Calculus of Variations and PDE manuscript No.
(will be inserted by the editor)
A Local Saddle Point Theorem and an Application
to a nonlocal PDE
James Bisgard
Received: date / Accepted: date
Abstract We prove a local saddle point theorem that can be viewed as a
generalization of the saddle point theorem of Rabinowitz. A difficulty to overcome is that there isn’t any linking. We then apply the theorem to show the
existence of solutions of a nonlocal partial differential equations. MSC: 58E05,
47J30, 35A01
Keywords Critical Point Theory · Variational Methods · Nonlocal PDE
1 Introduction
Suppose E is a Hilbert space, T ∈ C 1 (E, R) has locally Lipschitz derivative,
and suppose further that
(SP1) E = H ⊕ C, where C is finite dimensional, and
(SP2) there exist
r1 > 0 and r2 > 0 such
that
(a) inf T > max T B (0)∩C and
r2
∂Br1 (0) ∩H
.
(b) inf T > max T Br1 (0)∩H
∂Br2 (0) ∩C
The main goal of the paper is to prove the following
Theorem 1.1. Suppose T ∈ C 1 (E, R) has locally Lipschitz derivative, and
satisfies (SP1) and (SP2). Then, there exists
a Palais-Smale sequence xn of
T such that T (xn ) → α, where α ≥ inf T Br (0)∩H .
1
James Bisgard
Central Washington University, 400 E. University Way, Ellensburg, WA 98926, USA
Tel.: +1-509-963-2823
Fax: +1-509-963-3226
E-mail: [email protected]
2
James Bisgard
Fig. 1 Geometric View of Assumptions
Notice that replacing r1 with ∞ makes (SP2)(a) vacuous, and turns (SP2)(b)
into the standard assumption in the saddle point theorem (see e.g. [3] or [6]),
and so Theorem 1.1 can be viewed as a generalization of the saddle
point
theorem. If r1 < ∞, a major difficulty is that the sets ∂Br2 (0) ∩ C and
Br1 (0) ∩ H don’t link in the sense of Silva (see [1]): there exist
functions
h : Br2 (0) ∩ C → E such that h(x) = x for all x ∈ ∂Br2 (0) ∩ C, and yet
h Br2 (0) ∩ C ∩ Br1 (0) ∩ H = ∅. In fact, these sets don’t link in the sense
of Schechter and Tintarev as in [5].
The conditions (SP2) are reminiscent of saying that T has a local linking
at 0 in the sense of Li and Willem in [2]:
(LL1) E = H ⊕ C
(LL2) there exist r > 0 such that
(a) T (u) ≥ 0 for all u ∈ H with kuk ≤ r and
(b) T (u) ≤ 0 for all u ∈ C with kuk ≤ r.
An immediate consequence of local linking in this sense is that 0 is a critical
point, which may not be the case under assumptions (SP1) and (SP2). On the
other hand, (LL1) is a weaker assumption, since it is not assumed that C is
finite dimensional. Note that (SP2) allows the possibility that in Br2 (0) ∩ C,
T may be larger than some points in Br1 (0) ∩ H - so long as those “large”
values are smaller than the smallest T is in ∂Br1 (0) ∩ H.
In Figure 1, H is the one-dimensional horizontal subspace and C is its
two dimensional complement. Shading represents the value of T , with black
representing small values and gray to white representing progressively larger
values. Finally, the thick white represents the largest values of T . That means
A Local Saddle Point Theorem and an Application to a nonlocal PDE
the diagram above represents the following:
>
inf
T
>
max
T
inf T ∂Br1 (0) ∩H
Br2 (0)∩C
Br1 (0)∩H
In local linking, the inequalities would be
≥ max T ≥ inf T inf T ∂Br1 (0) ∩H
Br2 (0)∩C
Br1 (0)∩H
> max T ≥ max T 3
.
∂Br2 (0) ∩C
.
