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On a semilinear elliptic equation from the sigma model of quantum field theory
Nonlinear Analysis: Real World Applications 12 (2011) 885–894
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On a semilinear elliptic equation from the sigma model of quantum field theory Janpou Nee ∗ ChienKuo Technology University, Chung-Hua, Taiwan, ROC
article
abstract
info
Article history: Received 7 January 2010 Accepted 9 August 2010
In this paper, we consider both the existence and the non-existence of solutions to the generalized semilinear elliptic equations from the sigma model of quantum field theory of condense matter. The interesting part of the result is that the conditions to prove the existence and the non-existence of the solution are not determined by the nonlinearity of the nonlinear function but rather the quantitative value of its source term. © 2010 Elsevier Ltd. All rights reserved.
Keywords: Elliptic equations Existence of solutions Super-solution Sub-solution Minimization of variational method
1. Introduction In this paper, we consider both the existence and the non-existence problem to the following type of the nonlinear elliptic equation,
1u =
f (u) 1 + g ( u)
− h(x),
x ∈ R2 .
(1)
This type of equation appears frequently in the problem of the prescribed curvature or the sigma model of the quantum field theory of condensed matter [1]. For the results of the prescribed curvature, the reader may refer to [2–7], and the references for the quantum field theory of condensed matter, in particular, the sigma model of Heisenberg ferromagnetic can be found in [1,8]. The sigma model of quantum field theory relates to the existence of Higgs which concern Lepton and Quark, thus the source of all matters. Therefore, it is important to understand the behavior of the solution of the sigma model. Schroers [8] proposed a model of the sigma field as follows:
1u =
4eu 1 + eu
− 4π Σjk=1 nj δpj + 4π Σjl=1 mj δqj ,
x ∈ R2 ,
(2)
where δp is the delta function at p ∈ R2 and p is the location of the singularity of the map. Yang [1] shows that for each natural number, N, there are solutions to saturate the classical energy lower bound for the field configurations in the topological family deg(0) = N, if and only if N ̸= 1. The unexpected results of [1] are that both the existence and the non-existence of the solution of Eq. (2) depend fully on the quantitative value of source function h. In this paper, we will extend results of Yang [1] to Eq. (1). The main assumptions of this paper are as follows: E-1
h ∈ L1c (R2 ), and 4 >
R2
hdx = H > 2,
where L1c (R2 ) = {u ∈ L1 |u has compact support}.
∗
Tel.: +886 4 7111111x2556. E-mail addresses: [email protected], [email protected].
1468-1218/$ – see front matter © 2010 Elsevier Ltd. All rights reserved. doi:10.1016/j.nonrwa.2010.08.013
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J. Nee / Nonlinear Analysis: Real World Applications 12 (2011) 885–894
For the nonlinear functions f (·) and g (·), we assume that
µf eu ≤ f (u) ≤ νf eu , µg eu ≤ g (u) ≤ νg eu where 0 < µf < νf , 0 < µg < νg . f , g both are of C 2 satisfying Cf > f ′ > 0, Cg > g ′ > 0 for some constant Cf and Cg .
A-1 A-2
We will show that under assumptions (E-1), and (A-1), and (A-2), the results of [1] remain true. For the non-existence problem, the assumption of symmetry is essential in general. However, we shall apply the method as that of Yang [1] to show that Eq. (1) has no solution under a weaker assumption of symmetry together with the complementary condition of (E-1). 2. Construction of super-solution and sub-solutions It is difficult to exhibit the behavior of the solution of Eq. (1) directly. Therefore, we introduce an auxiliary equation which helps us to get a super-solution and a sub-solution to Eq. (1). The auxiliary equation that we insert is of the following form:
1u =
beu 1 + aeu
− h,
(3)
bx where the reaction term 1+ is monotone. Therefore, by the condition (A-1) the solution of Eq. (3) is a super-solution of ax Eq. (1) when a = µg , b = νf and is a sub-solution when a = νg , b = µf . The monotonic property of the reaction term will be frequently applied throughout this paper. Before we start to prove our results, we introduce the following notations. First, the weighted function:
K = eβ v¯ ≥ 0,
K = O|x|−β ,
β > 0,
(4)
where the function v¯ and the parameter β of Eq. (4) above will be given later. We denote the usual Lp space over R2 by Lp and its norm by ‖ · ‖p . We define the weighted measure dµ = K dx and we use Lp (dµ) to denote the induced Lp space. In order to demonstrate the asymptotic behavior of the solution, we consider the subspace Ξ of L2 with norm:
‖w‖2Ξ = ‖∇w‖2 + ‖w‖2L2 (dµ) ≤ ∞,
for w ∈ Ξ .
(5)
Note that Ξ contains those constant functions defined on R2 . Thus, to exclude the constant functions from Ξ , we consider ˜ of Ξ where the closed subspace Ξ
∫ ˜ = w ∈ Ξ Ξ
R2
w(x)dµ = 0 .
(6)
˜ then satisfies the following Trudinger–Moser inequality: The function w ∈ Ξ ∫ R2
e|w| dµ ≤ C e
‖∇w‖2 2 4γ
,
w ∈ Ξ˜
(7)
where
γ ≤ min{2, β − 2},
and C = C (γ ).
(8)
The constant β is chosen according to function h which will be given later. The other constant γ we choose in (8) relates to the asymptotic behavior of the solution while the range of γ in (7) is not an issue that we concerned at here. Note that the embedding Ξ → L2 (dµ) is completely continuous. To prove our results, we first construct a sub-solution to Eq. (1). To this end, we consider the regularization of h and we denote D1 = {x ∈ R2 |h(x) > 0}, D2 = {x ∈ R2 |h(x) ≤ 0}, and h(x), 0,
h1 =
x ∈ D1 otherwise
−h(x), h2 = 0,
x ∈ D2 otherwise.
By the assumption (E-1), Di are bounded domains where supp(hi ) = Di for i = 1, 2. Let Di ⊃ Dϵi = {x ∈ Di |dist(x, ∂ Di ) ≤ ϵ} and G˜ i ∈ C ∞ (R2 ) be the regularization of hi respectively. We define
1vi =
hi − G˜ i , 0,
x ∈ Di x ∈ R2 \ Di ,
(9)
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and thus vi are non-positive functions for i = 1, 2. Furthermore, we choose v3 ∈ C ∞ (R2 ), so that
−1 ln |x|, 2π
v3 =
|x| ≥ 1, x ∈ R2 ,
(10)
and then we have
∫
(−1v3 ) = 1.
R2
(11)
We choose β > 0 such that 2<β
(12)
βv3
and set K = e which is the same as the weighted function K (x) of Eq. (4) of the weighted Sobolev space that we mentioned previously. By taking the transformation: u = −v1 + v2 + βv3 + w,
(13)
Eq. (3) becomes bK ev2 +w
1w =
ev1 + aK ev2 +w
− G,
(14)
where G = G˜ 1 − G˜ 2 + βv3 is of compact support satisfying
∫
G(x)dx = H − β > 0.
R2
(15)
From (15) we choose h¯ ∈ Cc∞ (R2 ), such that h¯ > 0, and it satisfies
∫
h¯ (x)dx = H − β.
