Electronic Journal of Differential Equations, Vol. 2013 (2013), No. 204, pp. 1–9.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
THE (n − 1)-RADIAL SYMMETRIC POSITIVE CLASSICAL
SOLUTION FOR ELLIPTIC EQUATIONS WITH GRADIENT
YONG ZHANG, QIANG XU, PEIHAO ZHAO
Abstract. In this article, we study the existence of the (n − 1)-radial symmetric positive classical solution for elliptic equations with gradient. By some
special techniques in two variables, we show a priori estimates, and then show
the existence of a solution using a fixed point theorem.
1. Introduction
In this article, we consider the following boundary-value problem of a secondorder elliptic equation,
−∆u = f (x, u, ∇u) in Ω,
(1.1)
u(x) = 0, on ∂Ω,
where Ω is a bounded domain in Rn , n ≥ 3.
This type of equations have been studied by several authors. As the nonlinearity
f depends on the gradient of the solution, solving (1.1) is not variational and the
well developed critical point theory can not be applied directly. But if f has a
special form, by changing variables, (1.1) can be transformed into a boundaryvalue problem which is independent of ∇u. For example, When f (x, u, ∇u) =
g(u) + λ|∇u|2 + η, Ghergu and Rădulescu [8] used the above method to show the
existence of positive classical solution under the assumption that g is decreasing
and unbounded at the origin. A similar method appears in [1], where f (x, u, ∇u)
has critical growth with respect to ∇u; see also [9, 20]. In addition, Chen and Yang
[5] considered the existence of positive solutions for (1.1) on a smooth compact
Riemannian manifold. As far as we know, the methods used to solve (1.1) are mainly
sub and super-solution, fixed point theorems, Galerkin method, and topological
degree, see, for instance, [2, 3, 7, 13, 17, 18, 19].
It is worth mentioning that de Figueiredo, Girardi and Matzeu [6] developed
a quite different method of variational type. Firstly, for each ω ∈ H01 (Ω), they
considered the boundary problem
−∆u = f (x, u, ∇ω)
u(x) = 0,
in Ω,
on ∂Ω.
(1.2)
2000 Mathematics Subject Classification. 35J60, 35B09.
Key words and phrases. Elliptic equations; symmetric; positive solution; a priori estimates;
fixed point theorem.
c 2013 Texas State University - San Marcos.
Submitted June 14, 2013. Published Spetember 16, 2013.
1
2
Y. ZHANG, Q. XU, P. ZHAO
EJDE-2013/204
which is a variational problem. Under the assumptions that f has a superlinear
subcritical growth at zero and at infinity with respect to the second variable, and
f is locally Lipschitz continuous with the third variable, they proved that a weak
solution uω of (1.2) exists by mountain-pass theorem. Then they have constructed
a sequence {uk } ⊂ H01 (Ω) as solutions of
−∆un = f (x, un , ∇un−1 )
un (x) = 0,
in Ω,
on ∂Ω,
(1.3)
and verified that {uk } converges to a solution of (1.1). However, this solution is
just in H01 (Ω).
Additionally, the existence of classical solutions for (1.1) has been obtained by
mountain-pass lemma and a suitable truncation method in [11], but the conditions
imposed on f are very strong:
(1) f is locally Lipschitz continuous on Ω̄ × R × Rn ,
converges to zero uniformly with respect to x ∈ Ω, ξ ∈ Rn as t tends
(2) f (x,t,ξ)
t
to zero,
n+2
) and r ∈ (0, 1) such that
(3) there exist a1 > 0, p ∈ (1, n−2
|f (x, t, ξ)| ≤ a1 (1 + |t|p )(1 + |ξ|r ),
∀x ∈ Ω̄, t ∈ R, ξ ∈ Rn ,
(4) there exist ϑ > 2 and a2 , a3 , t0 > 0 such that
0 < ϑF (x, t, ξ) ≤ tf (x, t, ξ),
∀x ∈ Ω̄, t ≥ t0 , ξ ∈ Rn , F (x, t, ξ) ≥ a2 |t|ϑ − a3 ;
F (x, t, ξ) ≥ a2 |t|ϑ − a3 ,
Rt
where F (x, t, ξ) = 0 f (x, s, ξ)ds.
