Directory UMM :Journals:Journal_of_mathematics:EJQTDE:

Electronic Journal of Qualitative Theory of Differential Equations
2005, No. 19, 1-12; http://www.math.u-szeged.hu/ejqtde/

REMARKS ON INHOMOGENEOUS ELLIPTIC
PROBLEMS ARISING IN ASTROPHYSICS
MARCO CALAHORRANO & HERMANN MENA
Abstract. We deal with the variational study of some type of
nonlinear inhomogeneous elliptic problems arising in models of solar flares on the halfplane Rn+ .

1. Introduction
In this paper we study a boundary value problem of type

−∆u + c(x)u = λm(y)f (u)
Rn+
(1.1)
n−1
u(z, 0) = h(z)
∀z ∈ R
where x = (z, y) ∈ Rn−1 × R+ ≡ Rn+ with R+ = {y ∈ R : y > 0} and
n ≥ 2, f :] − ∞, +∞[→ R is a function satisfying:
(f-1) There exists s0 > 0 such that f (s) > 0 for all s ∈]0, s0 [.

(f-2) f (s) = 0 for s ≤ 0 o s ≥ s0 .
n+2
(f-3) f (s) ≤ asσ , a is a positive constant and 1 < σ < n−2
if n > 2
or σ > 1 if n = 2.
(f-4) There exists l > 0 such that |f (s1 ) − f (s2 )| ≤ l|s1 − s2 |, for all
s1 , s2 ∈ R.
h is a non-negative
bounded smooth function, h 6= 0, min h < s0 , c ≥ 0,
T
∞
c ∈ L (Ω) C(Ω) and mes{x ∈ Ω : c(x) = 0} = 0.
The problem (1.1) is a generalization of an astrophysical gravity model
of solar flares in the half plane R2+ , given in [1], namely:

−∆u = λe−βy f (u)
R2+
(1.2)
u(x, 0) = h(x)
∀x ∈ R

besides the above mentioned conditions for f , h and β > 0. See [1], [8]
and [6] for a detailed description and related problems.
By this, we study the problem (1.1) with m : R+ → R+ a C 1 function
such that
Z
+∞

ym(y)dy < +∞

0

more general than e−βy .
1991 Mathematics Subject Classification. 35J65, 85A30.
Key words and phrases. Solar flares, variational methods, inhomogeneous semilinear elliptic problems.

EJQTDE, 2005 No. 19, p. 1

We shall follow the ideas of F. Dobarro and E. Lami Dozo in [8].
The authors prove the existence of solutions of (1.1) in the special case
c(x) = 0. In fact, the result presented here follows from the one obtained by the authors.

First of all we note that problem (1.1) is equivalent to

−∆ω + c(x)ω = λm(y)f (ω + τ )
Rn+
(1.3)
n−1
ω(z, 0) = 0
∀z ∈ R
where ω = u − τ and τ is solution of the problem

−∆τ + c(x)τ = 0
Rn+
(1.4)
τ (z, 0) = h(z)
∀z ∈ Rn−1
We will study (1.3) instead of (1.1).
The problem (1.1), or equivalently (1.3), is interesting not only on
whole Rn+ , but also in an arbitrary big but finite domain in Rn+ , for example for semidisks DR = {(x, y) ∈ Rn−1 ×R+ : |x|2 +y 2 < R2 , y > 0},
with R big enough.
Motivated by this observation in section 2, we will study the following

approximate problem

−∆ω + c(x)ω = λm(y)f (ω + τ )
DR
(1.5)
ω=0
∂DR
whose solutions are related to those of (1.3).
Using variational techniques we will prove the existence of an interval Λ ⊂ R+ such that for all λ ∈ Λ there exists at least three positive
solutions of (1.5), with R large enough.
Finally in section 3 we prove the existence of solutions of (1.3) as
limit of a special family of solutions of (1.5) obtained in theorem 5 and
its uniqueness to λ small enough.
2. Problem in DR
Letting Ω be either DR or Rn+ , we denote by Lpm (Ω) the usual weighted
L space on Ω for a suitable weight m and 1 ≤ p < ∞, and by Vm1,2 (Ω),
Vc1,2 (Ω) the completion of C0∞ (Ω) in the norm
Z
Z
2

