ON POSITIVE SOLUTIONS OF QUASILINEAR
ELLIPTIC EQUATIONS
NGUYEN HOANG LOC AND KLAUS SCHMITT
Abstract. In 1981 Peter Hess established a multiplicity result for
solutions of boundary value problems for nonlinear perturbations
of the Laplace operator. The sufficient conditions given were later
shown to be also necessary by Dancer and the second author. In
this paper we show that similar (and slightly more general) results
hold when the Laplace operator is replaced by the p− Laplacian.
1. Introduction.
Let Ω be an open bounded subset of RN with smooth boundary ∂Ω,
p ∈ (1, ∞), and E be the usual Sobolev space W01,p (Ω) with norm
p1
|∇u|p dx .
u =
Ω
Let f be a continuous function on R and assume throughout that f
satisfies:
(F ) f (0) ≥ 0, and there exist 0 < a1 < b1 < a2 < b2 < · · · < bm−1 <
am such that for all k = 1, · · · , m − 1
f (·) ≤ 0 on (ak , bk )
f (·) ≥ 0
on (bk , ak+1 ).
Motivated by the results in [1] (for the case p = 2), we establish
necessary and sufficient conditions in order that the problem
in Ω
−Δp u = λf (u)
(1.1)
u = 0
on ∂Ω
has, for λ >> 1, at least m − 1 nonnegative weak solutions
{u1 , · · · , um−1 } ⊂ E ∩ L∞ (Ω),
such that
ak < uk ∞ ≤ ak+1 , k = 1, · · · , m − 1.
1991 Mathematics Subject Classification. 35B45,35J60, 35,J65.
Key words and phrases. p-Laplacian, weak solutions, multiple solutions.
1
2
NGUYEN HOANG LOC AND KLAUS SCHMITT
Definition 1.1. The function f above is said to be in class (H) if
ak+1
ak
f (s)ds > 0 for all k ∈ {1, · · · , m − 1}.
The following is the result to be established in the paper:
Theorem 1.1. If f is in class (H), then for all λ, sufficiently large,
(1.1) has at least m − 1 non-negative solutions {u1 , · · · , um−1 } ⊂ E ∩
L∞ (Ω) such that ak < uk ∞ ≤ ak+1 for each k = 1, · · · , m − 1. It is
also the case that if (1.1) has a non-negative weak solution u such that
u∞ ∈ (ak , ak+1 ], then
ak+1
f (s)ds > 0,
ak
for k ∈ {1, · · · , m − 1}.
The proof of this theorem will be given in the next two sections and
follows ideas used in the proofs of the paper [1], suitably modified and
expanded for the case being considered.
2. Proof of the first part of Theorem 1.1
We establish first a result which is an extension of a theorem of Hess
[3] and which gives us the first part of Theorem 1.1. The proof follows
Hess’ arguments very closely and we give it here, for completeness’
sake.
Theorem 2.1. Suppose f is in class (H). Then there exists a number
λ̄ > 0 such that for all λ ≥ λ̄, problem (1.1) has at least m - 1 nonnegative weak solutions u1 , u2, · · · , um−1 ∈ E ∩ L∞ (Ω) and uk ∞ ∈
(ak , ak+1 ], k = 1, · · · , m − 1.
The following Lemma is a consequence of the weak maximum principle for the p− Laplace operator.
Lemma 2.1. Let g : R → R be a continuous function such that there
exists s0 ≥ 0 such that g(s) ≥ 0 if s ∈ (−∞, 0) and g(s) ≤ 0 if s ≥ s0 .
If u is a weak solution of
(2.1)
−p u = g(u) in Ω, u = 0 on ∂Ω,
then u is non-negative a.e. and belongs to L∞ (Ω). Moreover,
u∞ ≤ s0 .
3
Proof. We let v = u− = max{−u, 0} ∈ E, then
−∇u u < 0
∇v =
0
u ≥ 0.
Hence, since u is a weak solution of (2.1), we have
p−2
|∇u| ∇u · ∇vdx =
g(u)vdx.
Ω
Ω
Ω
Ω
This implies v ≤ 0 and therefore v = 0, and u ≥ 0 a.e. on Ω.