As far as we know, a few authors have paid attention to the radial solutions of
(1.1); see for example [4, 7]. So we will limit us to the radially symmetric case and
try to focus on some new methods to study (1.1). We consider the boundary-value
problem (1.1) and assume the following:
(D1) Ω is a so-called (n − 1)-symmetric domain in Rn (n ≥ 3), that is, Ω is
symmetric with respect to x1 , x2 , · · · , xn−1 and 0 ∈
/ Ω;
(F1) f (x, u, η) isqa nonnegative function satisfying f (x, u, η) = f (r, xn , u, |η|),
where r =
x21 + x22 + · · · + x2n−1 ;
2p
) such that
(F2) there exist c0 ≥ 1, M > 0, p > 1, τ ∈ (0, p+1
up − M |η|τ ≤ f (x, u, η) ≤ c0 up + M |η|τ ,
∀(x, u, η) ∈ Ω × R × Rn ;
(F3) f (x, u, η) ∈ C β (Ω, R, Rn ) for some β ∈ (0, 1).
2p
We remark that in [14], the constants p and τ belong to (1, 2(n−1)
n−2 ) and (1, p+1 )
respectively. Obviously, the conditions in (F2) are weaker than those in [14].
If the solution u(x) is (n − 1)-radial symmetric, that is u(x) = u(r, xn ), then by
(F1) Equation (1.1) can be transformed into the following elliptic equation in two
variables:
−(urr + uxn xn ) = H(r, xn , u, ur , uxn ), in Ω,
(1.4)
u(x) = 0, on ∂Ω,
where H(r, xn , u, ur , uxn ) = f (r, xn , u, |∇u|) + n−2
r ur . Motivated by the priori
estimates mentioned in [14] and special technique for the equation in two variables
developed in [10], we develop an approach which is distinct from the previous
EJDE-2013/204
THE (n − 1)-RADIAL SYMMETRIC SOLUTION
3
works, and shows the existence of the (n − 1)-radial symmetric positive classical
C 2,β -solutions of (1.1). Note that solution in [14] is just in C 1,α (Ω).
The rest of this work is organized as follows. Motivated by [14] we give a priori
estimates in section 2. In section 3 we show the existence of (n−1)-radial symmetric
positive classical solutions with the help of [10].
2. A priori estimates
Compared with the reference [14], we should deal with the second term n−2
r ur
of H(r, xn , u, ur , uxn ) in (1.4) additionally, it is necessary to give a brief proof of
the a priori estimates although the process is similar to that in [14].
Theorem 2.1. Assume that (D1) and (F2) hold, and that λ < λ0 for some λ0
fixed. Then, for any C 1 -solution u of the equation
−(urr + uxn xn ) = H(r, xn , u, ur , uxn ) + λ,
u(x) = 0,
in Ω,
on ∂Ω,
(2.1)
there exists a positive constant C such that supΩ u < C.
To prove this theorem, we need the following lemmas.
Lemma 2.2. Let (D1) hold and u(r, xn ) be a positive weak C 1 -solution of the
inequality
n−2
ur ,
(2.2)
− (urr + uxn xn ) ≥ up − M |∇u|τ +
r
2p
). Denote
where 1 < p and 0 < τ < 2p/(p + 1). Take γ ∈ (0, p) and µ ∈ (0, p+1
by B2R a ball of radius 2R contained in Ω, where R < R0 and R0 is a positive
constant. Then there exists a positive constant C = C(p, γ, µ, R0 ) such that
Z
uγ ≤ CR2−2γ/(p−1) ,
(2.3)
BR
Z
|∇u|µ ≤ CR2−(p+1)µ/(p−1) .
(2.4)
BR
Proof. We can assume that BR is centered at x0 ∈ Ω and first focus on proving
(2.3). Let ξ be a C 2 -cut-off function on B2 satisfying:
(1) ξ(x) = ξ(|x − x0 |), 0 ≤ |x − x0 | ≤ 2.
(2) ξ(x) = 1 for |x − x0 | ≤ 1.
