In this section we prove the exponential decay of ground states with the aid of Propo-sition 4.8, The approach of using the fundamental solution of Helmholtz equation is exactly taken from [14], Lemma A.5.
Proposition 4.13. Let u ∈ GM,q, for arbitrary0 ≤ M ≤ 1 and q ≥ 0. For any a > 0, there exist constantsUj(a) (j = 1, 0, −1) such that uj(x) ≤ Uj(a)e−a|x|.
Proof. (2.4b) can be arranged as (−∆ + a2)u0 = Q0u0, where
Q0 = a2+ µ − V − 2β0|u|2− 2β1(u1− u−1)2. (4.12) Thus
u0(x) = (Ya∗ (Q0u0))(x) = Z
Ya(x − y)Q0(y)u0(y)dy,
where Ya(x) = e−a|x|/(4π|x|) is the fundamental solution of the operator −∆ + a2. (Ya is also referred to as the Yukawa potential. See [13], 6.23.) By the assumption (A1), Q0 < 0 outside a bounded set, say B(R0), the open ball centered at the origin with radius R0. Thus we obtain
u0(x) ≤ Z
|y|<R0
Ya(x − y)Q0(y)u0(y)dy = e−a|x|
Z
|y|<R0
ea(|x|−|x−y|)
4π|x − y| Q0(y)u0(y)dy.
Thus u0(x) ≤ U0(a)e−a|x|, where (see also Lemma 4.14 below) U0(a) = sup
x∈R3
Z
|y|<R0
ea(|x|−|x−y|)
4π|x − y| Q0(y)u0(y)dy < ∞. (4.13) For uj, j = 1, −1, we similarly have
(−∆ + a2)uj = Qjuj − 2β1u20(uj − u−j)
from (2.4a) and (2.4c), where all the indices 0 replaced by 1. In contrast, the fact u−1 ≤ u1makes it difficult to apply the same argument to u−1. Nevertheless, also since u−1 ≤ u1, at least we can choose U−1(a) = U1(a).
For our next result, we give the following estimate of Uj(a).
Lemma 4.14. For j = 1 and 0,
where the last inequality is obtained by H¨older’s inequality and the fact Z
(uj)2 ≤ Z
|u|2 = 1.
We thus obtain the assertion of the lemma.
The assertion of exponential decay is indeed far stronger than what we need. On the other hand, we will consider sequences {un} of ground states corresponding to differ-ent values of q, and hence differdiffer-ent Lagrange multipliers, where estimates independdiffer-ent of n are required. We give what we really need in the following.
Corollary 4.15. Given a sequence un = (un1, un0, un−1) ∈ GM,qn. Letµnandλnbe the Lagrange multipliers corresponding toun. If the sequences{µn} and {λn} are both bounded, then for anyε > 0, there is rj > 0 (j = 1, 0, −1) independent of n such that uj(x) ≤ ε for |x| ≥ rj.
Proof. The assertion is easily seen by repeating the proof of Proposition 4.13 for un. It suffices to give Ujn(a) (the analogue of Uj(a) for un) an upper bound independent of n. Take U0n(a) for example. By assumption, there is c > 0 such that µn < c for every n. From (4.12) we have
Qn0 = a2+ µn− V − 2β0|un|2 − 2β1(un1 − un−1)2
≤ a2+ µn− V
< a2+ c − V.
Hence we can find R0 independent of n so that Qn0 < 0 outside B(R0). Then, by Lemma 4.14, we have
U0n(a) ≤ eaR0 4π sup
x∈R3
Z
|y|<R0
Qn0(y)2
|x − y|2dy
12
≤ eaR0 4π sup
x∈R3
Z
|y|<R0
(a2+ c − V (y))2
|x − y|2 dy
12 ,
which is independent of n. U1n(a) can be estimated similarly, and again for U−1n (a) we use the fact U−1n (a) ≤ U1n(a).
Chapter 5
The Bifurcation Phenomenon
We begin our proof of the bifurcation phenomenon. According to Remark 4.3, it suf-fices to consider 0 < M < 1.
