Schechter Minimax Systems and Critical Point Theory (Springer, 2009)
.pdfChapter 8
Semilinear Problems
8.1 Introduction
In the present chapter we study several nonlinear boundary value problems that arise frequently in applications and illustrate the techniques described in the book.
8.2 Bounded domains
We assume that is a bounded domain in Rn with boundary ∂ sufficiently regular so that the Sobolev inequalities hold and the embedding of H m,2( ) in L2( ) is compact (cf., e.g., [1]). Let A be a self-adjoint operator on L2( ). We assume that A ≥ λ0 > 0
and that
C0∞( ) D := D( A1/2) H m,2( )
for some m > 0, where C0∞( ) denotes the set of test functions in (i.e., infinitely differentiable functions with compact supports in ) and H m,2( ) denotes the Sobolev space. If m is an integer, the norm in H m,2( ) is given by
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1/2 |
(8.1) |
u m,2 := |
Dμu 2 |
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| μ|≤m |
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Here Dμ represents the generic derivative of order |μ| and the norm on the right-hand side of (8.1) is that of L2( ). We shall not assume that m is an integer. Let q be any number satisfying
(8.2) |
2 |
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≤ 2n/(n − 2m), |
2m < n |
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≤ |
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< ∞, |
n ≤ 2m |
and let f (x, t) be a Carath´eodory function on × R. This means that f (x, t) is continuous in t for a.e. x and measurable in x for every t R. We make the following assumption.
M. Schechter, Minimax Systems and Critical Point Theory, DOI 10.1007/978-0-8176-4902-9_8, © Birkhäuser Boston, a part of Springer Science+Business Media, LLC 2009
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8. Semilinear Problems |
(A) |
The function f (x, t) satisfies |
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| f (x, t)| ≤ V (x)q (|t|q−1 + W (x)) |
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and |
f (x, t)/ V (x)q = o(|t|q−1) as |t| → ∞, |
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where V (x) > 0 is a function in Lq ( ) such that |
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(8.3) |
V u q ≤ C u D , u D, |
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and W is a function in L∞( ). Here |
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1/q |
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u q := |
|u(x)|q dx |
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and |
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(8.4) |
u D := A1/2u . |
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With the norm (8.4), D becomes a Hilbert space. Define |
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F(x, t) := 0 |
f (x, s) ds |
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and |
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(8.5) |
G(u) := u 2D − 2 |
F(x, u) dx. |
It is readily shown that G is a continuously differentiable functional on the whole of D (cf., e.g., [122]). Since the embedding of D in L2( ) is compact, the spectrum of A consists of isolated eigenvalues of finite multiplicity:
0 < λ0 < λ1 < · · · < λ < · · · .
Let λ , > 0, be one of these eigenvalues. We assume that the eigenfunctions of λ are in L∞( ) and that the following hold:
(8.6) |
2F(x, t) ≤ λ t2 + W1(x), |
x , t R |
for some W1(x) L1(R), |
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(8.7) |
λ t2 ≤ 2F(x, t), |
|t| ≤ δ, |
for some δ > 0, |
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(8.8) |
νt2 ≤ 2F(x, t), x , t R, |
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for some ν > λ −1, |
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(8.9) |
H (x, t) := 2F(x, t) − t f (x, t) ≤ C(|t| + 1), |
8.2. Bounded domains |
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and |
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(8.10) |
σ (x) := lim sup H (x, t)/|t| < 0 a.e. |
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|t|→∞ |
We wish to obtain a solution of |
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(8.11) |
Au = f (x, u), u D. |
By a solution of (8.11) we shall mean a function u D such that
(8.12) |
(u, v)D = ( f (·, u), v), v D. |
If f (x, u) is in L2( ), then a solution of (8.12) is in D( A) and solves (8.11) in the classical sense. Otherwise we call it a weak (or semi-strong) solution. We have
Theorem 8.1. Under the above hypotheses,
(8.13) |
Au = f (x, u), u D |
has at least one nontrivial solution.
Proof. Let N denote the subspace of L2 responding to the eigenvalues λ0, . . . , λ This time we take
( ) spanned by the eigenfunctions of A cor- , and let M = N ∩ D. Thus, D = M N.
G(u) = 2 F(x, u) dx − u 2D ,
the negative of (8.5). We are therefore looking for solutions of G (u) = 0. Let N be the set of those functions in N that are orthogonal to E(λ ). It is spanned by those eigenfunctions corresponding to λ0, . . . , λ −1. Let v0 be an eigenfunction of λ with norm 1. Let M1 = M E(λ ) {v0}. We can write
E = M1 {v0} N .
Consider the mapping
F(v + w + sv0) = w + [s + ρ − ρϕ( v 2/ρ2)]v0, v N, w M1, s R, where ϕ satisfies the hypotheses of Proposition 7.3 and ρ > 0 is to be chosen. We take
A = M1 {v0}, B = F−1(ρv0).
