φa −φa0|(cid:12)(cid:12)]
Z0
+C(|a−a |tm+1+tm+3|φ −φ |)}dt
0 a a0
1 1
≤ |a−a | mrm+C rm+2
r 0 (m+2)2
(cid:26) (cid:18) (cid:19)
m2 C
+kφ −φ k ( rm+2+ rm+4)
a a0 C[0,δ] m+2 m+4
(cid:27)
≤ C|a−a |, (2.15)
0
where C is a positive constant depending only on m and a . Thus, we obtain the desired
0
estimate:
kh (r)−h (r)k ≤ C|a−a |.
a a0 C1[0,δ] 0
Remark 4. By the standard elliptic regularity theory, it is not difficult for us to see that, the
conclusions in Theorem 2.1 and Theorem 2.2 also hold true, if g ∈ Cα(m)(R), where α(m) =
max{1,|m|−2}.
7
3 Qualitative Analysis of the O.D.E
In this section, we will establish a series of lemmas to characterize the behavior of solutions to
(2.1) under some suitable assumptions on the function g.
First, let’s recall the Pohozaev identity of (2.1). By multiplying the both sides of equation
(2.3) by rh′(r) and integrating from s to r, we obtain the Pohozaev identity
r
rh′(r) 2− sh′(s) 2 = m2[sin2h(r)−sin2h(s)]+2 g(h(t))h′(t)t2dt, (3.1)
Zs
(cid:0) (cid:1) (cid:0) (cid:1)
or
rh′(r) 2− sh′(s) 2 = m2[sin2h(r)−sin2h(s)]+2[G(h(r))r2 −G(h(s))s2]
r
(cid:0) (cid:1) (cid:0) (cid:1) −4 G(h(t))tdt, (3.2)
Zs
where
π
G(x) = − g(t)dt.
Zx
Lemma 3.1. Assume that g(x) ∈ C∞([0, π]) satisfies (i) − (iii) and h(r) satisfies (2.1). If
there exists r ∈ (0,+∞) such that
0
0 ≤ h(r )≤ min{π−ξ, ξ} and h′(r )> 0,
0 0
then there exists r > r such that h′(r)> 0 for any r ∈ [r , r ] and h(r ) = ξ.
1 0 0 1 1
Proof. By the Pohozaev identity, we have
r
rh′(r) 2 = r h′(r ) 2+m2[sin2h(r)−sin2h(r )]+2 g(h(t))h′(t)t2dt. (3.3)
0 0 0
Zr0
(cid:0) (cid:1) (cid:0) (cid:1)
Let
r∗ = sup{s ∈ [r ,+∞) | h′(r)> 0, r ∈ [r ,s)}.
0 0
It is easy to see that r < r∗ ≤ +∞, since h′(r ) > 0.
0 0
We claim that
h(r∗) > ξ. (3.4)
If this was false, then there existed some r ∈ [r ,r∗) such that h(r) ≤ ξ. Hence, from the
0
assumptions on g(x) we have, for r ∈ [r , r∗), there holds
0
sin2h(r) ≥ sin2h(r ) and g(h(r)) ≥ 0. (3.5)
0
Combining (3.3) and (3.5) we obtain
rh′(r) ≥ r h′(r )> 0, ∀r ∈ [r , r∗). (3.6)
0 0 0
It follows that
r r 1
h(r)= h′(t)dt+h(r ) ≥ r h′(r ) dt+h(r ).
0 0 0 0
t
Zr0 Zr0
8
This implies that r∗ < +∞. Otherwise, we would deduce that h(r) is unboundedin the interval
[r ,+∞), a contradiction.
0
By the definition of r∗, we get h′(r∗) = 0 which contradicts (3.6). Thus, we show the
assertion. Therefore, we can choose r ∈ (r ,r∗) such that h(r ) = ξ. By the definition of r∗,
1 0 1
we get
h′(r)> 0, ∀r ∈ [r ,r ].
0 1
Corollary 3.2. Suppose that g(x) satisfies (i)−(iii). If h (r) is the solution of (2.1)-(2.2) with
a
a > 0, then there exists s ∈(0,+∞) such that h(r) increases monotonically from 0 to ξ on the
a
interval [0,s ].
a
Lemma 3.3. Suppose that g(x) satisfies (i)−(iii). If h(r), which is not a constant, satisfies
(2.1) on the interval (r ,+∞), ξ ≤ h(r) ≤ π for any r ∈ [r ,+∞) and lim h(r) = l > ξ, then,
0 0
r→∞
there hold that for any r ∈ [r ,+∞)
0
h′(r) > 0 and l = π.