Our main theorem is the following.
Theorem 5.1. For fixed 0 < M < 1, there is a qc > 0 such that for q > qc,u ∈ GM,q satisfiesu0 > 0, while for 0 ≤ q < qc,z is the unique element in GM,q.
Remark 5.1. We do not know what happens at the critical qc. Since Eg is continuous with respect to q, z is of course an element in GM,qc. However, since we do not prove the uniqueness of ground state, we are not sure if it’s possible that there are other three-component ground states at qc. We give more detailed discussions in §7.1.
The proof idea of Theorem 5.1 is to use (2.7) to derive some conditions on the situation to be excluded. More precisely, to prove that u ∈ GM,q cannot have some property, we assume the opposite, then exploit the fact that any redistribution v ∈ AM
of u (in particular those not having the property) satisfy (2.7). How this idea works will be clear in the proof.
We regard M as a fixed number in (0, 1) in the following. The proof is divided into three claims.
Claim 1. For q large enough, z is not an element in GM,q.
Proof. Assume z ∈ GM,qfor some M, q. Since z is independent of q, it’s quite easy to prove the claim by (2.7). For example, consider v to be the redistribution of z defined
by
v21 = (1 − σ)z12 v20 = σz12+ z2−1 v2−1= 0 ,
(5.1)
where σ = (1 − M )/(1 + M ), which is just the constant making v ∈ AM. Then (2.7) implies
EZee[z] − EZee[v] ≤ E1[v] − E1[z]. (5.2)
It’s easy to check that the left-hand side of (5.2) equals (1 − M )q. In contrast, the right-hand side, no matter what it is, is independent of q. Thus, since M < 1, (5.2) gives an upper bound of q. That is there is an upper bound of q for z to be in GM,q.
Since three-component elements in GM,q in general depend on q, it’s far more difficult to prove that there is a positive lower bound of q for the existence of u ∈ GM,q
with u0 > 0. We leave this to the last claim. We shall first give the observation that the two-component regime and the three-component regime are really separated by a specific qc. That is, we exclude by the next claim the possibility that two-component and three-component ground states will alternately be the case (in any range on the q-axis).
Claim 2. Assume for some q there exists u ∈ GM,qwithu0 > 0, then for every q0 > q, z /∈ GM,q0.
Proof. Let’s here write E [u, q] instead of E [u] to specify the value of q. Since u ∈ GM,q, E [u, q] ≤ E[z, q]. Thus, by the assumption u0 > 0, for q0 > q we have
E[u, q0] = E [u, q] + (q0− q) Z
u21+ u2−1
< E [z, q] + (q0− q) Z
z12+ z−12 = E[z, q0].
Hence z /∈ GM,q0.
Now define
qc= infq
z /∈ GM,q0 for q0 > q .
From Claim 1, qc < ∞. By definition of qc, for any q > qc and v ∈ GM,q, we have v0 > 0. Moreover, Claim 2 implies that for any 0 ≤ q < qc, z is the unique element in GM,q. To complete the proof of Theorem 5.1, it remains to show qc > 0. That is we have to prove the following assertion.
Claim 3. There exists q > 0 such that z ∈ GM,q.
Since the proof of Claim 3 requires much more effort, we give it in a separate section.
5.1 Proof of Claim 3
Let u ∈ GM,q. To give a restriction on the presence of u0, we consider the redistribu-tion v ∈ AM of u defined by
v21 = u21+1 2u20 v02 = 0
v−12 = u2−1+1 2u20.
(5.3)
Then (2.7) implies
q Z
u20 ≥ 2β1 Z
u20(u1− u−1)2. (5.4)
From (5.4), it’s easy to see u0 ≡ 0 if q = 0 (we can not have u1 − u−1 ≡ 0 since M > 0). This is exactly the argument used in the proof of Theorem 3.3. For q > 0, however, no matter how small it is, whether u0 ≡ 0 is not so obvious. We shall prove that, for q small enough, there does exist a positive constant c independent of q, such that the right-hand side of (5.4) is no less than cR u20, and hence obtain a lower bound of q for u0 > 0. This is made possible by the assertions of the following lemma.
