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On noncoercive semilinear equations
Nonlinear Analysis: Hybrid Systems 2 (2008) 344–358 www.elsevier.com/locate/nahs
On noncoercive semilinear equations Kamal N. Soltanov 1 F. Agaev Street 9, Institute of Mathematics and Mechanics of National Academy of Sciences of Azerbaijan, Baku, Az 1141, Azerbaijan Received 14 April 2006; accepted 17 April 2006 Dedicated to Prof. Dr. V. Lakshmikantham on the occasion of his 80th birthday.
Abstract In this article some noncoercive semilinear equations are investigated. Here, a subset of a right side for which the corresponding boundary value problems are uniquely solvable is described. c 2006 Elsevier Ltd. All rights reserved.
Keywords: Semilinear equations; Noncoercive operator; Biorthogonal system; Solvability theorem; Eigenfunction; Uniqueness
1. Introduction We consider the problem −∆u − f (u) = h(x),
x ∈ Ω ⊂ Rn , u |∂ Ω = 0,
(1.1)
where Ω ⊂ Rn , n ≥ 1, is a bounded domain with sufficiently smooth boundary ∂Ω , and f is a continuous mapping such that the problem (1.1) generates a noncoercive operator A: 0
A : W2` (Ω ) ∩ W 12 (Ω ) −→ W2`−2 (Ω ),
(` ≥ 1)
0
(in particular, A : W 12 (Ω ) → W2−1 (Ω )). Here W2` (Ω ), ` ≥ 1, is the Sobolev space which is defined as ( ) Z X α 2 2 ` D u dx < ∞, α = (α1 , . . . , αn ) , W2 (Ω ) ≡ u(x) kukW ` = 2 Ω |α|≤` 0
W 12 (Ω ) ≡ W21 (Ω ) ∩ {u(x) |u|∂ Ω = 0} ,
D α ≡ D1α1 D2α2 · · · Dnαn ,
Di ≡
∂ , ∂ xi
∗ 0 and W2−1 (Ω ) ≡ W 12 (Ω ) [1]. E-mail addresses: [email protected], [email protected]. 1 Current address: Department of Mathematics, Faculty of Sciences, Hacettepe University, Beytepe, Ankara, TR-06532, Turkey. c 2006 Elsevier Ltd. All rights reserved. 1751-570X/$ - see front matter doi:10.1016/j.nahs.2006.04.010
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We recall that A : X → X ∗ is a coercive operator if the following relation is valid for a dual form h·, ·i with respect to the pair of spaces (X, X ∗ ): hA(x), xi % ∞, kxk X
under kxk X % ∞,
where X is a Banach space and X ∗ is its dual (see [8]). And A : S ⊆ X → Y is a “coercive” operator in a generalized sense if there exists a bounded operator B : X 0 ⊆ S −→ Y ∗ that satisfies certain continuity conditions, X 0 = S, Im B = Y ∗ and a continuous function µ : R+ → R non-decreasing for τ ∈ [τ0 , Υ0 ] ⊆ R+ such that the following relation is valid for a dual form h·, ·i with respect to the pair of spaces (Y, Y ∗ ): hA(x), B(x)i ≥ µ ([x] S ) ,
for x ∈ X 0 and ∃r > 0 : µ (r ) ≥ 0
where S is an pn-space with a p-norm [•] S (see [12,14]). In this case it is said that the mappings A and B generate a “coercive” pair on X 0 . 0
Let the mapping f be such that the operator A generated by the problem (1.1) is acting from W 12 (Ω ) to W2−1 (Ω ). It is known ([1,10,12] etc.) that if A is noncoercive operator then, generally speaking, the problem (1.1) is not solvable for arbitrary right hand side unlike in the coercive case (in particular, in this case the problem is normally solvable). Moreover, if the problem (1.1) is solvable then the solution may be non-single (and usually not unique); furthermore there may be an infinite number of solutions ([10,9,6,3,15] etc.). It should be noted that some parts of the results relating to this problem were published previously in the paper [14]. Here some properties of noncoercive problems are investigated. And for the problems considered it is shown that there exist subsets M from the image of the operator A defined by the problem such that for any h of M the problem (1.1) is uniquely solvable in corresponding subsets from the domain. To be more exact, it is shown that there exists a 0
subset G of W 12 (Ω ) on which A : G −→ A (G) ⊂ W2−1 (Ω ) is injective. At beginning the problem (1.1) with nonlinearity of functional type is studied, and after that the one with nonlinearity of exponential type is studied. 2. Some general results At beginning some general results are given on which the proofs of solvability of the problems considered are based. Let X, Y be Banach spaces and X ∗ , Y ∗ be their dual spaces; let S be a weakly complete pn-space (see [12,13], where other variants of the following theorem are given). Let X 0 be a topological space such that X 0 ⊆ S ⊆ X , and let A be a nonlinear mapping acting from X to Y , where Y is a reflexive space such that both Y and Y ∗ are strictly convex. We consider the equation A(x) = y,
y ∈ Y.
(2.1)
This equation is considered in the case where the operator A is not coercive in the generalized sense. But it can be investigated locally as in [14], i.e. in the sense described in this section. It is known that in the general case a definition of a solution of the equation A(x) = y has the form [13]: Definition 1 (Generalized Solution). Let M ∗ be a subset of Y ∗ and y be an element of Y . Then an element x ∈ S satisfying hA(x), y ∗ i = hy, y ∗ i , ∀y ∗ ∈ M ∗ ⊂ Y ∗ , is called an M ∗ -solution of equation A(x) = y. We will consider the following conditions: (1) A : S −→ Y is a weakly continuous mapping and there exists a closed linear subspace Y1 of Y such that A : S −→ Y1 ⊂ Y ; (2) there exists a mapping B : X 0 −→ Y ∗ such that Im B contains a linear manifold from Y ∗ which is dense in a closed linear subspace Y1∗ of Y ∗ and generates a “coercive” pair with A on X 0 in a generalized sense; moreover one of the following conditions ((α) or (β)) holds: (α) if B is a linear continuous operator then S is a “reflexive” space (see [12,13]), X 0 is a separable vector topological space which is dense in S and ker B ∗ = {0};
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(β) if B is a nonlinear operator then Y1∗ is a separable subspace of Y ∗ and B −1 is weakly continuous from Y ∗ to S. Theorem 1. Let conditions (1) and (2) hold. Then Eq. (2.1) is Y1∗ -solvable in S for every y ∈ Y1 satisfying the following condition: There exists r > 0 such that µ ([x] S ) ≥ hy, B(x)i ,
for ∀x ∈ X 0 with [x] S ≥ r,
(2.2)
where h·, ·i is the dual form of the pair (Y, Y ∗ ). Notation 1. It is clear that the Eq. (2.1) is equivalent to the following functional equation:
A(x), y ∗ = y, y ∗ , ∀y ∗ ∈ Y ∗ . It should be noted that if the space Y satisfies the above-stated conditions then Y1∗ ≡ (Y1 )∗ ; see Proposition 1. From here we obtain that the solution in the general sense which is the Y1∗ -solution of Eq. (2.1) is actually the solution in the usual sense. Consequently, in this case, the solution of the equation A(x) = y for any y ∈ Y1 is the solution in the general sense and vice versa. 2 Proof (Short Scheme k ∞). Let the conditions (1) and (2) (α) be satisfied. We will use the Galerkin approximation method. So, as usual let x k=1 be a complete system in the (separable) space X 0 . We then look for approximate solutions P k in the form xm = m k=1 cmk x . Here cmk are unknown coefficients which have to be determined from the system of algebraic equations
8k (cm ) := hA (xm ) , Bxk i − hy, Bxk i = 0,
k = 1, 2, . . . , m
(∗)
where cm ≡ (cm1 , cm2 , . . . , cmm ). Now we observe that the mapping 8 (cm ) := (81 (cm ) , 82 (cm ) , . . . , 8m (cm )) is continuous by virtue of condition 1. From (2.2) it follows that there exists r > 0 such that the “acute angle” condition is satisfied for all xm with [xm ] S ≥ r , i.e. the following inequality holds for any x ∈ Sr1 (0) ⊂ R m , r1 ≥ r : m X
h8k (cm ), cmk i ≥ 0,
∀cm ∈ Rm , kcm kRm = r1 .
k=1
Then the solvability of the system (∗) follows from the well-known lemma on the “acute angle” which is equivalent to Brouwer’s fixed-point theorem (see Lemma 2 and, e.g., [8,9,15]). Thus we obtain a sequence which is contained in a bounded subset of S. Further arguments are analogous to the arguments from [13,14]; therefore we omit them. Remark 1. It is obvious that if there exists a function ψ ∈ C 0 such that ψ (ξ ) = 0 ⇐⇒ ξ = 0 and if the following inequality is fulfilled: kx1 − x2 k X ≤ ψ (kA (x1 ) − A (x2 )kY ) for all x1 , x2 ∈ S, then a Y1∗ -solution of Eq. (2.1) is unique. The following corollary follows from this theorem. Corollary 1. Let the conditions (1) and (2) hold, and the operators A and B generate a “coercive” pair in the generalized sense on a space X 0 ; furthermore let µ be such that µ (τ ) % +∞ as τ % +∞, and let Y be a reflexive Banach space which, together with its dual space Y ∗ , has a strictly convex norm, and say there exists a ) function ϕ1 : R+ −→ R+ , ϕ1 ∈ C 0 such that ϕµ(τ % +∞ as τ % +∞; then for every y ∈ Y1 such that 1 (τ ) n o |hy,B(x)i| sup ϕ1 ([x] S ) x ∈ X 0 < +∞, Eq. (2.1) is solvable in S. The problem considered can be solved by applying the above theorem. But we shall construct the proofs of the solvability theorems for the problems considered without directly applying the general theorem, for better 2 For the proof see also [13,14].
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visualization of the proof of existence and uniqueness. The general concept of a solution coincides with the usual concept of a solution in the case considered, by virtue of a result which was proved in the article “Remarks on separation of convex sets, fixed-point theorem and applications in theory of linear operators” (which will be published).3 3 Proposition 1. Let X and its dual space X ∗ be strictly convex reflexive Banach spaces, and let X 0 ⊂ X be a subspace of X . Then the dual space of a subspace X 0 ⊂ X is equivalent to a subspace of X ∗ which is determined by the subspace X 0 ; i.e. X ∗ has a subspace X 0∗ ⊂ X ∗ defined by the unit sphere of the subspace X 0 such that (X 0 )∗ ≡ X 0∗ . Consequently, X 0 and its dual space (X 0 )∗ are strictly convex reflexive Banach spaces under the norms induced by the norms of X and X ∗ , respectively. Proof. It is known from [4,5,7] that the dual space of the subspace X 0 ⊂ X is equivalent to a factor (quotient) space of the form X ∗ / X 0⊥ , where X 0⊥ ⊂ X ∗ is the annihilator of X 0 ⊂ X : X 0⊥ ≡ x ∗ ∈ X ∗ |hx, x ∗ i = 0, ∀x ∈ X 0 . (Here the expression h·, ·i denotes the dual form for the pair X, X ∗ , or an inner product if X is a Hilbert space.) It is also known from [11,16] that the subspace X 0 ⊂ X is a reflexive Banach space under the norm induced from X and that its dual space (X 0 )∗ is also reflexive. Moreover, if X is a strictly convex reflexive Banach space then so is X 0 . In addition, by [4], if X is a strictly convex reflexive Banach space then an arbitrary element of the unit sphere is an extremal point and the dual space (X 0 )∗ is equivalent to a subspace of X ∗ . It remains, therefore, to identify this subspace. = In order to construct a subspace dual to X 0 we will consider the duality mapping = : X −→ X ∗ for the pair X ; X ∗ , i.e. X ←→ X ∗ (see [4,8, 16] and the references therein). In the case under consideration, the duality mapping is bijective and, together with its inverse mapping, is strictly
monotone, surjective, odd, demicontinuous, bounded and coercive. Hence we have x, x ∗ ≡ x ∗ , x for any x ∈ X , x ∗ ∈ X ∗ , and in particular for any x ∈ X we have x ←→ x ∗ = =(x), i.e. it is an equivalence relation [4,8,16]. It follows from this that it will be enough to consider these mappings on the unit spheres of X and X ∗ . ∗ ∗ We shall denote the unit spheres of X and X ∗ by S1X and S1X , respectively. Then we have = S1X ≡ S1X . In addition, the following relations hold: ∗
(∀x)(x ∈ S1X ⇐⇒ =(x) = x ∗ ∈ S1X ).
(∀x)(x ∈ S1X ⇐⇒ hx, =(x)i = x, x ∗ = kxk X · x ∗ X ∗ = 1 · 1), and conversely D E
∗ (∀x ∗ )(x ∗ ∈ S1X ⇐⇒ x ∗ , =−1 x ∗ = x ∗ , x = x ∗ X ∗ · kxk X = 1 · 1), since the duality mapping is a homeomorphism, and by virtue of the conditions of the proposition (see [8,16] and the references therein). Moreover the following relation holds: ∀x ∈ X,
∃e x ∈ S1X
so that x = e x kxk X .
Hence we have that the unit sphere S1X defines the whole space X in the sense that X ≡ S1X × R+ ; R+ ≡ {τ ∈ R : τ ≥ 0}. X Hence, if X 0 ⊂ X is a subspace of X then X 0 can be defined through a subset of the unit sphere of the form S1 0 ≡ S1X ∩ X 0 ≡ S1X (0)∩ X 0 . Here, X0 S1 denotes the unit sphere of X 0 with the norm induced from X . Thus the space X 0 is a strictly convex reflexive Banach space. Consequently there exists an equivalent norm such that X 0 , together with its dual space, is a strictly convex reflexive Banach space. Under the induced topology X – which we obtain by virtue of the duality mapping = from X onto X ∗ – the sphere S1 0 will be transformed onto a subset which can be expressed in the form ∗ X e S1∗ ≡ {x ∗ ∈ S1X | h=−1 (x ∗ ), x ∗ i = kxk X · kx ∗ k X ∗ = 1, =−1 (x ∗ ) = x ∈ S1 0 }, because we have ∗ X X =−1 e S1∗ ≡ S1 0 ; ∀x ∗ ∈ e S1∗ ⊂ S1X ⇐⇒ x = =−1 x ∗ ∈ S1 0 .