(3) ξ has compact support in B2 and 0 ≤ ξ ≤ 1.
(4) |∇ξ| ≤ 2.
k −d
0
Let d = p − γ > 0 and φ = [ξ( x−x
as a test function for (2.2) (k to be fixed
R )] u
later). We obtain
Z
Z
n−2
k −d
− (urr + uxn xn )ξ u ≥ (up − M |∇u|τ +
ur )ξ k u−d .
r
Ω
Ω
2k
Integrating by parts and using that |∇ξ k | = kξ k−1 |∇ξ| ≤ ξ k Rξ
, we obtain
Z
Z
d
ξ k uγ−p−1 |∇u|2 +
ξ k uγ
Ω
Ω
Z
Z
Z
n−2
−d
k
τ k −d
ur ξ k u−d
≤
u |∇u||∇ξ | + M
|∇u| ξ u −
r
Ω
Ω
Ω
4
Y. ZHANG, Q. XU, P. ZHAO
Z
≤
u−d |∇u|ξ k
Ω
2k
+M
Rξ
Z
EJDE-2013/204
|∇u|τ ξ k u−d +
Ω
n−2
dist(0, ∂Ω)
Z
|∇u|ξ k u−d .
Ω
Applying the Young inequality to the first right term, we have
Z
Z
Z
d
2k
≤
ξ k uγ−p−1 |∇u|2 + CR−2
u−d |∇u|ξ k
ξ k−2 uγ−p+1 ,
Rξ
4 Ω
Ω
Ω
so
Z
Z
3
k γ−p−1
2
d
ξ u
|∇u| +
ξ k uγ
4 Ω
Ω
Z
Z
Z
n−2
|∇u|ξ k u−d .
≤ CR−2
ξ k−2 uγ−p+1 + M
|∇u|τ ξ k u−d +
dist(0,
∂Ω)
Ω
Ω
Ω
2γ
. By using the Young
Next we focus on the case of γ > p − 1. Take k = p−1
inequality again, we have
Z
Z
1
CR−2
ξ k−2 uγ−p+1 ≤
ξ k uγ + CR2−2γ/(p−1)
4 Ω
Ω
and
Z
τ k −d
|∇u| ξ u
M
Ω
Z
Z
d
k γ−p−1
2
ξ u
|∇u| + C
ξ k ut
≤
4 Ω
Ω
Z
Z
d
1
≤
ξ k uγ−p−1 |∇u|2 +
ξ k uγ + CR−2 ,
4 Ω
4 Ω
2
the second inequality holds becasue t = (−d − τ γ−p−1
) 2−τ
< γ, and
2
Z
Z
Z
n−2
d
|∇u|ξ k u−d ≤
ξ k uγ−p−1 |∇u|2 + C
ξ k uγ−p+1
dist(0, ∂Ω) Ω
4 Ω
Z
ZΩ
1
d
k γ−p−1
2
ξ u
|∇u| +
ξ k uγ + CR−2 .
≤
4 Ω
4 Ω
So
Z
Z
1
d
k γ−p−1
2
ξ u
|∇u| +
ξ k uγ ≤ CR2−2γ/(p−1) ,
(2.5)
4 Ω
4 Ω
which gives (2.3).
If γ = p − 1, (2.3) is obvious by the above arguments. For the case of γ < p − 1,
the following Höder inequality
Z
Z
γ/(p−1)
uγ ≤ CR2(1−γ)/(p−1)
up−1
BR
BR
and the above argument yields to (2.3).
To prove (2.4), we use Höder inequality:
Z
Z
µ/2 Z
|∇u|µ ≤
uγ−p−1 |∇u|2
BR
BR
us
1− µ2
,
BR
where s = (p + 1 − γ)/(2 − µ). We can choose γ close enough to p − 1 such that
s < p, and then obtain (2.4) by combining (2.3) and (2.5). Thus we complete the
proof.