Next, choosing the test function v = (u − s0 )+ = max{u − s0, 0} ∈ E
in the equation
p−2
|∇u| ∇u · ∇vdx =
g(u)vdx,
we have v ≤ 0 and therefore u ≤ s0 a.e., i.e., u∞ ≤ s0 .
For k = 2, · · · , m, define fk as follows:
⎧
f (0) s ≤ 0
⎪
⎪
⎨
f (s) 0 ≤ s ≤ ak
fk (s) :=
⎪
⎪
⎩
0
s > ak ,
and let
Fk (s) =
s
0
fk (σ)dσ.
For any λ ≥ 0, let the functional
Φk (λ, ·) : E → R
be defined by
1
Φk (λ, u) =
p
p
Ω
|∇u| − λ
Ω
Fk (u)dx.
Denote by Kk (λ) the set of critical points of Φk (λ, ·), then if u is in
Kk (λ), u is a weak solution of
−p u = λfk (u) in Ω
u = 0
on ∂Ω
and by Lemma 2.1, u is nonnegative and u∞ ≤ ak . We have thus
shown:
Lemma 2.2. u is in Kk (λ) if and only if u is a non-negative weak
solution of (1.1) and belongs to L∞ (Ω) with u∞ ≤ ak .
4
NGUYEN HOANG LOC AND KLAUS SCHMITT
We next claim that Kk (λ) is not empty. Since fk is bounded and
vanishes on (ak , ∞), Φk (λ, ·) is coercive. Futher, since the first summand defining Φk (λ, ·) is 1p · p , it is continuous and since it is a convex
functional it is weakly lower semi continuous. The second summand is
weakly continuous, as follows from the compact embedding of W01,p (Ω)
in Lp (Ω). Thus, there exists uk (λ) such that
Φk (λ, uk (λ)) = inf{Φk (λ, v) : v ∈ E}.
The following Lemma shows that for k = 2, · · · , m, ak−1 < uk ∞ ≤
ak and therefore, (1.1) has at least m − 1 solutions when λ > 0 sufficiently large.
Lemma 2.3. For each k = 2, · · · , m, there exists λk > 0, such that for
all λ > λk , uk (λ) ∈ Kk−1 (λ).
Proof. We show that there exist λk > 0 and w ∈ E, w ≥ 0, and
w∞ ≤ ak , such that
Φk (λ, w) < Φk−1 (λ, uk−1(λ)),
∀λ > λk .
This will imply the assertion.
Since f is in class (H),
0 < α := F (ak ) − max{F (s) : 0 ≤ s < ak−1 },
s
where F (s) = 0 f (σ)dσ. Then for all u in E satisfying 0 ≤ u ≤ ak−1
a.e.,
Ω
F (u)dx ≤
Ω
F (ak )dx − α|Ω|,
where |Ω| is the Lebesgue measure of Ω.
For δ > 0, let Ωδ := {x ∈ Ω : dist(x, ∂Ω) < δ}. By Lebesgue’s
Theorem, |Ωδ | → 0 as δ → 0. On the other hand, for each δ > 0, there
exists wδ ∈ C0∞ (Ω) with 0 ≤ wδ ≤ ak , wδ (x) = ak ∀x ∈ Ω \ Ωδ . Thus
Ω
F (ak )dx −
F (wδ )dx
Ωδ
F (ak )dx −
(F (ak ) + F (wδ ))dx
=
Ω
Ωδ
≥
F (ak )dx − 2C|Ωδ |,
F (wδ )dx =
Ω\Ωδ
Ω
where C = max{|F (s)| : 0 ≤ s ≤ ak }. Hence, for all u in E, 0 ≤ u ≤
ak−1 ,
F (wδ )dx ≥
F (u)dx + α|Ω| − 2C|Ωδ |.
Ω
Ω
5
Fix δ > 0 such that η := α|Ω| − 2C|Ωδ | > 0 and set w := wδ . Then
for all u, 0 ≤ u ≤ ak−1 ,
1
p
p
(w − u ) − λ (F (w) − F (u))dx
Φk (λ, w) − Φk−1 (λ, u) =
p
Ω
1
wp − λη < 0,
≤
p
provided λ > 0 is chosen sufficiently large.