By Proposition 7.3, A, B form a sandwich pair.
For v N, we write v = v + y, where v N and y E(λ ). Since E(λ ) is finite-dimensional and contained in L∞( ), there is a ρ > 0 such that
(8.14) |
y D ≤ ρ implies y ∞ ≤ δ/2, |
where δ is given by (8.7). Thus, if |
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(8.15) |
v D ≤ ρ and |v(x)| ≥ δ, |
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8. Semilinear Problems |
then |
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δ ≤ |v(x)| ≤ |v (x)| + |y(x)| ≤ |v (x)| + δ/2. |
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Hence, |
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|v(x)| ≤ 2|v (x)| |
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holds for all x |
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satisfying (8.15). Thus by (8.7) |
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G(v) ≥ λ |
v2dx − 2 |
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{|V v|q + |V q v|W }dx − v 2D |
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|v |<δ |
|v |>δ |
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≥ λ v 2 − λ |
v2dx − v 2D − C |
{|V v |q + δ1−q |V v |q }dx |
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|v |>δ |
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2|v |>δ |
≥ λ v 2 − v 2D − C |
{|V v |q + δ1−q |V v |q + δ2−q |v |q }dx |
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2|v |>δ |
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≥λ v 2 − v 2D − C v qD
≥ |
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C v |
q−2 |
v |
2 . |
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−1 |
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To see this, note that when (8.15) holds, we have |V v| ≤ 2|V v | and
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V q v |
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v |
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|v|q−1 |
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δ1−q V q |
2v |
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q . |
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δq 1 ≤ |
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Moreover,
v q ≤ C v m,2 ≤ C v D
by the Sobolev inequality and the embedding of D in H m,2( ). From this we see that there are positive constants , ρ such that
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G(v) ≥ v 2D , v D ≤ ρ, v N . |
Moreover, this shows that for each positive ρ1 ≤ ρ, |
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(8.16) |
G(v) ≥ 1, v D = ρ1, v N , |
for some positive 1 unless there is a solution of
Ay = λ y = f (x, y), y E(λ )\{0}
(cf. [122]). Since such a solution would solve (8.13), we may assume that (8.16) holds.
Since
v 2D ≤ λ v 2, v N,
and
λ +1 w 2 ≤ w 2D , w M,
we have, by (8.6),
G(w) ≤ λ w 2 + B1 − w 2D ≤ B1, w A,
8.2. Bounded domains |
67 |
where B1 = W1(x) dx. Moreover, (8.8) implies
G(v ) ≥ (ν − λ −1) v 2, v N .
Hence, there is an ε > 0 such that
G(v) ≥ ε, v B.
In view of these inequalities, we can now apply Proposition 7.3 to conclude that there is a sequence {uk } D such that
(8.17) |
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G(uk ) → c, |
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ε ≤ c ≤ B1, G (uk ) → 0. |
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Let ρk = uk D . If ρk → ∞, then |
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(8.18) |
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G(uk ) = 2 |
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F(x, uk ) dx − ρk2 → c |
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and |
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(G (uk ), uk )/2 = f (x, uk )uk dx − ρk2 = o(ρk ). |
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Hence, |
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˜k |
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H (x, uk) dx = o(ρk ). |
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= uk /ρk . Then u˜k D = |
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Let u |
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1. Thus, there is a renamed subsequence such that |
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u˜ k → u˜ weakly in D, strongly in L ( ), and a.e. in . By (8.9) and (8.10), |
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lim sup |
H (x, uk) dx/ρk ≤ |
lim sup[H (x, uk)/|uk |]|u˜k | dx |
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= |
σ (x)|u˜| dx. |
Since σ (x) < 0 a.e. in , the last two statements imply that u˜ ≡ 0. However, we see from (8.18) that
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F(x, uk ) dx/ρk2 → 1, |
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while (8.6) implies |
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lim sup 2 |
F(x, uk ) dx/ρk2 ≤ λ |
u˜ 2 dx, |
showing that u˜ ≡ 0. This contradiction tells us that the ρk must be bounded. We can now apply Theorem 3.4.1 of [122] to conclude that there is a u D satisfying
(8.19) |
G(u) = c, G (u) = 0. |
Since c ≥ ε > 0, we see that u = 0, and the proof is complete.
68 |
8. Semilinear Problems |
The proof of Theorem 8.1 implies
Corollary 8.2. If λ is a simple eigenvalue, then hypothesis (8.6) in Theorem 8.1 can be weakened to
(8.20) 2F(x, t) ≤ λ +1t2 + W1(x), x , t R,
for some W1(x) L1(R).
Remark 8.3. The proof of Theorem 8.1 is much simpler if = 0. In this case N = {0} and (8.14) immediately implies (8.16). The rest of the proof is unchanged.