Lemma 2.3. Let u(r, xn ) be a nonnegative weak solution of the following inequality,
in a domain Ω,
|urr + uxn xn | ≤ c(x)|∇u| + d(x)u + f (x),
THE (n − 1)-RADIAL SYMMETRIC SOLUTION
EJDE-2013/204
5
0
where c(x) ∈ Lq (Ω), d, f ∈ Lq (Ω), q 0 > 2 and q ∈ (1, 2). Then for every R such
2
1− 2
that B2R ⊂ Ω, there exists a constant C = C(q, q 0 , R q0 kckLq0 , R2− q kdkLq ) such
that
2
sup u ≤ C(inf u + R2− q kf kLq ).
BR
BR
Note that this lemma is of Harnack type; see [15] for more information on this
type of inequalities. The next theorem is similar to [14, Theorem 2.3].
Theorem 2.4. Let (D) hold and R ≤ R0 such that B2R ⊂ Ω. Suppose u(r, xn ) is
a positive weak solution of the inequality
n−2
n−2
up − M |∇u|τ +
ur ≤ −(urr + uxn xn ) ≤ c0 up + M |∇u|τ +
ur + λ,
r
r
2p
where p > 1, 0 < τ < p+1
, λ > 0. Then there exists a constant C = C(p, τ, R0 , M )
such that
sup u ≤ C(inf u + λR2 ).
BR
BR
Proof. From (2.4), we obtain
n−2
|∇u| + λ.
r
and d = c0 up−1 . To prove this theorem, we only
|urr + uxn xn | ≤ c0 up + M |∇u|τ +
Take f = λ, c = M |∇u|τ −1 + n−2
r
need to verify that
0
c(x) ∈ Lq (B2R ),
d ∈ Lq (B2R ).
0
q
τ −1
Note that n−2
∈
r obviously belongs to L (B2R ), so we only need to prove M |∇u|
q0
L (B2R ). By lemma 2.1, we have
Z
1/q0
2−(p+1)µ/(p−1)
τ −1
q0
kM |∇u|
kLq0 = M
|∇u|µ
,
≤ CR
B(2R)
2p
where µ = q (τ − 1) should satisfy q 0 (τ − 1) < p+1
for some q 0 > 2. Since τ <
and q 0 > 2 can be close enough to 2, so we just need to verify
2p
2p
2
−1 <
.
p+1
p+1
0
2p
p+1
0
The above inequality is obvious, that is to say, c(x) ∈ Lq (B2R ).
For d = c0 up−1 , by lemma 2.1 we have
Z
1/q
kdkLq (B2R ) = c0
uγ
≤ CR(2−2q)/q ,
B(2R)
where γ = (p − 1)q should satisfy (p − 1)q < p. By choosing q > 1 close enough to
1, we can get (p − 1)q < p, that is, d ∈ Lq (B2R ). The proof is complete.
For completeness, we sketch the proof of Theorem 2.1 which is similar as the
proof of [14, Proposition 3.3].
Proof of Theorem 2.1. Suppose, by contradiction, that there exist λn < λ0 , un > 0
such that un is solution of (2.1) with λ substituted by λn and maxΩ un → ∞. Let
zn be a point in Ω such that un (zn ) = maxΩ un , Sn . Denote δn = dist(zn , ∂Ω).
In order to prove there exists a y0 ∈ Ω such that un (y0 ) → ∞, we proceed in three
steps:
6
Y. ZHANG, Q. XU, P. ZHAO
EJDE-2013/204
(p−1)/2
Step 1: There exists c > 0 such that c < δn Sn
. Define w(x) = Sn−1 un (y),
(1−p)/2
where y = Mn x + zn , Mn = Sn
. By easy computation and condition (F2), we
obtain
−∆wn (x) = Sn−1 Mn2 (H(Mn x + zn , Sn wn (x), Sn Mn−1 ∇wn (x)) + λn )
n−2
τ p+1
|∇wn | + λn Sn−p .
≤ c0 wnp + M Sn−p Sn 2 |∇wn |τ +
dist(0, ∂Ω)
τ p+1
2
Notice that M Sn−p Sn
so
and λn Sn−p tend to zero respectively as n tends to infinity,
−∆wn (x) ≤ c0 wnp + |∇wn |τ +
n−2
|∇wn | + 1.
dist(0, ∂Ω)
By the regularity result in [12], there exists a constant C independent of n such
that supΩ wn ≤ C. Let yn ∈ ∂Ω such that d(zn , yn ) = δn ; then, by the mean value
theorem, we have
1 = wn (0) = wn (0) − wn (Mn−1 (yn − zn )) ≤ sup wn Mn−1 δn ≤ CMn−1 δn .