∂Br2 (0) ∩C
2 Preliminaries
Throughout this section, we will assume that T ∈ C 1 (E, R) has locally Lip .
schitz derivative and satisfies (SP1) and (SP2). Let β := max T ∂Br2 (0) ∩C
By (SP2), there is an ε > 0 such that
η := inf T > ε + β.
Br1 (0)∩H
Let T β := {x ∈ E : T (x) ≤ β} and Tη := {x ∈ E : T (x) ≥ η} and define
g(x) :=
dist(x, T β )
.
dist(x, T β ) + dist(x, Tη )
Since T β ∩ Tη = ∅, the denominator in the definition of g is never zero. We
also have g(x) = 0 for all x ∈ T β and g(x) = 1 for all x ∈ Tη . Moreover, g is
locally Lipschitz. Let ∇T := D ◦ T 0 , where D : E 0 → E is the duality operator.
Suppose now that ϕt (x) satisfies:
d
g(ϕt (x))
ϕt (x) = −
∇T (ϕt (x))
dt
1 + k∇T (ϕt (x))k
ϕ0 (x) = x.
(1)
Because the right side of the first equation in (1) is bounded, the flow exists
for all time. Next, notice that ϕt (x) = x for all x ∈ T β . However, for any
x∈
/ T β , ϕt (x) = x if and only if T 0 (x) = 0.
Lemma 2.1. For any x ∈ E, t 7→ T (ϕt (x)) is a decreasing function.
Proof. Let x ∈ E be arbitrary. Then
d
d
T (ϕt (x)) = T 0 (ϕt (x)) ϕt (x)
dt
dt
g(ϕt (x))
∇T (ϕt (x))i
1 + k∇T (ϕt (x))k
g(ϕt (x))
=−
k∇T (ϕt (x))k2 ≤ 0,
1 + k∇T (ϕt (x))k
= h∇T (ϕt (x)), −
and so T decreases along the flow of ϕt (x).
(2)
4
James Bisgard
Lemma 2.2. Suppose now that ϕt (x) ∈ Tη for all t ≥ 0. Then there exists a
sequence tn → ∞ such that ϕtn (x) is a Palais-Smale sequence for T .
Proof. Since ϕt (x) ∈ Tη for all t ≥ 0, we know that g(ϕt (x)) = 1 for all t ≥ 0.
Therefore
Z s
d
T (ϕs (x)) − T (x) =
T (ϕt (x)) dt
dt
0
Z s
g(ϕt (x))
−
k∇T (ϕt (x))k2 dt
=
1 + k∇T (ϕt (x))k
0
Z s
k∇T (ϕt (x))k2
=−
dt.
0 1 + k∇T (ϕt (x))k
Since T (ϕs (x)) is bounded from below and decreasing in s, we know that
lims→∞ T (ϕs (x)) exists. In fact, we have
Z ∞
k∇T (ϕt (x))k2
lim T (ϕs (x)) = T (x) −
dt.
s→∞
1 + k∇T (ϕt (x))k
0
Z s
k∇T (ϕt (x))k2
dt converges, and so there must be a sequence
Therefore,
0 1 + k∇T (ϕt (x))k
k∇T (ϕtn (x))k2
tn with tn → ∞ such that
→ 0. But that implies that
1 + k∇T (ϕtn (x))k
T 0 (ϕtn (x)) → 0, and so ϕtn (x) is a Palais-Smale sequence for T .
Thus, to prove Theorem 1.1, it suffices to find an x such that ϕt (x) ∈ Tη
for all t ≥ 0. The next lemma shows that deformations of Br2 (0) ∩ C by the
negative gradient flow must still intersect Br1 (0) ∩ H.
Lemma 2.3. For every t ≥ 0, there exists an x ∈ Br2 (0) ∩ C such that
ϕt (x) ∈ Br1 (0) ∩ H.
Proof. The statement is clearly true for t = 0, since ϕ0 (x) = x for all x ∈ E.