Moreover, h(r) converges to π exponentially as r → +∞.
Proof. As π is a trivial solution of equation (2.1) and h(r) is not a constant function, then we
have
h(r) < π, r ∈ (r ,+∞). (3.7)
0
In fact, if there exists a r ∈(r ,+∞) such that h(r ) = π, by the condition of ξ ≤ h(r)≤ π for
1 0 1
any r ∈ [r ,+∞), we get h′(r ) = 0. The uniqueness of solutions to initial value problem tells
0 1
that h(r) ≡ π, which contradicts the fact that h(r) is not a constant function.
Since lim h(r) = l > ξ, then, there exists r ∈ [r ,+∞) such that, for any r ∈ [r ,+∞), we
1 0 1
r→∞
have
m2
sinh(r)cosh(r)+g(h(r)) < 0.
r2
From (2.3) we have that, for r ≤ s≤ r < +∞,
1
r m2
rh′(r) = sh′(s)+ sinhcosh+g(h) tdt (3.8)
t2
Zs (cid:20) (cid:21)
and
r
h(r) = h(s)+ h′(t)dt. (3.9)
Zs
Since ξ ≤ h(r) ≤ π for any r ∈ [r ,+∞), (3.9) implies that there exists C > 0 and r → +∞
0 k
such that
C
−h′(r ) ≤ .
k
r
k
Set
r m2
A(r) = sinhcosh+g(h) tdt,
t2
Zs (cid:20) (cid:21)
9
it follows from (3.8) that A(r ) ≥ −C. Onthe other hand, A(r)is decreasing and negative since
k
the integrand is negative on the interval [r ,+∞), we see that lim A(r) exists. It implies that
1
r→∞
m2
lim sinhcosh+g(h) = g(l) = 0.
r→∞ r2
Then we get l = π.
By (3.8) we infer that lim rh′(r) = l exists. Then, we can easily see that l is zero.
0 0
r→∞
Otherwise, h(r) would be unbounded by (3.9). Let r → +∞ in (3.8) and replace s by r, we get
+∞ m2
rh′(r) = − sinhcosh+g(h)t dt. (3.10)
t
Zr (cid:20) (cid:21)
From (3.10) and (3.7) we can deduce that for any r ∈[r ,+∞)
1
h′(r)> 0.
By the Pohozaev identity (3.2), we have that, for r ≤ s ≤ r < +∞,
0
rh′(r) 2− sh′(s) 2 = m2[sin2h(r)−sin2h(s)]+2[G(h(r))r2 −G(h(s))s2]
r
(cid:0) (cid:1) (cid:0) (cid:1) −4 G(h(t))tdt. (3.11)
Zs
Since lim rh′(r)= 0 and lim h(r) = π, we get
r→∞ r→∞
+∞
0 < G(h(t))tdt < +∞.
Zs
Replacing s by r and then letting r → +∞ in the above identity, we obtain
+∞
rh′(r) 2 = m2sin2h(r)+2G(h(r))r2 +4 G(h(t))tdt. (3.12)
Zr
(cid:0) (cid:1)
Since ξ ≤ h(r) < π for r ∈ [r ,+∞), from the above identity, we get h′(r) 6= 0 for any
0
r ∈ [r ,+∞). Ash′(r)iscontinuous ontheinterval[r ,+∞)andh′(r) > 0foranyr ∈ [r ,+∞),
0 0 1
we obtain h′(r) > 0 for any r ∈ [r ,+∞).
0
Now, we are in the position to prove that h(r) converges to π exponentially as r → +∞.
Let
h(r) = π−h(r).
Then, h(r) >0 on (0,+∞) satisfies the following equation
e
e 1 m2sin2h
h′′+ h′ = −g(π−h). (3.13)
r 2r2
e
Let f(r)= be−ǫr. Then, it iseeasy toeverify that f(r) satisfiees the following equation
1 ǫ
f′′+ f′ = (ǫ2− )f. (3.14)
r r
10