Ω
Thus, the first step is complete.
Step 2: There exists γ > 0 such that
Z
|un |γ → ∞.
B(zn ,δn /2)
By Theorem 2.4, we obtain
Sn =
max
B(zn ,δn /2)
un ≤ C
min
B(zn ,δn /2)
un + λn
δn2 .
4
Since λn and δn are bounded, we obtain that minB(zn ,δn /2) un ≥ cSn for some c > 0.
So
Z
|un |γ ≥ cSnγ δn2 ≥ cSnγ Sn1−p .
B(zn ,δn /2)
We can choose a γ > p − 1 such that cSnγ Sn1−p → +∞. The proof of step 2 is
complete.
Step 3: There exists a y0 ∈ Ω such that un (y0 ) → ∞. Notice that ∂Ω is C 2 and
compact boundary , so we can find ε > 0 independent of n and yn ∈ Ω such that:
• d(yn , ∂Ω) = 2ε, for all n ∈ N.
• B(zn , δ2n ) ⊂ B(yn , 2ε), for all n ∈ N.
By the weak Harnack inequality in [16] and step 2, we conclude that
Z
1/γ
min un ≥ c
|un |γ
→ +∞.
B(yn ,ε)
B(yn ,2ε)
Taking a subsequence {ỹn } ⊂ {yn } such that ỹn → y0 ∈ Ω. For n large enough, we
have y0 ∈ B(ỹn , ε) and un (y0 ) → ∞, which contradicts with Theorem 2.4. Thus
we obtain a priori estimate of solutions.
THE (n − 1)-RADIAL SYMMETRIC SOLUTION
EJDE-2013/204
7
3. Existence of positive classical C 2,β -solutions
Theorem 3.1. Assume (D1), (F1)–(F3) hold. Then (1.1) admits an (n − 1)-radial
symmetric positive classical solution u(r, xn ) ∈ C 2,β (Ω) ∩ C 0 (Ω).
The following lemma mentioned in [10] will be used in our proof.
Lemma 3.2 ([10, Theorem 12.4]). Let u be a bounded C 2 (Ω) solution of
Lu = a(x, y)uxx + 2b(x, y)uxy + c(x, y)uyy = f (x, y),
where L is uniformly elliptic in a domain Ω ⊂ R2 , satisfying
λ(ξ 2 + η 2 ) ≤ aξ 2 + 2bξη + cη 2 ≤ Λ(ξ 2 + η 2 ), ∀(ξ, η) ∈ R2 ,
Λ
≤γ
λ
for some constant γ ≥ 1. Then for some α = α(γ) > 0, we have
[u]∗1,α = sup d1+α
1,2
z1 ,z2 ∈Ω
where C = C(γ),
(2)
| λf |0
f (2)
|Du(z2 ) − Du(z1 )|
≤ C(|u|0 + | |0 ),
α
|z2 − z1 |
λ
= supz∈Ω d2z | λf |, dz = dist(z, ∂Ω) and d1,2 = min{dz1 , dz2 }.
Since the conditions imposed on f in Theorem 3.1 are different from those in
[10, Theorem 12.5], it is necessary to give the proof, although similar to that of [10,
Theorem 12.5].
Proof of Theorem 3.1. We now proceed by truncation of H to reduce (1.4) to the
case of bounded H. Namely, let ψN denote the function given by
(
t,
|t| ≤ N
ψN (t) =
N sign t, |t| > N,
and define the truncation of H by
HN (r, xn , u, ur , uxn ) = H(r, xn , ψN (u), ψN (ur ), ψN (uxn )).
From (F2), we have |HN | ≤ c0 N p + M N τ +
family of problems
n−2
dist(0,∂Ω) N
−(urr + uxn xn ) = HN (r, xn , u, ur , uxn )
u(x) = 0,
on ∂Ω.