2
kukV 1,2 (Ω) =
u (z, y)m(y)dzdy +
|∇u|2 dzdy
p

m

Ω

Ω

EJQTDE, 2005 No. 19, p. 2

and
kuk2V 1,2 (Ω)
c

=

Z

2

u (x)c(x)dx +
Ω

Z

|∇u|2dx

Ω

Let m : R+ → R+ be such that
Z +∞
(2.1)
0 0 such that
′

(2.4)

|(log m) | ≤ k

then the identity map is an immersion from Vm1,2 (Ω) into Lp p (Ω) for
m2

1 0, ΘR (Vc1,2 (Dr )) ⊂ Vc1,2 (θR
Dr ) and if R ≥ 1, Vc1,2 (θR
Dr ) ⊂
1,2
Vc (DRr ).
ii. If η ∈ C0∞ (Rn+ ), is non identically 0, then
(2.7)

k∇ηR kL2 (

n)
+

→ +∞

R → +∞

as

iii. Let f be defined before and m such that verifies (2.1). Then there
exists 0 < λ < ∞ such that if λ > λ, η ∈ C0∞ (Rn+ ), η ≥ 0, non
identically 0 and
R
1
2
2
n {|∇η| + kckL∞ η }
2
+
R
(2.8)
λ ≤ Q(η) ≡
0 : ηR ∈
for all non negative function τ .

′
′
a. Ψλ,τ,R′ (ηR ) < 0, ∀R , R: R ≥ Rrn ≥ rn .
b. Ψλ,τ,Rrn (ηR ) → −∞ as R → +∞
EJQTDE, 2005 No. 19, p. 4

Proof.- This proof follows almost directly from lemma 6 in [8]. However, by completeness we present all the proof.
i. It is immediate from the definition of ΘR .
ii. We observe


1
1
2
2
|∇ηR | (z, y) = 2 |∇η|θR + 1 − 2 |∂y η|2θR (z,y)
R
R
thus, changing variables
(2.9)

k∇ηR k2L2 (

so, since
iii. Set
(2.10)

R

n
+

n)
+

=R

n−1



1
R2

Z



1
|∇η| + 1 − 2
n
R
+
2

Z

|∂y η|
n
+

2



|∂y η|2 > 0, (2.9) implies (2.7).
λ ≡ inf {Q(η) : η ∈ C0∞ (Rn+ ), η ≥ 0, η 6= 0}

by (f-3) and since F is bounded
b
F (s)
≡ sup 2 < +∞
2
s>0 s

(2.11)

so, by (2.2) and since c(x) ≥ 0
Z
Z
bM
m(y)F (η) ≤
n
2
+

|∇η|2 + kckL∞ η 2
n
+

hence

1
≤λ Q(η) be, since η ∈ C0∞ (Rn+ ), there exists rn > 0 such that supp
−1
η ⊂ DRrn , for all R ≥ 1. Then by i. and (2.3) ηR ∈ Vc1,2 (θR
Dr n ) ⊂
′
1,2
1,2
Vc (DRrn ) ⊂ Vc (DR′ ) for all R ≥ Rrn ≥ rn .
For simplicity from now on we call Rrn ≡ Rn , where R ≥ 1.
Then, by remark 1
0<

(2.12)

Ψλ,τ,R′ (ηR ) ≤ Ψλ,τ,Rn (ηR ) ≤ Ψλ,0,Rn (ηR )