(2.3)
It is known that if X and X ∗ are strictly convex reflexive spaces, then the duality mapping = : X X ∗ : =−1 is the Gateaux-differential of =
a strictly convex functional z and =−1 is the Gateaux-differential of a strictly convex functional z∗ , i.e. the duality mapping X ←→ X ∗ is a positively homogeneous potential operator with strictly convex potential. In addition, there is a strongly monotone increasing continuous function
∗ Φ : R+ −→ R+ , Φ (0) = 0, Φ (τ ) % +∞ when τ % +∞ such that = (τ x) = Φ (τ ) x ∗ for any x ∈ S1X and x ∗ ∈ S1X , where x, x ∗ ≡ 1 X S ∗ defined by (2.3) is the unit sphere of the and τ ∈ R+ [8]. Consequently, = B 0 is a convex subset X ∗ (see also [16]). Thus we obtain that e 1
1
subspace (X 0 )∗ from X ∗ , which we can denote by X 0∗ (i.e. (X 0 )∗ ≡ X 0∗ ) that also is equivalent to X ∗ / X 0⊥ . In other words we have obtained that X 0∗ is equivalent to the dual space of the subspace X 0 of X , and so a subspace X 0 of a reflexive Banach space X is a reflexive Banach space under the induced topology under the conditions of the proposition.
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3. Solvability of the problem (1.1) in the case of functional nonlinearity Let Ω ⊂ Rn , n ≥ 1, be a bounded domain with sufficiently smooth boundary ∂Ω . Let the eigenfunctions of the problem −∆wk = λk wk ,
wk |∂ Ω = 0,
(3.1)
generate an orthogonal basis in space L 2 (Ω ). We denote a system of eigenfunctions of problem (3.1) by {wk (x)}∞ k=1 where wk (x) corresponds to the eigenvalue λk . (Usually an eigenvalue of problem (3.1) is considered in the form λk = µ2k .) We will arrange eigenvalues in the order of growth, i.e. 0 < λ1 ≤ λ2 ≤ · · ·. Further we denote a subspace of the space L 2 (Ω ) with codimension 0
m − 1 ≥ 0 generated by elements {wk (x)} for k ≥ m by E m0 and the corresponding subspace of W12 (Ω ) by E m1 . Now we will consider 0 W 12 (Ω ) ≡ W21 (Ω ) ∩ u(x) | u |∂ Ω = 0 , with a norm which is equivalent to the previous one and is defined by
2
2 kuk2 ≡ kuk20 ≡ n D j u L (Ω ) ≡ n D j u 2 . W 12 (Ω )
2
j=1
j=1
0
i (i = 0, 1) are subspaces of L (Ω ) and As E m W 12 (Ω ) respectively, with induced topologies generated by a norm in 2 the corresponding space, we have that E m−1 is a dual space of E m1 and it is a subspace of W2−1 (Ω ) (see, for example, [8, 5,16]) with the codimension m − 1 (i.e. codim E m−1 = m − 1).
Note. It follows from Proposition 1 that the dual space of E m1 is a space E m−1 which is a subspace of the space W2−1 (Ω ). Lemma 1. Let X be a strictly convex reflexive Banach space together with its dual space X ∗ . If X is a space with a basis and X 1 is a subspace of X with codimension m (i.e. codim X 1 = m) then X 1∗ is the dual space of X 1 , and X 1∗ is a subspace of X ∗ with the same codimension (i.e. codim X 1∗ = m). This assertion follows from the Proposition 1 and results devoted to the properties of biorthogonal bases (see, for example, [5, Chapter 2 & 6] etc.). Moreover since the space considered is a Hilbert space we obtain immediately that this property remains. Thus the spaces scale framed on L 2 (Ω ) are transferred to spaces scale framed on E m0 (subspace). Therefore, it is possible to define the biorthogonal base in E m−1 . So, we consider the problem (1.1) −∆u − f (u) = h(x),
x ∈ Ω ⊂ Rn , u |∂ Ω = 0
and assume that the following condition holds: 0
(ι) the mapping f : W 12 −→ W2−1 from the problem (1.1) is such that f : E m0 −→ E m−1 , f ∈ C 0 , as f : E m1 −→ E m−1 , f ∈ C 1 and for any u ∈ E m1 the following inequalities hold:
0
−1
ρ ρ+1 k f (u)kW −1 ≤ c1 λm 2 kuk2 , ≤ c0 λ−1
f (u) 0 1 m kuk2 , −1 W 2 −→W2
2
where ρ, c1 ≥ 0 are constants, and c0 = c0 (ρ) > 0. (It is clear that if c0 = 0 then the mapping f does not depend on u; therefore we assume that c0 = c0 (ρ) > 0.) Remark 2. The assumption that f : E m0 → E m−1 ,
f ∈ C 0 (is continuous) in condition (ι) may be superfluous.
For example the operator f may have the following form: Z ρ 2 ρ 2 |u| dx u, a > 0 − const. f (u) ≡ a kuk2 u ≡ a × Ω
K.N. Soltanov / Nonlinear Analysis: Hybrid Systems 2 (2008) 344–358
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This operator satisfies the condition (ι). In fact for any u ∈ E m1 we have
−1 ρ+1 ρ ρ k f (u)kW −1 = a kuk2 u W −1 = a kuk2 kukW −1 ≤ aλm 2 kuk2 . 2
2
2
− 21
Also we have kukW −1 ≤ λm kuk2 for any u ∈ E m1 . 2
0
W 12 (Ω )
For any u ∈ we have Z Z ∆u u dx = ∇u(x)∇u(x)dx = kuk20 . − Ω
W 12
Ω
0
From here we obtain that, if the λi ’s are eigenvalues of the operator −∆ : W 12 (Ω ) −→ W2−1 (Ω ) and λ1 = min {λi |i ∈ N }, then 1
kuk20 ≥ λ1 kuk22 ,
i.e. kuk 0 1 ≥ λ12 kuk2 .
W 12
W2
Now, if an u ∈ E m1 is an arbitrary element, then we have 1
−1
kuk 0 1 ≥ λm2 kuk2 H⇒ λm 2 kuk 0 1 ≥ kuk2 . W2
W2
0
Further it is well known that the operator −∆ : W 12 (Ω ) −→ W2−1 (Ω ) is an isomorphism. Then for any 0
v(x) ∈ W2−1 (Ω ) there exists a function ψ(x) ∈ W 12 (Ω ) such that −∆ψ(x) = v(x), and ψ(x) = (−∆)−1 v(x). Consequently, from the definition we have 0 kvkW −1 = sup hv, gi | g ∈ S1 (0) ⊂ W 12 (Ω ) 2
0
for any v(x) ∈ W2−1 (Ω ). Let a function ψ(x) ∈ W 12 (Ω ) be such as the above: hv, ψi =
h−∆ψ(x), ψi = h∇ψ, ∇ψi H⇒ kvkW −1 = k∇ψk2 = kψk 0 1 2
1
H⇒ (−∆)− 2
W2
−1
v ≤ λ1 2 kvk2 . 2
−1
Consequently we have kukW −1 ≤ λm 2 kuk2 for any u ∈ E m1 . 2
Notation 2. Let the system of eigenfunctions {wk }∞ k=1 of the problem (3.1) be an orthogonal basis in L 2 (Ω ), where wk (x) is the eigenfunction corresponding to the eigenvalue λk . Let the eigenvalues λk be arranged in order of growth, 1 i.e. 0 < λ1 ≤ λ2 ≤ · · ·. Then, from the above reasoning, we will obtain that the subspace P∞E m is generated by this ∞ 1 system over the basis {wk }k=m , m ≥ 1, and consequently for any u ∈ E m we have u(x) = k=m u k wk (x). Moreover we have ∞ ∞ X X −∆u(x) = −∆ u k wk (x) = λk u k wk (x). k=m
k=m
Consequently, h−∆u(x), u(x)i =
∞ X
λk u 2k kwk (x)k22 ≥ λm kuk22 ,
k=m
as λm is a minimum of λk at k ≥ m. We interpret the solution of problem (1.1) in a somewhat general sense as in Section 2. Therefore using Proposition 1 this definition for problem (1.1) can be formulated as follows.
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Definition 2. A function u(x) ∈ E m1 is called an E m1 -solution of the problem (1.1) if u(x) satisfies Eq. (1.1) in the sense of the space E m−1 , i.e. the equality Z Z Z hA (u) , vi ≡ h(x)v(x)dx, (3.2) f (u(x)) v(x)dx = ∇u(x)∇v(x)dx − holds for any v(x) ∈
Ω
Ω
Ω
E m1 ,
where the operator A is generated by problem (1.1).
Remark 3. As is known, E m1 is a Hilbert space with dual space E m−1 and when h(x) is contained in E m−1 , the solvability of problem (1.1) in the usual sense follows from the definition. 0 0 1 Now, let Br (0) ≡ u(x) ∈ W 2 (Ω ) | kuk ≤ r be the closed ball of W 12 (Ω ) and Sr (0) ≡ 0 u(x) ∈ W 12 (Ω ) | kuk = r be the sphere (i.e. the boundary of the ball Br (0)). Theorem 2. Let all above-mentioned conditions of problem (1) be fulfilled, let c1 = 1, and let b be a fixed number from (c0 , 2c0 ), and b0 = min b−1 , 2−1 . Then for each integer m ≥ 1 there exists an integer m 0 ≥ 1 such that for any h(x) ∈ E m−10 satisfying the inequality hh, ui ≤ hA (u) , ui ≡ h−∆u − f (u) , ui ,
∀u ∈ Srm 0 (0) ∩ E m1 0 ,
(3.3) 0
1
ρ+2
the problem (1) is uniquely solvable on the ball Brm 0 (0) ∩ E m1 0 ⊂ W 12 (Ω ) for rm 0 = (1 − b0 ) ρ λm2ρ0 , where m 0 = inf {e m |λm e = λm }. (Here for simplicity we assume that c1 = 1.) Remark 4. It is known that λm % +∞ when m % +∞, so we have rm 0 and the corresponding norm of h(x) ∈ E m−10 can be chosen sufficiently large; m 0 = m 0 (m). Proof. First of all, we shall show that the set of the right side — h(x) defined by (3.3) — is a nonempty subset of 0
E m−10 . In fact, for any u(x) ∈ Brm 0 (0) ∩ E m1 0 ⊂ W 12 (Ω ) and for any m 0 ≥ 1 we have Z Z hA (u) , ui ≡ A (u(x)) u(x)dx ≡ [−∆u − f (u)] u(x)dx = kuk2 − h f (u) , ui Ω
Ω
≥ kuk2 − k f (u)kW −1 kuk , 2
and taking into account the first inequality for f from condition (ι) we obtain h i −1 ρ+1 ρ kuk hA (u) , ui ≥ kuk2 − λm 02 kuk2 kuk ≥ kuk2 1 − λ−1 m0 2 . Further we will consider this inequality for the functions u(x) from Srm 0 (0) ∩ E m1 0 as in the condition of the theorem. Then we obtain the inequality using the condition (i), selecting u(x) from Srm 0 (0) ∩ E m1 0 and using kuk2 −1
≤ λm 02 kuk (as c1 = 1) −1− ρ2 −1− ρ2 ρ ρ 2 hA (u) , ui ≥ kuk 1 − λm 0 kuk ≥ rm 0 1 − λm 0 rm 0 . 2
Hence we obtain by substituting the value rm 0 from the theorem 1 1 ρ 1 + −1− ρ2 2 +1 hA (u) , ui ≥ rm 0 1 − λm 0 (1 − b0 ) λm (1 − b0 ) ρ λmρ 0 2 1
1
+1
= rm 0 b0 (1 − b0 ) ρ λmρ 0 2 . The inequality obtained (3.4) shows that the subset of right sides satisfying the inequality (3.3) is nonempty.
(3.4)
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Now, we shall proceed with the proof of the theorem. For the proof we use a variant of the Galerkin method. We k will search for an approximate solution as an element of the linear variety generated by {wi (x)}i=m , i.e. in the form 0 Pk u k (x) = i=m 0 aki wi (x), where aki (i = m 0 , m 0 + 1, . . . , k) are unknown coefficients which will be determined from the system of algebraic equations hA (u k ) , wi i = hh, wi i ,
i = m 0 , m 0 + 1, . . . , k.
(3.5)
The solvability of system (3.5) follows from a known result, which is equivalent to a variant of the Brouwer theorem on a fixed point [8,9,15] and which can be formulated as follows: Lemma 2 ([8]). Let the mapping F acting on Rn (n ≥ 1) satisfy the following conditions on the closed ball Br (0) for some number r > 0: (α) F is continuous; (β) the inequality hF(x), xi ≥ 0, holds for any x ∈ Sr (0). Then there exists a point x0 ∈ Br (0) such that F (x0 ) = 0. If we denote the mapping generated by problem (3.5) in the form F (ak ) ≡ Fm 0 (ak ) , Fm 0 +1 (ak ) , . . . , Fk−m 0 (ak )
k where F j (ak ) ≡ A (u k ) − h, wm 0 + j , and ak ≡ {aki }i=m , then obviously F is continuous on R k−m 0 +1 and from 0 inequality (3.4) it follows that the condition (β) holds. Therefore we obtain the solvability of the system (3.5) for each k : k = m 0 , m 0 + 1, . . . by applying Lemma 2. From here we can say that all approximate solutions u k (x) are contained in the ball Brm 0 (0) ∩ E m1 0 . Thus we obtain a sequence of approximate solutions {u k (x)} of problem (1.1) which is contained in a bounded subset of E m1 0 . Therefore it is possible to select a subsequence of the sequence {u k (x)} weakly converging in E m1 0 by virtue of the reflexivity of the E m1 0 (for any m 0 = 0, 1, . . .) which we shall designate through {u k (x)}, for simplicity. 0
Then u k * u weakly in E m1 0 (and in W 12 (Ω )), where u(x) is an element of E m1 0 . Consequently u k → u strongly in E m0 0 (also in L 2 (Ω )) and almost everywhere in Ω when k → +∞. It is clear that ∆u k * ∆u
weakly in E m−10 (and in W2−1 (Ω )) as k → +∞.
Now it is necessary to show that weakly in E m−10 (and in W2−1 (Ω )) as k −→ +∞.
f (u k ) * f (u)
This is true, since f is a continuous mapping from E m0 0 in E m−10 and u k → u strongly in E m0 0 (also in L 2 (Ω )) and almost everywhere in Ω . Thus we can pass to the limit in (3.5) for k → +∞ for each i = m 0 , m 0 + 1, . . . . Then passing to the limit in (3.5) for k → +∞ we have hA (u) , wi i = hh, wi i ,
i = m 0 , m 0 + 1, . . . .