Thus, for all λ large enough, there are m−1 solutions u2 (λ), · · · , um (λ)
as asserted by Theorem 2.1.
3. Proof of the second part of Theorem 1.1
It follows easily that it will suffice to consider the case that λ = 1
and k = 1. We also assume f (0) > 0 and remove this condition later.
Thoughout this section, all weak solutions u of (1.1) are in class C 1
because of the Theorem 1.7 in [5]. We first establish a strong maximum
principle for weak solutions of equation (1.1).
Lemma 3.1. Let u ∈ C 1 (Ω) be a nonnegative weak solution of (1.1).
If f (0) > 0 then u is positive in Ω.
Proof. Assume there exists x0 in Ω such that u(x0 ) = 0. Let D be a
ball contained in Ω such that x0 ∈ ∂D. Denote by y0 and r the center
and the radius of D, respectively, and let g ≤ f be a strictly decreasing
continuous function defined on [0, ∞) such that g(0) = f (0) > 0 and
γ := g( a21 ) = inf{g(s) : 0 ≤ s ≤ a21 } > 0. Now, define a function b on
D as
b(x) = e−|
x−y0 2
|
r
− e−1 ,
where is sufficiently small such that
sup |div(|∇b(x)|p−2 ∇b(x))| ≤ γ.
x∈D
Then b is a subsolution of
−div(|∇u|p−2∇u) = g(u)
u=0
in D
on ∂D.
It follows that for all ϕ ≥ 0 in W01,p (D),
p−2
p−2
(|∇b| ∇b − |∇u| ∇u) · ∇ϕdx ≤
(g(b) − g(u))ϕdx.
D
D
6
NGUYEN HOANG LOC AND KLAUS SCHMITT
Choosing ϕ = (b−u)+ , and using the fact that the p-Laplace operator
is monotone and g is strictly decreasing, we obtain
0 ≤
≤
D
(|∇b|p−2∇b − |∇u|p−2∇u) · ∇(b − u)dx
+
(g(b) − g(u))(b − u)dx ≤ 0,
D+
where D + = {x ∈ D : b(x) > u(x)}. Therefore, D + is empty
equivalently u ≥ b in D. Since u(x0 ) = b(x0 ) = 0, and b > 0 in
guarantee that ∂ν u(x0 ) ≤ ∂ν b(x0 ) < 0 implying that |∇u(x0 )| =
which contradicts that u(x0 ) = 0 is a minimum value of u in Ω.
or
D
0,
Let B be an open ball centered at 0 and containing Ω. Define the
function α on Ω by
α(x) =
u(x) x ∈ Ω̄
0
x ∈ B̄ \ Ω.
Since Ω is a domain with smooth boundary, α ∈ W01,p (B). We also
have:
Lemma 3.2. α is a subsolution of
(3.1)
−Δp u = f (u)
in B
u = 0
on ∂B.
Proof. For each n in N, define vn (x) = n min{u(x), n1 }, x ∈ Ω, where
u is as in Lemma 3.1. Then ∇u · ∇vn is nonnegative and by Lemma
3.1, {vn } converges to 1 pointwise in Ω. Let w ≥ 0 be in C0∞ (B). Then
wvn is in W01,p (Ω) and since u is a weak solution of (1.1),
Ω
or
w|∇u|
Ω
p−2
|∇u|
p−2
∇u · ∇(wvn )dx =
∇u · ∇vn dx +
Ω
vn |∇u|
p−2
Ω
f (u)wvndx,
∇u · ∇wdx =
Ω
f (u)wvn dx.
7
Now, applying Lebesgue’s dominated convergence Theorem and noting that 0 ≤ vn ≤ 1 for all n, we have
p−2
|∇α| ∇α · ∇wdx =
|∇u|p−2∇u · ∇wdx
B
Ω
vn |∇u|p−2∇u · ∇wdx
= lim
n→∞ Ω
f (u)wvn dx − w|∇u|p−2∇u · ∇vn dx
= lim
n→∞ Ω
f (u)wdx
≤
Ω
f (α)wdx,
≤
B
proving the Lemma.
The following demonstrates the second part of Theorem 1.1, which
is an extention of a result of [1].