On the other hand, if we define the function ξ : R+ → R
Z

Z
1
1
1
2
2
ξ(R) ≡ n−1 k∇ηR kL2 ( n+ ) =
|∇η| + 1 − 2
|∂y η|2
n
R
R2 n+
R
+
Z
 R |∇ η|2
n
z
R+
+1
=
|∂y η|2
2
2
n
R
|∂
η|
n
y
+
+

is non increasing. So applying ξ(R) ≤ ξ(1) to (2.9)
Z
Z
Z
2
n−1
2
|∇ηR | ≤ R
|∇η|2
|∇ηR | =
DRn

n
+

n
+

EJQTDE, 2005 No. 19, p. 5

furthermore
Z

c(x)ηR2

=

DRn

and

Z

m(y)F (ηR ) =

DRn

so
Ψλ,0,Rn ≤ R

n−1

Z

Z

 Z
1
2

n
+

Ψλ,0,Rn

≤R

n−1

kckL∞

m(y)F (ηR ) = R

n−1

n
+

2

2

|∇η| + kckL∞ η − λ
n
+

then
(2.13)

c(x)ηR2

Rn−1
≤
2

Z

Z

η2
n
+

m(y)F (η)
n
+

m(y)F (η)
n
+


|∇η| + kckL∞ η 1 −
2

2

n
+

Z

Z

λ
Q(η)





thus, from (2.12) and (2.13) we obtain immediately a and b.

Remark 3. i. Let m : R+ → R+ be a bounded C 1 function and let
2
m
b ≡ m σ+1 . It is easy to prove that m verifies (2.4) if and only if m
b
′
does it. Furthermore, given a positive constant k, |(log m) | ≤ k if and
′
2
k.
only if |(log m)
b | ≤ σ+1
ii. If there exists a non negative value m1 ≥ 0 such that {m > 1} ⊂
[0, m1 ] and
Z +∞
c≡
0 1 and 0 <

2
σ+1

1
m≤1
Z

Z

c)
Lemma 4. There exists a positive constant C = C(a, σ, k, sup m, M
such that for all λ < λ(kτ kLσ+1
n ) and for all u : kuk 1,2
Vc ( n ) =
m ( +)
+

kτ kLσ+1
m (

n ) , Ψλ,τ (u) > 0 where λ(kτ kLσ+1 (
m
+

Moreover λ(kτ k

n
Lσ+1
m ( +)

) → +∞ as kτ k

1−σ
n ) ) ≡ Ckτ k σ+1
L
(
+
m

n ).
+

→0
EJQTDE, 2005 No. 19, p. 6

n
Lσ+1
m ( +)

Proof.- Let u ∈ Vc1,2 (Rn+ ) be, using (f-3) and Minkowsky inequality
with respect to measure m(y)dxdy and (2.2), (2.5) we obtain
Z
Z
Z
Z u+τ
a
0≤
mF (u + τ ) =
m
m(u + τ )σ+1
f (t)dt ≤
n
n
σ + 1 n+
0
+
+
a
(kukLσ+1
)σ+1
+ kτ kLσ+1
≤
m
σ+1
σ+1
m 2
a
c) 12 k∇ukL2( n ) + kτ k σ+1 )σ+1
(Cs K(1 + M
≤
Lm
+
σ+1
a
c) 12 kuk 1,2 n + kτ k σ+1 )σ+1
≤
(Cs K(1 + M
Lm
Vc ( + )
σ+1
then
(2.14)
1
a
c 21 kuk 1,2 n +kτ k σ+1 )σ+1
(Cs K(1+ M)
Ψλ,τ (u) ≥ kuk2V 1,2 ( n ) −λ
Lm
Vc ( + )
c
+
2
σ+1
then, if we define
σ+1
c) 12 + 1)−σ−1
(Cs K(k, sup m)(1 + M
C≡
2a
then Ψλ,τ (u) > 0 for all λ < λ ≡ Ckτ k1−σ
n , and since σ > 1. The
Lσ+1
m ( +)
lemma is proved.