R2
(16)
To get a proper sub-solution of (1), we study the following equation: bK ew
1w =
ev1 + aK ew
− h¯ ,
(17)
which in term is a sub-solution of Eq. (14). To get the solution of Eq. (17), we study the solution of the following functional: I (w) =
∫
1 R2
2
|∇w|2 +
b a
ln(ev1 + aK ew ) − h¯ w
(18)
in the admissible space: bK ew
∫
A = w ∈ Ξ
R2
ev1 + aK ew
∫ = R2
h¯ .
Our goal is to show that I (u) is a weakly lower semi-continuous functional on Ξ . Lemma 1. I ≥ C1 ‖∇w‖2 − C2 is bounded below over A. Proof. To achieve our goal, we use the fact that supp(v1 ) ⊂ D1 . We rewrite I (w) of (18) as I (w) =
∫
1 R2
2
|∇w|2 +
b a
ln(1 + aK ew ) − h¯ w dx +
b
∫
a
ln D1
ev1 + aK ew 1 + aew
.
t Since the function F (t ) = α+ increases for α < 1, we have F (t ) ≥ F (0) for all t ≥ 0. Therefore, 1 +t
∫ ln
ev1 + aK ew
D1
1 + aew
∫
(19) ev1 +aK ew 1+aK ew
> ev1 and
v1 dx > −C∗ > −∞.
≥ D1
Substituting the above into (19), we have I (w) ≥
[
∫ R2
1 2
|∇w|2 +
b a
]
ln(1 + aK ew ) − h¯ w − C∗ .
(20)
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Since ln(1 + aK ew ) = βv3 + w + ln(a + e−(βv3 +w) ), we have W =
b a
ln(1 + aK ew ).
(21)
Thus, from (20) we derive I (w) ≥
∫
1 R2
2
|∇w|2 +
∫
∫
¯ + hW
W− R2
∫
R2
R2
h¯ (βv3 ) +
b a
∫ R2
h¯ ln(a + e−(βv3 +w) ) − C∗ .
(22)
From (21), it yields
∇W =
bK ew
· (β∇v3 + ∇w),
1 + aK ew
and we have bK ew
∫
2
1 + aK ew
R2
bK ew
∫ ≤
1 + aK ew
R2
bK ew
∫ ≤ R2
ev1 + aK ew
= H − β.
Applying Trudinger–Moser inequality (7), it yields
‖∇ W ‖2 ≤ ‖∇w‖2 + C . Substitute the above inequality into (22) and by the fact that W > 0, we obtain I (w) ≥
1 4
1
2 2
2 2
W2
∫
‖∇w‖ + ‖∇ W ‖ + 4
R2
1+W
∫ dx − R2
¯ − C1 . hW
(23)
To complete the proof, we need the following interpolation inequality over R2 :
∫
∫
4
W dx ≤ 2 R2
∫
2
W dx R2
R2
|∇ W |2 .
(24)
Since h¯ is of compact support, we induced from (24) that
∫
¯ dx ≤ ‖h¯ ‖ 4 ‖W ‖4 hW 3
R2
≤ ϵ‖W ‖2 + C (ϵ)‖∇ W ‖2 + C 1
≤ ϵ‖W ‖2 + ‖∇ W ‖22 + C (ϵ),
(25)
8
where ϵ is small. From (24) and W > 0, we obtain W2
∫ |W |2 ≤ 1 + R2
1+W
+ ‖∇ W ‖22 .
(26)
Substituting (24), (26) into (23) yields I (w) ≥
1 4
‖∇w‖22 + C1 ‖∇ W ‖22 +
where Ci > 0 are constant for i = 1, 2.
W2
∫ R2
1+W
− C2 ,
(27)
Lemma 2. If I (w) satisfies (18), then min I (w) ≡ γ0 ,
(28)
w∈A
has a solution. Proof. Let wj be a minimizing sequence of (28) and let Wj =
b a
ln(1 + aK ewj ),
j = 1, 2, . . . ,
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then with (23) and (26) imply that {Wj } is bounded in L2 and thus bounded in L1 . Applying I (wj ) = γ0 and inequality (27), we get ‖∇wj ‖2 bounded. Let wj = w ¯j+w ˜ j where w ¯ j is the constant part and w ˜ j ∈ Ξ˜ , and we will show that w ¯ j is bounded. Let supp(v1 ) ⊂ D0 . By (19) and v1 < 0, we have
γ0 ≤ I (wj ) 1
=
2 1
≤
2
b
‖∇wj ‖22 + ‖Wj ‖1 − a
∫ R2
b
‖∇wj ‖22 + ‖Wj ‖1 − w ¯j
∫
a
∫
b
h¯ wj +
ln
a
h¯ −
1 + aewj
D0
∫
R2
ev1 + aK ewj
R2
h¯ w ˜ j,
thus
γ0 ≤
1 2
b
‖∇w ‖ + ‖Wj ‖1 − w ¯j 2 j 2
∫
a
h¯ −
∫
R2
R2
h¯ w ˜ j.
(29)
Inserting the following Poincaré type of inequality
∫
w2 dµ ≤
R2
∫ R2
|∇wj |2 w ∈ Ξ˜
(30)
into inequality (29), we obtain
w ¯j
∫ R2
1 h¯ ≤ |γ0 | + ‖∇wj ‖22 + ‖Wj ‖1 − 2
∫ R2
h¯ w ˜ j,
and hence
w ¯j
∫
1
h¯ ≤ |γ0 | +
2
R2
‖∇wj ‖22 + ‖Wj ‖1 + C .
(31)
Therefore, w ¯ j is bounded from above. Next, we will show that w ¯ j is bounded from below. To this end, we will use the constraint of the definition of the admissible space that
∫
∫ bK ewj + w v j 1 + aK ewj R2 \D0 1 + aK e D e ∫ ∫ 0 wj bK e bK ewj < + , v1 + aK ewj v1 + aK ewj R2 e D0 e
h¯ = R2
∫
bK ewj
and thus
∫ R2
h¯ < |D0 | + µew¯ j
∫ R2
ew˜ j dµ.
(32)
Without lost of generality, we shall assume that |D0 | ≤
∫ w ¯ j > ln |D0 | −
h¯
∫
R2
¯ Substituting (7) into (32), we obtain h.
− ln
R2
R2
ew˜ j dµ ,
and hence
w ¯ j > −C1 − C2 ‖∇wj ‖22 .
(33)
Since ‖∇wj ‖ is bounded, w ¯ j is bounded from below. Similarly {w ˜ j } is bounded in Ξ , there exists a w ∈ Ξ and a subsequence which converges weakly to w in Ξ . Thus
wj = w ¯j +w ˜ j ⇀ w,
in Ξ .
Since the embedding Ξ → L (dµ) is completely continuous, we may assume without generality that wj → w strongly in L2 (dµ). Next, we will show that w ∈ A, or equivalently 2
bK ew
∫ R2
ev1 + aK ew
= H.