Let Π : E → C be projection onto C, and notice that Π ◦ ϕt (x) = 0 if and
only if ϕt (x) ∈ H. By definition, (∂Br2 (0)) ∩ C ⊆ T β , and so by definition of g
and ϕt , Π ◦ ϕt (x) = Πx = x for all x ∈ (∂Br2 (0)) ∩ C and all t ≥ 0. Therefore,
by standard properties of degree (and using I to mean the identity function
on Br2 (0))
d(Π ◦ ϕt (·), Br2 (0) ∩ C, 0) = d(I, Br2 (0) ∩ C, 0) = 1
for each fixed t ≥ 0. Thus, for every t ≥ 0, there must be an x ∈ Br2 (0) ∩ C
such that Π ◦ ϕt (x) = 0, which in turn implies that for every t ≥ 0, there is
an x ∈ Br2 (0) ∩ C such that ϕt (x) ∈ H. We next show for at least one such x
that kϕt (x)k ≤ r1 .
Suppose then that there exists a τ > 0 such that kϕτ (x)k > r1 whenever
x ∈ Br2 (0) ∩ C and ϕτ (x) ∈ H. Let
Z := {[0, τ ] × (Br2 (0) ∩ C) : Π ◦ ϕt (x) = 0} .
A Local Saddle Point Theorem and an Application to a nonlocal PDE
5
Because (t, x) 7→ ϕt (x) is jointly continuous in t and x, the properties of
degree imply there is a connected component K of Z that intersects both
{0} × (Br2 (x) ∩ C) and {τ } × (Br2 (x) ∩ C). For any (t, x) ∈ [0, τ ] × Br2 (0),
let P (t, x) = kϕt (x)k. Since P is continuous and K is connected, P (K) is an
interval in R. Because ϕ0 (x) = x for all x ∈ Br2 (0), Z ∩({0} × (Br2 (x) ∩ C)) =
(0, 0), and so 0 ∈ P (K). Next, by assumption, if (τ, x) ∈ Z, kxk > r1 , and
so P (K) contains numbers larger than r1 . Because P (K) is an interval, there
must be a (t, x) ∈ K such that P (t, x) = r1 , i.e. kϕt (x)k = r1 . For such an x,
Lemma 2.1 implies
,
≥ T (x) ≥ T (ϕt (x)) ≥ inf T max T Br2 (0)∩C
(∂Br1 (0))∩H
which contradicts assumption (SP2a). Thus, for every t > 0, there is an x ∈
Br2 (0) ∩ C such that ϕt (x) ∈ Br1 (0) ∩ H.
3 Proof of Theorem 1.1
We can now prove Theorem 1.1:
Theorem. Suppose T ∈ C 1 (E, R) has locally Lipschitz derivative, and satisfies (SP1) and (SP2). Then, there exists
a Palais-Smale sequence xn of T
such that T (xn ) → α, where α ≥ inf T Br (0)∩H .
1
Proof. Since η := inf T B
r1 (0)∩H
, Lemma 2.2 implies it suffices to find an
x ∈ Tη such that ϕt (x) ∈ Tη for all t ≥ 0. For each n ∈ N, there is an
xn ∈ Br2 (0) ∩ C such that
T (ϕn (xn )) =
max
T (ϕn (x)).
x∈Br2 (0)∩C
By Lemma 2.3, for each n ∈ N, there is a yn ∈ Br2 (0) ∩ C such that ϕn (yn ) ∈
Br1 (0) ∩ H. Therefore, we know
= η.
(3)
T (ϕn (xn )) ≥ T (ϕn (yn )) ≥ inf T Br1 (0)∩H
Since Br2 (0) ∩ C is compact, along a subsequence xnj → x ∈ Br2 (0) ∩ C. To
finish the proof, we show that T (ϕt (x)) ≥ η for all t ≥ 0. Suppose this is false.
Then there is a τ ≥ 0 such that
T (ϕτ (x)) < η.
