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4.4 Some Applications 161

) There exist p ∈(2, 2

∗

) and C>0 such that for all s, t ∈R,

|f(t)−f(s)|≤C|t −s|

(

|s|+|t|+1

)

p−2

) lim

t→0

f(t)

=0.

) There exist M>0 and μ>2 such that f(t)t ≥μF (t) when |t|≥M.

) There exists t

∈R with |t

|≥M such that F(t

)>0.

Remark 4.4.1 The meaning of assumption (f

) is twofold. First of all, it is a growth

condition: setting s =0 one sees that (f

) implies

|f(t)|≤K(|t|

p−1

+1) (4.13)

for all t and for a suitable positive constant K.

Next, (f

) implies that f is locally Lipschitz continuous on R. These two facts

allow one to prove that the functional G :H

() →R deﬁned by

G(u) =





F(u)dx

is of class C

1,1

and G



(u)v =





f(u)vdx for all v ∈H

().

Also, assumption (f

) means that the function t →

F(t)

|t|

is increasing for t ≥M

and decreasing for t ≤−M, as one can immediately see by differentiation.

Remark 4.4.2 Problem (4.1), which we used as a motivation for the construction of

the minimax methods, is a particular case of Problem (4.12).

If, as usual, we equip H

() with the norm

u





|∇u|

+q(x)u

dx, (4.14)

we can say that weak solutions of Problem (4.12) are critical points of the functional

J :H

() →R deﬁned as

J(u)=





|∇u|

+q(x)u

dx −





F(u)dx =

u

−G(u),

which is in C

()).

We will prove the following theorem, where assumption (h

) is the one intro-

duced at the beginning of Sect. 2.1 (recall also Remark 2.1.1).

Theorem 4.4.3 Assume that (h

) and (f

)–(f

) hold. Then Problem (4.12) has at

least one non trivial solution.

In view of Remark 4.1.16, the abstract minimax principle of Sect. 4.2 holds by

merely assuming J of class C

, and so do the Mountain Pass and Saddle Point

theorems. Therefore, in the previous result, assumption (f

) could be replaced by

the continuity of f and (4.13).

162 4 Introduction to Minimax Methods

Remark 4.4.4 Theorem 4.4.3 gives a stronger result than those we proved in

Sect. 2.5. Indeed, it is easy to see that hypotheses (f

)–(f

) imply (f

)–(f

) (with

>M); therefore Theorem 2.5.2 is a special case of Theorem 4.4.3, except for the

positivity of the solution. However, if t

>M this can be recovered by the method

of Example 1.7.10 after extending f to zero for negative arguments. Moreover, the

minimax method allows one to drop the hypothesis of C

nonlinearities and the

assumption (f

). Also the hypotheses of Theorem 2.5.11 imply (f

)–(f

We are going to show that the Mountain Pass Theorem is applicable. First, a tech-

nical result.

Lemma 4.4.5 The functional J is in C

1,1

()).

Proof We already know that J is C

; of course, since the differential of the

quadratic term is (globally) Lipschitz continuous, we only have to check the local

Lipschitz continuity for the differential of the functional G(u).

To this aim, ﬁx R>0 and take u

∈H

() such that u

, u

≤R. Then,

for all v ∈H

() we have, by (f

|(G



) −G



))v|=







(f (u

) −f(u

))v dx



≤









f(u

) −f(u

)



p−1



p−1







|v|



1/p

≤C







−u

p−1

(1 +|u

|+|u

p(p−2)

p−1



p−1

v.

Applying the Hölder inequality with exponents p −1 and

p−1

p−2

we obtain





−u

p−1

(

1 +|u

|+|u

)

p(p−2)

p−1

≤







−u



p−1







(1 +|u

|+|u



p−2

p−1

≤u

−u



p−1

K(R),

where K(R) depends on R but not on u

. This shows that

|(G



) −G



))v|≤u

−u

K(R)

p−1

v,

and therefore, maximizing over all v=1,

G



) −G



)≤C(R)u

−u

,

where C(R) is a constant depending on R.SowehaveprovedthatG



(and hence J )

is locally Lipschitz continuous. 

Now we turn to the compactness properties of J .

4.4 Some Applications 163

Lemma 4.4.6 The functional J satisﬁes (PS)

for every c ∈R.

Proof Take c ∈ R and assume that {u

} is a Palais–Smale sequence at level c,

namely such that

J(u

) →c and J



) →0(in(H

())



This implies of course that there is a constant C>0 such that

|J(u

)|≤C and |J



|≤Cu

 (4.15)

for every k. The following is a standard trick in superlinear problems to show that

Palais–Smale sequences are bounded, and is the main motivation for the use of

assumption (f

).By(4.15) we can write

C(1 +u

) ≥ J(u

) −





−



u









f(u

−F(u

)



dx.