Let
∫ aj =
bK ewj bK ew . − w v v w j 1 1 + aK e e + aK e R2 e
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For any ϵ > 0, we choose an open set D satisfying |D| ≤ 2ϵ , then
∫ wj bK ewj bK ew bK ew + bK e − − w w w w v v v v j j 1 + aK e 1 + aK e e 1 + aK e e 1 + aK e R 2 \D e D e ∫ bK e|w|+|wj | ≤ 2|D| + |wj − w|; v ∫
aj =
e
R2 \D
1
thus, ¯ C (‖∇wj ‖2 +‖∇w‖2 ) ). aj < ϵ + C (ϵ)‖wj − w‖L2 (dµ) sup(ew¯ j +w+ 2
2
(34)
j
Let j → ∞ and we have a = lim sup aj ≤ ϵ; j→∞
thus, a = 0 and hence w ∈ A. Next, we will show that I (·) is weakly lower semi-continuous on Ξ . To this end, we let η = t wj + (1 − t )w where 0 ≤ t ≤ 1; then,
∫
wj
w
| ln(ev1 +aK e ) − ln(ev1 +aK e )| ∫ ∫ K eη |wj − w| ≤ |wj − w| + v
ϵj =
R2
e
R2 \D
D
≤ C1 ‖wj − w‖L2 (dµ) +
1
K e|wj |+|w|
∫ R2 \D
ev1
|wj − w|
and ϵj → 0 as j → ∞. This shows that I (·) is weakly lower semi-continuous on Ξ and I (w) ≤ lim inf I (wj ) = a thus w ∈ A that minimize I (·). Lemma 3. The optimization problem (28) of Eq. (17) has a solution. Proof. The Lagrange multiplier implies that bK ew
∫ ∇w · ∇η + R2
ev1
+
aK ew
∫ − h¯ ηdx = λ
bK ev1 +w R2
(ev1
+ aK ew )2
η,
(35)
where λ ∈ R is a constant and η ∈ Ξ is an arbitrary trail function. Let η ≡ 1 and by definition w ∈ A, we see that λ = 0. Inserting λ = 0 into (35), we see that w is a weak solution of (17). By the standard elliptic theory, w is a classical solution of (17). To demonstrate the asymptotic behavior, we now introduce notations that will appear later. For δ ∈ R and s ∈ N, we let Ws2,δ to be the closure of Cc∞ (R2 ) under the norm:
‖η‖2W 2 = s,δ
−
‖(1 + |x|)δ+|α| Dα η‖22 ,
|α|≤s
where Cc∞ (R2 ) contains the functions defined over R2 with compact supports and C0 (R2 ) denote the set of continuous functions over R2 vanishing at infinity. The following results of weighted Sobolev space will be used to exhibit the asymptotic behavior of the solution of Eq. (14). Lemma 4.
(i) If s > 1 and δ > −1, then Ws2,δ ⊂ C0 (R2 ).
(ii) For −1 < δ < 0, the Laplace operator ∆ : W22,δ → W02,δ+2 is 1 − 1 and
∫ ∆(W22,δ ) = F ∈ W02,δ+2
R2
F =0 .
(iii) If η ∈ Ξ and 1η = 0, then η = k where k is a constant. Using β < 4 and the above results, we obtain the following lemma. Lemma 5. If c < 0, then Eq. (14) has a sub-solution w − satisfying w − < c in R2 .
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Proof. We will first consider the solution of the following equation:
1w = h¯ − G,
(36)
by finding the minimizer w1 of functional J (w) =
∫
1 R2
2
‖∇w‖2 + (h¯ − G)w.
(37)
˜ . To show that w1 → k as |x| → ∞, we By condition (E-1), the solution of Eq. (37) is unique over the admissible space Ξ choose 0 > δ > −1 and since h¯ − G ∈ L(R2 ) ∩ W02,δ+2 and R2 h¯ − G = 0, (i) and (ii) of Lemma 4 of weighted Sobolev
space give η ∈ W22,δ satisfying Eq. (36) and η = 0 as ‖x‖ → ∞, moreover η ∈ L2 (dµ). Since ∇η ∈ W02,δ+1 and δ > −1, we deduce that η ∈ L2 (R2 ). Hence η ∈ Ξ . Since η − w1 ∈ Ξ and ∆(η − w1 ) = 0, (iii) of the previous lemma then implies that η − w1 = k where k is a constant, and we have that w1 → k at infinity. Let w2 be the solution of Eq. (17) and in fact by the constrain of admissible space, we have bK ew
∫
ev1 + aK ew
R2
− h¯ = 0.
Let supp(v1 ) ⊂ D; then,
∫ R2
(1 + |x|δ+2 )2
bK ew2
ev1 + aK ew2
2
∫
(1 + |x|δ+2 )2 + C0
≤ D
∫ R2
e2w2 dµ,
where C0 = sup {K (x)(1 + |x|δ+2 )2 }, x∈R2
and C0 is finite if 2(δ + 2) ≤ β . Thus, we choose β > 0 such that δ = −2 + bK ex ev1 + aK ex
− h¯ = ρ(x);
β 2
, and we let (38)
then, ρ(x) ∈ W02,δ+2 and condition β < 4 is equivalent to −1 < δ < 0. Now by (i) and (ii) of the previous lemma, there exists η′ ∈ W22,δ satisfying
1η′ = ρ and hence η′ − w2 = k′ where k′ is constant and hence w2 → k′ at infinity. x Since wi are bounded, we may choose C such that w ¯ = w1 − C +w2 < c < 0 and w1 − C < 0. Let function F (x) = 1+bKaKe ex ; then by the monotonicity of F (x) we have F (w) ¯ = F (w1 − C + w2 ) ≤ F (w2 ). Hence
1w ¯ = F (w2 ) − G ≥ F (w) ¯ −G bK ev2 +w¯
− G, + aK ev2 +w¯ and by transformation (13), w ¯ is a sub-solution to Eq. (14). ≥
ev 1
Next, we shall construct the super-solution of Eq. (1). Similar to the construction of the sub-solution, we consider the modified equation:
1w =
bK ew 1 + aK ew
− G¯ ,
(39)
where G¯ = G1 + β 1v3 . Let h˜ ∈ Cc∞ (R2 ) such that h˜ ≥ 0 and
∫
˜ = hdx R2
∫ R2
G¯ dx = 2π (2H + − β).
(40)
We consider the following equation:
1w =
bK ew 1 + aK ew
− h˜ .
(41)
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Since the function h˜ is chosen as similar to the h¯ of the modified Eq. (17), we may apply Lemmas 2–4 to get a solution w1 of Eq. (41) satisfying w1 → C as ‖x‖ → ∞. Let w2 be a bounded solution of
1w = h˜ − G¯ .
(42)
We may assume that w2 ≥ 0. Let w ˜ = w1 + w2 ; then, from (41) and (42) we have
1w ˜ =
bK ew 2
1 + aK ew 2
− G¯ .
x By monotonicity of function 1+ , we derive x
1w ˜ ≤
≤
bK ew1 +w2
− G¯
1 + aK ew1 +w2 bK ew˜ 1 + aK ew˜
− G¯ + h+
in the distribution sense. Let w ∗ = w ˜ − v2 ; then, the above inequality implies that
1w ≤ ∗
∗ bK ew +v2
− G. ∗ ev1 + aK ew +v2
(43)
Thus, w ∗ is a super-solution to Eq. (3) in the distribution sense. Since w ∗ is bounded from below, we may assume w ¯ < w∗ 2 everywhere in R , where w ¯ is the sub-solution to Eq. (14) that we obtained previously. 3. Existence of the solution In this section, we shall prove the existence of a solution to Eq. (1). First, we use the monotone iteration method to prove that problem (1) has a unique solution wr on Br which satisfies w ¯ < wr < w ∗ when r is sufficiently large and Br = {x ∈ R2 | |x| ≤ r }. For any r > 0, we consider the following boundary value problem:
1w =
f (w)
w = w∗1, + g (w)
− G,
x ∈ Br
(44)
for |x| = r .