Taking N ∈ N such that N > τ , Lemma 2.1 implies
T (ϕN (x)) ≤ T (ϕτ (x)) < η.
6
James Bisgard
Since xnj → x, we then have for all sufficiently large j that
T (ϕN (xnj )) < η.
Since nj → ∞, we may assume nj > N . By (3) and Lemma 2.1, we have
η ≤ T (ϕnj (xnj )) ≤ T (ϕN (xnj )) < η,
which is a contradiction. Thus, T (ϕt (x)) ≥ η for all t ≥ 0, and so Lemma 2.2
implies that there is a subsequence tn → ∞ such that ϕtn (x) is a Palais-Smale
sequence for T . Moreover, we will have T (ϕtn (x)) ≥ η = inf T Br (0)∩H , which
1
finishes the proof.
4 An Application to a nonlocal PDE
Suppose now that Ω ⊂ Rn is a bounded open set with smooth boundary. We
will now show
Theorem 4.1. For any λ > 0 and f ∈ L2 (Ω), there is a µ0 > 0 such that for
all 0 < µ < µ0 , there is a weak solution of
2
−∆u = λu + f + 4µ kukL2 u for x ∈ Ω
u = 0 for x ∈ ∂Ω.
(4)
Equations such as (4) arise as stationary solutions of reaction-diffusion
equations, with nonlocal reaction terms (see [4] and its references). Solutions
of (4) correspond to critical points of Tµ : W01,2 (Ω) → R given by
Z
λ
1
|∇u|2 − u2 − f u dx − µkuk4L2
Tµ (u) =
2
Ω 2
λ
1
4
2
2
= k∇ukL2 − kukL2 − hf, uiL2 − µ kukL2 .
2
2
Notice that whenever f is not zero, solutions of (4) are nontrivial. We prove
(4.1) by showing that Tµ satisfies the Palais-Smale condition and the requirements of Theorem 1.1. For the norm on W01,2 , we use kukW 1,2 := k∇ukL2 . It
0
is straightforward to show that Tµ is (Frechet-)differentiable, and
Z
Z
Tµ0 (u)h =
∇u · ∇h − λuh − f h dx − 4µ kukL2
uh dx
Ω
= h∇u, ∇hiL2 − hλu + f, hiL2 −
Ω
2
4µ kukL2
hu, hiL2
for any h ∈ W01,2 . Thus, for all h ∈ W01,2 , we have
2
h∇Tµ (u), hiW 1,2 = h∇u, ∇hiL2 − hλu + f, hiL2 − 4µ kukL2 hu, hiL2 .
0
Lemma 4.2. For any µ > 0, any λ > 0 and any f ∈ L2 (Ω), Tµ satisfies the
Palais-Smale condition.
A Local Saddle Point Theorem and an Application to a nonlocal PDE
7
Proof. Notice that ∇Tµ = Id − K(u), where K : W01,2 (Ω) → W01,2 (Ω) is
uniquely determined by requiring that
2
hK(u), hiW 1,2 = hλu + f, hiL2 + 4µ kukL2 hu, hiL2
0
for all h ∈ W01,2 (Ω). For any h ∈ W01,2 (Ω), we then have
hK(u)− K(w), hiW01,2 ≤ λ |hu − w, hiL2 |
2
2
+ 4µ kukL2 hu, hiL2 − kwkL2 hw, hiL2 ≤ λ ku − wkL2 khkL2
2
2
+ 4µ kukL2 hu, hiL2 − kwkL2 hw, hiL2 ≤ λ ku − wkL2 khkL2
2
2
+ 4µ kukL2 hu, hiL2 − kwkL2 hu, hiL2 2
2
+ 4µ kwkL2 hu, hiL2 − kwkL2 hw, hiL2 ≤ λ ku − wkL2 khkL2
2
2 + 4µ kukL2 − kwkL2 hu, hiL2 2 + 4µ kwk 2 hu − w, hi 2 L
L
≤ λ ku − wkL2 khkL2
+ 4µ(kukL2 + kwkL2 ) · kukL2 − kwkL2 · kukL2 khkL2
2
+ 4µ kwkL2 ku − wkL2 khkL2
≤ λ ku − wkL2 khkL2
+ 4µ(kukL2 + kwkL2 ) ku − wkL2 kukL2 khkL2
2
+ 4µ kwkL2 ku − wkL2 khkL2
2
≤ λ + 4µ kukL2 + kwkL2 kukL2 + 4µ kwkL2 ku − wkL2 khkL2 .