Theorem 5. Let us assume (f-1-2-3-4), let m:R+ → R+ be a C 1
2
function such that m and m
b ≡ m σ+1 verify (2.1) and (2.4), and let
τ : Rn+ → R+ be a C 1 function, non identically 0. So there exists
c) and λ such that if
positive constants C = C(a, σ, k, sup m, M
1
  σ−1
c
(2.15)
kτ kLσ+1
n <
m ( +)
λ
then
∀λ : λ < λ < λ ≡ Ckτ k1−σ
Lσ+1 (
m

n)
+

there exists a positive R0 = R0 (λ) such that for all R ≥ R0 , (1.5) has
at least three strictly positive solutions.
c) and λ be the positive constant
Proof.- Let C = C(a, σ, k, sup m, M
defined in lemmas 4 and 2 respectively . Since τ verifies (2.15), by
lemma 4 and remark 1, for all λ ∈]λ, λ[ and for all R ≥ 1
(2.16)

Ψλ,τ,R (u) > 0

∀u ∈ Vc1,2 (DR ) : kukVc1,2 (DR ) = kτ kLσ+1
n
m (R+ )

On the other hand, fixed λ ∈]λ, λ[, η ∈ C0∞ (Rn+ ), and letting rn > 0, the
radius of any semidisk Drn such that supp η ⊂ Drn , by lemma 2 there
EJQTDE, 2005 No. 19, p. 7

exists R1 ≥ 1 such that for all R ≥ R1 rn , we have ηR1 ∈ Vc1,2 (DR ),
furthermore
(2.17) kτ kLσ+1
m (

n)
+

< k∇ηR1 kL2 (DR ) = k∇ηR1 kL2 (

n)
+

< kηR1 kVc1,2 (

n)
+

and
(2.18)

Ψλ,τ,R (ηR1 ) < µ < 0

where µ ∈ R defined as
1 2
a
c) 12 t + kτ k σ+1 )σ+1
µ≡
min
t −λ
(Cs K(1 + M
Lm
0≤t≤kτ kLσ+1 ( n ) 2
σ+1
m
+

Let R ≥ R1 , we divide the proof in three steps.
1. Local minimum.- Let
νR ≡ inf Ψλ,τ,R (u)
BΓ

where BΓ = {u ∈ Vc1,2 (DR ) : kukVc1,2 (DR ) < Γ ≡ kτ kLσ+1
n }.
m ( +)
Since Ψλ,τ,R (0) < 0, νR < 0. Furthermore µ ≤ νR < 0, by (2.14) and
remark 1 . Therefore inf ∂BΓ Ψλ,τ,R > νR .
Now we will prove that νR is achieved in BΓ . Using a modification in
the proof of proposition 5 and corollaries 6 and 7 in [3], we can obtain
a sequence (un )n in BΓ such that
Ψλ,τ,R (un ) → νR
′

Ψλ,τ,R (un ) → 0
since Ψλ,τ,R verifies Palais-Smale condition, there exists a subsequence
(unk )k such that unk → u1,R in Vc1,2 (DR ) and u1,R 6= 0 because 0 it is
not a critical point of Ψλ,τ,R .
2. Absolute minimum.- Let
uR ≡

inf

Vc1,2 (DR )

Ψλ,τ,R

Then uR < µ, by (2.17). Now using similar arguments to local minimum, but without any modification, we have that uR is achieved in
Vc1,2 (DR ) at a function u2,R .
3.Mountain pass.- Let
cR ≡ inf sup Ψλ,τ,R (u)
δ∈ΛR u∈δ

where ΛR is the set of paths
ΛR = {γ : γ ∈ C([0, 1], Vc1,2 (DR )), γ(0) = 0, γ(1) = ηR1 }
Since Ψλ,τ,R (0) < 0, by (2.15), (2.16) and (2.17), cR > 0.
Then by the mountain pass theorem, see [4], cR is achieved in Vc1,2 (DR )
at a function u3,R .
EJQTDE, 2005 No. 19, p. 8

