⎩
u(rˆ )=r2(N1+N2)eu+ˆ vˆ v(r)ˆ =r2(N1+N2)eu+ˆ vˆ
ˆ
u(0)=0,uˆ(0)=0,v(ˆ 0)=0,vˆ(0)=0.
(5.5)
Since U(t,s,ds)and V(t,s,ds)are both decreasing in t, we haveu andˆ vˆare nonin-creasing in r . But, any solution pair of (5.5) must be innonin-creasing in r . Actually, bothuˆ or vˆmust blow up at finite r . This contradiction shows that there exists s0 > 0 such that u(r,s,ds)or v(r,s,ds)blows up∀s > s0, and hence, by Lemma 2.3, so does u(r, α1, α2)orv(r, α1, α2)for anyα1≥s0≡ ˜α1orα2≥ds0≡ ˜α2. This proves (B).
Step 4. In this step, we prove the result of (C). For this purpose we claim the following statements.
(a) Letα∈and(u(r), v(r))=(u(r, α), v(r, α)). Then the corresponding(β1, β2) satisfies eitherβ1<∞orβ2<∞.
(b) =N T
u
v. (c) ∂⊂Su
T Sv.
(d) Ifα=(α1, α2)∈∂and(α1−, α2)∈for some >0, thenα∈ Su. (e) Ifα=(α1, α2)∈∂and(α1, α2−)∈for some >0, thenα∈Sv.
(f) Let1=∂
Suand2=∂
Sv. Then1
2=T . (g) is a simply connected and unbounded region.
(h) Ifα=(α1, α2)∈Su, then(α1+, α2)∈Su∀ >0.
(i) Ifα=(α1, α2)∈Sv, then(α1, α2+)∈Sv∀ >0.
(a) Suppose the result is not true. Thenβ1 = ∞, β2 = ∞and lim
r→∞r u(r)= −∞.
Hence, for each M >2 there exists rM >0 such that r u(r) < −M ∀r >rM. Furthermore we get eu(r)<C·r−M for large r and
β2= ∞
0
r eu(1−ev)dr ≤ ∞
0
r eudr <∞.
This contradiction proves (a).
(b) By (a) and the definitions ofuandv, we easily have (b).
(c) Let E=Su
T
Sv, α∈∂and(u, v)be the respective solution. Then, by the Hopf lemma we have u(r) < 0, v(r) < 0∀r > 0. By Proposition4.1, we have
∂⊂E. This proves (c).
(d) Ifα = (α1, α2) ∈ ∂andα = (α1−, α2) ∈ for some > 0, then, by Lemma2.3,v(r, α)→ −∞andv(r, α) < v(r, α)∀r>0. This provesα∈Su. (e) The proof is similar to (d).
(f) From (c)-(e),1 = ∅and2 = ∅, then by the continuity w.r.t. initial data, we obtain that
1∩Sv= ∅ and 2∩Su= ∅.
Therefore, (f) is proved.
(g) Supposeis not connected. Then there exist two disjoint open sets O1and O2
satisfying O1
O2 = . From (d)–(f), O1 and O2 possess one initial data of topological solutions at least, respectively, and by (A)–(B), there exist two distinct initial data of topological solutions T1and T2such that T1∈∂O1and T2 ∈∂O2. This contradicts the uniqueness of the topological solution. Henceis connected.
By using the similar arguments, we getis unbounded. From (A), we obtain that the setdoes not have a hole, that is,is a simply connected set. This shows (g).
(h) First we show that ifα=(α1, α2)∈ Su, then∀ >0 we haveα =(α1+, α2) and u(r0, α)≥0 for some r0=r0() >0, that is,α ∈/ Su∀ >0 by Lemma4.1.