Theorem 3.1. Assume that f (0) is positive. If (1.1) has a nonnegative
weak solution u in L∞ (Ω) such that u∞ ∈ (a1 , a2 ], then
a2
f (s)ds > 0.
a1
We remark that the condition f (0) > 0 will be removed later.
Proof. Define β(x) = a2 for all x in B. Since β and α are a supersolution and subsolution, respectively, (3.1) has a maximum solution ū
such that α(x) ≤ ū(x) ≤ a2 for all x in B (see Remark 1.5 in [?]). This
means, for all solution v of (3.1) with α(x) ≤ v(x) ≤ a2 , v ≤ ū. This
will imply that ū is radially symmetric i.e. ū(x1 ) = ū(x2 ) for all x1 , x2
in B such that |x1 | = |x2 |. Assuming this is not the case, then there
exist x1 and x2 in B with |x1 | = |x2 | such that ū(x1 ) < ū(x2 ). Let P be
a N × N matrix in SO(N, R), the special orthogonal group, such that
x2 = P x1 . Note that the transpose matrix P T of P is also its inverse
matrix. Let u1 (x) = ū(P x). Since for all x in Ω
∇u1 (x) = P ∇ū(P x),
and the map x → P x is an isometry, it follows that
|∇u1 (x)| = |P ∇ū(P x)| = |∇ū(P x)|.
8
NGUYEN HOANG LOC AND KLAUS SCHMITT
We next show that u1 is a weak solution of (3.1). That is, we need
to verify for all ϕ in W01,p (B)
p−2
|∇u1 (x)| ∇u1 (x) · ∇ϕ(x)dx =
f (u1(x))ϕ(x)dx.
(3.2)
B
B
T
W01,p (B).
Let ψ(x) = ϕ(P x) ∈
The left hand side of (3.2) becomes
|∇ū(P x)|p−2 P ∇ū(P x) · ∇ϕ(x)dx
B
|∇ū(P x)|p−2P ∇ū(P x) · (P ∇ψ(P x)) dx
=
B
|∇ū(y)|p−2∇ū(y) · ∇ψ(y) det P dy
=
B
f (ū(y))ψ(y)dy
=
B
f (ū(P P T y))ψ(P P T y)dy
=
B
f (ū(P x))ψ(P x) det P T dx
=
B
=
f (u1(x))ϕ(x)dx.
B
Hence (3.2) holds.
Now, (3.1) has two subsolutions, α and u1 . It follows from Theorem
1.4 in [4] that (3.1) has another solution u2 such that
max{α, u1} ≤ u2 ≤ β.
Since ū is the maximum solution with respect to the pair of subsupersolutions (α, β), we have
ū(x1 ) ≥ u2 (x1 ) ≥ u1 (x1 ) = ū(x2 ) > ū(x1 ).
This contradiction shows that ū is radially symmetric.
Next, define a C 1 −function u : [0, R) → R+ by u(|x|) = ū(x) for all
x in B, where R is the radius of B. Using the chain rule for classical
differentiation, we have for all r in (0, R)
xi
du ∂r
∂ ū
=
= u , i = 1, ..., N
∂xi
dr ∂xi
r
12
N
x2
i
|∇ū| = |u |
= |u |.
2
r
i=1
9
For any v in C0∞ (0, R), put
w(r) =
v(r)
r N −1
∀r ∈ (0, R), w(0) = 0
and
v̄(x) = v(|x|), w̄(x) = w(|x|)
∀x ∈ B.
Now, as a weak solution of (3.1), ū satisfies
p−2
|∇ū| ∇ū · ∇w̄dx =
f (ū)w̄dx.
B
But,
∂ w̄(x)
∂xi
=
B
xi
w |x|
,
R
0
and thus,
p−2 |u |
N −1
uwr
dr =
R
f (u)wr N −1dr.
0
v
v
and w = rN−1
− Nr−1
Substituting w = rN−1
N v into this equation, we
obtain
R
R
v
N −1
v
p−2 N −1
|u | u
−
v r
dr =
f (u) N −1 r N −1 dr,
N
−1
N
r
r
r
0
0
or
R
R
R
N − 1 p−2 p−2 |u | u vdr =
|u | u v dr −
f (u)vdr,
r
0
0
0
for all v in C0∞ (0, R). This implies that u is a C 1 weak solution of the
equation
N − 1 p−2 |u | u + f (u),
r
and by the continuity of the right hand side, the distributional derivative ∂ above becomes a classical derivative and hence u is a classical
solution of (3.3).