Let w0 = w ∗ , and wi satisfies
1wi − C0 wi =
w = w, ˜ i
f (wi−1 ) 1 + g (wi−1 )
− C0 wi−1 − G,
x ∈ Br
|x| = r ,
(45)
f (w)
where C0 is chosen so that 1+g (w) − C0 w is a decreasing function. The existence of C0 is according to conditions (A-1) and (A-2). wi then satisfies
w ¯ < · · · < wn < · · · < w2 < w1 < w∗ .
(46)
Since w1 satisfies
1w1 − C0 w1 =
f (w ∗ ) 1 + g (w ∗ )
− C0 w ∗ − G,
and the right-hand side is of Ls where s > 2; thus, w1 ∈ C 1,α (B¯r ) for some 0 < α < 1. In particular, w1 < w ∗ for x ∈ supp h2 ∩ Br and ∆(w1 − w ∗ ) ≥ C0 (w1 − w ∗ ). Hence, the maximum principle implies that w1 < w ∗ . Similarly, w ¯ < w∗ ¯ and hence by the mean value theorem and conditions (A-1), (A-2), there exists a constant C = C (f , g ) such that
(∆ − C0 )(w ¯ − w1 ) ≥ (C¯ − C0 )(w ¯ − w1 ) ≥ 0. By induction, let w ¯ < wk and if wk < wk−1 , then again by the mean value theorem:
(∆ − C0 )(wk+1 − wk ) ≥ (C¯ − C0 )(wk − wk−1 ) ≥ 0, we conclude that wk+1 < wk . Similarly
(∆ − C0 )(w ¯ − wk+1 ) ≥ (C¯ − C0 )(w ¯ − wk ) ≥ 0. Thus, {wi } is indeed a monotonic decreasing sequence.
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Since w ¯ is a bounded function, we have
wr = lim wk .
(47)
k→∞
By standard elliptic embedding theory, (47) exists in the strong sense, and hence wr is a unique classical solution on Br . Let r = n, and denote wr by wn ; then for x ∈ ∂ Bn , we have wn = w ∗ > wn+1 . By the mean value theorem (∆ − C0 )(wn − wn+1 ) ≥ 0 in Bn ; thus, the point-wise limit w = limn→∞ wn exists. Moreover, w ¯ ≤ w ≤ w∗ . We obtain the existence theorem. Theorem 6. If assumptions (E − 1) and (A − 1), (A − 2) hold then Eq. (1) has a solution. 4. Non-existence of the solution In this section, we consider the non-existence of the solution to Eq. (1). For the existence problem, we assume that the source function satisfies 4>
∫
hdx = H > 2.
R2
For the non-existence problem, we consider the complementary cases of the above condition and we assume that
N-1
hdx = H > 4.
R2
We use wµ to denote the solution of Eq. (3) for a = µg , b = νf ; then, wµ satisfies f (wµ )
1wµ ≤
1 + g (wµ )
− h,
and is a sub-solution to Eq. (1). Moreover, wµ ≤ u where u satisfies Eq. (1). Thus, problem (1) has no solution, provided that wµ → ∞ when |x| ≥ R0 for some R0 . Let vi be defined as in the beginning of Section 2 for i = 1, 2, 3 and u = v1 + v2 + w, K (x) = ev1 +v2 ; then, bK (x)ew
1w =
≡ h(x).
1 + aK (x)ew
(48)
Since vi = 0 for |x| > R0 , we have K (x)ew = eu ≤ 1,
| x| > R 0 ,
and we derive the following result. Theorem 7. If f , h satisfies conditions (N − 1), (A − 1), and (A − 2), then Eq. (1) admits no solution. Proof. We integrate (48) over region 0 ≤ ρ ≤ r , 0 ≤ θ ≤ 2π , then π
∫ r∫ 0
1wρ dθ dρ =
0
∫ r∫ 0
π
hρ dθ dρ.
(49)
0
Let W (r ) =
2π
∫
1 2π
w(r , θ )dθ ;
(50)
0
then, from (49) we have rWr =
=
1 2π 1 2π
2π
∫ r∫ 0
(ρ Wρ )ρ +
0 2π
∫ r∫ 0
1 ∂ 2W
ρ ∂θ 2
dθ dρ
hρ dθ dρ.
0
Taking the derivation of (51) with respect to r, we obtain 1 r
(rWr )r =
1
2π
∫
2π
hdθ .
0
By the assumptions (N-1) and (A-1), there exists a sufficiently large R0 such that ev1 +v2 ≥ |x|−2H ,
for |x| > R0 .
(51)
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J. Nee / Nonlinear Analysis: Real World Applications 12 (2011) 885–894
From (48), h(x) ≥ |x|−2H ew ,
for |x| > R0 .
By Jensen’s inequality, we obtain 1 r
(rWr )r ≥
eW r 2H
,
for r ≥ R0 .
(52)
Let t = ln r and φ(t ) = W (et ) = W (r ); then, (52) yields
φtt ≥ e2(1−H )t +φ ≥ C0 eφ t ≥ t0 ,
and C0 = e2(1−H )t0
(53)
by taking the initial point at t0 = ln(r0 ). By assumption (N-1) and (50), we have φt > 0 for t > t0 . Multiplying φt to (53) and integrating over (t0 , t1 ), we obtain
φt2 (t1 ) − φt2 (t0 ) ≥ C0 (eφ(t1 ) − eφ(t0 ) ). Let τ0 = φt2 (t0 ) − C0 eφ(t0 ) ; then, by the fact that φt ≥ 0 when t is sufficiently large we have
φt (t ) ≥
τ0 + C0 eφ(t ) ,
t ≥ t0 ,
or equivalently
∫
φ(t ) φ(t0 )
dφ ≥ t − t0 , τ0 + C0 eφ(t )
t ≥ t0 .
(54)
The left-hand side of inequality (54) is bounded above but the right-hand side is not, since
∫
dφ
∞ φ(t0 )
τ0 + C0 eφ(t )
we derive a contradiction.
<∞
References [1] [2] [3] [4] [5] [6] [7] [8]
Y. Yang, Solitons in Field Theory and Nonlinear Analysis, Springer-Verlag, New York, 2001. M. Berger, Riemannian structures of prescribed Gaussian curvature for compact 2-manifolds, J. Differential Geom. 5 (1971) 325–332. R. McOwen, Conformal metrics in R2 with prescribed Gaussian curvature and positive total curvature, Indiana Univ. Math. J. 34 (1984) 97–104. R. McOwen, On the equation 1u + K e2u = f and prescribed negative curvature in R2 , J. Math. Anal. Appl. 103 (1984) 365–370. W.M. Ni, On the elliptic equation 1u + k(x)e2u = 0 and comformal metrics with prescribed Gaussian curvatures, Invent. Math. 66 (1982) 343–352. W.M. Ni, On the elliptic equation 1u + k(x)u(n+2)/(n−2) = 0, its generalizations, and applications in geometry, Indiana Univ. Math. J. 31 (1982) 493–529. D.H. Sattinger, Conformal metrics in R2 with prescribed curvature, Indiana Univ. Math. J. 22 (1972) 1–4. B.J. Schroers, Bogomol’nyi solitons in a gauged 0(3) sigma model, Phys. Lett. B 356 (1995) 291–296.