= C0 . Consider now the
in Ω,
(3.1)
By Theorem 2.1, any solution u of (3.1) is subject to the bound M̃ , independent
of N ,
sup |u| ≤ M̃ .
(3.2)
Ω
Now we make the following observation. Let v be any bounded function with locally
Hölder continuous first derivatives in Ω and H˜N = HN (r, xn , v, vr , vxn ). Then the
following linear problem
−(urr + uxn xn ) = H˜N in Ω,
(3.3)
u(x) = 0, on ∂Ω,
has a unique solution u ∈ C 2 (Ω)∩C 0 (Ω̄). We observe from classical priori estimates
that
|u|0 = sup |u| ≤ M0 .
Ω
8
Y. ZHANG, Q. XU, P. ZHAO
EJDE-2013/204
Furthermore, if supΩ |v| ≤ M0 , from lemma 3.1, we have
|u|∗1,α ≤ C(|u|0 + C0 (diam(Ω))2 ) ≤ C(M0 + C0 (diam(Ω))2 ) = K,
where C, α depend on M0 . So K depends on M0 , N and Ω.
Next, define the Banach space
C∗1,α (Ω) = {u ∈ C 1,α (Ω)||u|∗1,α;Ω < +∞}
and define a mapping T on the set
S = {v ∈ C∗1,α : |v|∗1,α ≤ K, |v|0 ≤ M0 }.
So u = T v is the unique solution of the linear Dirichlet problem (3.3). It is easy
to show that S is convex and closed in the Banach space, and T is continuous in
C∗1 = {u ∈ C 1 (Ω)||u|∗1;Ω < +∞ and T S is precompact. So we may conclude from
the Schauder fixed point theorem and Schauder estimates that T has a fixed point,
uN = T uN , uN ∈ C∗1,α (Ω) ∩ C 2,β (Ω) ∩ C o (Ω̄). This will provide a solution of the
problem (3.1).
Furthermore, from lemma 3.1 we infer the estimate
(2)
[uN ]∗1,α ≤ C(|u|0 + |GHN |0 ).
By (F2) and (3.2), we obtain
[uN ]∗1,α ≤ C(1 + [uN ]∗1 ),
where C = C(M̃ , M, c0 , p, τ, diam(Ω)). Furthermore, the interpolation inequality
yields the uniform bound which is independent of N ,
[uN ]∗1,α ≤ C = C(M̃ , M, c0 , p, τ, diam(Ω)).
By similar arguments as in the proof of [10, Theorem 12.5], it is easy to show
there is a subsequence {un } of {uN } which converges to a solution u of (1.4), and
u also satisfies the boundary condition u = 0 on ∂Ω. Since f is nonnegative, by
comparison principles, u is positive. This completes the proof.
Remark 3.3. If Ω = Ω1 × Ω2 ⊂ Rk × Rn−k , Ω1 and
p Ω2 are symmetric and
Ω,
f
(x,
u,
|∇u|)
=
f
(r
,
r
,
u,
|∇u|),
where
r
=
x21 + x22 + · · · + x2k , r2 =
0
∈
/
1
2
1
q
x2k+1 + x2k+2 + · · · + x2n . Under the conditions of (F2) and (F3), (1.1) admits an
(n − 1)-radial symmetric positive classical solution u(r1 , r2 ) ∈ C 2,β (Ω) ∩ C 0 (Ω).
The proof is left to readers.
Acknowledgments. This work is partly supported by the National Natural Science Foundation of China (10971088) and Natural Science Foundation of Chizhou
College (2013ZRZ002).
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Yong Zhang
Department of Mathematics, Lanzhou University, Lanzhou, Gansu, 730000, China
Department of Mathematics, Chizhou College, Chizhou, Anhui, 247000, China
E-mail address: zhangy12@lzu.edu.cn
Qiang Xu
Department of Mathematics, Lanzhou University, Lanzhou, Gansu, 730000, China
E-mail address: xuqiang09@lzu.edu.cn
Peihao Zhao
Department of Mathematics, Lanzhou University, Lanzhou, Gansu, 730000, China
E-mail address: zhaoph@lzu.edu.cn