Suppose there exists0 >0 such that u(r, α0) < 0∀r >0. By Lemma2.3, we have
u(r, α) <u(r, α) <u(r, α0) <0
v(r, α0) < v(r, α) < v(r, α) <0 ∀r ∈(0,∞). (5.6) From (5.6) andα∈Su, we obtain u(r, α0)is bounded below for large r . Then, by (ii) of Lemma4.1, we have u(r, α0) > 0∀r >0, and hence lim
r→∞u(r, α0)= −cu ≤ 0.
Furthermore, by Lemma4.2, we get lim
r→∞r u(r, α)= lim
r→∞r u(r, α0)=0. Combining these facts, we attainβ1(α)=2N1=β1(α0). But, from (5.6) we haveβ1(α) > β1(α0).
This contradiction proves our assertion and thus (h) holds.
(i) The proof is similar to (h).
By above claims (b)–(i) and the existence of the topological solution, we finally obtain (C) and the proof is complete.
In the following, we letu andv be defined in Sect.1. Then the corresponding (β1, β2)of each solution can be classified as follows.
Proposition 5.2. Let(u, v)be an entire solution of (1.13)–(1.14) and the corresponding (β1, β2)be defined in(1.20). Then the following statements are valid:
(a) β1 is continuous w.r.tα1 and α2 for all(α1, α2) ∈ N T ∪u. Similarly,β2 is continuous w.r.tα1andα2for all(α1, α2)∈N T ∪v.
(b) If(u, v)is a topological solution, then(β1, β2)=(2N1,2N2).
(c) If(u, v)is a non-topological solution, then the respective(β1, β2)satisfies (β1−2(N1+ 1))(β2−2(N2+ 1)) >4(N1+ 1)(N2+ 1).
(d) For anyα ∈ Su (resp.α ∈ Sv),(u, v)is a Type (IV) solution with(β1, β2) = (2N1,∞)(resp.(β1, β2)=(∞,2N2)).
(e) For anyα∈u(resp.α∈v),(u, v)is a Type (III) solution with 2N1< β1 ≤ 2N1+ 2 andβ2= ∞(resp.β1= ∞and 2N2< β2≤2N2+ 2).
Proof. (a) We prove the case ofβ1. The case ofβ2is similar. Let D=N T
u, α0= (α10, α20) ∈ D. First, we want to show that D = N T
u is open. Because of β1(α0) <∞, we obtain that
rlim→∞rv(r;α0) <−2−ε
for someε >0, by continuity and (1.13), there exist r0, δ >0 such that for all|α−α0|<
δ,
rv(r;α) <−2−ε
2 on [r0,∞),
which implyβ1(α) <∞, that is,α ∈ D. Thus the set D is open. Now ifδn >0 and δn→0 as n→ ∞such thatα1n =α10+δn, (α1n, α20)∈D andβ1,n =β1(α1n, α20)∀n.
We want to proveβ1n →β1(α0)as n → ∞. By monotone property, Lemma2.3, and the continuity of(u, v)w.r.t. the initial data, we obtain that
u(r, α1n, α20)u(r, α0)
v(r, α1n, α20)v(r, α0) pointwise in r as n→ ∞. (5.7) By (5.7), the definition ofβ1and the monotone convergence theorem, we obtainβ1n → β1(α0)as n→ ∞. The case ofδn<0 is similar. Soβ1is continuous w.r.t.α1. By using the same arguments, we getβ1is continuous w.r.t.α2. This proves (a).
(b) By (i) of Lemma2.2we obtain the result.
(c) Since (u, v)is a non-topological solution of (1.13), the respective β1 < ∞and β2<∞. Hence there exists a sequence{rj}such that rj → ∞and
r2jeu(rj)(1−ev(rj))→0, r2jev(rj)(1−eu(rj))→0 as j→ ∞.