Contents lists available at ScienceDirect
Nonlinear Analysis: Real World Applications journal homepage: www.elsevier.com/locate/nonrwa
On a semilinear elliptic equation from the sigma model of quantum field theory Janpou Nee ∗ ChienKuo Technology University, Chung-Hua, Taiwan, ROC
article
abstract
info
Article history: Received 7 January 2010 Accepted 9 August 2010
In this paper, we consider both the existence and the non-existence of solutions to the generalized semilinear elliptic equations from the sigma model of quantum field theory of condense matter. The interesting part of the result is that the conditions to prove the existence and the non-existence of the solution are not determined by the nonlinearity of the nonlinear function but rather the quantitative value of its source term. © 2010 Elsevier Ltd. All rights reserved.
Keywords: Elliptic equations Existence of solutions Super-solution Sub-solution Minimization of variational method
1. Introduction In this paper, we consider both the existence and the non-existence problem to the following type of the nonlinear elliptic equation,
1u =
f (u) 1 + g ( u)
− h(x),
x ∈ R2 .
(1)
This type of equation appears frequently in the problem of the prescribed curvature or the sigma model of the quantum field theory of condensed matter [1]. For the results of the prescribed curvature, the reader may refer to [2–7], and the references for the quantum field theory of condensed matter, in particular, the sigma model of Heisenberg ferromagnetic can be found in [1,8]. The sigma model of quantum field theory relates to the existence of Higgs which concern Lepton and Quark, thus the source of all matters. Therefore, it is important to understand the behavior of the solution of the sigma model. Schroers [8] proposed a model of the sigma field as follows:
1u =
4eu 1 + eu
− 4π Σjk=1 nj δpj + 4π Σjl=1 mj δqj ,
x ∈ R2 ,
(2)
where δp is the delta function at p ∈ R2 and p is the location of the singularity of the map. Yang [1] shows that for each natural number, N, there are solutions to saturate the classical energy lower bound for the field configurations in the topological family deg(0) = N, if and only if N ̸= 1. The unexpected results of [1] are that both the existence and the non-existence of the solution of Eq. (2) depend fully on the quantitative value of source function h. In this paper, we will extend results of Yang [1] to Eq. (1). The main assumptions of this paper are as follows: E-1
h ∈ L1c (R2 ), and 4 >
R2
hdx = H > 2,
where L1c (R2 ) = {u ∈ L1 |u has compact support}.
∗
Tel.: +886 4 7111111x2556. E-mail addresses: [email protected], [email protected].
1468-1218/$ – see front matter © 2010 Elsevier Ltd. All rights reserved. doi:10.1016/j.nonrwa.2010.08.013
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J. Nee / Nonlinear Analysis: Real World Applications 12 (2011) 885–894
For the nonlinear functions f (·) and g (·), we assume that
µf eu ≤ f (u) ≤ νf eu , µg eu ≤ g (u) ≤ νg eu where 0 < µf < νf , 0 < µg < νg . f , g both are of C 2 satisfying Cf > f ′ > 0, Cg > g ′ > 0 for some constant Cf and Cg .
A-1 A-2
We will show that under assumptions (E-1), and (A-1), and (A-2), the results of [1] remain true. For the non-existence problem, the assumption of symmetry is essential in general. However, we shall apply the method as that of Yang [1] to show that Eq. (1) has no solution under a weaker assumption of symmetry together with the complementary condition of (E-1). 2. Construction of super-solution and sub-solutions It is difficult to exhibit the behavior of the solution of Eq. (1) directly. Therefore, we introduce an auxiliary equation which helps us to get a super-solution and a sub-solution to Eq. (1). The auxiliary equation that we insert is of the following form:
1u =
beu 1 + aeu
− h,
(3)
bx where the reaction term 1+ is monotone. Therefore, by the condition (A-1) the solution of Eq. (3) is a super-solution of ax Eq. (1) when a = µg , b = νf and is a sub-solution when a = νg , b = µf . The monotonic property of the reaction term will be frequently applied throughout this paper. Before we start to prove our results, we introduce the following notations. First, the weighted function:
K = eβ v¯ ≥ 0,
K = O|x|−β ,
β > 0,
(4)
where the function v¯ and the parameter β of Eq. (4) above will be given later. We denote the usual Lp space over R2 by Lp and its norm by ‖ · ‖p . We define the weighted measure dµ = K dx and we use Lp (dµ) to denote the induced Lp space. In order to demonstrate the asymptotic behavior of the solution, we consider the subspace Ξ of L2 with norm:
‖w‖2Ξ = ‖∇w‖2 + ‖w‖2L2 (dµ) ≤ ∞,
for w ∈ Ξ .
(5)
Note that Ξ contains those constant functions defined on R2 . Thus, to exclude the constant functions from Ξ , we consider ˜ of Ξ where the closed subspace Ξ
∫ ˜ = w ∈ Ξ Ξ
R2
w(x)dµ = 0 .
(6)
˜ then satisfies the following Trudinger–Moser inequality: The function w ∈ Ξ ∫ R2
e|w| dµ ≤ C e
‖∇w‖2 2 4γ
,
w ∈ Ξ˜
(7)
where
γ ≤ min{2, β − 2},
and C = C (γ ).
(8)
The constant β is chosen according to function h which will be given later. The other constant γ we choose in (8) relates to the asymptotic behavior of the solution while the range of γ in (7) is not an issue that we concerned at here. Note that the embedding Ξ → L2 (dµ) is completely continuous. To prove our results, we first construct a sub-solution to Eq. (1). To this end, we consider the regularization of h and we denote D1 = {x ∈ R2 |h(x) > 0}, D2 = {x ∈ R2 |h(x) ≤ 0}, and h(x), 0,
h1 =
x ∈ D1 otherwise
−h(x), h2 = 0,
x ∈ D2 otherwise.
By the assumption (E-1), Di are bounded domains where supp(hi ) = Di for i = 1, 2. Let Di ⊃ Dϵi = {x ∈ Di |dist(x, ∂ Di ) ≤ ϵ} and G˜ i ∈ C ∞ (R2 ) be the regularization of hi respectively. We define
1vi =
hi − G˜ i , 0,
x ∈ Di x ∈ R2 \ Di ,
(9)
J. Nee / Nonlinear Analysis: Real World Applications 12 (2011) 885–894
887
and thus vi are non-positive functions for i = 1, 2. Furthermore, we choose v3 ∈ C ∞ (R2 ), so that
−1 ln |x|, 2π
v3 =
|x| ≥ 1, x ∈ R2 ,
(10)
and then we have
∫
(−1v3 ) = 1.
R2
(11)
We choose β > 0 such that 2<β
(12)
βv3
and set K = e which is the same as the weighted function K (x) of Eq. (4) of the weighted Sobolev space that we mentioned previously. By taking the transformation: u = −v1 + v2 + βv3 + w,
(13)
Eq. (3) becomes bK ev2 +w
1w =
ev1 + aK ev2 +w
− G,
(14)
where G = G˜ 1 − G˜ 2 + βv3 is of compact support satisfying
∫
G(x)dx = H − β > 0.