∞ We obtain the solvability of problem (1.1) taking into account the completeness of the system {wi (x)}i=m in E m1 0 ; 0 1 i.e. the function u(x) satisfies the equality (3.2) for any v(x) of E m 0 . From here we obtain the first part of the statement of the theorem by virtue of Remark 3. Now to complete the proof of the theorem it is necessary to show the uniqueness of the solution. For this purpose it is sufficient to prove the following inequality for any u(x), v(x) ∈ Brm 0 (0) ∩ E m1 0 : kA (u) − A (v)kW −1 ≥ K ku − vk , K = K rm 0 , λm 0 , ρ > 0, (3.6) 2
where K is some number which is independent of u(x) and v(x). We set w(x) = u(x) − v(x). Then Z kA (u) − A (v)kW −1 kwk ≥ [A (u) − A (v)] w(x)dx 2
Ω
= |hA (u) − A (v) , w(x)i| = |h−∆w − f (u) + f (v) , wi|
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K.N. Soltanov / Nonlinear Analysis: Hybrid Systems 2 (2008) 344–358
≥ kwk2 − |h f (u) − f (v) , wi| ≥ kwk2 − k f (u) − f (v)kW −1 kwk 2
0
2
kwk , = kwk − f (e u) 0 1 −1 W 2 (Ω )→W2
where e u (x) is an element of the “interval” [u(x), v(x)] ⊂ Brm 0 (0) ∩ E m1 0 . Hence we will obtain by virtue of condition (ι) ρ kA (u) − A (v)kW −1 kwk ≥ kwk2 1 − c0 λ−1 ke k u m0 2 2 ρ −1− ≥ kwk2 1 − c0 λm 0 2 ke u kρ . Taking ke u k ≤ rm 0 , ρ > 0 and substituting the expression for rm 0 , −1− ρ kA (u) − A (v)kW −1 ≥ kwk 1 − c0 λm 0 2 rmρ 0 ≥ kwk [1 − c0 (1 − b0 )] .
(3.7)
2
The correctness of inequality (3.6) and consequently uniqueness of a solution of the problem in a ball Brm 0 (0) ∩ 0
E m1 0 ⊂ W 12 (Ω ) under the conditions of Theorem 2 follows from the last inequality by virtue of the condition on 1 c0 > 1 − b0 > 0. Thereby the theorem is proved completely. The following statement follows from Theorem 2. Corollary 2. Let the conditions of Theorem 2 be fulfilled, let c1 = 1, and let b be a fixed number in (c0 , 2c0 ), and b0 = min b−1 , 2−1 . Then for each integer m ≥ 1 there exists an integer m 0 ≥ 1 such that the problem 0
(1.1) is uniquely solvable in a ball Brm 0 (0) ∩ E m1 0 ⊂ W 12 (Ω ) for any h(x) ∈ B R∗ m (0) ∩ E m−10 ⊂ W2−1 (Ω ), where 0 m 0 = inf {e m | λm e = λm }, B R∗ m (0) ≡ {h(x) ∈ W2−1 (Ω ) | khkW −1 (Ω ) ≤ Rm 0 } 0
is the closed ball of following form:
2
W2−1 (Ω )
centered at the origin with radius Rm 0 > 0, and rm 0 and Rm 0 are determined in the
2+ρ
1
1
2+ρ
rm 0 = (1 − b0 ) ρ λm2ρ0 , Rm 0 = b0 (1 − b0 ) ρ λm2ρ0 ! 1 1 1 − b0 ρ 2+ρ 1 − b0 ρ ρ + b0 2+ρ 2ρ 2ρ or rm 0 = λm 0 , R m 0 = λm 0 . 1+ρ 1+ρ 1+ρ Proof. For this purpose it is enough to show that the inequality (3.4) holds for any h(x) ∈ B R∗ m (0)∩ E m−10 ⊂ W2−1 (Ω ). 0 The fulfillment of the conditions of Theorem 2 for an h(x) follows from the proof of Theorem 2 under the condition on the radius Rm 0 . The following inequalities follow from the inequalities (3.4) and (3.7): 1
2+ρ
hA (u) , ui ≥ rm 0 b0 (1 − b0 ) ρ λm2ρ0 and −1− ρ kA (u) − A (v)kW −1 ≥ kwk 1 − c0 λm 0 2 rmρ 0 ≥ kwk [1 − c0 (1 − b0 )] , 2
which shows that the problem (1.1) is uniquely solvable in Brm 0 (0) ∩ E m1 0 for any h(x) ∈ E m−10 satisfying the 1
2+ρ
inequality hh, ui ≤ rm 0 b0 (1 − b0 ) ρ λm2ρ0 for any u(x) ∈ Srm 0 (0) ∩ E m1 0 .
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Further, the estimation hh, ui gives, under the conditions of the corollary, 2+ρ
1
|hh, ui| ≤ khkW −1 (Ω ) kuk = khkW −1 (Ω ) rm 0 ≤ rm 0 b0 (1 − b0 ) ρ λm2ρ0 . 2 2
So the corollary follows from here. Pm∗ :
W2−1 (Ω )
Now, let −→ following statements are true.
E m−1
be a canonical projector. From Theorem 2 and Corollary 2 it follows that the
Corollary 3. Let all conditions of Corollary 2 be fulfilled. Then for any h(x) ∈ W2−1 (Ω ) there exists an integer 2+ρ
1 m 0 ≥ 1 such that for Pm∗ 0 h ∈ E m−10 , Pm∗ 0 h W −1 ≤ b0 (1 − b0 ) ρ λm2ρ0 is fulfilled, in other words, for Pm∗ 0 h ∈ E m−10 the 2
1
2+ρ
problem (1.1) is uniquely solvable in the ball Brm 0 (0) ∩ E m1 0 where rm 0 = (1 − b0 ) ρ λm2ρ0 . 0
Corollary 4. Let the operator f : W 12 (Ω ) → W2−1 (Ω ) be such that f ∈ C 1 ∩ Cw0 and there exist constants α ≥ 12 , β = α + 12 , 1 > γ > 21 , ρ > 0 such that the following inequalities hold:
−β ρ ρ+1 k f (u)kW −1 (Ω ) ≤ λ−α ≤ λ1 kuk2 , f 0 (u) 0 1 1 kuk2 −1 W 2 →W2
2
0
for any u(x) ∈ W 12 (Ω ). Then for any h(x) ∈ W2−1 (Ω ), khkW −1 (Ω ) ≤ γ · r1 , the problem (1.1) is uniquely solvable in 2 0
1
ρ+1+2α 2ρ
the ball Br1 (0) ⊂ W 12 (Ω ), r1 = (1 − γ ) ρ λ1
. (We denote the space of weakly continuous mappings by Cw0 .) 0
Note that in the case m = 1 we obtain E m−1 = W2−1 (Ω ) and E m1 = W 12 (Ω ), i.e. P1 = id. Remark 5. Assertions similar to the statements of the problem (1.1) can also be proved when appropriate spaces and 0
operator A are determined as follows: the expression for the operator A does not vary but now A : W22 (Ω )∩ W 12 (Ω ) → 0
L 2 (Ω ), and E m2 is the corresponding subspace of the space W22 (Ω ) ∩ W 12 (Ω ). The necessary a priori estimations will be obtained by multiplication with −∆u. In this case the condition (ι) will be replaced by an appropriate condition. 4. Solvability of the problem (1.1) with polynomial nonlinearity 0
i i be a subspace such as the above, and E i Let E m −m be the respective complementary subspace, so that W 2 (Ω ) ≡ i ⊕ E i , i = 0, 1. Let us designate through P , Q the projectors in L (Ω ) such that for each m ∈ N the following Em m m 2 −m relations apply:
kPm k ≤ 1, kQ m k ≤ 1,
P1 ≡ 1 ≡ id,
0
i Q m ≡ (1 − Pm ) : W i2 (Ω ) −→ E −m ,
0
i Pm : W i2 (Ω ) −→ E m ,
0
i i ⊕ E −m , i = 0, 1 W i2 (Ω ) ≡ E m
and −1 Pm∗ Q ∗m : W2−1 (Ω ) −→ E m−1 E −m ,
m = 1, 2, . . . .
We will consider the problem (1.1) in the case when f (u) ≡ |u|ρ u, i.e. −∆u − f (u) ≡ −∆u − |u|ρ u = h(x), x ∈ Ω ; u |∂ Ω = 0, 0 ≤ ρ < which may be rewritten in the following form: ∗ Pm A (u) ≡ −∆u m − Pm∗ |u|ρ u = h m (x) ≡ Pm∗ h(x), (1) Q ∗m A (u) ≡ −∆u −m − Q ∗m |u|ρ u = h −m (x) ≡ Q ∗m h(x), (2) ,
4 , n−2
(4.1)
u |∂ Ω = 0,
(4.2)
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K.N. Soltanov / Nonlinear Analysis: Hybrid Systems 2 (2008) 344–358
where u m ≡ Pm u ∈ E m1 ;
1 u −m ≡ Q m u ∈ E −m ,
Pm∗ ∆u ≡ ∆Pm u ≡ ∆u m .
u ≡ u m + u −m ;
Let us introduce the following notation: ρ ρ 1 1 rm 0 = 1 − bm 0 ρ 2−1 C p λνm 0 ; r−m 0 = 1 − bm 0 ρ 2−1 C p λν1 ; [4 − ρ (n − 2)] ν≡ , m 0 ≥ 2; 4ρ if n ≥ 3 and ρ > 0 is an arbitrary number if n = 1, 2, h σ i ρ 1 , m0 ≥ 2 Rm 0 ≡ 1 − bm 0 ρ 2−µ C p λνm 0 bm 0 − 1 − bm 0 λ1 λ−1 m0 h i 1 ρ σ , m0 ≥ 2 R−m 0 ≡ 1 − bm 0 ρ 2−µ C p λνm 0 bm 0 − 1 − bm 0 λm 0 λ−1 1
0<ρ<
σ =
4 n−2
2 (ρ + 1) ν (ρ + 1) [4 − ρ (n − 2)] = , 2 pρ p
where bm 0 > 0 is a number which satisfies the inequalities 1 > bm 0 > 1 −
2λ1 ; p λm 0 + λ1
p = ρ + 2,
m 0 = m 0 (m) and C > 0 is a constant of the Gagliardo–Nirenberg inequality [8] n n n kukW ps (Ω ) ≤ C kukθW l (Ω ) kuk1−θ − s = − θ) + θ − l , , (1 L p2 (Ω ) p1 0 p0 p2 p1 where C ≡ C ( p0 , p1 , p2 , l, s) > 0; if s = 0, l = 1, p1 = p2 = 2, p0 = ρ + 2 = p, then θ =
(4.3) nρ 2p .
4 if n ≥ 3 and 0 < p − 2 = ρ be an arbitrary number if n = 1, 2, and let Theorem 3. Let 0 < ρ < n−2 i i E m 0 , E −m 0 (i = −1, 0, 1) be the above-determined subspaces: 0
0
0
1 . W 12 (Ω ) ≡ Pm 0 W 12 (Ω ) ⊕ Q m 0 W 12 (Ω ) ≡ E m1 0 ⊕ E −m 0 0
1 Then the problem (4.1) ((4.2)) is uniquely solvable in the “ball” Brm 0 (0) ∩ E m1 0 ⊕ Br−m 0 (0) ∩ E −m ⊂ W 12 (Ω ) 0 for any h(x) ∈ W2−1 (Ω ) satisfying the inequality
1 hh, ui ≤ − −∆u − |u|ρ u, u , for ∀u ∈ Srm 0 (0) ∩ E m1 0 ⊕ Sr−m 0 (0) ∩ E −m . (4.4) 0
Proof. We remark that the solution of problem (4.1) (or (4.2)) will be understood in the generalized sense; it is similar to Definition 1. We denote by A the operator generated by problem (4.1). At the beginning one needs to show that the subset of the right sides h(x) determined by the inequality (4.3) is nonempty. It is enough to show that the inequality 1 hA (u) , ui ≥ µ (kuk) kuk applies on the subset Brm 0 (0) ∩ E m1 0 ⊕ Br−m 0 (0) ∩ E −m with a function µ (τ ), τ ≥ 0, such 0 that µ rm 0 ≥ δ0 > 0 for some number δ0 . These allow us to use Lemma 1. For this purpose we consider the problem (4.1) in the form of the system (4.2), as the problem (4.2) is equivalent to the problem (4.1) by construction. From the problem (4.2)(1) we have
2
∗
ρ+1 Pm 0 A (u) , u m 0 ≡ −∆u m 0 − Pm∗ 0 |u|ρ u, u m 0 ≥ u m 0 − kuk L p (Ω ) u m 0 L (Ω ) p
2
p
ρ+1
ρ ρ
u m ≥ u m − 2 u m − 2 u −m 0
0
0
L p (Ω )
0
L p (Ω )
0
L p (Ω )
for any u(x) ∈ W 12 (Ω ) (here, if m 0 = 1 we will obtain the whole space; therefore we suppose that m 0 ≥ 2).
K.N. Soltanov / Nonlinear Analysis: Hybrid Systems 2 (2008) 344–358
355
Now we assume that ρ > 0, since when −1 < ρ < 0 this problem turns into the known one which has already been investigated well enough. We will obtain by using the inequality (4.3) in the last inequality
2
2νρ
2νρ
ρn
ρ+1
ρn Pm∗ 0 A (u) , u m 0 ≥ u m 0 − 2ρ C p u m 0 2 u m 0 L (Ω ) − 2ρ C u −m 0 L (Ω ) u m 0 2 p u m 0 Lp2 (Ω ) p
2
2 − ρν
u m p − 2ρ C u −m ρ+1 λm p u m ≥ u m 0 − 2ρ C p λ−ρν m0 0 0 0 L (Ω ) 0 p
(where ν =
1 ρ
−
n−2 4 ).
(4.5)
From here, correctness of the indicated inequality follows.
1 Now assuming u(x) ∈ Srm 0 (0) ∩ E m1 0 ⊕ Sr−m 0 (0) ∩ E −m in the last inequality, we obtain 0
ρ+1
2 − ρν p ρ
um0 Pm∗ 0 A (u) , u m 0 ≥ u m 0 1 − 2ρ C p rmρ 0 λ−ρν − 2 C u λ −m 0 L (Ω ) m 0 m0 p
=
rm2 0
ρ
1−2
C p rmρ 0 λ−ρν m0
ρ+1 − ρν − 2 C u −m 0 L (Ω ) λm 0 p rm 0 ρ
p
ρν ρµ ρ p ρ+1 − p − 2 rm 0 ≥ rm2 0 1 − 2ρ C p rmρ 0 λ−ρν λ − 2 C r −m 0 m 0 λ1 m0 ρν ρµ ρ p ρ −ρν ρ p ρ+1 − p − 2 = rm 0 rm 0 − 2 C rm 0 λm 0 − 2 C r−m 0 λm 0 λ1 .