The Poincaré inequality and the inequalities above imply
kK(u)−K(v)kW 1,2 =
sup hK(u) − K(v), hiW 1,2 0
khk
1,2 =1
W0
0
2
≤ C λ + 4µ kukL2 + kwkL2 ) kukL2 + 4µ kwkL2 ku − wkL2 ,
and so K(u) is locally Lipschitz. Moreover, whenever un → u in the L2 norm,
the inequalities above imply that K(un ) → K(u) in the W 1,2 norm. Thus, K is
a compact operator when Ω is bounded. By standard arguments, to show that
Tµ satisfies the Palais-Smale condition, it suffices to show that a Palais-Smale
sequence is bounded.
Suppose then that un is a sequence such that Tµ (un ) is bounded and for
which kTµ0 (un )k → 0. We must now show that un is bounded. Without loss of
8
James Bisgard
generality, we may assume there is a δ > 0 such that kun kW 1,2 = k∇un kL2 ≥ δ.
0
(Otherwise, there is a subsequence of un that converges to 0.) By definition,
we have
Tµ (un ) =
λ
1
2
2
4
k∇un kL2 − kun kL2 − hf, uiL2 − µ kun kL2 ,
2
2
and so we must have
2
2
4
k∇un kL2 = λ kun kL2 + 2 hf, un iL2 + 2µ kun kL2 + 2Tµ (un ).
(5)
Notice that (5) implies that if un is bounded in the L2 norm, then un will
be bounded in the W01,2 norm. If there is a subsequence of un that converges
to 0 in the L2 norm, we will be finished. Suppose then that there is no such
subseqence. There must be a δ1 > 0 such that kun kL2 ≥ δ1 > 0 for all n.
Dividing by k∇un kL2 , we see that
4
2
k∇un kL2 = λ
Similarly, we have
0
Tµ (un )
hf, un iL2
kun kL2
kun kL2
Tµ (un )
+2
+ 2µ
+2
.
k∇un kL2
k∇un kL2
k∇un kL2
k∇un kL2
un
k∇un kL2
(6)
2
= k∇un kL2 − λ
kun kL2
hf, un iL2
−
k∇un kL2
k∇un kL2
2
2
− 4µ kun kL2
kun kL2
,
k∇un kL2
and so in particular
4
2
k∇un kL2 = λ
hf, un iL2
kun kL2
kun kL2
+
+ 4µ
+ Tµ0 (un )
k∇un kL2
k∇un kL2
k∇un kL2
un
k∇un kL2
.
(7)
Subtracting (7) from (6), we see
4
kun kL2
hf, un iL2
Tµ (un )
− 2µ
+2
− Tµ0 (un )
0=
k∇un kL2
k∇un kL2
k∇un kL2
un
k∇un kL2
.
Rearranging then yields
4
kun kL2
hf, un iL2
Tµ (un )
2µ
=
+2
− Tµ0 (un )
k∇un kL2
k∇un kL2
k∇un kL2
un
k∇un kL2
.
The Cauchy-Schwarz inequality, the Poincaré inequality and our assumption
that k∇un kL2 ≥ δ > 0 for all n then imply
4
kun kL2
2
2µ
≤ C kf kL2 + Tµ (un ) − Tµ0 (un )
k∇un kL2
δ
un
k∇un kL2
.