R2
(15)
From (15) we choose h¯ ∈ Cc∞ (R2 ), such that h¯ > 0, and it satisfies
∫
h¯ (x)dx = H − β.
R2
(16)
To get a proper sub-solution of (1), we study the following equation: bK ew
1w =
ev1 + aK ew
− h¯ ,
(17)
which in term is a sub-solution of Eq. (14). To get the solution of Eq. (17), we study the solution of the following functional: I (w) =
∫
1 R2
2
|∇w|2 +
b a
ln(ev1 + aK ew ) − h¯ w
(18)
in the admissible space: bK ew
∫
A = w ∈ Ξ
R2
ev1 + aK ew
∫ = R2
h¯ .
Our goal is to show that I (u) is a weakly lower semi-continuous functional on Ξ . Lemma 1. I ≥ C1 ‖∇w‖2 − C2 is bounded below over A. Proof. To achieve our goal, we use the fact that supp(v1 ) ⊂ D1 . We rewrite I (w) of (18) as I (w) =
∫
1 R2
2
|∇w|2 +
b a
ln(1 + aK ew ) − h¯ w dx +
b
∫
a
ln D1
ev1 + aK ew 1 + aew
.
t Since the function F (t ) = α+ increases for α < 1, we have F (t ) ≥ F (0) for all t ≥ 0. Therefore, 1 +t
∫ ln
ev1 + aK ew
D1
1 + aew
∫
(19) ev1 +aK ew 1+aK ew
> ev1 and
v1 dx > −C∗ > −∞.
≥ D1
Substituting the above into (19), we have I (w) ≥
[
∫ R2
1 2
|∇w|2 +
b a
]
ln(1 + aK ew ) − h¯ w − C∗ .
(20)
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J. Nee / Nonlinear Analysis: Real World Applications 12 (2011) 885–894
Since ln(1 + aK ew ) = βv3 + w + ln(a + e−(βv3 +w) ), we have W =
b a
ln(1 + aK ew ).
(21)
Thus, from (20) we derive I (w) ≥
∫
1 R2
2
|∇w|2 +
∫
∫
¯ + hW
W− R2
∫
R2
R2
h¯ (βv3 ) +
b a
∫ R2
h¯ ln(a + e−(βv3 +w) ) − C∗ .
(22)
From (21), it yields
∇W =
bK ew
· (β∇v3 + ∇w),
1 + aK ew
and we have bK ew
∫
2
1 + aK ew
R2
bK ew
∫ ≤
1 + aK ew
R2
bK ew
∫ ≤ R2
ev1 + aK ew
= H − β.
Applying Trudinger–Moser inequality (7), it yields
‖∇ W ‖2 ≤ ‖∇w‖2 + C . Substitute the above inequality into (22) and by the fact that W > 0, we obtain I (w) ≥
1 4
1
2 2
2 2
W2
∫
‖∇w‖ + ‖∇ W ‖ + 4
R2
1+W
∫ dx − R2
¯ − C1 . hW
(23)
To complete the proof, we need the following interpolation inequality over R2 :
∫
∫
4
W dx ≤ 2 R2
∫
2
W dx R2
R2
|∇ W |2 .
(24)
Since h¯ is of compact support, we induced from (24) that
∫
¯ dx ≤ ‖h¯ ‖ 4 ‖W ‖4 hW 3
R2
≤ ϵ‖W ‖2 + C (ϵ)‖∇ W ‖2 + C 1
≤ ϵ‖W ‖2 + ‖∇ W ‖22 + C (ϵ),
(25)
8
where ϵ is small. From (24) and W > 0, we obtain W2
∫ |W |2 ≤ 1 + R2
1+W
+ ‖∇ W ‖22 .
(26)
Substituting (24), (26) into (23) yields I (w) ≥
1 4
‖∇w‖22 + C1 ‖∇ W ‖22 +
where Ci > 0 are constant for i = 1, 2.
W2
∫ R2
1+W
− C2 ,
(27)
Lemma 2. If I (w) satisfies (18), then min I (w) ≡ γ0 ,
(28)
w∈A
has a solution. Proof. Let wj be a minimizing sequence of (28) and let Wj =
b a
ln(1 + aK ewj ),
j = 1, 2, . . . ,
J. Nee / Nonlinear Analysis: Real World Applications 12 (2011) 885–894
889
then with (23) and (26) imply that {Wj } is bounded in L2 and thus bounded in L1 . Applying I (wj ) = γ0 and inequality (27), we get ‖∇wj ‖2 bounded. Let wj = w ¯j+w ˜ j where w ¯ j is the constant part and w ˜ j ∈ Ξ˜ , and we will show that w ¯ j is bounded. Let supp(v1 ) ⊂ D0 . By (19) and v1 < 0, we have
γ0 ≤ I (wj ) 1
=
2 1
≤
2
b
‖∇wj ‖22 + ‖Wj ‖1 − a
∫ R2
b
‖∇wj ‖22 + ‖Wj ‖1 − w ¯j
∫
a
∫
b
h¯ wj +
ln
a
h¯ −
1 + aewj
D0
∫
R2
ev1 + aK ewj
R2
h¯ w ˜ j,
thus
γ0 ≤
1 2
b
‖∇w ‖ + ‖Wj ‖1 − w ¯j 2 j 2
∫
a
h¯ −
∫
R2
R2
h¯ w ˜ j.
(29)
Inserting the following Poincaré type of inequality
∫
w2 dµ ≤
R2
∫ R2
|∇wj |2 w ∈ Ξ˜
(30)
into inequality (29), we obtain
w ¯j
∫ R2
1 h¯ ≤ |γ0 | + ‖∇wj ‖22 + ‖Wj ‖1 − 2
∫ R2
h¯ w ˜ j,
and hence
w ¯j
∫
1
h¯ ≤ |γ0 | +
2
R2
‖∇wj ‖22 + ‖Wj ‖1 + C .
(31)
Therefore, w ¯ j is bounded from above. Next, we will show that w ¯ j is bounded from below. To this end, we will use the constraint of the definition of the admissible space that
∫
∫ bK ewj + w v j 1 + aK ewj R2 \D0 1 + aK e D e ∫ ∫ 0 wj bK e bK ewj < + , v1 + aK ewj v1 + aK ewj R2 e D0 e
h¯ = R2
∫
bK ewj
and thus
∫ R2
h¯ < |D0 | + µew¯ j
∫ R2
ew˜ j dµ.
(32)
Without lost of generality, we shall assume that |D0 | ≤
∫ w ¯ j > ln |D0 | −
h¯
∫
R2
¯ Substituting (7) into (32), we obtain h.
− ln
R2
R2
ew˜ j dµ ,
and hence
w ¯ j > −C1 − C2 ‖∇wj ‖22 .
(33)
Since ‖∇wj ‖ is bounded, w ¯ j is bounded from below. Similarly {w ˜ j } is bounded in Ξ , there exists a w ∈ Ξ and a subsequence which converges weakly to w in Ξ . Thus
wj = w ¯j +w ˜ j ⇀ w,
in Ξ .