This shows that the subset of elements h(x) which satisfy the inequality (4.4) is a nonempty subset of W2−1 (Ω ) by virtue of the determination of rm 0 . Thus if we are searching for approximate solutions as in Section 3, repeating the same arguments we will obtain 1 that for arbitrary fixed u −m 0 ∈ Br−m 0 (0) ∩ E −m and for any h m 0 (x) ≡ Pm 0 h(x) ∈ E m−10 satisfying the following 0 inequality:
ρ+1
− ρν h m 0 , u m 0 + 2ρ C u −m 0 L (Ω ) λm 0 p rm 0 ≤ bm 0 rm2 0 , p
∀u m 0 ∈ Srm 0 (0) ∩ E m1 0
(4.6)
∞ 0 the sequence of approximate solutions u m 0 k k=1 are contained in a bounded subset of E m1 0 ⊂ W 12 (Ω ), more precisely in Brm 0 (0) ∩ E m1 0 . 0
Then we have obtained that u m 0 k * u m 0 weakly in E m1 0 (and in W 12 (Ω )) just as in the previous section from the 0 0 reflexivity of the spaces E m1 0 W 12 (Ω ) and L p (Ω ) and because in the conditions of the theorem W 12 (Ω ) ⊂ L p (Ω ) is compact, where u m 0 (x) is an element of E m1 0 . Consequently, u m 0 k → u m 0 strongly in E m0 0 (and also in L p (Ω )) and almost everywhere in Ω as k → +∞. From here it follows that the following relation is correct: −∆u m 0 k * −∆u m 0 weakly in E m−10 and in W2−1 (Ω ) , Pm∗ 0 |u k |ρ u k * Pm∗ 0 |u|ρ u weakly in E m−10 and in W2−1 (Ω ) , and, furthermore, Pm∗ 0 |u k |ρ u k → Pm∗ 0 |u|ρ u
strongly in L q (Ω ) q = p 0 under k → +∞.
So it is sufficient to repeat the arguments of the proof of Theorem 2 to finish the proof of the solvability for problem (4.2)(1) . Thereby the solvability of the problem (4.2)(1) is proved. Now we move on to the proof of uniqueness of the solution of this problem. For the proof, we assume the contrary. Therefore, first we show that the following sequences of inequalities are correct:
∗
P A (u) − P ∗ A (v) −1 ku − vk ≥ Pm∗ 0 (A (u) − A (v)) , u m 0 − vm 0 m0 m0 W2 (Ω )
2
= u m 0 − vm 0 − Pm∗ 0 |u|ρ u − |v|ρ v , u m 0 − vm 0
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K.N. Soltanov / Nonlinear Analysis: Hybrid Systems 2 (2008) 344–358
2
2 ρ u k L p (Ω ) u m 0 − vm 0 L (Ω ) ≥ u m 0 − vm 0 − (ρ + 1) ke p 2νρ
2
2 − ρ u k L p (Ω ) u m 0 − vm 0 ≥ u m 0 − vm 0 − C 2 (ρ + 1) λm 0 p ke
2 − 2νρ ρ ≥ u m 0 − vm 0 × 1 − 2ρ−1 C p (ρ + 1) λm 0 p rmρ 0 + r−m 0 . 0
1 for any u(x) and v(x) of Brm 0 (0) ∩ E m1 0 ⊕ Br−m 0 (0) ∩ E −m ⊂ W 12 (Ω ) when u −m 0 ≡ v−m 0 , where 0 e e u (x) ≡ e u m 0 (x) + u −m 0 (x), u m 0 (x) ∈ u m 0 , vm 0 ⊂ Brm 0 (0) ∩ E m1 0 .
We get, taking into account the values of rm 0 and r−m 0 , " !# 2νρ
∗
p ∗ −1 −1
P A (u) − P A (v) −1 . ≥ u m 0 − vm 0 × 1 − 2 1 − bm 0 (ρ + 1) 1 + λ1 λm 0 m0 m0 W (Ω ) 2
Hence we obtain the uniqueness of the solution using the conditions on bm 0 and u −m 0 (x) = v−m 0 (x). We have 1 proved that for each fixed u −m 0 ∈ Br−m 0 (0) ∩ E −m the problem (4.2)(1) is uniquely solvable in Brm 0 (0) ∩ E m1 0 0 under the conditions of Theorem 3. Thereby the unique solvability of problem (4.2)(1) in Brm 0 (0) ∩ E m1 0 for any h(x) ∈ W2−1 (Ω ) as h m 0 (x) = Pm 0 h(x) ∈ B R∗ m (0) ∩ E m−10 is proved. 0 With the help of the reasoning indicated for problem (4.2)(1) we shall obtain that problem (4.2)(2) is uniquely 0
−1 1 solvable in Br−m 0 (0) ∩ E −m ⊂ W 12 (Ω ) for any h(x) ∈ W2−1 (Ω ) for h −m 0 = Q ∗m 0 h ∈ E −m satisfying the inequality 0 0
ρ+1 − νρ 2 , h −m 0 , u −m 0 + 2ρ u m 0 L (Ω ) λm 0p r−m 0 ≤ bm 0 r−m 0 p
1 u −m 0 ∈ Sr−m 0 (0) ∩ E −m 0
for each fixed u m 0 ∈ Brm 0 (0) ∩ E m1 0 . Thus we show that problem (4.1) is uniquely solvable in 0
1 Brm 0 (0) ∩ E m1 0 ⊕ Br−m 0 (0) ∩ E −m ⊂ W 12 (Ω ) 0
for any h(x) ∈ W2−1 (Ω ) which satisfies the inequality from Theorem 3. The validity of the following statement is easily seen from the proof of Theorem 3. Corollary 5. Let all conditions of Theorem 3 be fulfilled. Then the problem (4.1) ((4.2)) is uniquely solvable in 0
1 Brm 0 (0) ∩ E m1 0 ⊕ Br−m 0 (0) ∩ E −m ⊂ W 12 (Ω ) for any 0 −1 h(x) ∈ B R∗ m (0) ∩ E m−10 ⊕ B R∗ −m (0) ∩ E −m ⊂ W2−1 (Ω ), 0 0
0
i.e. h(x) ≡ Pm∗ 0 h + Q ∗m 0 h
and Pm∗ 0 h ∈ B R∗ m (0) ∩ E m−10 , 0
−1 Q ∗m 0 h ∈ B R∗ −m (0) ∩ E −m . 0 0
5. Some remarks From the proof of Theorem 3 it follows that a statement similar to Corollary 5 is valid under the condition m = 1, namely: Corollary 6. Let the conditions of the Theorem 3 be fulfilled. Then the problem (4.1) is uniquely solvable in 0
1
Br1 (0) ⊂ W 12 (Ω ) for any h(x) ∈ B R∗ 1 (0) ⊂ W2−1 (Ω ), where R1 ≡ b1 (1 − b1 ) ρ C and b1 is a fixed number of 2−1 , 1 .
− ρp ν λ1
1
and r1 ≡ (1 − b1 ) ρ C
− ρp ν λ1 ,
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K.N. Soltanov / Nonlinear Analysis: Hybrid Systems 2 (2008) 344–358 j
j
Now we shall lead to remarks which are associated with the use of spaces E m , E −m ( j = −1, 0, 1). This allows us to obtain additional information about the problems considered. P In fact, we have from the definition of u −m (x) that u −m (x) ≡ m−1 k=1 ak wk where ak are unknown coefficients and Pm−1 ek . If w ek = wk then (4.2)(2) has the form h −m ≡ k=1 bk w ρ ! m−1 ∞ ∞ m−1 X X X X λk ak wk − Q ∗m ak wk ak wk = bk wk , k=1 k=1 k=1 k=1 which generates a system of algebraic equations for determining the unknown coefficients ak , k = 1, m − 1. (It is 0
1 known that [5], since {wk }∞ k=1 is a basis in W 2 (Ω ) (and in L 2 (Ω )) and as these spaces satisfy the corresponding −1 conditions by construction, it is possible to define a biorthogonal basis {e w k }∞ k=1 in W2 (Ω ).) We shall consider the case when problem (1.1) has functional nonlinearity since we can give a better illustration in this case. To be exact, we consider the case when f (u) = kuk2L 2 (Ω ) u, i.e. ρ = 2. Thus, we consider the problem
−∆u − kuk2L 2 (Ω ) u = h(x),
x ∈ Ω ⊂ Rn ,
(5.1)
u |∂ Ω = 0,
(5.2)
and we shall use Corollary 5. In order to make use of Corollary 5 we consider the representation of the type (4.2) for problem (5.1) and (5.2): (
Pm∗ A (u) ≡ −∆u m − Pm∗ kuk2L 2 (Ω ) u = h m (x) ≡ Pm∗ h(x), (1) Q ∗m A (u) ≡ −∆u −m − Q ∗m kuk2L 2 (Ω ) u = h −m (x) ≡ Q ∗m h(x), (2)
u |∂Ω = 0.
(5.3)
e > 0 for some number R. e Then if we take Rm ≥ R, e there Now let h(x) ∈ W2−1 (Ω ) be preassigned and khkW −1 = R 0 2
0
exists m 0 such that the problem (5.3)(1) in the conditions indicated is uniquely solvable in Brm 0 (0) ∩ E m1 0 ⊂ W 12 (Ω ) 0
1 for any fixed u −m 0 (x) ∈ Br−m 0 (0) ∩ E −m ⊂ W 12 (Ω ) by virtue of Theorem 3 (or Corollary 5). 0 In this case we obtain for problem (5.3)(2) under the same conditions the following equation which is similar to the equation stated above: ! m −1 mX mX ∞ 0 −1 0 0 −1 X X 2 2 λk ak wk − ak kwk k L 2 (Ω ) ak wk = bk wk , (wk (x) = w ek (x)) . k=1
k=1
k=1
k=1
If we assume kwk k L 2 (Ω ) = 1 then we obtain the system of algebraic equations λk a k −
∞ X
! ak2
a k = bk ,
k = 1, 2, . . . , m 0 − 1.
(5.4)
k=1
e which, generally Further, we need to study the system (5.4)(2) for h −m 0 = Q ∗m 0 h in the case when R−m 0 ≥ R, speaking, does not satisfy the conditions of Corollary 3. Therefore we need to consider the operator A generated by 0
the system (5.4) and must investigate Q m 0 A on the ball Br (0) ⊂ W 12 (Ω ), namely in the case when u −m 0 (x) belongs 0
1 to Br (0) ∩ E −m ⊂ W 12 (Ω ) for various numbers r > 0. 0 2 Let 0 < r ≤ rm 0 ; then we obtain that there exists k0 : 1 ≤ k0 ≤ m 0 − 1 such that λk0 ≤ λ−1 1 r , where ∞ X k=1
ak2 ≡ kuk2L 2 (Ω ) ,
−1
kuk L 2 (Ω ) ≤ λ1 2 kuk .
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K.N. Soltanov / Nonlinear Analysis: Hybrid Systems 2 (2008) 344–358
It is clear that, for such λk , the statement on the unique solvability of the problem considered does not apply. In fact, it is enough to consider the following part of the system (5.4)(2) : ! ∞ X λk − a 2j ak − ak3 = bk , k = 1, 2, . . . , k0 . j=1, j6=k
Each equation in this system can have three solutions. If the system (5.4)(2) is solvable, the number of solutions in 2 e the ball indicated depends on the number k0 : λk0 ≤ λ−1 1 r which in turn depends on R > 0 and ρ ≥ 0 (ρ = 2 in this case). Thus we obtain the following proposition. Proposition 2. Let all conditions of the problem considered be fulfilled. If this problem is solvable in the case considered, then the number of solutions will be increased with magnification of the number m 0 (and consequently e and ρ ≥ 0. rm 0 ) which depends on R Since many works (see, in particular, [2,10,3,6] and their references) are devoted to the study of the number of solutions of such problems and in these works various results on numbers of solutions are obtained, we will not continue these discussions. Let us just remark that the indicated reasoning rather simply illustrates the local connection between the number of solutions and the diameter of the considered domain from the range of definition of the operator generated for the problem considered. References [1] O.V. Besov, V.P. Il‘in, S.M. Nikolskiy, Integral Representations of Functions and Imbedding Theorem, M., Nauka, 1996. [2] H. Brezis, Points critiques dans les problemes variationnels sans compacite, in: Sem. Burbaki 1987–88, in: Asterisque, vol. 161–162, 1988, pp. 239–256. n. 698. [3] R.D. Benguria, J. Dolbeault, M. Esteban, Classification of the solutions of semilinear elliptic problems in a ball, J. Differential Equations 167 (2000) 2. [4] J. Diestel, Geometry of Banach Spaces. Selected Topics, Springer-Verlag, Berlin, Heidelberg, 1975. [5] N. Dunford, J.T. Schwartz, Linear Operators, Part I : General Theory, Interscience Publishers, New York, London, 1958. With the assistance W.G. Bade, R.G. Bartle. [6] F. Huang, Existence of two solutions nonlinear elliptic equations with critical Sobolev exponents and mixed boundary conditions, in: Proceed. Royal Soc. Edinburgh, vol. 126A, n. 1, 1996. [7] E. Hille, R.S. Phillips, Functional Analysis and Semi-groups, vol. 31, AMS, Colloq. Publ., 1957. [8] J.-L. Lions, Quelques methodes de resolution des problemes aux limites non lineares, Dunod, Gauthier-Villars, Paris, 1969. [9] L. Nirenberg, Topics in nonlinear functional analysis, New York, 1974. [10] S.I. Pohozhaev, On equations of the form ∆u = f (x, u, Du), Math. Sb. 113 (2) (1980). [11] H.H. Schaefer, Topological Vector Spaces, The Macmillan Company, Collier–Macmillan Limited, London, New York, 1966. [12] K.N. Soltanov, To the theory of normal solvability of nonlinear equations, Dokl. Ac. Sci. USSR 278 (1) (1984). [13] K.N. Soltanov, Solvability of nonlinear equations with operators of the form of the sum of a pseudomonotone and weakly compact operator, Russian Ac. Sci Dokl. Math. 45 (3) (1992). [14] K.N. Soltanov, Some Applications of the Nonlinear Analysis to the Differential Equations, ELM, Baku, 2002, p. 292. [15] K.N. Soltanov, On nonlinear equations with continuous operators in reflexive Banach spaces, Funct. Anal. Appl. 33 (1) (1999). [16] E. Zeidler, Nonlinear functional analysis and its applications II/b, in: Nonlinear Monotone Operators, Springer-Verlag, Berlin, 1990.