Then from (f

) we obtain







f(u

−F(u

)



dx =



{|u

|≤M}



f(u

−F(u

)





{|u

|>M}



f(u

−F(u

)



≥



{|u

|≤M}



f(u

−F(u

)



dx ≥−C

and hence

C(1 +u

) ≥



−



u



−C

which of course implies that {u

} is bounded.

By usual arguments we can assume that, up to a subsequence, there exists

u ∈H

() such that

• u

uin H

(),

• u

→u in L

() for all q ∈[2, 2

∗

• u

(x) →u(x) for almost every x ∈.

We now show that the convergence of u

to u is strong, and this is a consequence

of the fact that p<2

∗

First of all, from the above convergence properties, we obtain by usual argu-

ments,





f(u

dx →





f(u)udx and





f(u

)u dx →





f(u)udx. (4.16)

As J



) → 0 and u

u,wealsohaveJ



)(u

− u) → 0 and obviously



(u)(u

−u) →0. Then, as k →∞,

164 4 Introduction to Minimax Methods

o(1) =





) −J



(u)



−u) =u

−u

−







f(u

) −f(u)



−u) dx

=u

−u

+o(1)

by (4.16). This shows that u

→u in H

(), and proves that J satisﬁes (PS)

for

every c ∈R. 

End of the proof of Theorem 4.4.3 We just have to check that the geometrical as-

sumptions of the Mountain Pass Theorem are satisﬁed.

First of all, as F(0) =0, we have J(0) = 0. We now prove that hypothesis 1 of

Theorem 4.3.1 holds.

Let λ

> 0 be the ﬁrst eigenvalue of the operator − + q on .Fixε>0so

small that

−

≥

By (f

) we have lim

t→0

F(t)

=0, and therefore there exists δ>0 such that

F(t)≤εt

if |t|≤δ.By(f

), as we remarked above, we deduce that there are a,b > 0 such

that |f(t)|≤a +b|t|

∗

−1

, and then also

|F(t)|≤˜a +

b|t|

∗

for some positive ˜a,

b and all t ∈R. Collecting these inequalities, it is not difﬁcult

to see that there is a constant C

> 0 such that

|F(t)|≤εt

|t|

∗

for all t. Hence

J(u)=

u

−





F(u)dx ≥

u

−ε





|u|

dx −C





|u|

∗

≥

u

−

u

−Cu

∗

≥

u

−Cu

∗

from which the assumption 1. of Theorem 4.3.1 easily follows, for example taking

as ρ any number so small that

−Cρ

∗

is positive.

To check that assumption 2 also holds, we notice that it is easy, using the same

arguments of Lemma 2.5.3, to show that there is a constant C>0 such that F(t)≥

C|t|

for all t ≥t

,ift

>M, and for all t ≤t

,ift

< −M. To proceed, we assume

that t

≥M (the other case being similar). Take ϕ ∈C

∞

() such that ϕ(x) ≥0for

all x ∈ and ϕ(x) ≥1 on some ball B =B

) ⊂. Then

tϕ(x) ≥t

, and hence F (tϕ(x)) ≥Ct

(x) for all t ≥t

and all x ∈B,

while |F(tϕ)|≤C if 0 ≤tϕ(x)≤t

. So we obtain

4.4 Some Applications 165





F(tϕ)dx =



{0≤tϕ≤t

}

F(tϕ)dx +



{tϕ>t

}

F(tϕ)dx

≥−C +C



{tϕ>t

}

≥−C +C



dx ≥−C +Ct

and then

J(tϕ)=

ϕ

−





F(tϕ)dx ≤t

+C −Ct

As μ>2, it is now obvious that for large t we have J(tϕ)<0, so also assumption 2

is proved. We then apply Theorem 4.3.1 to obtain a critical point at level c.Asc>0,

this critical point cannot be zero, and therefore the solution of Problem (4.12)is

nontrivial. 

We now give another application of the Mountain Pass Theorem. On a bounded

open set  ⊂R

, we look for a solution u of



−u +q(x)u =λu +|u|

p−2

u in ,

u =0on∂,

(4.17)

with λ ∈ R. As usual, q is bounded and nonnegative on , and we assume

2 <p<2

∗

. Weak solutions of (4.17) are critical points of the differentiable func-

tional J : H

() →R deﬁned by

J(u)=

u

−





dx −





|u|

dx,

where the norm of u is the one deﬁned in (4.14).

Remark 4.4.7 It is easy to see that if we deﬁne f(u)=λu+|u|

p−2

u, then f satisﬁes

), (f

), but not (f

). Therefore we cannot use directly Theorem 4.4.3 to

solve Problem (4.17). Actually one could reenter in the frame of Theorem 4.4.3 by

assuming that λ

(− +q(x)−λ) > 0 and use as an equivalent norm in H

() the

ﬁrst two terms of J . We prefer not do so because in the applications one encounters

frequently problems written as (4.17).