A Local Saddle Point Theorem and an Application to a nonlocal PDE
9
Since Tµ (un ) is assumed to be bounded and Tµ0 (un ) → 0, this last inequality
implies that there is a constant M such that
4
kun kL2
≤M
k∇un kL2
for all n. In particular, we then have
1
4
1
kun kL2 ≤ M k∇un kL2 and hence kun kL2 ≤ M 4 k∇un kL4 2 .
Therefore,
1
λ
2
2
4
k∇un kL2 − kun kL2 − hf, un iL2 − µ kun kL2
2
2
1
1
1
λM 2
2
≥ k∇un kL2 −
k∇un kL2 2 − kf kL2 kun kL2 − M µ k∇un kL2
2
2
1
1
1
1
λM 2
1
2
k∇un kL2 2 − M 4 kf kL2 k∇un kL4 2 − M µ k∇un kL2 .
≥ k∇un kL2 −
2
2
Tµ (un ) =
Since Tµ (un ) is bounded, the inequality above implies that k∇un kL2 is bounded.
Thus, any Palais-Smale sequence for Tµ is bounded, and so Tµ satisfies the
Palais-Smale condition.
Lemma 4.3. For given λ > 0 and f ∈ L2 (Ω), there is a µ0 > 0 such that for
all 0 < µ < µ0 , Tµ satisfies (SP1) and (SP2).
Proof. Let λn and un be the eigenvalues and corresponding eigenfunctions of
−∆u, normed so that kun kW 1,2 = 1 and assume that λn is increasing. We then
2
2
0
2
2
have λn kun kL2 = k∇un kL2 = kun kW 1,2 = 1, and so kun kL2 =
0
W01,2 ,
1
λn .
Notice that
for any v ∈
if we multiply −∆un = λun by v and integrate by parts,
we see that hun , viW 1,2 = λn hun , viL2 . Taking v = uj , we see that the un are
0
P∞
orthogonal in L2 . Thus, for any v ∈ W01,2 , if v = j=1 aj uj , we must then
have
2
kvkL2
=
∞
X
j=1
a2j
2
kuj kL2
∞
X
a2j
.
=
λ
j=1 j
Suppose now λ > 0 is fixed, and let N := max{n ∈ N : λn ≤ λ}. In particular,
this tells us that λN ≤ λ < λN +1 . Let C := span{u1 , u2 , . . . , uN }, and let
H := span{uN +1 , uN +2 , . . . }. Notice that W01,2 (Ω) = H ⊕ C, and so (SP1) is
satisfied. We now show that there is a µ0 > 0 such that (SP2) is satisfied for
all 0 < µ ≤ µ0 .
The Poincaré inequality implies that kukL2 ≤ √1λ k∇ukL2 for all u ∈
1
PN
W01,2 (Ω). Moreover, if u ∈ C, then u = j=1 aj uj and (since λn is increasing)
2
kukL2 =
N
N
N
X
X
a2j
a2j
1 X 2
1
2
≥
=
aj =
kuj kW 1,2 .
0
λ
λ
λ
λ
j
N
N
N
j=1
j=1
j=1
10
James Bisgard
p
λN kukL2 .
0
PN
Suppose now that u ∈ C is arbitrary. Using u = j=1 aj uj and the fact
that 0 < λj ≤ λN ≤ λ for all j ≤ N , we will have
Thus, for u ∈ C, we have
p
λ1 kukL2 ≤ kukW 1,2 ≤
1
λ
2
2
4
k∇ukL2 − kukL2 − hf, uiL2 − µ kukL2
2
2
N
N
1 X 2 λ X a2j
4
=
aj −
− hf, uiL2 − µ kukL2
2 j=1
2 j=1 λj


N 1 X
λ
4
=
1−
a2j  − hf, uiL2 − µ kukL2
2 j=1
λj
Tµ (u) =
4
4
≤ − hf, uiL2 − µ kukL2 ≤ kf kL2 kukL2 − µ kukL2 .