Lemma 5.2. Given q ∈ [0, ∞). Let un ∈ GM,qn and u∞ ∈ GM,q be as claimed in Corollary 4.6, then the following assertions hold.
(a) There exists a large enough R such that 1
We first prove Claim 3 by this lemma.
Proof of Claim 3. Let un = (un1, un0, un−1) ∈ GM,qn be as claimed in Corollary 4.6 for q = 0. Then un → z in B since z is the unique element in GM,0for 0 < M < 1. For this sequence, let R be the corresponding radius asserted in (a) of Lemma 5.2, and let k = infB(R)(z1− z−1). Note that k > 0 by Corollary 4.12. Now by (b) of Lemma 5.2, un → z uniformly, and hence (un1 − un−1) ≥ k/2 on B(R) for n large enough. From this fact and (5.5) we obtain
Z
for n large enough. On the other hand, for any n, (5.4) implies
qn
Now we prove Lemma 5.2. The proofs of both assertions need the following ob-servation.
Lemma 5.3. Given q ∈ [0, ∞). Let un ∈ GM,qn and u∞ ∈ GM,q be as claimed in Corollary 4.6, then the Lagrange multipliersµn, λn corresponding to un converge respectively to those corresponding tou∞, denoted byµ∞, λ∞.
Proof. Multiply (2.4a) by u1 and multiply (2.4c) by u−1, and take integration, we obtain un0, and take integration, we obtain
µn On the other hand, by the assumption (A1), there exists R > 0 such that V (x) ≥ 2C for |x| > R, and hence
From (5.10) and (5.11), we obtainR (un0)2 ≥ 2R
B(R)c(un0)2, which is easily checked to be equivalent to (5.5).
Proof of Lemma 5.2 (b). The idea is that, if the GP system (2.4) for un tends to that for u∞ in a suitable sense, then uniform convergence can be obtained by the global boundedness result for elliptic operators. We take (2.4a) for example.
Let vn1 = un1 − u∞1 . Subtract (2.4a) for u∞from (2.4a) for un, we obtain
∆v1n− V (x)v1n= Pn− P∞+ Sn− S∞, (5.12)
where
Pn = −(µn+ λn− qn)un1,
Sn = 2β0|un|2un1 + 2β1(un0)2(un1 − un−1) + un1 (un1)2− (un−1)2,
and P∞and S∞are given by the same expressions with n replaced by ∞ (q∞is under-stood to be q). Apply global boundedness theorem for elliptic operators (see e.g. [8], Theorem 8.16) to (5.12), we obtain for every r > 0
sup
B(r)
|v1n| ≤ sup
∂B(r)
|v1n| + CkPn− P∞+ Sn− S∞kL2, (5.13)
where C > 0 depends only on the radius r and supB(r)V . Now since qn → q, µn → µ∞, λn → λ∞ (by Lemma 5.3), and un → u∞ in B, we see Pn − P∞ → 0 in L2. Also, Sn− S∞ → 0 in L2 since H1 is continuously embedded in L6. On the other hand, µn → µ∞ and λn → λ∞ also implies µn, λn, µ∞ and λ∞ all lie in a bounded set. By Corollary 4.15, given ε > 0, we can find r1such that each un1 as well as u∞1 are bounded above by ε outside B(r1). In particular, we have
sup
|x|≥r1
|vn1(x)| ≤ 2ε for all n. (5.14)
Let r = r1in (5.13), and let n → ∞, we obtain
lim sup
n→∞
sup
x∈B(r1)
|v1n(x)|
≤ 2ε. (5.15)
From (5.14) and (5.15) we have sup
x∈R3
|v1n(x)| ≤ 3ε for n large enough.
Since ε > 0 is arbitrary, we conclude that v1n → 0 uniformly on R3. Similarly v0nand v−1n converge to zero uniformly, which complete the proof.