We are going to prove the following result; in its statement we denote by λ

(q)

the ﬁrst eigenvalue of the operator − +q(x).

Theorem 4.4.8 Assume that (h

) holds and that 2 <p<2

∗

. If λ<λ

(q), then

Problem (4.17) has at least one non trivial solution.

Proof The argument is essentially the same as that of the previous theorem. The

nonlinearity f(u)=λu +|u|

p−2

u in (4.17)isC

and satisﬁes (f

), so the proof of

Lemma 4.4.5 works in the same way and we obtain that J ∈ C

1,1

()).

166 4 Introduction to Minimax Methods

Let us check that the Palais–Smale condition holds at every level. Let {u

} be a

sequence such that

J(u

) →c and J



) →0.

As in the previous theorem, this implies that there is a constant C>0 such that

C(1 +u

) ≥ J(u

) −





−



u



−



−







dx −











−



u



−λ



−







≥



−



u



−

(q)



−



u





−



1 −

(q)



u



Since λ<λ

(q), this implies that {u

} is bounded and, as in the previous theorem,

we deduce that there is u ∈H

() such that

• u

uin H

(),

• u

→u in L

() for all q ∈[2, 2

∗

• u

(x) →u(x) for almost all x ∈.

Then we easily see that as k →∞,





p−2

−u) dx =o(1).

We conclude by computing

o(1) =





) −J



(u)



−u)

=u

−u

−λ





−u|

dx −







p−2

−|u|

p−2



−u) dx

=u

−u

+o(1),

which shows that u

→u strongly. Condition (PS)

holds for every c.

Finally we check that the geometric assumptions of Theorem 4.3.1 are satisﬁed.

It is obvious that J(0) =0. For every u ∈H

() we have

J(u)=

u

−





dx −





|u|

≥

u

−

2λ

(q)

u

−Cu



1 −

(q)



u

−Cu

4.4 Some Applications 167

As λ<λ

(q) and p>2, it is enough to choose any positive number ρ so small that

(1 −

(q)

)ρ

−Cρ

> 0. Hypothesis 1 of Theorem 4.3.1 is satisﬁed.

Finally for every u ∈H

()\{0} and t>0wehave

J(tu)=

u

−

λt





dx −





|u|



u

−λ







−





|u|

dx.

Therefore J(tu)→−∞as t →+∞, and we can certainly ﬁnd v such that v≥ρ

and J(v) < 0. The mountain Pass Theorem then yields the existence of a critical

point u for J at a positive level. This critical point is then a nontrivial solution of

Problem (4.17). 

Remark 4.4.9 If λ ≥λ

(q), the Mountain Pass Theorem cannot be applied, because

the “geometry around zero” is lost, in the sense that zero is no longer a local min-

imum for J . To see this, let ϕ

be the eigenfunction associated to λ

(q). For every

t =0wehave

J(tϕ

) =

ϕ



−

λt





dx −

|t|





|ϕ

≤

ϕ



−

(q)t





dx −

|t|





|ϕ

=−

|t|





|ϕ

dx < 0,

and J is strictly negative on all the line tϕ

, except t =0.

4.4.2 Asymptotically Linear Problems

We now present some applications of the Saddle Point Theorem. The most frequent

situation where this theorem is applicable is that of asymptotically linear problems,

namely those where the nonlinearity grows linearly at inﬁnity.

To illustrate the method, we consider an open and bounded subset  of R

, and

we look for a solution of



−u +q(x)u =λu +f(u)+h(x) in ,

u =0on∂.

(4.18)

Throughout this section, in addition to (h

) and (h

), we will use the following

assumption:

)f∈C

(R), f is bounded and |f



(t)|≤C(1 +|t|

∗

−2

168 4 Introduction to Minimax Methods

The requirement that f be C

can be weakened, and we use it only to avoid some

technicalities (for example we could take f bounded and satisfying (f

)). If we

deﬁne

f : ×R →R as

f(x,t)=λt +f(t)+h(x),

we see that

lim

t→±∞

f(t,x)

=λ

almost everywhere in . Thus

f grows linearly in t at inﬁnity, and this characterizes

(4.18) as an asymptotically linear problem.

Once H

() is equipped with the usual norm deﬁned by (4.14), the weak so-

lutions of (4.18) are the critical points of the functional J : H

() → R deﬁned

J(u)=

u

−





dx −





F(u)dx −





hu dx,

where of course F(t) =



f(s)ds.Dueto(f

), repeating the arguments of

Lemma 4.4.5, the functional J is readily seen to be in C

1,1

()), and we will

not discuss this point anymore.