√
Because kukW 1,2 ≤ λN kukL2 for any u ∈ C, we then have for any u ∈ C
0
4
2
that kukW 1,2 ≤ λ2N kukL2 and hence (by the Poincaré inequality)
0
kf k 2
µ
4
Tµ (u) ≤ √ L kukW 1,2 − 2 kukW 1,2
0
0
λN
λ1
Therefore, we must have
max Tµ Br2 (0)∩C
≤ max
t∈[0,r2 ]
for any u ∈ C.
kf kL2
µ
√
t − 2 t4
λN
λ1
and
(8)
max Tµ (∂Br2 (0))∩C
kf k 2
µ
≤ √ L r2 − 2 r24
λN
λ1
We nowPget an estimate from below on Tµ restricted to H. Suppose u ∈ H.
∞
Then u = j=N +1 aj uj , and so we must have
1
λ
2
2
4
k∇ukL2 − kukL2 − hf, uiL2 − µ kukL2
2
2

∞
1 X
λ
4
=
1−
a2j  − hf, uiL2 − µ kukL2 .
2
λj
Tµ (u) =
j=N +1
Since λ < λN +1 ≤ λj for all j ≥ N + 1, we have 1 − λλj ≥ 1 − λNλ+1 > 0 for all
j ≥ N + 1. Thus, for any u ∈ H, we will have


∞
1 X
λ
4
Tµ (u) ≥
1−
a2j  − hf, uiL2 − µ kukL2
2
λN +1
j=N +1


∞
X
1
λ
4
2

=
1−
aj  − hf, uiL2 − µ kukL2
2
λN +1
j=N +1
λ
1
2
4
=
1−
kukW 1,2 − hf, uiL2 − µ kukL2 .
0
2
λN +1
A Local Saddle Point Theorem and an Application to a nonlocal PDE
11
Next, the Cauchy-Schwarz and Poincaré inequalities imply that for every u ∈
kf k
W01,2 , | hf, uiL2 | ≤ kf kL2 kukL2 ≤ √λL2 kukW 1,2 , and so for any u ∈ H
1
Tµ (u) ≥
1
2
1−
λ
λN +1
0
kf k 2
4
2
kukW 1,2 − √ L kukW 1,2 − µ kukL2 .
0
0
λ1
4
4
Finally, since kukL2 ≤ √1λ kukW 1,2 , we will have µ kukL2 ≤ λµ2 kukW 1,2 , and
1
0
0
1
therefore for any u ∈ H
kf k 2
1
µ
λ
4
2
Tµ (u) ≥
1−
kukW 1,2 − √ L kukW 1,2 − 2 kukW 1,2
0
0
0
2
λN +1
λ1
λ1
Therefore, we must have
inf Tµ (∂Br1 (0))∩H
≥
1
2
1−
λ
λN +1
kf k 2
µ
r12 − √ L r1 − 2 r14 and
λ1
λ1
(9)
inf Tµ ≥ min
Br1 (0)∩H
t∈[0,r1 ]
1
2
1−
λ
λN +1
kf k 2
µ
t2 − √ L t − 2 t4 .
λ1
λ1
Therefore, in order to show that (SP2) is satisfied, (8) and (9) imply that it
is sufficient to show there is a µ0 > 0 such that for all µ ∈ (0, µ0 ), there exist
r1 > 0 and r2 > 0 such that
kf k 2
kf kL2
λ
µ
µ
1
√
1−
t − 2 t4 and
r12 − √ L r1 − 2 r14 > max
(a)
2
λ1
λN
t∈[0,r2 ]
λ1
λN+1
λ1
kf kL2
kf kL2
µ 4
µ
1
λ
2
(b) min
t − 2t > √
r2 − 2 r24 ,
1−
t − √
λN +1
λ1
λN
t∈[0,r1 ] 2
λ1
λ1
which follows from Propositon 4.4.