On noncoercive semilinear equations Kamal N. Soltanov 1 F. Agaev Street 9, Institute of Mathematics and Mechanics of National Academy of Sciences of Azerbaijan, Baku, Az 1141, Azerbaijan Received 14 April 2006; accepted 17 April 2006 Dedicated to Prof. Dr. V. Lakshmikantham on the occasion of his 80th birthday.
Abstract In this article some noncoercive semilinear equations are investigated. Here, a subset of a right side for which the corresponding boundary value problems are uniquely solvable is described. c 2006 Elsevier Ltd. All rights reserved.
Keywords: Semilinear equations; Noncoercive operator; Biorthogonal system; Solvability theorem; Eigenfunction; Uniqueness
1. Introduction We consider the problem −∆u − f (u) = h(x),
x ∈ Ω ⊂ Rn , u |∂ Ω = 0,
(1.1)
where Ω ⊂ Rn , n ≥ 1, is a bounded domain with sufficiently smooth boundary ∂Ω , and f is a continuous mapping such that the problem (1.1) generates a noncoercive operator A: 0
A : W2` (Ω ) ∩ W 12 (Ω ) −→ W2`−2 (Ω ),
(` ≥ 1)
0
(in particular, A : W 12 (Ω ) → W2−1 (Ω )). Here W2` (Ω ), ` ≥ 1, is the Sobolev space which is defined as ( ) Z X α 2 2 ` D u dx < ∞, α = (α1 , . . . , αn ) , W2 (Ω ) ≡ u(x) kukW ` = 2 Ω |α|≤` 0
W 12 (Ω ) ≡ W21 (Ω ) ∩ {u(x) |u|∂ Ω = 0} ,
D α ≡ D1α1 D2α2 · · · Dnαn ,
Di ≡
∂ , ∂ xi
∗ 0 and W2−1 (Ω ) ≡ W 12 (Ω ) [1]. E-mail addresses: [email protected], [email protected]. 1 Current address: Department of Mathematics, Faculty of Sciences, Hacettepe University, Beytepe, Ankara, TR-06532, Turkey. c 2006 Elsevier Ltd. All rights reserved. 1751-570X/$ - see front matter doi:10.1016/j.nahs.2006.04.010
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We recall that A : X → X ∗ is a coercive operator if the following relation is valid for a dual form h·, ·i with respect to the pair of spaces (X, X ∗ ): hA(x), xi % ∞, kxk X
under kxk X % ∞,
where X is a Banach space and X ∗ is its dual (see [8]). And A : S ⊆ X → Y is a “coercive” operator in a generalized sense if there exists a bounded operator B : X 0 ⊆ S −→ Y ∗ that satisfies certain continuity conditions, X 0 = S, Im B = Y ∗ and a continuous function µ : R+ → R non-decreasing for τ ∈ [τ0 , Υ0 ] ⊆ R+ such that the following relation is valid for a dual form h·, ·i with respect to the pair of spaces (Y, Y ∗ ): hA(x), B(x)i ≥ µ ([x] S ) ,
for x ∈ X 0 and ∃r > 0 : µ (r ) ≥ 0
where S is an pn-space with a p-norm [•] S (see [12,14]). In this case it is said that the mappings A and B generate a “coercive” pair on X 0 . 0
Let the mapping f be such that the operator A generated by the problem (1.1) is acting from W 12 (Ω ) to W2−1 (Ω ). It is known ([1,10,12] etc.) that if A is noncoercive operator then, generally speaking, the problem (1.1) is not solvable for arbitrary right hand side unlike in the coercive case (in particular, in this case the problem is normally solvable). Moreover, if the problem (1.1) is solvable then the solution may be non-single (and usually not unique); furthermore there may be an infinite number of solutions ([10,9,6,3,15] etc.). It should be noted that some parts of the results relating to this problem were published previously in the paper [14]. Here some properties of noncoercive problems are investigated. And for the problems considered it is shown that there exist subsets M from the image of the operator A defined by the problem such that for any h of M the problem (1.1) is uniquely solvable in corresponding subsets from the domain. To be more exact, it is shown that there exists a 0
subset G of W 12 (Ω ) on which A : G −→ A (G) ⊂ W2−1 (Ω ) is injective. At beginning the problem (1.1) with nonlinearity of functional type is studied, and after that the one with nonlinearity of exponential type is studied. 2. Some general results At beginning some general results are given on which the proofs of solvability of the problems considered are based. Let X, Y be Banach spaces and X ∗ , Y ∗ be their dual spaces; let S be a weakly complete pn-space (see [12,13], where other variants of the following theorem are given). Let X 0 be a topological space such that X 0 ⊆ S ⊆ X , and let A be a nonlinear mapping acting from X to Y , where Y is a reflexive space such that both Y and Y ∗ are strictly convex. We consider the equation A(x) = y,
y ∈ Y.
(2.1)
This equation is considered in the case where the operator A is not coercive in the generalized sense. But it can be investigated locally as in [14], i.e. in the sense described in this section. It is known that in the general case a definition of a solution of the equation A(x) = y has the form [13]: Definition 1 (Generalized Solution). Let M ∗ be a subset of Y ∗ and y be an element of Y . Then an element x ∈ S satisfying hA(x), y ∗ i = hy, y ∗ i , ∀y ∗ ∈ M ∗ ⊂ Y ∗ , is called an M ∗ -solution of equation A(x) = y. We will consider the following conditions: (1) A : S −→ Y is a weakly continuous mapping and there exists a closed linear subspace Y1 of Y such that A : S −→ Y1 ⊂ Y ; (2) there exists a mapping B : X 0 −→ Y ∗ such that Im B contains a linear manifold from Y ∗ which is dense in a closed linear subspace Y1∗ of Y ∗ and generates a “coercive” pair with A on X 0 in a generalized sense; moreover one of the following conditions ((α) or (β)) holds: (α) if B is a linear continuous operator then S is a “reflexive” space (see [12,13]), X 0 is a separable vector topological space which is dense in S and ker B ∗ = {0};
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(β) if B is a nonlinear operator then Y1∗ is a separable subspace of Y ∗ and B −1 is weakly continuous from Y ∗ to S. Theorem 1. Let conditions (1) and (2) hold. Then Eq. (2.1) is Y1∗ -solvable in S for every y ∈ Y1 satisfying the following condition: There exists r > 0 such that µ ([x] S ) ≥ hy, B(x)i ,
for ∀x ∈ X 0 with [x] S ≥ r,
(2.2)
where h·, ·i is the dual form of the pair (Y, Y ∗ ). Notation 1. It is clear that the Eq. (2.1) is equivalent to the following functional equation:
A(x), y ∗ = y, y ∗ , ∀y ∗ ∈ Y ∗ . It should be noted that if the space Y satisfies the above-stated conditions then Y1∗ ≡ (Y1 )∗ ; see Proposition 1. From here we obtain that the solution in the general sense which is the Y1∗ -solution of Eq. (2.1) is actually the solution in the usual sense. Consequently, in this case, the solution of the equation A(x) = y for any y ∈ Y1 is the solution in the general sense and vice versa. 2 Proof (Short Scheme k ∞). Let the conditions (1) and (2) (α) be satisfied. We will use the Galerkin approximation method. So, as usual let x k=1 be a complete system in the (separable) space X 0 . We then look for approximate solutions P k in the form xm = m k=1 cmk x . Here cmk are unknown coefficients which have to be determined from the system of algebraic equations
8k (cm ) := hA (xm ) , Bxk i − hy, Bxk i = 0,
k = 1, 2, . . . , m
(∗)
where cm ≡ (cm1 , cm2 , . . . , cmm ). Now we observe that the mapping 8 (cm ) := (81 (cm ) , 82 (cm ) , . . . , 8m (cm )) is continuous by virtue of condition 1. From (2.2) it follows that there exists r > 0 such that the “acute angle” condition is satisfied for all xm with [xm ] S ≥ r , i.e. the following inequality holds for any x ∈ Sr1 (0) ⊂ R m , r1 ≥ r : m X
h8k (cm ), cmk i ≥ 0,
∀cm ∈ Rm , kcm kRm = r1 .
k=1
Then the solvability of the system (∗) follows from the well-known lemma on the “acute angle” which is equivalent to Brouwer’s fixed-point theorem (see Lemma 2 and, e.g., [8,9,15]). Thus we obtain a sequence which is contained in a bounded subset of S. Further arguments are analogous to the arguments from [13,14]; therefore we omit them. Remark 1. It is obvious that if there exists a function ψ ∈ C 0 such that ψ (ξ ) = 0 ⇐⇒ ξ = 0 and if the following inequality is fulfilled: kx1 − x2 k X ≤ ψ (kA (x1 ) − A (x2 )kY ) for all x1 , x2 ∈ S, then a Y1∗ -solution of Eq. (2.1) is unique. The following corollary follows from this theorem. Corollary 1. Let the conditions (1) and (2) hold, and the operators A and B generate a “coercive” pair in the generalized sense on a space X 0 ; furthermore let µ be such that µ (τ ) % +∞ as τ % +∞, and let Y be a reflexive Banach space which, together with its dual space Y ∗ , has a strictly convex norm, and say there exists a ) function ϕ1 : R+ −→ R+ , ϕ1 ∈ C 0 such that ϕµ(τ % +∞ as τ % +∞; then for every y ∈ Y1 such that 1 (τ ) n o |hy,B(x)i| sup ϕ1 ([x] S ) x ∈ X 0 < +∞, Eq. (2.1) is solvable in S. The problem considered can be solved by applying the above theorem. But we shall construct the proofs of the solvability theorems for the problems considered without directly applying the general theorem, for better 2 For the proof see also [13,14].
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visualization of the proof of existence and uniqueness. The general concept of a solution coincides with the usual concept of a solution in the case considered, by virtue of a result which was proved in the article “Remarks on separation of convex sets, fixed-point theorem and applications in theory of linear operators” (which will be published).3 3 Proposition 1. Let X and its dual space X ∗ be strictly convex reflexive Banach spaces, and let X 0 ⊂ X be a subspace of X . Then the dual space of a subspace X 0 ⊂ X is equivalent to a subspace of X ∗ which is determined by the subspace X 0 ; i.e. X ∗ has a subspace X 0∗ ⊂ X ∗ defined by the unit sphere of the subspace X 0 such that (X 0 )∗ ≡ X 0∗ . Consequently, X 0 and its dual space (X 0 )∗ are strictly convex reflexive Banach spaces under the norms induced by the norms of X and X ∗ , respectively. Proof. It is known from [4,5,7] that the dual space of the subspace X 0 ⊂ X is equivalent to a factor (quotient) space of the form X ∗ / X 0⊥ , where X 0⊥ ⊂ X ∗ is the annihilator of X 0 ⊂ X : X 0⊥ ≡ x ∗ ∈ X ∗ |hx, x ∗ i = 0, ∀x ∈ X 0 . (Here the expression h·, ·i denotes the dual form for the pair X, X ∗ , or an inner product if X is a Hilbert space.) It is also known from [11,16] that the subspace X 0 ⊂ X is a reflexive Banach space under the norm induced from X and that its dual space (X 0 )∗ is also reflexive. Moreover, if X is a strictly convex reflexive Banach space then so is X 0 . In addition, by [4], if X is a strictly convex reflexive Banach space then an arbitrary element of the unit sphere is an extremal point and the dual space (X 0 )∗ is equivalent to a subspace of X ∗ . It remains, therefore, to identify this subspace. = In order to construct a subspace dual to X 0 we will consider the duality mapping = : X −→ X ∗ for the pair X ; X ∗ , i.e. X ←→ X ∗ (see [4,8, 16] and the references therein). In the case under consideration, the duality mapping is bijective and, together with its inverse mapping, is strictly
monotone, surjective, odd, demicontinuous, bounded and coercive. Hence we have x, x ∗ ≡ x ∗ , x for any x ∈ X , x ∗ ∈ X ∗ , and in particular for any x ∈ X we have x ←→ x ∗ = =(x), i.e. it is an equivalence relation [4,8,16]. It follows from this that it will be enough to consider these mappings on the unit spheres of X and X ∗ . ∗ ∗ We shall denote the unit spheres of X and X ∗ by S1X and S1X , respectively. Then we have = S1X ≡ S1X . In addition, the following relations hold: ∗
(∀x)(x ∈ S1X ⇐⇒ =(x) = x ∗ ∈ S1X ).
(∀x)(x ∈ S1X ⇐⇒ hx, =(x)i = x, x ∗ = kxk X · x ∗ X ∗ = 1 · 1), and conversely D E
∗ (∀x ∗ )(x ∗ ∈ S1X ⇐⇒ x ∗ , =−1 x ∗ = x ∗ , x = x ∗ X ∗ · kxk X = 1 · 1), since the duality mapping is a homeomorphism, and by virtue of the conditions of the proposition (see [8,16] and the references therein). Moreover the following relation holds: ∀x ∈ X,
∃e x ∈ S1X
so that x = e x kxk X .
Hence we have that the unit sphere S1X defines the whole space X in the sense that X ≡ S1X × R+ ; R+ ≡ {τ ∈ R : τ ≥ 0}. X Hence, if X 0 ⊂ X is a subspace of X then X 0 can be defined through a subset of the unit sphere of the form S1 0 ≡ S1X ∩ X 0 ≡ S1X (0)∩ X 0 . Here, X0 S1 denotes the unit sphere of X 0 with the norm induced from X . Thus the space X 0 is a strictly convex reflexive Banach space. Consequently there exists an equivalent norm such that X 0 , together with its dual space, is a strictly convex reflexive Banach space. Under the induced topology X – which we obtain by virtue of the duality mapping = from X onto X ∗ – the sphere S1 0 will be transformed onto a subset which can be expressed in the form ∗ X e S1∗ ≡ {x ∗ ∈ S1X | h=−1 (x ∗ ), x ∗ i = kxk X · kx ∗ k X ∗ = 1, =−1 (x ∗ ) = x ∈ S1 0 }, because we have ∗ X X =−1 e S1∗ ≡ S1 0 ; ∀x ∗ ∈ e S1∗ ⊂ S1X ⇐⇒ x = =−1 x ∗ ∈ S1 0 .