By the Pohozaev identity, Lemma2.1, we easily obtain r u(r)·rv(r)−2
r 0
ev(s)(1−eu(s))s ds−2 r
0
eu(s)(1−ev(s))s ds +r2eu(r)(1−ev(r))+ r2ev(r)(1−eu(r))
=4N1N2+ 6 r
0
seu(s)+v(s)ds ∀r ∈(0,∞). (5.8) Taking r =rj on both sides of (5.8) and then letting j → ∞, we have
(2N1−β1)(2N2−β2)−2β1−2β2=4N1N2+ 6 ∞
0
r eu(r)+v(r)dr which implies
(β1−2(N1+ 1))(β2−2(N2+ 1))=4(N1+ 1)(N2+ 1)+ 6 ∞
0
r eu(r)+v(r)dr.
This proves (c).
(d) We prove the case of Su. The proof for Svis similar. Letα∈ Suand(u(r), v(r))= (u(r, α), v(r, α)). By Lemma 4.2 we have lim
r→∞r u(r) = 0 = 2N1−β1(α). This proves (d).
(e) We prove the case ofu. The proof forvis similar. Letα∈uand(u(r), v(r))= (u(r, α), v(r, α)). By (1.21), (1.22) and the definition ofu, we have lim
r→∞r u(r;α) <0 and lim
r→∞r u(r)=2N1−β1(α). Henceβ1(α) >2N1. Now we claimβ1(α)≤2N1+ 2 ∀α ∈ u. Suppose not, then there existsα ∈ u such thatβ1(α) > 2N1+ 2 and β2(α)= ∞. Then we obtain lim
r→∞r u(r)=2N1−β1(α) <−2, and thus∞
0 r eudr<
∞. From this we deduce
∞>
∞
0
r eudr>
∞
0
r eu(1−ev)dr =β2(α)= ∞.
This contradiction proves (e) and the proof is complete.
The following results describe the existence and properties of Type (V) solution.
Proposition 5.3. Wuand Wvare open subsets of R2. Furthermore, the following
In order to prove Proposition5.3, we need the following lemmas.
Lemma 5.1. Suppose(u(r), v(r))is a radial solution satisfying u(r0) >0, v(r0) <0
Proof. Suppose (u, v)is not an entire solution. Then there exists R0 > 0 such that u(r)→ ∞as r → R−0. Then we claim that
By (a), there exists 0< R1 < R0such that 2ev(r)−1 ≤ −12 ∀r ∈ (R1,R0). By (1.13) we easily have
r(u(r)+v(r))=R1(u(R1)+v(R1))+ r
R1
seu(2ev−1−ev−u)ds
≤C−1 2
r R1
seuds<0 for r close to R0−. This proves(u +v)(r) <0 for r near R0−and thus (b) follows.
Now by (a)-(b) we deduce
∞ = lim
r→R0−
r u(r)=R1u(R1)+ R0
R1
sev(eu−1)ds<∞.
This contradiction proves the first result. From Lemma4.1, we obtain the second result and the proof is complete.
The following lemma depicts the asymptotic behaviors of Type (V) solution at infin-ity.
Lemma 5.2. Let(u, v)be an entire solution of (1.13)–(1.14) on(0,∞). If u(r0)≥0 for some r0>0, then
(a) lim
r→∞r u(r)=λand lim
r→∞(u(r)−λlog r)=cufor some constantsλ >0 and cu. (b) rpeu(r)+v(r)∈ L1(0,∞)for any p≥0.
(c) lim
r→∞
rv(r)
r2+λ = −2+ecuλand lim
r→∞
v(r)
r2+λ = −(2+ecuλ)2, whereλand cuare the constants in (a).
Proof. (a) By Lemma4.1, we see that
u(r)≥C and 1−ev(r)≤ 1
2 on[r0,∞) for some C,r0>0, then
rlim→∞rv(r)= −∞ and v(r)≤ −Cr2 ∀r≥r0. From (4.1) we obtain(r u(r))>0 on[r0,∞)and
rlim→∞r u(r)=λ, 0< λ≤ ∞. (5.11) To complete this proof, we need the following fact.