Proposition 4.4. Suppose that b, c and d are fixed strictly positive constants,
and let h(t) = bt2 − at − cµt4 and k(t) = at − dµt4 . Given any a > 0, there is
a µ0 > 0 such that for any µ ∈ (0, µ0 ), there exist r1 > 0 and r2 > 0 such that
(i) h(r1 ) > max k(t) and
t∈[0,r2 ]
(ii)
min h(t) > k(r2 ).
t∈[0,r1 ]
b3/2
1
Proof. Let a > 0 be given. Pick µ0 sufficiently small that both √ · √ > a
µ0
2c
r
2
4/3
b √
1
1
b
a
2/3
>a·
· µ0 + 1/3
− 4/3 · µ0 . Suppose now that 0 <
and
4c
2c
d
41/3
4
b3/2
1
µ < µ0 . Then, we will have √ · √ > a and (since the right side is increasing
µ
2c
r
2
4/3
b
a
b √
1
1
in µ)
>a·
· µ + 1/3
−
· µ2/3 .
4c
2c
d
41/3
44/3
12
James Bisgard
b3/2
1
Notice that the inequality √ · √ > a then implies
µ
2c
that r1 b > a, i.e. r1 > ab . We now show that
Let r1 :=
q
b
2cµ .
b
(10)
whenever t ∈ [0, r1 ], then −at + t2 ≤ h(t).
2
q
b
b
Equality clears holds when t = 0. If 0 < t ≤ r1 = 2cµ
, then 0 < t2 ≤ 2cµ
,
and so 0 < cµt2 ≤ 2b . Thus, multiplying through by t2 , we have 0 < cµt4 ≤ 2b t2 .
Thus, if 0 < t ≤ r1 , we will have − 2b t2 ≤ −cµt4 . Adding −at + bt2 , then yields
−at + 2b t2 ≤ −at + bt2 − cµt4 , as desired. Next, a straightforward calculation
shows that −at + 2b t2 is minimized when t = ab . Therefore, since ab ∈ [0, r1 ],
(10) implies
−
a2
b
= min −at + t2 ≤ min h(t).
2b t∈[0,r1 ]
2
t∈[0,r1 ]
(11)
Next, since k(t) = at − dµt4 → −∞ as t → ∞, we may choose r2 > 0 such
1/3
2
a
that both − a2b > k(r2 ) = ar2 − dµr24 and r2 > 4dµ
are true. (11) then
2
implies min h(t) ≥ − a2b > k(r2 ), and so (ii) is satisfied. It remains to show
t∈[0,r1 ]
that (i) is satisfied for this r2 .
A straightforward calculation shows that the maximum of k(t) on [0, ∞)
1/3
1/3
a
a
. Since r2 > 4dµ
, we then know that
occurs when t = 4dµ
!
1/3
a
1
a4/3
1
max k(t) = k
=
1/3 41/3 − 44/3 .
4dµ
t∈[0,r2 ]
dµ
Therefore, to finish the proof of Proposition 4.4, it suffices to show that h(r1 ) >
1
1
a4/3
− 44/3
. We have
(dµ)1/3 41/3
s
s
s
!
b2
b2
b2
b
b
b
= −a ·
+
− cµ 2 2 = −a ·
+
.
h(r1 ) = h
2cµ
2cµ 2cµ
4c µ
2cµ 4cµ
By assumption on µ, we have
r
b2
b √
a4/3
1
1
>a·
· µ + 1/3
−
· µ2/3 ,
4c
2c
d
41/3
44/3
q
b
and therefore, upon dividing by µ > 0 and subtracting a · 2c
· √1µ from both
sides, we will have
r
b
1
b2
a4/3
1
1
1
−a ·
·√ +
> 1/3
− 4/3 · 1/3 ,
1/3
2c
µ 4cµ
d
4
4
µ
i.e. h(r1 ) > max k(t) as desired.
t∈[0,r2 ]
A Local Saddle Point Theorem and an Application to a nonlocal PDE
13
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