Since ū is radially symmetric, u(0) = 0. Hence, u is a solution of
(3.3) subject to the condition u (0) = 0 = u(R). Let r0 ∈ [0, R) such
that umax = u(r0 ) = max{u(r) : r ∈ [0, R)}. Multiplying both sides of
(3.3) by u and intergrating it, we obtain
r
r p r
|u |
p−2 dτ =
(p − 1)|u | u u dτ + (N − 1)
f (u)udτ,
−
τ
r0
r0
r0
(3.3)
−∂(|u |p−2 u) =
for all 0 < r < R. Now, since umax = u(r0 ) is greater than a1 , we can
choose r ∈ (0, R) such that u(r) = a1 . The above equality becomes
u (r)
r p
a1
|u |
p−2
f (s)ds = −(p − 1)
|s| sds − (N − 1)
dτ < 0.
τ
0
r0
umax
10
NGUYEN HOANG LOC AND KLAUS SCHMITT
u
This equation shows a1max f (s)ds > 0. Because f ≤ 0 in (a1 , b1 ],
umax ∈ (b1 , a2 ] and f is nonnegative in [umax , a2 ], we get the desired
result because
umax
a2
f (s)ds ≥
f (s)ds > 0.
a1
s0
4. Some remarks
Remark 1. We can remove the condition f (0) > 0 in Theorem
3.1. In fact, assume that f (0) ≤ 0 and again assume that (1.1) has a
nonnegative solution u in W01,p (Ω) ∩ L∞ (Ω) satisfying u∞ ∈ (a1 , a2 ].
Define f˜ so that f˜(0) > 0, f˜(s) ≥ f (s) when 0 ≤ s ≤ a1 and f˜(s) =
f (s) on [a1 , ∞). Then u is a subsolution of
−p u = f˜(u) x ∈ Ω
(4.1)
u = 0
x ∈ ∂Ω
and as before, we may use β(x) ≡ a2 as a supersolution for (4.1).
Hence, (4.1) has a solution ũ satisfying u ≤ ũ ≤ a2 . We now proceed
as in the first part of the proof with f˜ in place of f and obtain
a2
a2
˜
f (s)ds =
f(s)ds
> 0.
a1
a1
The previous remark and Theorem 3.1 consider the case 0 < a1 <
b1 < a2 . Now, we study the case 0 = a1 < b1 < a2 .
Remark 2 Let a < b be two positive numbers and let f be a continuous function on [0, b] such that f (0) = 0, f < 0 on (0, a), f > 0 on
(a, b) and f (a) = f (b) = 0. Assume the problem
−p u = f (u) x ∈ Ω
(4.2)
u=0
x ∈ ∂Ω
has a nonnegative weak solution u in E ∩ L∞ (Ω) such that u∞ is in
(a, b] then
b
f (s)ds ≥ 0.
0
Proof. Let < a be an arbitrarily small positive number. For each n in
N, define a continuous function g satisfying g(0) = 1, g > 0 on (0, ),
11
a
a
f ≤ g < 0 on (, a), g = f on [a, b] and g(s)ds < 0 f (s)ds + n1 .
Since u is a solution of (4.2) and g ≥ f , u is a subsolution of
−p u = g(u) x ∈ Ω
(4.3)
u=0
x ∈ ∂Ω
Now (4.3) has a pair of sub-supersolution (u, b), as in the proof of
Theorem 3.1, (4.3) has a weak solution u1 such that u ≤ u1 ≤ b a.e. in
Ω. Using Theorem 3.1 and the assumption on g, we have
b
b
1
f (s)ds + >
g(s)ds > 0.
n
0
It follows the assertion.
The problem (1.1) is said to be infinite semipositone if f (0) is −∞.
In recent years, it was concerned by some mathematicians [2, 6, 7]. One
of their results, the Theorem 1.1 in [2], can be deduced by Theorem 3.1
and Remark 2 in the case all functions in the semipositone problem in
[2] do not depend on x. However, our result will be more general than
Theorem 1.1 in [2] because we replace the problem of Laplacian type
by a p−Laplacian one and remove some smooth and growth conditions
required in that paper. This idea will be given in Remark 3.