We now show that we can essentially reverse the inequalities (8.6)–(8.10) and obtain the same results. In fact, we have
Theorem 8.4. Equation (8.13) has at least one nontrivial solution if we assume > 0 and
(8.21) |
λ t2 ≤ 2F(x, t) + W1(x), |
x , t R, |
for some W1(x) L1(R), |
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(8.22) |
2F(x, t) ≤ λ t2, |
|t| ≤ δ, |
for some δ > 0, |
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(8.23) |
2F(x, t) ≤ νt2, x , t R, |
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for some ν < λ +1, |
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(8.24) |
H (x, t) ≥ −C(|t| + 1), |
x , t R, |
and |
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(8.25) |
limt inf H (x, t)/|t| > 0 a.e. |
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Proof. In this case we take G to be the functional (8.5). We take A = N, N = N {v0}, and consider the mapping
F(v + w + sv0) = v + [s + δ − δϕ( w 2/δ2)]v0, v N , w M, s R, where ϕ satisfies the hypotheses of Proposition 7.3. By (8.21), we have
G(v) ≤ B1, v N.
For w M1, we write w = w + y, where w M and y E(λ ). Then (8.22) implies
G(w) ≥ ε1, w D = ρ, w M1,
unless (8.13) has a nontrivial solution. Hence, by the argument given in the proof of Theorem 8.1 we have a sequence satisfying (8.17). If u˜ k and u˜ are as in the proof of
8.3. Some useful quantities |
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Theorem 8.1, then (8.24), (8.25) imply that u˜ ≡ 0 as in that proof. However, (8.18) implies
2 |
F(x, uk ) dx/ρk2 → 1, |
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while (8.23) implies |
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lim sup 2 |
F(x, uk ) dx/ρk2 ≤ ν |
u˜2dx, |
showing that u˜ ≡ 0. This contradiction proves the theorem as in the case of Theorem 8.1.
The proof of Theorem 8.4 implies
Corollary 8.5. If λ is a simple eigenvalue, then hypothesis (8.21) in Theorem 8.4 can be weakened to
(8.26) λ −1t2 ≤ 2F(x, t) + W1(x), x , t R, for some W1(x) L1(R).
8.3 Some useful quantities
We now show how we can improve the results of the last section. For each fixed k, let
Nk denote the subspace of D := D( A1/2) spanned by the eigenfunctions corresponding to λ0, . . . , λk , and let Mk = Nk ∩ D. Then D = Mk Nk . We define
(8.27) αk := max{( Av, v) : v Nk , v ≥ 0, v = 1},
where v denotes the L2( )-norm of v. We assume that A has an eigenfunction ϕ0 of constant sign a.e. on corresponding to the eigenvalue λ0.
Next we define for a R |
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(8.28) |
γk (a) := max{( Av, v) − a v− 2 : v Nk , v+ = 1} |
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and |
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(8.29) |
k (a) := inf{( Aw, w) − a w− 2 : w Mk , w+ = 1}, |
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where u± = max{±u, 0}. |
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We take any integer ≥ 0 and let N denote the subspace of L |
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( ) spanned |
by the eigenspaces of A corresponding to the eigenvalues λ0, λ1, . . . , λ . We take M = N ∩ D, where D = D( A1/2). We assume that F(x, t) satisfies
(8.30) |
a1(t−)2 + γ (a1)(t+)2 |
− W1(x) ≤ 2F(x, t) |
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≤ a2(t−)2 + ν(t+)2, x , t R, |
for numbers a1, a2 satisfying α |
< a1 ≤ a2, where W1 is a function in L1( ) and |
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ν < (a2). We also assume that |
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(8.31) |
2F(x, t) ≤ λ +1t2, |t| ≤ δ |
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8. Semilinear Problems |
for some δ > 0, |
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(8.32) |
| f (x, t)| ≤ C|t| + W (x), W L2( ), |
(8.33) |
f (x, t)/t → α±(x) a.e. as t → ±∞, |
and the only solution of |
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(8.34) |
Au = α+(x)u+ − α−(x)u− |
is u ≡ 0. We have
Theorem 8.6. Under the above hypotheses, (8.13) has a nontrivial solution. Proof. By (8.28),
(8.35) |
v 2D ≤ a1 v− 2 + γ (a1) v+ 2, |
v N, |
and by (8.29), we have |
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(8.36) |
a2 w− 2 + (a2) w+ 2 ≤ w 2D , |
w M. |
Hence, |
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(8.37) |
G(v) ≤ B1, v N. |
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Since ν < (a2), we see by continuity that there is an ε > 0 such that
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ν < (1 − ε) |
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Hence, |
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(8.38) |
G(w) ≥ ε w 2D + (1 − ε) |
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ε |
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≥ ε w 2D , w M,
by (8.30).