On the other hand it is clear that u1,R , u2,R and u3,R are different,
indeed
Ψλ,τ,R (u2,R ) = uR < µ ≤ νR = Ψλ,τ,R (u1,R ) < 0 < cR = Ψλ,τ,R (u3,R )

Remark 6. When λ is small enough it is easy to prove uniqueness for
(1.5), so u1,R = u2,R , and the local minimum in BΓ od Ψλ,τ,R is the
absolute in Vc1,2 (DR ).
n
3. Problem in R+

Ψλ,τ does not verifies Palais-Samale condition, furthermore by lemma
2 and remark 1 Ψλ,τ is not coercive and not bounded from below.
However for λ small enough:
Proposition 7. Let f be as above, let b be given by (2.11) and suppose
m verifies (2.1). Then
1
i. For all λ < bM
, Ψλ,τ is coercive and bounded from below.
1
ii. For all λ < lM , (1.3) has at most one solution in Vc1,2 (Rn+ ).
λ < λ holds in both cases.
Proof.- i. By (2.11), (2.2) and Cauchy-Schwartz for the measure
mdxdy
Z
1
λb
2
Ψλ,τ (u) ≥
m(u + τ )2
kukV 1,2 ( n ) −
c
+
n
2
2
+
1
1
λb
kuk2V 1,2 ( n ) − (M 2 k∇ukL2 ( n+ ) + kτ kL2m ( n+ ) )2
c
+
2
2
1
1
≥
(1 − λbM)kuk2V 1,2 ( n ) − (λbM 2 kτ kL2m ( n+ ) )kukVc1,2 (
c
+
2 

λb
−
kτ k2L2m ( n+ )
2

≥

n)
+

−

so, i. is proved.

ii. Uniqueness is proved as in [1] using the inequality (2.2) and (f-4).
Indeed: if u1 and u2 are two solutions of (1.3) then
Z
Z
Z
2
2
2
(u1 −u2 ) m ≤ M
|∇(u1 −u2 )| +c(x)(u1 −u2 ) ≤ Mlλ
(u1 −u2 )2 m
n
+

n
+

n
+


Now we will prove a sufficient condition to approximate solutions of
(1.3) with solutions of (1.5) with R large enough.
EJQTDE, 2005 No. 19, p. 9

Lemma 8. Let f and τ be as above and λ ∈ R+ . Suppose (Rn )n
is a sequence R+ such that Rn → +∞ and (un )n is a sequence of
positive solutions of (1.5) with Rn instead of R, such that for all n,
un ∈ Vc1,2 (DRn ) and (un )n is bounded in Vc1,2 (Rn+ ), i.e. there exists
′
′
Γ > 0 such that for all n, kun kVc1,2 (DRn ) < Γ . Then, there exists a
subsequence (called again (un )n )) and a function u ∈ Vc1,2 (Rn+ ) such
that un → u weakly in Vc1,2 (Rn+ ) and u is a classical solution (1.3).
Proof.- By the Calder´on-Zygmund1 inequality for all n, un ∈ H01 (DRn )
′
′
2,p
H (DRn ) and fixed R > 0, for any Ω ⊂⊂ DR′
(3.1)

kun kH 2,p (Ω′ ) ≤ C(kun kLp (DR′ ) + kλm(y)f (un + τ )kLp (DR′ ) )
′

for all n such that Rn > R . The constant C depends on DR′ , n, p and
′
Ω . Since m is decreasing and strictly positive, and (un )n is bounded
in Vc1,2 (Rn+ ), by (2.2), (2.5), (3.1) and the hypothesis of f and m, we
obtain
′

1

1

1

′

kun kH 2,p (Ω′ ) ≤ C(m(R )− 2 Cs K(1 + M) 2 Γ + λ sup m sup f |DR′ | p )
for p such that
1