(2.3)
It is known that if X and X ∗ are strictly convex reflexive spaces, then the duality mapping = : X X ∗ : =−1 is the Gateaux-differential of =
a strictly convex functional z and =−1 is the Gateaux-differential of a strictly convex functional z∗ , i.e. the duality mapping X ←→ X ∗ is a positively homogeneous potential operator with strictly convex potential. In addition, there is a strongly monotone increasing continuous function
∗ Φ : R+ −→ R+ , Φ (0) = 0, Φ (τ ) % +∞ when τ % +∞ such that = (τ x) = Φ (τ ) x ∗ for any x ∈ S1X and x ∗ ∈ S1X , where x, x ∗ ≡ 1 X S ∗ defined by (2.3) is the unit sphere of the and τ ∈ R+ [8]. Consequently, = B 0 is a convex subset X ∗ (see also [16]). Thus we obtain that e 1
1
subspace (X 0 )∗ from X ∗ , which we can denote by X 0∗ (i.e. (X 0 )∗ ≡ X 0∗ ) that also is equivalent to X ∗ / X 0⊥ . In other words we have obtained that X 0∗ is equivalent to the dual space of the subspace X 0 of X , and so a subspace X 0 of a reflexive Banach space X is a reflexive Banach space under the induced topology under the conditions of the proposition.
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3. Solvability of the problem (1.1) in the case of functional nonlinearity Let Ω ⊂ Rn , n ≥ 1, be a bounded domain with sufficiently smooth boundary ∂Ω . Let the eigenfunctions of the problem −∆wk = λk wk ,
wk |∂ Ω = 0,
(3.1)
generate an orthogonal basis in space L 2 (Ω ). We denote a system of eigenfunctions of problem (3.1) by {wk (x)}∞ k=1 where wk (x) corresponds to the eigenvalue λk . (Usually an eigenvalue of problem (3.1) is considered in the form λk = µ2k .) We will arrange eigenvalues in the order of growth, i.e. 0 < λ1 ≤ λ2 ≤ · · ·. Further we denote a subspace of the space L 2 (Ω ) with codimension 0
m − 1 ≥ 0 generated by elements {wk (x)} for k ≥ m by E m0 and the corresponding subspace of W12 (Ω ) by E m1 . Now we will consider 0 W 12 (Ω ) ≡ W21 (Ω ) ∩ u(x) | u |∂ Ω = 0 , with a norm which is equivalent to the previous one and is defined by
2
2 kuk2 ≡ kuk20 ≡ n D j u L (Ω ) ≡ n D j u 2 . W 12 (Ω )
2
j=1
j=1
0
i (i = 0, 1) are subspaces of L (Ω ) and As E m W 12 (Ω ) respectively, with induced topologies generated by a norm in 2 the corresponding space, we have that E m−1 is a dual space of E m1 and it is a subspace of W2−1 (Ω ) (see, for example, [8, 5,16]) with the codimension m − 1 (i.e. codim E m−1 = m − 1).
Note. It follows from Proposition 1 that the dual space of E m1 is a space E m−1 which is a subspace of the space W2−1 (Ω ). Lemma 1. Let X be a strictly convex reflexive Banach space together with its dual space X ∗ . If X is a space with a basis and X 1 is a subspace of X with codimension m (i.e. codim X 1 = m) then X 1∗ is the dual space of X 1 , and X 1∗ is a subspace of X ∗ with the same codimension (i.e. codim X 1∗ = m). This assertion follows from the Proposition 1 and results devoted to the properties of biorthogonal bases (see, for example, [5, Chapter 2 & 6] etc.). Moreover since the space considered is a Hilbert space we obtain immediately that this property remains. Thus the spaces scale framed on L 2 (Ω ) are transferred to spaces scale framed on E m0 (subspace). Therefore, it is possible to define the biorthogonal base in E m−1 . So, we consider the problem (1.1) −∆u − f (u) = h(x),
x ∈ Ω ⊂ Rn , u |∂ Ω = 0
and assume that the following condition holds: 0
(ι) the mapping f : W 12 −→ W2−1 from the problem (1.1) is such that f : E m0 −→ E m−1 , f ∈ C 0 , as f : E m1 −→ E m−1 , f ∈ C 1 and for any u ∈ E m1 the following inequalities hold:
0
−1
ρ ρ+1 k f (u)kW −1 ≤ c1 λm 2 kuk2 , ≤ c0 λ−1
f (u) 0 1 m kuk2 , −1 W 2 −→W2
2
where ρ, c1 ≥ 0 are constants, and c0 = c0 (ρ) > 0. (It is clear that if c0 = 0 then the mapping f does not depend on u; therefore we assume that c0 = c0 (ρ) > 0.) Remark 2. The assumption that f : E m0 → E m−1 ,
f ∈ C 0 (is continuous) in condition (ι) may be superfluous.
For example the operator f may have the following form: Z ρ 2 ρ 2 |u| dx u, a > 0 − const. f (u) ≡ a kuk2 u ≡ a × Ω
K.N. Soltanov / Nonlinear Analysis: Hybrid Systems 2 (2008) 344–358
349
This operator satisfies the condition (ι). In fact for any u ∈ E m1 we have
−1 ρ+1 ρ ρ k f (u)kW −1 = a kuk2 u W −1 = a kuk2 kukW −1 ≤ aλm 2 kuk2 . 2
2
2
− 21
Also we have kukW −1 ≤ λm kuk2 for any u ∈ E m1 . 2
0
W 12 (Ω )
For any u ∈ we have Z Z ∆u u dx = ∇u(x)∇u(x)dx = kuk20 . − Ω
W 12
Ω
0
From here we obtain that, if the λi ’s are eigenvalues of the operator −∆ : W 12 (Ω ) −→ W2−1 (Ω ) and λ1 = min {λi |i ∈ N }, then 1
kuk20 ≥ λ1 kuk22 ,
i.e. kuk 0 1 ≥ λ12 kuk2 .
W 12
W2
Now, if an u ∈ E m1 is an arbitrary element, then we have 1
−1
kuk 0 1 ≥ λm2 kuk2 H⇒ λm 2 kuk 0 1 ≥ kuk2 . W2
W2
0
Further it is well known that the operator −∆ : W 12 (Ω ) −→ W2−1 (Ω ) is an isomorphism. Then for any 0
v(x) ∈ W2−1 (Ω ) there exists a function ψ(x) ∈ W 12 (Ω ) such that −∆ψ(x) = v(x), and ψ(x) = (−∆)−1 v(x). Consequently, from the definition we have 0 kvkW −1 = sup hv, gi | g ∈ S1 (0) ⊂ W 12 (Ω ) 2
0
for any v(x) ∈ W2−1 (Ω ). Let a function ψ(x) ∈ W 12 (Ω ) be such as the above: hv, ψi =
h−∆ψ(x), ψi = h∇ψ, ∇ψi H⇒ kvkW −1 = k∇ψk2 = kψk 0 1 2
1
H⇒ (−∆)− 2
W2
−1
v ≤ λ1 2 kvk2 . 2
−1
Consequently we have kukW −1 ≤ λm 2 kuk2 for any u ∈ E m1 . 2
Notation 2. Let the system of eigenfunctions {wk }∞ k=1 of the problem (3.1) be an orthogonal basis in L 2 (Ω ), where wk (x) is the eigenfunction corresponding to the eigenvalue λk . Let the eigenvalues λk be arranged in order of growth, 1 i.e. 0 < λ1 ≤ λ2 ≤ · · ·. Then, from the above reasoning, we will obtain that the subspace P∞E m is generated by this ∞ 1 system over the basis {wk }k=m , m ≥ 1, and consequently for any u ∈ E m we have u(x) = k=m u k wk (x). Moreover we have ∞ ∞ X X −∆u(x) = −∆ u k wk (x) = λk u k wk (x). k=m
k=m
Consequently, h−∆u(x), u(x)i =
∞ X
λk u 2k kwk (x)k22 ≥ λm kuk22 ,
k=m
as λm is a minimum of λk at k ≥ m. We interpret the solution of problem (1.1) in a somewhat general sense as in Section 2. Therefore using Proposition 1 this definition for problem (1.1) can be formulated as follows.
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Definition 2. A function u(x) ∈ E m1 is called an E m1 -solution of the problem (1.1) if u(x) satisfies Eq. (1.1) in the sense of the space E m−1 , i.e. the equality Z Z Z hA (u) , vi ≡ h(x)v(x)dx, (3.2) f (u(x)) v(x)dx = ∇u(x)∇v(x)dx − holds for any v(x) ∈
Ω
Ω
Ω
E m1 ,
where the operator A is generated by problem (1.1).
Remark 3. As is known, E m1 is a Hilbert space with dual space E m−1 and when h(x) is contained in E m−1 , the solvability of problem (1.1) in the usual sense follows from the definition. 0 0 1 Now, let Br (0) ≡ u(x) ∈ W 2 (Ω ) | kuk ≤ r be the closed ball of W 12 (Ω ) and Sr (0) ≡ 0 u(x) ∈ W 12 (Ω ) | kuk = r be the sphere (i.e. the boundary of the ball Br (0)). Theorem 2. Let all above-mentioned conditions of problem (1) be fulfilled, let c1 = 1, and let b be a fixed number from (c0 , 2c0 ), and b0 = min b−1 , 2−1 . Then for each integer m ≥ 1 there exists an integer m 0 ≥ 1 such that for any h(x) ∈ E m−10 satisfying the inequality hh, ui ≤ hA (u) , ui ≡ h−∆u − f (u) , ui ,
∀u ∈ Srm 0 (0) ∩ E m1 0 ,
(3.3) 0
1
ρ+2
the problem (1) is uniquely solvable on the ball Brm 0 (0) ∩ E m1 0 ⊂ W 12 (Ω ) for rm 0 = (1 − b0 ) ρ λm2ρ0 , where m 0 = inf {e m |λm e = λm }. (Here for simplicity we assume that c1 = 1.) Remark 4. It is known that λm % +∞ when m % +∞, so we have rm 0 and the corresponding norm of h(x) ∈ E m−10 can be chosen sufficiently large; m 0 = m 0 (m). Proof. First of all, we shall show that the set of the right side — h(x) defined by (3.3) — is a nonempty subset of 0
E m−10 . In fact, for any u(x) ∈ Brm 0 (0) ∩ E m1 0 ⊂ W 12 (Ω ) and for any m 0 ≥ 1 we have Z Z hA (u) , ui ≡ A (u(x)) u(x)dx ≡ [−∆u − f (u)] u(x)dx = kuk2 − h f (u) , ui Ω
Ω
≥ kuk2 − k f (u)kW −1 kuk , 2
and taking into account the first inequality for f from condition (ι) we obtain h i −1 ρ+1 ρ kuk hA (u) , ui ≥ kuk2 − λm 02 kuk2 kuk ≥ kuk2 1 − λ−1 m0 2 . Further we will consider this inequality for the functions u(x) from Srm 0 (0) ∩ E m1 0 as in the condition of the theorem. Then we obtain the inequality using the condition (i), selecting u(x) from Srm 0 (0) ∩ E m1 0 and using kuk2 −1
≤ λm 02 kuk (as c1 = 1) −1− ρ2 −1− ρ2 ρ ρ 2 hA (u) , ui ≥ kuk 1 − λm 0 kuk ≥ rm 0 1 − λm 0 rm 0 . 2
Hence we obtain by substituting the value rm 0 from the theorem 1 1 ρ 1 + −1− ρ2 2 +1 hA (u) , ui ≥ rm 0 1 − λm 0 (1 − b0 ) λm (1 − b0 ) ρ λmρ 0 2 1
1
+1
= rm 0 b0 (1 − b0 ) ρ λmρ 0 2 . The inequality obtained (3.4) shows that the subset of right sides satisfying the inequality (3.3) is nonempty.
(3.4)
K.N. Soltanov / Nonlinear Analysis: Hybrid Systems 2 (2008) 344–358
351
Now, we shall proceed with the proof of the theorem. For the proof we use a variant of the Galerkin method. We k will search for an approximate solution as an element of the linear variety generated by {wi (x)}i=m , i.e. in the form 0 Pk u k (x) = i=m 0 aki wi (x), where aki (i = m 0 , m 0 + 1, . . . , k) are unknown coefficients which will be determined from the system of algebraic equations hA (u k ) , wi i = hh, wi i ,
i = m 0 , m 0 + 1, . . . , k.
(3.5)
The solvability of system (3.5) follows from a known result, which is equivalent to a variant of the Brouwer theorem on a fixed point [8,9,15] and which can be formulated as follows: Lemma 2 ([8]). Let the mapping F acting on Rn (n ≥ 1) satisfy the following conditions on the closed ball Br (0) for some number r > 0: (α) F is continuous; (β) the inequality hF(x), xi ≥ 0, holds for any x ∈ Sr (0). Then there exists a point x0 ∈ Br (0) such that F (x0 ) = 0. If we denote the mapping generated by problem (3.5) in the form F (ak ) ≡ Fm 0 (ak ) , Fm 0 +1 (ak ) , . . . , Fk−m 0 (ak )
k where F j (ak ) ≡ A (u k ) − h, wm 0 + j , and ak ≡ {aki }i=m , then obviously F is continuous on R k−m 0 +1 and from 0 inequality (3.4) it follows that the condition (β) holds. Therefore we obtain the solvability of the system (3.5) for each k : k = m 0 , m 0 + 1, . . . by applying Lemma 2. From here we can say that all approximate solutions u k (x) are contained in the ball Brm 0 (0) ∩ E m1 0 . Thus we obtain a sequence of approximate solutions {u k (x)} of problem (1.1) which is contained in a bounded subset of E m1 0 . Therefore it is possible to select a subsequence of the sequence {u k (x)} weakly converging in E m1 0 by virtue of the reflexivity of the E m1 0 (for any m 0 = 0, 1, . . .) which we shall designate through {u k (x)}, for simplicity. 0
Then u k * u weakly in E m1 0 (and in W 12 (Ω )), where u(x) is an element of E m1 0 . Consequently u k → u strongly in E m0 0 (also in L 2 (Ω )) and almost everywhere in Ω when k → +∞. It is clear that ∆u k * ∆u
weakly in E m−10 (and in W2−1 (Ω )) as k → +∞.
Now it is necessary to show that weakly in E m−10 (and in W2−1 (Ω )) as k −→ +∞.
f (u k ) * f (u)
This is true, since f is a continuous mapping from E m0 0 in E m−10 and u k → u strongly in E m0 0 (also in L 2 (Ω )) and almost everywhere in Ω . Thus we can pass to the limit in (3.5) for k → +∞ for each i = m 0 , m 0 + 1, . . . . Then passing to the limit in (3.5) for k → +∞ we have hA (u) , wi i = hh, wi i ,
i = m 0 , m 0 + 1, . . . .