Claim. λ <∞.
Proof of Claim. Supposeλ= ∞, then, using (1.13)–(1.14), we obtain ∞
0
r eu(r)+v(r)dr≥ lim
r→∞r u(r)−2N1= ∞, (5.12)
and by lim From (5.14) we see that r2+ peu+v is bounded from above by a positive constant and hence r eu+v ≤ Cr−(1+ p) for all large r and∞ C >0 and R>0. From this inequality and the claim in the proof of (a), we have that rpeu(r)+v(r)<Cr−2for all large r>0 for any p>0. Thus the result (b) is valid.
Thus (c) is true and the proof is complete.
Remark 5.2. If we replace u by v in the condition of Lemma5.2, we can obtain the following respective results. The proof is similar.
Lemma 5.3. Let(u, v)be an entire solution of (1.13)– (1.14) on(0,∞). Ifv(r0)≥0
Now we are in the position to prove Proposition5.3.
Proof of Proposition5.3. We prove the case of(Wu,Su). The case of(Wv,Sv)is simi-lar. We divide the proof into the following steps.
Step 1. Wuis open.
Letα¯ ∈ Wu. Then there exists r0 > 0 such that u(r0,α) >¯ 0, v(r0,α) <¯ 0 and v(r0,α) <¯ 0. By the continuity of(u, v), (u, v)w.r.tα, there existsδ >0 such that
u(r0, α) >0, v(r0, α) <0 and v(r, α) <0 ∀α∈Bδ(α).¯ (5.16) By (5.16) and Lemma5.1, we obtain(u(r, α), v(r, α))→(∞,−∞)as r → ∞ ∀α∈
Bδ(¯α). This proves Bδ(α)¯ ⊂Wuand hence Wuis open.
Step 2. Let α˜ = (θ, η) ∈ Su. Then there exists r0 > 0 such thatv(r0,α) <˜ 0 and v(r0,α) <˜ 0. By continuity, there exists >0 such thatv(r0, α) <0 andv(r0, α) <
0∀α=(α1, η)withη < α1< η+. By (h) of Step 4 in the proof of Proposition5.1 and (1.13), we have u(r1, α) >0 andv(r1, α) <0 for some r1>r0. From Lemma5.1, we obtain that(α1, η) ∈ Wu ∀η < α1 < η+. By Lemma 5.2, we also obtain the asymptotic behavior of(u, v)at∞and the corresponding(β1, β2)which satisfies
β1=2N1− lim
r→∞r u(r)=2N1−λ and β2=2N2− lim
r→∞rv(r)= ∞.
These prove (i) and (ii).
Step 3. Wuand Wvare all nonempty.
Since, by (C) of Proposition5.1, Suand Svare all nonempty, we get Wuand Wvare also all nonempty from (i)-(ii). This completes the proof.
Now, we give the proof of Theorems1.3and1.2in the following.
Proof of Theorem1.3. We divide the proof into the following steps.
Step 1. By (C) of Proposition5.1, (c) of Proposition5.2and (i)-(ii) of Proposition5.3, we obtain (i), (iii) and (iv)-(v) respectively.
Step 2. First we claim the following statements.
(a) For eachα∈ (Sv
v)(resp.α ∈(Su
u)), there does not exist{αk} ⊂u
(resp.{αk} ⊂v) such thatαk →αas k→ ∞.
(b) For eachα = (α1, α2) ∈ Sv (resp.α ∈ Su), there does not existk 0 with {αk = (α1+k, α2)} ⊂ N T (resp.{αk = (α1−k, α2)} ⊂ N T) such that αk→αas k→ ∞.
(c) For eachα = (α1, α2) ∈ Sv (resp.α ∈ Su), there does not existk 0 with {αk = (α1, α2−k)} ⊂ N T (resp.{αk = (α1, α2+k)} ⊂ N T) such that αk→αas i → ∞.