Let g be continuous on (0, ∞) and lim+ g(s)ds = ∞. Let h : [0, ∞) →
s→0
[0, ∞) be continuous. Assume there exists uniquely s0 > 0 such that
f = −g + h is negative
on (0, s0 ) and positive on (s0 , ∞). We have
s0
Remark 3 If 0 g(s)ds = ∞ then the problem
⎧
⎨ −Δp u + g(u) = h(u) in Ω,
u>0
in Ω,
(4.4)
⎩
u=0
on ∂Ω
has no nonnegative weak solution in L∞ (Ω).
Proof. Assume by contrary that u is a nonnegative weak solution of
(4.4) such that M = u∞ < ∞. It is easy to see M > s0 . In fact, if
M ≤ s0 then
p
0≤
|∇u| dx = (−g(u) + h(u))udx ≤ 0
Ω
and thus u = 0. It is impossible.
Ω
12
NGUYEN HOANG LOC AND KLAUS SCHMITT
Define a continuous function f on (0, ∞) and a sequence of continuous functions {fn }n∈N on [0, ∞) such that
⎧
⎨ −g(s) + h(s) s ∈ (0, M),
≥0
s ∈ (M, M + 1),
f (s) =
⎩
0
s>M +1
and
⎧
⎨ fn (0) = 0,
f (s) ≤ fn (s) ≤ 0 s ∈ (0, n1 ],
⎩
fn (s) = f (s)
s ∈ ( n1 , ∞).
Since u ≤ M a.e. in Ω and fn ≥ f for all n ∈ N, u is a subsolution
of
(4.5)
−Δp u = fn (u) in Ω,
u=0
on ∂Ω
for all n ∈ N. Applying Theorem 1.4 in [4], we have (4.5) has a weak
solution un such that u ≤ un ≤ M + 1 a.e. in Ω. Now, Remark 2
implies
M +1
fn (s)ds ≥ 0
0
and
0
s0
−fn (s)ds ≤
M +1
s0
fn (s)ds =
M +1
f (s)ds
s0
for all n ∈ N.
On the other hand, −fn (s) converges to −f (s) = g(s)−h(s) for every
s in (0, s0 ) as n approaches to ∞. Noting that −fn ≥ 0 on (0, s0 ), we
can use the Fatou Lemma to get
s0
s0
(g(s) − h(s))ds ≤ lim inf
−fn (s)ds ≤ C < ∞
0
M +1
n→∞
0
where C = s0 f (s)ds is a constant. It follows that
and this is a contradiction to the hypothesis.
1
0
g(s)ds < ∞
References
[1] E. Dancer and K. Schmitt, On positive solutions of semilinear elliptic equations, Proceedings of the American Mathematical Society, 101 (1987), pp. 445–
452.
[2] M. Ghergu and V. Radulescu, Sublinear singular elliptic problems with two
parameters, J. Differential Equations, 195 (2003), pp. 520–536.
[3] P. Hess, On multiple positive solutions of nonlinear elliptic eigenvalue problems, Commun. Partial Differential Equations, 6 (1981), pp. 951–961.
13
[4] V. K. Le and K. Schmitt, Some general concepts of sub-supersolutions
for nonlinear elliptic problems, Topological Methods in Nonlinear Analysis, 28
(2006), pp. 87–103.
[5] G. M. Lieberman, The natural generalization of the natural conditions of
Ladyzhenskaya and Ural’tseva for elliptic equations, Comm. Partial Differential
Equations, 16 (1991), pp. 311–361.
[6] M. Ramaswamy, R. Shivaji and J. Ye, Positive solutions for a class of
infinite semipositone problems, Differential and Integral Equations, 20 (2007),
p. 14231433.
[7] Z. Zhang, On a Dirichlet problem with a singular nonlinearty, J. Math. Anal.
Appl., 194 (1995), pp. 103–113.
Department of Mathematics, University of Utah, 155 South 1400
East, Salt Lake City, UT 84112, USA
E-mail address: [email protected]
E-mail address: [email protected]