∞ We obtain the solvability of problem (1.1) taking into account the completeness of the system {wi (x)}i=m in E m1 0 ; 0 1 i.e. the function u(x) satisfies the equality (3.2) for any v(x) of E m 0 . From here we obtain the first part of the statement of the theorem by virtue of Remark 3. Now to complete the proof of the theorem it is necessary to show the uniqueness of the solution. For this purpose it is sufficient to prove the following inequality for any u(x), v(x) ∈ Brm 0 (0) ∩ E m1 0 : kA (u) − A (v)kW −1 ≥ K ku − vk , K = K rm 0 , λm 0 , ρ > 0, (3.6) 2
where K is some number which is independent of u(x) and v(x). We set w(x) = u(x) − v(x). Then Z kA (u) − A (v)kW −1 kwk ≥ [A (u) − A (v)] w(x)dx 2
Ω
= |hA (u) − A (v) , w(x)i| = |h−∆w − f (u) + f (v) , wi|
352
K.N. Soltanov / Nonlinear Analysis: Hybrid Systems 2 (2008) 344–358
≥ kwk2 − |h f (u) − f (v) , wi| ≥ kwk2 − k f (u) − f (v)kW −1 kwk 2
0
2
kwk , = kwk − f (e u) 0 1 −1 W 2 (Ω )→W2
where e u (x) is an element of the “interval” [u(x), v(x)] ⊂ Brm 0 (0) ∩ E m1 0 . Hence we will obtain by virtue of condition (ι) ρ kA (u) − A (v)kW −1 kwk ≥ kwk2 1 − c0 λ−1 ke k u m0 2 2 ρ −1− ≥ kwk2 1 − c0 λm 0 2 ke u kρ . Taking ke u k ≤ rm 0 , ρ > 0 and substituting the expression for rm 0 , −1− ρ kA (u) − A (v)kW −1 ≥ kwk 1 − c0 λm 0 2 rmρ 0 ≥ kwk [1 − c0 (1 − b0 )] .
(3.7)
2
The correctness of inequality (3.6) and consequently uniqueness of a solution of the problem in a ball Brm 0 (0) ∩ 0
E m1 0 ⊂ W 12 (Ω ) under the conditions of Theorem 2 follows from the last inequality by virtue of the condition on 1 c0 > 1 − b0 > 0. Thereby the theorem is proved completely. The following statement follows from Theorem 2. Corollary 2. Let the conditions of Theorem 2 be fulfilled, let c1 = 1, and let b be a fixed number in (c0 , 2c0 ), and b0 = min b−1 , 2−1 . Then for each integer m ≥ 1 there exists an integer m 0 ≥ 1 such that the problem 0
(1.1) is uniquely solvable in a ball Brm 0 (0) ∩ E m1 0 ⊂ W 12 (Ω ) for any h(x) ∈ B R∗ m (0) ∩ E m−10 ⊂ W2−1 (Ω ), where 0 m 0 = inf {e m | λm e = λm }, B R∗ m (0) ≡ {h(x) ∈ W2−1 (Ω ) | khkW −1 (Ω ) ≤ Rm 0 } 0
is the closed ball of following form:
2
W2−1 (Ω )
centered at the origin with radius Rm 0 > 0, and rm 0 and Rm 0 are determined in the
2+ρ
1
1
2+ρ
rm 0 = (1 − b0 ) ρ λm2ρ0 , Rm 0 = b0 (1 − b0 ) ρ λm2ρ0 ! 1 1 1 − b0 ρ 2+ρ 1 − b0 ρ ρ + b0 2+ρ 2ρ 2ρ or rm 0 = λm 0 , R m 0 = λm 0 . 1+ρ 1+ρ 1+ρ Proof. For this purpose it is enough to show that the inequality (3.4) holds for any h(x) ∈ B R∗ m (0)∩ E m−10 ⊂ W2−1 (Ω ). 0 The fulfillment of the conditions of Theorem 2 for an h(x) follows from the proof of Theorem 2 under the condition on the radius Rm 0 . The following inequalities follow from the inequalities (3.4) and (3.7): 1
2+ρ
hA (u) , ui ≥ rm 0 b0 (1 − b0 ) ρ λm2ρ0 and −1− ρ kA (u) − A (v)kW −1 ≥ kwk 1 − c0 λm 0 2 rmρ 0 ≥ kwk [1 − c0 (1 − b0 )] , 2
which shows that the problem (1.1) is uniquely solvable in Brm 0 (0) ∩ E m1 0 for any h(x) ∈ E m−10 satisfying the 1
2+ρ
inequality hh, ui ≤ rm 0 b0 (1 − b0 ) ρ λm2ρ0 for any u(x) ∈ Srm 0 (0) ∩ E m1 0 .
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K.N. Soltanov / Nonlinear Analysis: Hybrid Systems 2 (2008) 344–358
Further, the estimation hh, ui gives, under the conditions of the corollary, 2+ρ
1
|hh, ui| ≤ khkW −1 (Ω ) kuk = khkW −1 (Ω ) rm 0 ≤ rm 0 b0 (1 − b0 ) ρ λm2ρ0 . 2 2
So the corollary follows from here. Pm∗ :
W2−1 (Ω )
Now, let −→ following statements are true.
E m−1
be a canonical projector. From Theorem 2 and Corollary 2 it follows that the
Corollary 3. Let all conditions of Corollary 2 be fulfilled. Then for any h(x) ∈ W2−1 (Ω ) there exists an integer 2+ρ
1 m 0 ≥ 1 such that for Pm∗ 0 h ∈ E m−10 , Pm∗ 0 h W −1 ≤ b0 (1 − b0 ) ρ λm2ρ0 is fulfilled, in other words, for Pm∗ 0 h ∈ E m−10 the 2
1
2+ρ
problem (1.1) is uniquely solvable in the ball Brm 0 (0) ∩ E m1 0 where rm 0 = (1 − b0 ) ρ λm2ρ0 . 0
Corollary 4. Let the operator f : W 12 (Ω ) → W2−1 (Ω ) be such that f ∈ C 1 ∩ Cw0 and there exist constants α ≥ 12 , β = α + 12 , 1 > γ > 21 , ρ > 0 such that the following inequalities hold:
−β ρ ρ+1 k f (u)kW −1 (Ω ) ≤ λ−α ≤ λ1 kuk2 , f 0 (u) 0 1 1 kuk2 −1 W 2 →W2
2
0
for any u(x) ∈ W 12 (Ω ). Then for any h(x) ∈ W2−1 (Ω ), khkW −1 (Ω ) ≤ γ · r1 , the problem (1.1) is uniquely solvable in 2 0
1
ρ+1+2α 2ρ
the ball Br1 (0) ⊂ W 12 (Ω ), r1 = (1 − γ ) ρ λ1
. (We denote the space of weakly continuous mappings by Cw0 .) 0
Note that in the case m = 1 we obtain E m−1 = W2−1 (Ω ) and E m1 = W 12 (Ω ), i.e. P1 = id. Remark 5. Assertions similar to the statements of the problem (1.1) can also be proved when appropriate spaces and 0
operator A are determined as follows: the expression for the operator A does not vary but now A : W22 (Ω )∩ W 12 (Ω ) → 0
L 2 (Ω ), and E m2 is the corresponding subspace of the space W22 (Ω ) ∩ W 12 (Ω ). The necessary a priori estimations will be obtained by multiplication with −∆u. In this case the condition (ι) will be replaced by an appropriate condition. 4. Solvability of the problem (1.1) with polynomial nonlinearity 0
i i be a subspace such as the above, and E i Let E m −m be the respective complementary subspace, so that W 2 (Ω ) ≡ i ⊕ E i , i = 0, 1. Let us designate through P , Q the projectors in L (Ω ) such that for each m ∈ N the following Em m m 2 −m relations apply:
kPm k ≤ 1, kQ m k ≤ 1,
P1 ≡ 1 ≡ id,
0
i Q m ≡ (1 − Pm ) : W i2 (Ω ) −→ E −m ,
0
i Pm : W i2 (Ω ) −→ E m ,
0
i i ⊕ E −m , i = 0, 1 W i2 (Ω ) ≡ E m
and −1 Pm∗ Q ∗m : W2−1 (Ω ) −→ E m−1 E −m ,
m = 1, 2, . . . .
We will consider the problem (1.1) in the case when f (u) ≡ |u|ρ u, i.e. −∆u − f (u) ≡ −∆u − |u|ρ u = h(x), x ∈ Ω ; u |∂ Ω = 0, 0 ≤ ρ < which may be rewritten in the following form: ∗ Pm A (u) ≡ −∆u m − Pm∗ |u|ρ u = h m (x) ≡ Pm∗ h(x), (1) Q ∗m A (u) ≡ −∆u −m − Q ∗m |u|ρ u = h −m (x) ≡ Q ∗m h(x), (2) ,
4 , n−2
(4.1)
u |∂ Ω = 0,
(4.2)
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K.N. Soltanov / Nonlinear Analysis: Hybrid Systems 2 (2008) 344–358
where u m ≡ Pm u ∈ E m1 ;
1 u −m ≡ Q m u ∈ E −m ,
Pm∗ ∆u ≡ ∆Pm u ≡ ∆u m .
u ≡ u m + u −m ;
Let us introduce the following notation: ρ ρ 1 1 rm 0 = 1 − bm 0 ρ 2−1 C p λνm 0 ; r−m 0 = 1 − bm 0 ρ 2−1 C p λν1 ; [4 − ρ (n − 2)] ν≡ , m 0 ≥ 2; 4ρ if n ≥ 3 and ρ > 0 is an arbitrary number if n = 1, 2, h σ i ρ 1 , m0 ≥ 2 Rm 0 ≡ 1 − bm 0 ρ 2−µ C p λνm 0 bm 0 − 1 − bm 0 λ1 λ−1 m0 h i 1 ρ σ , m0 ≥ 2 R−m 0 ≡ 1 − bm 0 ρ 2−µ C p λνm 0 bm 0 − 1 − bm 0 λm 0 λ−1 1
0<ρ<
σ =
4 n−2
2 (ρ + 1) ν (ρ + 1) [4 − ρ (n − 2)] = , 2 pρ p
where bm 0 > 0 is a number which satisfies the inequalities 1 > bm 0 > 1 −
2λ1 ; p λm 0 + λ1
p = ρ + 2,
m 0 = m 0 (m) and C > 0 is a constant of the Gagliardo–Nirenberg inequality [8] n n n kukW ps (Ω ) ≤ C kukθW l (Ω ) kuk1−θ − s = − θ) + θ − l , , (1 L p2 (Ω ) p1 0 p0 p2 p1 where C ≡ C ( p0 , p1 , p2 , l, s) > 0; if s = 0, l = 1, p1 = p2 = 2, p0 = ρ + 2 = p, then θ =
(4.3) nρ 2p .
4 if n ≥ 3 and 0 < p − 2 = ρ be an arbitrary number if n = 1, 2, and let Theorem 3. Let 0 < ρ < n−2 i i E m 0 , E −m 0 (i = −1, 0, 1) be the above-determined subspaces: 0
0
0
1 . W 12 (Ω ) ≡ Pm 0 W 12 (Ω ) ⊕ Q m 0 W 12 (Ω ) ≡ E m1 0 ⊕ E −m 0 0
1 Then the problem (4.1) ((4.2)) is uniquely solvable in the “ball” Brm 0 (0) ∩ E m1 0 ⊕ Br−m 0 (0) ∩ E −m ⊂ W 12 (Ω ) 0 for any h(x) ∈ W2−1 (Ω ) satisfying the inequality
1 hh, ui ≤ − −∆u − |u|ρ u, u , for ∀u ∈ Srm 0 (0) ∩ E m1 0 ⊕ Sr−m 0 (0) ∩ E −m . (4.4) 0
Proof. We remark that the solution of problem (4.1) (or (4.2)) will be understood in the generalized sense; it is similar to Definition 1. We denote by A the operator generated by problem (4.1). At the beginning one needs to show that the subset of the right sides h(x) determined by the inequality (4.3) is nonempty. It is enough to show that the inequality 1 hA (u) , ui ≥ µ (kuk) kuk applies on the subset Brm 0 (0) ∩ E m1 0 ⊕ Br−m 0 (0) ∩ E −m with a function µ (τ ), τ ≥ 0, such 0 that µ rm 0 ≥ δ0 > 0 for some number δ0 . These allow us to use Lemma 1. For this purpose we consider the problem (4.1) in the form of the system (4.2), as the problem (4.2) is equivalent to the problem (4.1) by construction. From the problem (4.2)(1) we have
2
∗
ρ+1 Pm 0 A (u) , u m 0 ≡ −∆u m 0 − Pm∗ 0 |u|ρ u, u m 0 ≥ u m 0 − kuk L p (Ω ) u m 0 L (Ω ) p
2
p
ρ+1
ρ ρ
u m ≥ u m − 2 u m − 2 u −m 0
0
0
L p (Ω )
0
L p (Ω )
0
L p (Ω )
for any u(x) ∈ W 12 (Ω ) (here, if m 0 = 1 we will obtain the whole space; therefore we suppose that m 0 ≥ 2).
K.N. Soltanov / Nonlinear Analysis: Hybrid Systems 2 (2008) 344–358
355
Now we assume that ρ > 0, since when −1 < ρ < 0 this problem turns into the known one which has already been investigated well enough. We will obtain by using the inequality (4.3) in the last inequality
2
2νρ
2νρ
ρn
ρ+1
ρn Pm∗ 0 A (u) , u m 0 ≥ u m 0 − 2ρ C p u m 0 2 u m 0 L (Ω ) − 2ρ C u −m 0 L (Ω ) u m 0 2 p u m 0 Lp2 (Ω ) p
2
2 − ρν
u m p − 2ρ C u −m ρ+1 λm p u m ≥ u m 0 − 2ρ C p λ−ρν m0 0 0 0 L (Ω ) 0 p
(where ν =
1 ρ
−
n−2 4 ).
(4.5)
From here, correctness of the indicated inequality follows.
1 Now assuming u(x) ∈ Srm 0 (0) ∩ E m1 0 ⊕ Sr−m 0 (0) ∩ E −m in the last inequality, we obtain 0
ρ+1
2 − ρν p ρ
um0 Pm∗ 0 A (u) , u m 0 ≥ u m 0 1 − 2ρ C p rmρ 0 λ−ρν − 2 C u λ −m 0 L (Ω ) m 0 m0 p
=
rm2 0
ρ
1−2
C p rmρ 0 λ−ρν m0
ρ+1 − ρν − 2 C u −m 0 L (Ω ) λm 0 p rm 0 ρ
p
ρν ρµ ρ p ρ+1 − p − 2 rm 0 ≥ rm2 0 1 − 2ρ C p rmρ 0 λ−ρν λ − 2 C r −m 0 m 0 λ1 m0 ρν ρµ ρ p ρ −ρν ρ p ρ+1 − p − 2 = rm 0 rm 0 − 2 C rm 0 λm 0 − 2 C r−m 0 λm 0 λ1 .