Let 0 <λ

<λ

≤···be the usual sequence of the eigenvalues of the operator

− +q(x). The solvability of Problem (4.18) strongly depends on the location of

λ with respect to the eigenvalues λ

Deﬁnition 4.4.10 If λ =λ

for every k, Problem (4.18) is said to be nonresonant. If

λ = λ

for some k, the problem is said to be resonant or at resonance with the k-th

eigenvalue.

Resonant problems are much more difﬁcult to solve, and it may happen that they

do not have any solution at all, even in the linear case (compare with Theorem 1.7.8).

So, to illustrate the method we ﬁrst consider a nonresonant problem.

Theorem 4.4.11 Assume that (h

), (h

) and (f

) hold. If λ = λ

for every k, then

Problem (4.18) has at least one solution.

If λ<λ

, the functional J is coercive and can be shown to have a global min-

imum, with the methods of Sect. 2.1. The interesting case is thus λ>λ

, and we

deﬁne n ∈N to be the number such that

<λ<λ

n+1

. (4.19)

In this case we are going to show that the functional J satisﬁes the assumptions of

the Saddle Point Theorem. As usual we split the problem in two parts: compactness

and geometric assumptions.

In what follows, we always take for granted the assumptions of Theorem 4.4.11

and (4.19).

4.4 Some Applications 169

Lemma 4.4.12 The functional J satisﬁes (PS)

for every c ∈R.

Proof Let {u

} be a sequence such that

J(u

) →c and J



) →0.

We ﬁrst prove that {u

} is bounded.

Assume for contradiction that, up to a subsequence, u

→+∞. Since



) →0, for every v ∈H

() we can write J



)v =o(1)v, namely

|v) −λ





vdx−





f(u

)v dx −





hv dx =o(1)v. (4.20)

Deﬁne ψ

u



. Of course the sequence {ψ

} is bounded in H

(),sowecan

assume ψ

ψ∈H

(). Dividing by u

 the equation above we obtain

(ψ

|v) −λ





vdx−

u







f(u

)v dx −

u







hv dx =o(1)

v

u



As f is bounded, passing to the limit we get

(ψ|v) −λ





ψvdx =0. (4.21)

Let us now prove that ψ ≡ 0. Assume instead that ψ = 0. If this is the case, by



)ψ

=o(1) we ﬁrst obtain

|ψ

) −λ





dx −





f(u

)ψ

dx −





hψ

dx =o(1),

and then, dividing by u

,

ψ



−λ





dx =o(1).

We know that ψ

 0inH

(), which implies ψ

→ 0inL

(). Therefore we

obtain ψ

→0, a contradiction. So we have constructed a non zero function ψ

satisfying (4.21) for all v ∈H

(), namely a weak solution of



−ψ +q(x)ψ =λψ in ,

ψ =0on∂.

(4.22)

But this says that λ is an eigenvalue of − +q(x), which contradicts the nonreso-

nance assumption.

This proves that {u

}is a bounded sequence, so we can assume u

uin H

()

and u

→u in L

(). From this we obtain easily with the usual arguments

o(1) =





) −J



(u)



−u)

=u

−u

−λ





−u)

dx −





(f (u

) −f(u))(u

−u) dx

−





h(u

−u)

=u

−u

+o(1),

which shows that u

→u in H

; the functional J satisﬁes (PS)

for all c ∈R. 

170 4 Introduction to Minimax Methods

End of the proof of Theorem 4.4.11 We just have to check that J satisﬁes the geo-

metric assumption of the Saddle Point Theorem. Let {ϕ

}

be the sequence of the

eigenfunctions of − +q(x) (orthonormal in L

()), each associated to an eigen-

value λ

. Deﬁne

=span{ϕ

,...,ϕ

}, and V =E

⊥

=cl{span{ϕ

|k ≥ n +1}}.

These two subspaces are nothing else than the spaces X

and X

introduced in (2.8).

We recall from Lemma 2.2.5 that

u

≤λ

|u|

∀u ∈E

and u

≥λ

n+1

|u|

∀u ∈V. (4.23)

From this we deduce, for every u ∈E

J(u)=

u

−

|u|

−





F(u)dx −





hu dx

≤

u

−

2λ

u

+Cu=



1 −



u

+Cu.

Denoting by B

the ball in E

of center zero and radius R, the last inequality shows

that for every u ∈∂B

J(u)≤



1 −



+CR.

Since λ/λ

> 1, we have proved that

max

u∈∂B

J(u)→−∞

if R →∞.

We now prove that J is bounded from below on V . Still from (4.23), for every

u ∈V we have

J(u)=

u

−

|u|

−





F(u)dx −





hu dx

≥

u

−

2λ

n+1

u

−Cu



1 −

n+1



u

−Cu.

Since λ/λ

n+1

< 1, this implies inf

J>−∞. Hence for R large enough we can

guarantee that

max

u∈∂B

J(u)< inf

u∈V

J(u).

All the assumptions of the Saddle Point Theorem are satisﬁed and we apply it to

conclude. 