Since the embedding Ξ → L (dµ) is completely continuous, we may assume without generality that wj → w strongly in L2 (dµ). Next, we will show that w ∈ A, or equivalently 2
bK ew
∫ R2
ev1 + aK ew
= H.
Let
∫ aj =
bK ewj bK ew . − w v v w j 1 1 + aK e e + aK e R2 e
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J. Nee / Nonlinear Analysis: Real World Applications 12 (2011) 885–894
For any ϵ > 0, we choose an open set D satisfying |D| ≤ 2ϵ , then
∫ wj bK ewj bK ew bK ew + bK e − − w w w w v v v v j j 1 + aK e 1 + aK e e 1 + aK e e 1 + aK e R 2 \D e D e ∫ bK e|w|+|wj | ≤ 2|D| + |wj − w|; v ∫
aj =
e
R2 \D
1
thus, ¯ C (‖∇wj ‖2 +‖∇w‖2 ) ). aj < ϵ + C (ϵ)‖wj − w‖L2 (dµ) sup(ew¯ j +w+ 2
2
(34)
j
Let j → ∞ and we have a = lim sup aj ≤ ϵ; j→∞
thus, a = 0 and hence w ∈ A. Next, we will show that I (·) is weakly lower semi-continuous on Ξ . To this end, we let η = t wj + (1 − t )w where 0 ≤ t ≤ 1; then,
∫
wj
w
| ln(ev1 +aK e ) − ln(ev1 +aK e )| ∫ ∫ K eη |wj − w| ≤ |wj − w| + v
ϵj =
R2
e
R2 \D
D
≤ C1 ‖wj − w‖L2 (dµ) +
1
K e|wj |+|w|
∫ R2 \D
ev1
|wj − w|
and ϵj → 0 as j → ∞. This shows that I (·) is weakly lower semi-continuous on Ξ and I (w) ≤ lim inf I (wj ) = a thus w ∈ A that minimize I (·). Lemma 3. The optimization problem (28) of Eq. (17) has a solution. Proof. The Lagrange multiplier implies that bK ew
∫ ∇w · ∇η + R2
ev1
+
aK ew
∫ − h¯ ηdx = λ
bK ev1 +w R2
(ev1
+ aK ew )2
η,
(35)
where λ ∈ R is a constant and η ∈ Ξ is an arbitrary trail function. Let η ≡ 1 and by definition w ∈ A, we see that λ = 0. Inserting λ = 0 into (35), we see that w is a weak solution of (17). By the standard elliptic theory, w is a classical solution of (17). To demonstrate the asymptotic behavior, we now introduce notations that will appear later. For δ ∈ R and s ∈ N, we let Ws2,δ to be the closure of Cc∞ (R2 ) under the norm:
‖η‖2W 2 = s,δ
−
‖(1 + |x|)δ+|α| Dα η‖22 ,
|α|≤s
where Cc∞ (R2 ) contains the functions defined over R2 with compact supports and C0 (R2 ) denote the set of continuous functions over R2 vanishing at infinity. The following results of weighted Sobolev space will be used to exhibit the asymptotic behavior of the solution of Eq. (14). Lemma 4.
(i) If s > 1 and δ > −1, then Ws2,δ ⊂ C0 (R2 ).
(ii) For −1 < δ < 0, the Laplace operator ∆ : W22,δ → W02,δ+2 is 1 − 1 and
∫ ∆(W22,δ ) = F ∈ W02,δ+2
R2
F =0 .
(iii) If η ∈ Ξ and 1η = 0, then η = k where k is a constant. Using β < 4 and the above results, we obtain the following lemma. Lemma 5. If c < 0, then Eq. (14) has a sub-solution w − satisfying w − < c in R2 .
J. Nee / Nonlinear Analysis: Real World Applications 12 (2011) 885–894
891
Proof. We will first consider the solution of the following equation:
1w = h¯ − G,
(36)
by finding the minimizer w1 of functional J (w) =
∫
1 R2
2
‖∇w‖2 + (h¯ − G)w.
(37)
˜ . To show that w1 → k as |x| → ∞, we By condition (E-1), the solution of Eq. (37) is unique over the admissible space Ξ choose 0 > δ > −1 and since h¯ − G ∈ L(R2 ) ∩ W02,δ+2 and R2 h¯ − G = 0, (i) and (ii) of Lemma 4 of weighted Sobolev
space give η ∈ W22,δ satisfying Eq. (36) and η = 0 as ‖x‖ → ∞, moreover η ∈ L2 (dµ). Since ∇η ∈ W02,δ+1 and δ > −1, we deduce that η ∈ L2 (R2 ). Hence η ∈ Ξ . Since η − w1 ∈ Ξ and ∆(η − w1 ) = 0, (iii) of the previous lemma then implies that η − w1 = k where k is a constant, and we have that w1 → k at infinity. Let w2 be the solution of Eq. (17) and in fact by the constrain of admissible space, we have bK ew
∫
ev1 + aK ew
R2
− h¯ = 0.
Let supp(v1 ) ⊂ D; then,
∫ R2
(1 + |x|δ+2 )2
bK ew2
ev1 + aK ew2
2
∫
(1 + |x|δ+2 )2 + C0
≤ D
∫ R2
e2w2 dµ,
where C0 = sup {K (x)(1 + |x|δ+2 )2 }, x∈R2
and C0 is finite if 2(δ + 2) ≤ β . Thus, we choose β > 0 such that δ = −2 + bK ex ev1 + aK ex
− h¯ = ρ(x);
β 2
, and we let (38)
then, ρ(x) ∈ W02,δ+2 and condition β < 4 is equivalent to −1 < δ < 0. Now by (i) and (ii) of the previous lemma, there exists η′ ∈ W22,δ satisfying
1η′ = ρ and hence η′ − w2 = k′ where k′ is constant and hence w2 → k′ at infinity. x Since wi are bounded, we may choose C such that w ¯ = w1 − C +w2 < c < 0 and w1 − C < 0. Let function F (x) = 1+bKaKe ex ; then by the monotonicity of F (x) we have F (w) ¯ = F (w1 − C + w2 ) ≤ F (w2 ). Hence
1w ¯ = F (w2 ) − G ≥ F (w) ¯ −G bK ev2 +w¯
− G, + aK ev2 +w¯ and by transformation (13), w ¯ is a sub-solution to Eq. (14). ≥
ev 1
Next, we shall construct the super-solution of Eq. (1). Similar to the construction of the sub-solution, we consider the modified equation:
1w =
bK ew 1 + aK ew
− G¯ ,
(39)
where G¯ = G1 + β 1v3 . Let h˜ ∈ Cc∞ (R2 ) such that h˜ ≥ 0 and
∫
˜ = hdx R2
∫ R2
G¯ dx = 2π (2H + − β).
(40)
We consider the following equation:
1w =
bK ew 1 + aK ew
− h˜ .
(41)
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J. Nee / Nonlinear Analysis: Real World Applications 12 (2011) 885–894
Since the function h˜ is chosen as similar to the h¯ of the modified Eq. (17), we may apply Lemmas 2–4 to get a solution w1 of Eq. (41) satisfying w1 → C as ‖x‖ → ∞. Let w2 be a bounded solution of
1w = h˜ − G¯ .
