Theorem 39.2.1Let H be a Hilbert space and let I : H → ℝ be a C^{1}functional havingI^{′}Lipschitz continuous and such that I satisfies the Palais Smale condition. SupposeI
(0)
= 0 and I
(u)
≥ a > 0 for all
∥u∥
= r. Suppose also that there exists v,
∥v∥
> r suchthat I
(v)
≤ 0. Then define
Γ ≡ {g ∈ C ([0,1];H) : g(0) = 0,g(1) = v}
Let
c ≡ ingf∈Γ m0a≤xt≤1 I(g(t))
Then c is a critical value of I meaning that there exists u such that I
(u)
= c andI^{′}
(u)
= 0. In particular, there is u≠0 such that I^{′}
(u)
= 0.
This nice example is in Evans [?]. Let the Hilbert space be H_{0}^{1}
(U )
where U is a
bounded open set. To avoid cases, assume U is in ℝ^{3} or higher. The main results will
work in general but it would involve cases. Consider the functional
1 ∫
-∥u∥2H10 − F (u)dx ≡ I1(u)− I2(u)
2 U
where F^{′}
(u)
= f
(u)
,f
(0)
= 0. Here it is assumed that
p ′ ( p− 1) n+ 2
|f (u )| ≤ C (1+ |u|),|f (u)| ≤ C 1 + |u| , 1 < p < n−-2 (39.2.8)
(39.2.8)
Also suppose that
0 ≤ F (u) ≤ γf (u) u where 0 < γ < 1∕2 (39.2.9)
(39.2.9)
and finally that
α |u|p+1 ≤ F (u) ≤ A |u|p+1, α,A > 0 (39.2.10)
(39.2.10)
Let R : H_{0}^{1}
(U )
→ H^{−1}
(U )
be the Riesz map.
Showing Functional is C^{1,1}
Then it is not hard to verify that
(I1(u),v)
=
(u,v)
and so it is clearly the case
that I^{′}
(u)
exists and is a continuous function of u. In addition to this, it is
Lipschitz.
Next consider I_{2}.
∫
I2(u+ v)− I2(u) = F (u + v)− F (u) dx
U
∫
1 ′ 2
= U f (u )v+ 2f (ˆu)v dx, ˆu ∈ [u,u+ v]
Now H_{0}^{1}
(U )
embeds continuously into L^{2n∕(n−2)}
(U)
. Because of the estimate for f
(u)
,
we can regard f
(u)
as being in H^{−1}
(U)
as follows.
||∫ || ( ∫ 2n∕(n+2) ) (n+2)∕2n (∫ 2n∕(n−2))(n−2)∕2n
|| f (u)vdx|| ≤ |f (u)| dx |v|
U U U
(∫ ( ) )(n+2)∕2n
≤ C 1+ |u|2n∕(n− 2) dx ∥v∥H1
U 0
where C will be adusted as needed here and elsewhere. Thus, writing in terms of the
inner product on H_{0}^{1},
′ ( −1 )
(I2 (u),v)H10 = R f (u),v H10
This is so if the
1
2
f^{′}
(ˆu)
v^{2} term is as it should be. We need to verify that
∫ |1 ′ 2|
-U-|2f-(ˆu)v-|dx-→ 0
∥v∥H10
However, we can use the estimate and write that this is no larger than
∫ ( ) | |
U C 1+ |v|p−1 + |u|p−1 |v2|dx
---(∫----2n∕(n−2)--)(n−-2)∕2n--- (39.2.11)
U |v| dx
(39.2.11)
Then consider the term involving
|u |
.
∫ p− 1 2 (∫ p+1)pp−+11 ( ∫ p+1)2∕(p+1)
|u| |v| dx ≤ |u| |v|
U U U
Now p + 1 ≤ 2
-n-
n− 2
and so the first factor is finite. As to the second, it equals
Hence, for r large enough, the right side becomes negative because p + 1 > 2. Therefore,
r → I
(ru)
is positive for small r and is eventually negative as r gets larger. hence there is
some value of r where this equals 0. Then v = ru. This verifies the conditions for the
mountain pass theorem.
conclusions
It follows from the mountain pass theorem that there is some u≠0 such that
I^{′}
(u)
= 0. From the above computations,
u− R −1f (u) = 0
Now R = −Δ the Laplacian. In terms of weak derivatives,