(a) We prove the case of Sv
v. The case of Su
uis based on the same arguments.
Suppose there exists{αk} ⊂usuch thatαk →αas k→ ∞for someα∈(Sv v). Then, by (d)-(e) of Proposition5.2, we have
β1(α)= ∞ and β1(αk)≤2N1+ 2∀k.
Denote(u(r), v(r)) = (u(r, α), v(r, α))and(uk(r), vk(r)) = (u(r, αk), v(r, αk))∀k.
Then, by the definition ofβ1, we obtain that there exists R0>0 such thatR0
0 r ev(1− eu)dr >2N1+ 2, and
2N1+ 2≥ lim
k→∞
R0
0
r evk(1−euk)dr
= R0
0
r ev(1−eu)dr (by Bounded Convergence Theorem)
>2N1+ 2.
Fig. 1. Structure of all entire solutions
This contradiction proves (a).
(b) We prove the case of Sv. The case of Suis similar. Letα=(α1, α2)∈ Sv. Suppose there existsk 0 with{αk =(α1+k, α2)} ⊂N T such thatαk →αas k → ∞.
Then, by (c)–(d) of5.2, we have
β2(α)=2N2andβ2(αk) >2N2+ 2∀k.
Then by the continuity ofβ2w.r.t.α1, i.e., (a) of Proposition5.2, we obtainβ2(α) =
rlim→∞β2(αk)≥2N2+ 2. This contradiction shows (b).
(c) The proof is similar to (b). We omit the details.
Since Suand Svare all nonempty, by (a)-(c) above we obtain that∀α=(α1, α2)∈Sv and∀ ¯α=(¯α1,α¯2)∈Su, there respectively existsδ1=δ1(α) >0 andδ2=δ2(α) >¯ 0 such that
(α1+δ, α2), (α1, α2−δ)∈v ∀0< δ < δ1
(α¯1−δ,α¯2), (¯α1,α¯2+δ)∈u ∀0< δ < δ2. (5.17) By (5.17) we deduce bothu andv are nonempty and connected. Now, by (C) of Proposition5.1and (a) above, we obtainN T = ∅. By Proposition5.2,is simple and open connected. From Lemma2.3, we also haveu, N T andvare all simple.
SupposeN T is not open. Then, by (5.17) and Lemma2.3, w.l.o.g., there exists {αi =(αi 1, αi 2)} ⊂vsuch thatαi → ¯αas i → ∞for someα¯ =(¯α1,α¯2)∈ N T. Letαi =(α¯1, αi 2). Thenαi → ¯αand{αi} ⊂v. By (a),(c) and (e) of Proposition5.2, we finally obtain
2N2+ 2≥ lim
i→∞β2(αi)=β2(α) >¯ 2N2+ 2.
This contradiction proves thatN T is open.
Now, suppose Z = ∂N T
∂ = ∅, then there exist α ∈ u and a sequence {αi} ⊂vsuch thatαi →αas i→ ∞. This contradicts (a). Hence Z = ∅and Z =T . Furthermore, by (a) we also getN T is connected. The proof is complete.
Proof of Theorem1.2. Let(u, v)be a radial solution of (1.7). Then, by Propositions4.1 and 5.2, we obtain that (u, v) must be one of the Types (I)–(V). Conversely, by Theorem1.1, the Type (I) solution, i.e., topological solution, exists and is unique. Then,
by (C) of Proposition 5.1, we have that both ∂andare nonempty. Thus Types (II)–(IV) solutions all exist due to Proposition5.1and Theorem1.3. In particular, the non-topological solution exists. Furthermore, by Proposition5.3, the Type (V) solution exists. We complete the proof.
Remark 5.3. Combining the results of Theorems1.1,1.2and1.3, we can sketch the structure of entire solutions as in Fig.1.
Acknowledgement. The authors would like to express their gratitude to the referee for valuable comments and suggestions.
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Communicated by M. Aizenman