This shows that the subset of elements h(x) which satisfy the inequality (4.4) is a nonempty subset of W2−1 (Ω ) by virtue of the determination of rm 0 . Thus if we are searching for approximate solutions as in Section 3, repeating the same arguments we will obtain 1 that for arbitrary fixed u −m 0 ∈ Br−m 0 (0) ∩ E −m and for any h m 0 (x) ≡ Pm 0 h(x) ∈ E m−10 satisfying the following 0 inequality:
ρ+1
− ρν h m 0 , u m 0 + 2ρ C u −m 0 L (Ω ) λm 0 p rm 0 ≤ bm 0 rm2 0 , p
∀u m 0 ∈ Srm 0 (0) ∩ E m1 0
(4.6)
∞ 0 the sequence of approximate solutions u m 0 k k=1 are contained in a bounded subset of E m1 0 ⊂ W 12 (Ω ), more precisely in Brm 0 (0) ∩ E m1 0 . 0
Then we have obtained that u m 0 k * u m 0 weakly in E m1 0 (and in W 12 (Ω )) just as in the previous section from the 0 0 reflexivity of the spaces E m1 0 W 12 (Ω ) and L p (Ω ) and because in the conditions of the theorem W 12 (Ω ) ⊂ L p (Ω ) is compact, where u m 0 (x) is an element of E m1 0 . Consequently, u m 0 k → u m 0 strongly in E m0 0 (and also in L p (Ω )) and almost everywhere in Ω as k → +∞. From here it follows that the following relation is correct: −∆u m 0 k * −∆u m 0 weakly in E m−10 and in W2−1 (Ω ) , Pm∗ 0 |u k |ρ u k * Pm∗ 0 |u|ρ u weakly in E m−10 and in W2−1 (Ω ) , and, furthermore, Pm∗ 0 |u k |ρ u k → Pm∗ 0 |u|ρ u
strongly in L q (Ω ) q = p 0 under k → +∞.
So it is sufficient to repeat the arguments of the proof of Theorem 2 to finish the proof of the solvability for problem (4.2)(1) . Thereby the solvability of the problem (4.2)(1) is proved. Now we move on to the proof of uniqueness of the solution of this problem. For the proof, we assume the contrary. Therefore, first we show that the following sequences of inequalities are correct:
∗
P A (u) − P ∗ A (v) −1 ku − vk ≥ Pm∗ 0 (A (u) − A (v)) , u m 0 − vm 0 m0 m0 W2 (Ω )
2
= u m 0 − vm 0 − Pm∗ 0 |u|ρ u − |v|ρ v , u m 0 − vm 0
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K.N. Soltanov / Nonlinear Analysis: Hybrid Systems 2 (2008) 344–358
2
2 ρ u k L p (Ω ) u m 0 − vm 0 L (Ω ) ≥ u m 0 − vm 0 − (ρ + 1) ke p 2νρ
2
2 − ρ u k L p (Ω ) u m 0 − vm 0 ≥ u m 0 − vm 0 − C 2 (ρ + 1) λm 0 p ke
2 − 2νρ ρ ≥ u m 0 − vm 0 × 1 − 2ρ−1 C p (ρ + 1) λm 0 p rmρ 0 + r−m 0 . 0
1 for any u(x) and v(x) of Brm 0 (0) ∩ E m1 0 ⊕ Br−m 0 (0) ∩ E −m ⊂ W 12 (Ω ) when u −m 0 ≡ v−m 0 , where 0 e e u (x) ≡ e u m 0 (x) + u −m 0 (x), u m 0 (x) ∈ u m 0 , vm 0 ⊂ Brm 0 (0) ∩ E m1 0 .
We get, taking into account the values of rm 0 and r−m 0 , " !# 2νρ
∗
p ∗ −1 −1
P A (u) − P A (v) −1 . ≥ u m 0 − vm 0 × 1 − 2 1 − bm 0 (ρ + 1) 1 + λ1 λm 0 m0 m0 W (Ω ) 2
Hence we obtain the uniqueness of the solution using the conditions on bm 0 and u −m 0 (x) = v−m 0 (x). We have 1 proved that for each fixed u −m 0 ∈ Br−m 0 (0) ∩ E −m the problem (4.2)(1) is uniquely solvable in Brm 0 (0) ∩ E m1 0 0 under the conditions of Theorem 3. Thereby the unique solvability of problem (4.2)(1) in Brm 0 (0) ∩ E m1 0 for any h(x) ∈ W2−1 (Ω ) as h m 0 (x) = Pm 0 h(x) ∈ B R∗ m (0) ∩ E m−10 is proved. 0 With the help of the reasoning indicated for problem (4.2)(1) we shall obtain that problem (4.2)(2) is uniquely 0
−1 1 solvable in Br−m 0 (0) ∩ E −m ⊂ W 12 (Ω ) for any h(x) ∈ W2−1 (Ω ) for h −m 0 = Q ∗m 0 h ∈ E −m satisfying the inequality 0 0
ρ+1 − νρ 2 , h −m 0 , u −m 0 + 2ρ u m 0 L (Ω ) λm 0p r−m 0 ≤ bm 0 r−m 0 p
1 u −m 0 ∈ Sr−m 0 (0) ∩ E −m 0
for each fixed u m 0 ∈ Brm 0 (0) ∩ E m1 0 . Thus we show that problem (4.1) is uniquely solvable in 0
1 Brm 0 (0) ∩ E m1 0 ⊕ Br−m 0 (0) ∩ E −m ⊂ W 12 (Ω ) 0
for any h(x) ∈ W2−1 (Ω ) which satisfies the inequality from Theorem 3. The validity of the following statement is easily seen from the proof of Theorem 3. Corollary 5. Let all conditions of Theorem 3 be fulfilled. Then the problem (4.1) ((4.2)) is uniquely solvable in 0
1 Brm 0 (0) ∩ E m1 0 ⊕ Br−m 0 (0) ∩ E −m ⊂ W 12 (Ω ) for any 0 −1 h(x) ∈ B R∗ m (0) ∩ E m−10 ⊕ B R∗ −m (0) ∩ E −m ⊂ W2−1 (Ω ), 0 0
0
i.e. h(x) ≡ Pm∗ 0 h + Q ∗m 0 h
and Pm∗ 0 h ∈ B R∗ m (0) ∩ E m−10 , 0
−1 Q ∗m 0 h ∈ B R∗ −m (0) ∩ E −m . 0 0
5. Some remarks From the proof of Theorem 3 it follows that a statement similar to Corollary 5 is valid under the condition m = 1, namely: Corollary 6. Let the conditions of the Theorem 3 be fulfilled. Then the problem (4.1) is uniquely solvable in 0
1
Br1 (0) ⊂ W 12 (Ω ) for any h(x) ∈ B R∗ 1 (0) ⊂ W2−1 (Ω ), where R1 ≡ b1 (1 − b1 ) ρ C and b1 is a fixed number of 2−1 , 1 .
− ρp ν λ1
1
and r1 ≡ (1 − b1 ) ρ C
− ρp ν λ1 ,
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K.N. Soltanov / Nonlinear Analysis: Hybrid Systems 2 (2008) 344–358 j
j
Now we shall lead to remarks which are associated with the use of spaces E m , E −m ( j = −1, 0, 1). This allows us to obtain additional information about the problems considered. P In fact, we have from the definition of u −m (x) that u −m (x) ≡ m−1 k=1 ak wk where ak are unknown coefficients and Pm−1 ek . If w ek = wk then (4.2)(2) has the form h −m ≡ k=1 bk w ρ ! m−1 ∞ ∞ m−1 X X X X λk ak wk − Q ∗m ak wk ak wk = bk wk , k=1 k=1 k=1 k=1 which generates a system of algebraic equations for determining the unknown coefficients ak , k = 1, m − 1. (It is 0
1 known that [5], since {wk }∞ k=1 is a basis in W 2 (Ω ) (and in L 2 (Ω )) and as these spaces satisfy the corresponding −1 conditions by construction, it is possible to define a biorthogonal basis {e w k }∞ k=1 in W2 (Ω ).) We shall consider the case when problem (1.1) has functional nonlinearity since we can give a better illustration in this case. To be exact, we consider the case when f (u) = kuk2L 2 (Ω ) u, i.e. ρ = 2. Thus, we consider the problem
−∆u − kuk2L 2 (Ω ) u = h(x),
x ∈ Ω ⊂ Rn ,
(5.1)
u |∂ Ω = 0,
(5.2)
and we shall use Corollary 5. In order to make use of Corollary 5 we consider the representation of the type (4.2) for problem (5.1) and (5.2): (
Pm∗ A (u) ≡ −∆u m − Pm∗ kuk2L 2 (Ω ) u = h m (x) ≡ Pm∗ h(x), (1) Q ∗m A (u) ≡ −∆u −m − Q ∗m kuk2L 2 (Ω ) u = h −m (x) ≡ Q ∗m h(x), (2)
u |∂Ω = 0.
(5.3)
e > 0 for some number R. e Then if we take Rm ≥ R, e there Now let h(x) ∈ W2−1 (Ω ) be preassigned and khkW −1 = R 0 2
0
exists m 0 such that the problem (5.3)(1) in the conditions indicated is uniquely solvable in Brm 0 (0) ∩ E m1 0 ⊂ W 12 (Ω ) 0
1 for any fixed u −m 0 (x) ∈ Br−m 0 (0) ∩ E −m ⊂ W 12 (Ω ) by virtue of Theorem 3 (or Corollary 5). 0 In this case we obtain for problem (5.3)(2) under the same conditions the following equation which is similar to the equation stated above: ! m −1 mX mX ∞ 0 −1 0 0 −1 X X 2 2 λk ak wk − ak kwk k L 2 (Ω ) ak wk = bk wk , (wk (x) = w ek (x)) . k=1
k=1
k=1
k=1
If we assume kwk k L 2 (Ω ) = 1 then we obtain the system of algebraic equations λk a k −
∞ X
! ak2
a k = bk ,
k = 1, 2, . . . , m 0 − 1.
(5.4)
k=1
e which, generally Further, we need to study the system (5.4)(2) for h −m 0 = Q ∗m 0 h in the case when R−m 0 ≥ R, speaking, does not satisfy the conditions of Corollary 3. Therefore we need to consider the operator A generated by 0
the system (5.4) and must investigate Q m 0 A on the ball Br (0) ⊂ W 12 (Ω ), namely in the case when u −m 0 (x) belongs 0
1 to Br (0) ∩ E −m ⊂ W 12 (Ω ) for various numbers r > 0. 0 2 Let 0 < r ≤ rm 0 ; then we obtain that there exists k0 : 1 ≤ k0 ≤ m 0 − 1 such that λk0 ≤ λ−1 1 r , where ∞ X k=1
ak2 ≡ kuk2L 2 (Ω ) ,
−1
kuk L 2 (Ω ) ≤ λ1 2 kuk .
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K.N. Soltanov / Nonlinear Analysis: Hybrid Systems 2 (2008) 344–358
It is clear that, for such λk , the statement on the unique solvability of the problem considered does not apply. In fact, it is enough to consider the following part of the system (5.4)(2) : ! ∞ X λk − a 2j ak − ak3 = bk , k = 1, 2, . . . , k0 . j=1, j6=k
Each equation in this system can have three solutions. If the system (5.4)(2) is solvable, the number of solutions in 2 e the ball indicated depends on the number k0 : λk0 ≤ λ−1 1 r which in turn depends on R > 0 and ρ ≥ 0 (ρ = 2 in this case). Thus we obtain the following proposition. Proposition 2. Let all conditions of the problem considered be fulfilled. If this problem is solvable in the case considered, then the number of solutions will be increased with magnification of the number m 0 (and consequently e and ρ ≥ 0. rm 0 ) which depends on R Since many works (see, in particular, [2,10,3,6] and their references) are devoted to the study of the number of solutions of such problems and in these works various results on numbers of solutions are obtained, we will not continue these discussions. Let us just remark that the indicated reasoning rather simply illustrates the local connection between the number of solutions and the diameter of the considered domain from the range of definition of the operator generated for the problem considered. References [1] O.V. Besov, V.P. Il‘in, S.M. Nikolskiy, Integral Representations of Functions and Imbedding Theorem, M., Nauka, 1996. [2] H. Brezis, Points critiques dans les problemes variationnels sans compacite, in: Sem. Burbaki 1987–88, in: Asterisque, vol. 161–162, 1988, pp. 239–256. n. 698. [3] R.D. Benguria, J. Dolbeault, M. Esteban, Classification of the solutions of semilinear elliptic problems in a ball, J. Differential Equations 167 (2000) 2. [4] J. Diestel, Geometry of Banach Spaces. Selected Topics, Springer-Verlag, Berlin, Heidelberg, 1975. [5] N. Dunford, J.T. Schwartz, Linear Operators, Part I : General Theory, Interscience Publishers, New York, London, 1958. With the assistance W.G. Bade, R.G. Bartle. [6] F. Huang, Existence of two solutions nonlinear elliptic equations with critical Sobolev exponents and mixed boundary conditions, in: Proceed. Royal Soc. Edinburgh, vol. 126A, n. 1, 1996. [7] E. Hille, R.S. Phillips, Functional Analysis and Semi-groups, vol. 31, AMS, Colloq. Publ., 1957. [8] J.-L. Lions, Quelques methodes de resolution des problemes aux limites non lineares, Dunod, Gauthier-Villars, Paris, 1969. [9] L. Nirenberg, Topics in nonlinear functional analysis, New York, 1974. [10] S.I. Pohozhaev, On equations of the form ∆u = f (x, u, Du), Math. Sb. 113 (2) (1980). [11] H.H. Schaefer, Topological Vector Spaces, The Macmillan Company, Collier–Macmillan Limited, London, New York, 1966. [12] K.N. Soltanov, To the theory of normal solvability of nonlinear equations, Dokl. Ac. Sci. USSR 278 (1) (1984). [13] K.N. Soltanov, Solvability of nonlinear equations with operators of the form of the sum of a pseudomonotone and weakly compact operator, Russian Ac. Sci Dokl. Math. 45 (3) (1992). [14] K.N. Soltanov, Some Applications of the Nonlinear Analysis to the Differential Equations, ELM, Baku, 2002, p. 292. [15] K.N. Soltanov, On nonlinear equations with continuous operators in reflexive Banach spaces, Funct. Anal. Appl. 33 (1) (1999). [16] E. Zeidler, Nonlinear functional analysis and its applications II/b, in: Nonlinear Monotone Operators, Springer-Verlag, Berlin, 1990.