As in the proof of Theorem 8.1, we note that the following alternative holds: Either
(a) |
there is an infinite number of eigenfunctions y E(λ )\{0} such that |
(8.39) |
Ay = f (x, y) = λ y, |
or |
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(b) |
for each ρ > 0 sufficiently small, there is an ε > 0 such that |
(8.40) |
G(w) ≥ ε, w D = ρ, w M . |
8.4. Unbounded domains |
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Since option (a) solves our problem, we may assume that option (b) holds. |
Let |
v0 E(λ ), and let F be the mapping (7.10). Take A = N, B = F−1(δv0). By (8.37),
(8.38), and (8.40) we see that (7.3) holds with b0 |
> 0 and a0 |
= B1. By Proposition |
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7.3, we can conclude that there is a sequence {uk } D such that |
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(8.41) |
G(uk ) → c, b0 ≤ c ≤ B1, |
G (uk ) → 0. |
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Thus, |
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(8.42) |
G(uk ) = uk 2D − 2 |
F(x, uk )dx → c |
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and |
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(8.43) |
(G (uk ), v) = 2(uk , v)D − 2( f (uk ), v) → 0, |
v D. |
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If |
ρk = uk D |
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u˜ k = uk /ρk . Then u˜k D = |
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1. Thus, there is a renamed |
subsequence such that u˜k → u˜ weakly in D, strongly in L ( ), and a.e. in . Hence,
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G(uk )/ρk2 = u˜k 2D − 2 |
F(x, uk )dx/ρk2 → 0. |
Since |
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(8.44) |
|F(x, uk )|/ρk2 ≤ C(|u˜k (x)|2 + W3(x)/ρk2), W3 L1( ), |
by (8.30), and the right-hand side of (8.44) converges to C|u˜(x)|2 in L1( ) and
(8.45) |
2F(x, uk (x))/ρ2 |
→ |
α (x)(u+)2 |
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α |
(x)(u−)2 |
a.e., |
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we see that the convergence in (8.45) is not only pointwise a.e., but also in L1( ). Since u˜k D = 1, (8.44) implies
(8.46) |
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{ |
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α (u−)2 |
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dx |
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1. |
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Also, |
(u |
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(G |
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2(u |
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D − |
2( f (u |
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for each v D. This implies |
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(α u+ |
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α u−, v), v |
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˜ ˜ |
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Consequently, u˜ is a solution of (8.34). |
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By hypothesis, u˜ ≡ 0. |
But this contradicts |
(8.46). Hence, ρk ≤ C. The theorem now follows from Theorem 3.4.1 of [122].
8.4 Unbounded domains
Now we allow the domain Rn to be unbounded. Assume hypothesis (A), and assume that
(8.47) H (x, t) = 2F(x, t) − t f (x, t) ≥ −W3(x) L1( ), x , t R,
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8. Semilinear Problems |
and |
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(8.48) |
H (x, t) → ∞ a.e. as |t| → ∞. |
We have |
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Theorem 8.7. Assume that the spectrum of A consists of isolated eigenvalues of finite multiplicity
(8.49) |
0 < λ0 < λ1 < · · · < λk < · · · , |
and let be a nonnegative integer. Take N to be the subspace of D spanned by the eigenspaces of A corresponding to the eigenvalues λ0, λ1, . . . , λ . We take M = N ∩ D. Assume that there are functions W1, W2 L1( ) and numbers a1, a2 such that α < a1 ≤ a2 and
(8.50) |
a1(t−)2 + γ (a1)(t+)2 − W1(x) ≤ 2F(x, t) |
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≤ a2(t−)2 + (a2)(t+)2 + W2(x), |
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x , t R, |
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and that (8.47) and (8.48) hold. Then (8.13) has at least one solution. |
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Proof. First, we note that |
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(8.51) |
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G ≤ B1, |
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≥ −B2, |
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B j = |
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sup |
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inf G |
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To see this, note that by (8.28) we have |
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(8.52) |
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v 2D ≤ a1 v− 2 + γ (a1) v+ 2, |
v N. |
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By (8.29) we have |
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(8.53) |
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a2 w− 2 + (a2) w+ 2 ≤ w 2D , |
w M. |
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Hence, |
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G(v) ≤ B1, |
v N, |
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and |
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G(w) ≥ −B2, |
w M, |
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→ ∞, there is a |
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by (8.50). By Theorem 3.19, we conclude that for any sequence Rk |
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sequence {uk } D such that |
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(8.54) |
G(uk ) |
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c, |
− |
B2 |
≤ |
c |
≤ |
B1, ( Rk |
uk |
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D ) |
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G (uk ) |
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B1 + B2 |
. |
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+ |
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≤ ln(4/3) |
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In particular, we have |
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(8.55) |
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uk 2D − 2 |
F(x, uk ) dx → c |
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