(42)
We may assume that w2 ≥ 0. Let w ˜ = w1 + w2 ; then, from (41) and (42) we have
1w ˜ =
bK ew 2
1 + aK ew 2
− G¯ .
x By monotonicity of function 1+ , we derive x
1w ˜ ≤
≤
bK ew1 +w2
− G¯
1 + aK ew1 +w2 bK ew˜ 1 + aK ew˜
− G¯ + h+
in the distribution sense. Let w ∗ = w ˜ − v2 ; then, the above inequality implies that
1w ≤ ∗
∗ bK ew +v2
− G. ∗ ev1 + aK ew +v2
(43)
Thus, w ∗ is a super-solution to Eq. (3) in the distribution sense. Since w ∗ is bounded from below, we may assume w ¯ < w∗ 2 everywhere in R , where w ¯ is the sub-solution to Eq. (14) that we obtained previously. 3. Existence of the solution In this section, we shall prove the existence of a solution to Eq. (1). First, we use the monotone iteration method to prove that problem (1) has a unique solution wr on Br which satisfies w ¯ < wr < w ∗ when r is sufficiently large and Br = {x ∈ R2 | |x| ≤ r }. For any r > 0, we consider the following boundary value problem:
1w =
f (w)
w = w∗1, + g (w)
− G,
x ∈ Br
(44)
for |x| = r .
Let w0 = w ∗ , and wi satisfies
1wi − C0 wi =
w = w, ˜ i
f (wi−1 ) 1 + g (wi−1 )
− C0 wi−1 − G,
x ∈ Br
|x| = r ,
(45)
f (w)
where C0 is chosen so that 1+g (w) − C0 w is a decreasing function. The existence of C0 is according to conditions (A-1) and (A-2). wi then satisfies
w ¯ < · · · < wn < · · · < w2 < w1 < w∗ .
(46)
Since w1 satisfies
1w1 − C0 w1 =
f (w ∗ ) 1 + g (w ∗ )
− C0 w ∗ − G,
and the right-hand side is of Ls where s > 2; thus, w1 ∈ C 1,α (B¯r ) for some 0 < α < 1. In particular, w1 < w ∗ for x ∈ supp h2 ∩ Br and ∆(w1 − w ∗ ) ≥ C0 (w1 − w ∗ ). Hence, the maximum principle implies that w1 < w ∗ . Similarly, w ¯ < w∗ ¯ and hence by the mean value theorem and conditions (A-1), (A-2), there exists a constant C = C (f , g ) such that
(∆ − C0 )(w ¯ − w1 ) ≥ (C¯ − C0 )(w ¯ − w1 ) ≥ 0. By induction, let w ¯ < wk and if wk < wk−1 , then again by the mean value theorem:
(∆ − C0 )(wk+1 − wk ) ≥ (C¯ − C0 )(wk − wk−1 ) ≥ 0, we conclude that wk+1 < wk . Similarly
(∆ − C0 )(w ¯ − wk+1 ) ≥ (C¯ − C0 )(w ¯ − wk ) ≥ 0. Thus, {wi } is indeed a monotonic decreasing sequence.
J. Nee / Nonlinear Analysis: Real World Applications 12 (2011) 885–894
893
Since w ¯ is a bounded function, we have
wr = lim wk .
(47)
k→∞
By standard elliptic embedding theory, (47) exists in the strong sense, and hence wr is a unique classical solution on Br . Let r = n, and denote wr by wn ; then for x ∈ ∂ Bn , we have wn = w ∗ > wn+1 . By the mean value theorem (∆ − C0 )(wn − wn+1 ) ≥ 0 in Bn ; thus, the point-wise limit w = limn→∞ wn exists. Moreover, w ¯ ≤ w ≤ w∗ . We obtain the existence theorem. Theorem 6. If assumptions (E − 1) and (A − 1), (A − 2) hold then Eq. (1) has a solution. 4. Non-existence of the solution In this section, we consider the non-existence of the solution to Eq. (1). For the existence problem, we assume that the source function satisfies 4>
∫
hdx = H > 2.
R2
For the non-existence problem, we consider the complementary cases of the above condition and we assume that
N-1
hdx = H > 4.
R2
We use wµ to denote the solution of Eq. (3) for a = µg , b = νf ; then, wµ satisfies f (wµ )
1wµ ≤
1 + g (wµ )
− h,
and is a sub-solution to Eq. (1). Moreover, wµ ≤ u where u satisfies Eq. (1). Thus, problem (1) has no solution, provided that wµ → ∞ when |x| ≥ R0 for some R0 . Let vi be defined as in the beginning of Section 2 for i = 1, 2, 3 and u = v1 + v2 + w, K (x) = ev1 +v2 ; then, bK (x)ew
1w =
≡ h(x).
1 + aK (x)ew
(48)
Since vi = 0 for |x| > R0 , we have K (x)ew = eu ≤ 1,
| x| > R 0 ,
and we derive the following result. Theorem 7. If f , h satisfies conditions (N − 1), (A − 1), and (A − 2), then Eq. (1) admits no solution. Proof. We integrate (48) over region 0 ≤ ρ ≤ r , 0 ≤ θ ≤ 2π , then π
∫ r∫ 0
1wρ dθ dρ =
0
∫ r∫ 0
π
hρ dθ dρ.
(49)
0
Let W (r ) =
2π
∫
1 2π
w(r , θ )dθ ;
(50)
0
then, from (49) we have rWr =
=
1 2π 1 2π
2π
∫ r∫ 0
(ρ Wρ )ρ +
0 2π
∫ r∫ 0
1 ∂ 2W
ρ ∂θ 2
dθ dρ
hρ dθ dρ.
0
Taking the derivation of (51) with respect to r, we obtain 1 r
(rWr )r =
1
2π
∫
2π
hdθ .
0
By the assumptions (N-1) and (A-1), there exists a sufficiently large R0 such that ev1 +v2 ≥ |x|−2H ,
for |x| > R0 .
(51)
894
J. Nee / Nonlinear Analysis: Real World Applications 12 (2011) 885–894
From (48), h(x) ≥ |x|−2H ew ,
for |x| > R0 .
By Jensen’s inequality, we obtain 1 r
(rWr )r ≥
eW r 2H
,
for r ≥ R0 .
(52)
Let t = ln r and φ(t ) = W (et ) = W (r ); then, (52) yields
φtt ≥ e2(1−H )t +φ ≥ C0 eφ t ≥ t0 ,
and C0 = e2(1−H )t0
(53)
by taking the initial point at t0 = ln(r0 ). By assumption (N-1) and (50), we have φt > 0 for t > t0 . Multiplying φt to (53) and integrating over (t0 , t1 ), we obtain
φt2 (t1 ) − φt2 (t0 ) ≥ C0 (eφ(t1 ) − eφ(t0 ) ). Let τ0 = φt2 (t0 ) − C0 eφ(t0 ) ; then, by the fact that φt ≥ 0 when t is sufficiently large we have
φt (t ) ≥
τ0 + C0 eφ(t ) ,
t ≥ t0 ,
or equivalently
∫
φ(t ) φ(t0 )
dφ ≥ t − t0 , τ0 + C0 eφ(t )
t ≥ t0 .
(54)
The left-hand side of inequality (54) is bounded above but the right-hand side is not, since
∫
dφ
∞ φ(t0 )
τ0 + C0 eφ(t )
we derive a contradiction.
<∞
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