# On Second-Order Differential Equations with Nonsmooth Second Member

Authors: M. Milla Miranda; A. T. Lourêdo; L. A. Medeiros
Journal: ISRN Applied Mathematics
 (2014)

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## Abstract

In an abstract framework, we consider the following initial value problem:u′′ + μAu + F(u)u = f  in  (0,T), u(0)=u0,u′(0)=u1, where μ is a positive function and f a nonsmooth function. Given u0, u1, and f we determine Fu in order to have a solution u of the previous equation. We analyze two cases of Fu. In our approach, we use the Theory of Linear Operators in Hilbert Spaces, the compactness Aubin-Lions Theorem, and an argument of Fixed Point. One of our two results provides an answer in a certain sense to an open question formulated by Lions in (1981, Page 284).


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## Body

## 1. Introduction

LetV and H be two real separable Hilbert spaces with V dense in H and V continuously embedding in H. The scalar product and norms of V and H are represented, respectively, by
(1)(u,v),|u|,((u,v)),∥u∥.LetA be the self-adjoint operator of H defined by the triplet {V,H,((u,v))}. Consider α∈ℝ, α≥0. We denote by D(Aα) the Hilbert space
(2)D(Aα)={u∈H;Aαu∈H}
equipped with the scalar product
(3)(u,v)D(Aα)=(Aαu,Aαv)
(cf. Lions [1]).Consider the following initial value problem:(4)u′′+μAu+F(u)u=fin(0,T),u(0)=u0,u′(0)=u1,
where μ is a positive function and f a nonsmooth function.The objective of this work is to study the following inverse problem: givenu0, u1, and f∈L2(0,T;D(Aα)′) (D(Aα)′ dual space of D(Aα)) to determine F(u) such that Problem (4) has a solution u. We analyze two cases of F(u), more precisely, the cases
(5)F(u)=|A-θ/2u|2p,(6)F(u)=∥u∥C0([0,T];X),
where X is an appropriate Hilbert space.In Lions [2, Page 284], the following problem is formulated
(7)u′′-Δu+[∫0t‍(∫Ω‍u2dx)ds]u=v(t)δ(x-x0)inΩ×(0,T),u=0on⁡Γ×(0,T),u(x,0)=0,u′(x,0)=0,x∈Ω,
where Ω is an open bounded set of ℝn with boundary Γ, x0∈Ω, and δ(x-x0) being the Dirac mass supported at {x0}. He says not to know if this problem admits a solution. He say also that one of the difficulties in the study of existence of solutions of the nonlinear equations lies in the difficulty in defining weak solutions, since the transposition method is essentially a linear method. This is ultimately connected to the fact that one cannot multiply distributions.Problem (4) with F(u) of the form (6) is an abstract formulation of Problem (7) with a slight modification of the nonlinear term. Theorem 3 gives the existence of solutions of this problem. In applications we give examples of Problem (7), with the modification of the nonlinear term, for Ω an open bounded set of ℝn,  n=1,2,3.In Grotta Ragazzo [3] the following equation is studied:
(8)utt-uxx-au+(1π∫0π‍u2dx)αu=0in(0,π)×ℝ.
This equation is considered as a first approximation of the Klein-Gordon equation
(9)utt-uxx-au+u1+2α=0in(0,π)×ℝ.
Observe that (8) with a=0 and α=1 is the meson equation of Schiff [4] (cf. also Jörgens [5]).The physical motivation of (8) with α=1 can be seen in Lourêdo et al. [6].Problem (4) with F(u) of the form (5) generalizes (8) when a=0. The existence of solutions of this problem is studied in Theorem 1.In Louredo et al., loc.cit., is analyzed the equation(10)u′′-μ(t)Δu+a(∫Ω‍u2dx)u+b(∫Ω‍u′2dx)u′=0inΩ×(0,∞)
with nonlinear boundary condition. The F(u) given in (5) is different from the a(∫Ω‍u2dx) of this equation. The term F(u) is related to the nonsmoothness of f.


## 2. Main Results

We use the notationD(Aα)′=D(A-α),  α∈ℝ,  α≥0. Identifying H with H′, we have
(11)D(Aα)↪H↪D(A-α).
Here and in what follows the notation X↪Y means that the space X in dense in the space Y and the embedding of X in Y are continuous. Note that D(A-α)′=D(Aα). Also, if β, γ∈ℝ with β≥γ, we have
(12)D(Aβ)↪D(Aγ).
Assume that
(13)theembeddingofVinHiscompact.First we analyze Problem (2) with F(u)=|A-θ/2u|2p, that is, the problem
(14)u′′+μAu+|A-θ/2u|2pu=fin(0,T),u(0)=u0,u′(0)=u0.Theorem 1.
Assume condition (13). Let θ and p be real numbers with p≥1. Consider
(15)u0∈D(A(1-θ)/2),u1∈D(A-θ/2),μ∈W1,1(0,T),μ(t)≥μ0>0,hhhhhhhhh∀t∈[0,T](μ0constant),f∈W1,1(0,T;D(A-(1+θ)/2)).
Then there exists a function u in the class
(16)u∈L∞(0,T;D(A(1-θ)/2)),u′∈L∞(0,T;D(A-θ/2)),u′′∈L∞(0,T;D(A-(1+θ)/2))
such that u is solution of the equation
(17)u′′+μAu+|A-θ/2u|2pu=finL∞(0,T;D(A-(1+θ)/2))
and satisfies the initial conditions
(18)u(0)=u0,u′(0)=u1.Remark 2.
Whenμ=1 it is possible to obtain a solution u of Problem (14) by using the Theory of Semigroups (cf. Pazy [7]).To formulate the second problem, we introduce some notations. In fact, let us define(19)Y=D(A(1-θ)/2),Z=D(A-θ/2),X=D(Aλ),
where
(20)-θ2≤λ<1-θ2,λ∈ℝ.
By (12) we have
(21)Y↪X↪Z.Consider Problem (4) with F(u)=∥u∥C0([0,T];X) and u0=0, that is, the problem
(22)u′′+μAu+∥u∥C0([0,T];X)u=fin(0,T),u(0)=0,u′(0)=u1.Theorem 3.
Assume thatθ, u1, and f satisfy the hypotheses of Theorem 1 and u0=0. Then there exists a function u in the class (16) such that u is solution of the problem
(23)u′′+μAu+∥u∥C0([0,T];X)u=finL∞(0,T;D(A-(1+θ)/2)),u(0)=0,u′(0)=u1.Remark 4.
Note that ifu belongs to class (16) then u∈C0([0,T];X) (cf. Lions and Magenes [8]).Corollary 5.
Under the same hypotheses of Theorem1, there exists a function u in the class (16) such that u is solution of the problem
(24)u′′+μAu+∥u∥Lp(0,T;X)pu=finL∞(0,T;D(A-(1+θ)/2)),u(0)=0,u′(0)=u1.We analyze the uniqueness of solutions. Considerθ=0 in Theorem 1. Then the solution u gives by this theorem when θ=0 satisfies
(25)u∈L∞(0,T;D(A1/2)),u′∈L∞(0,T;H),u′′∈L∞(0,T;D(A-1/2)),(26)u′′+μAu+|u|2pu=finL∞(0,T;D(A-1/2)),u(0)=u0,u′(0)=u1.Theorem 6.
Letp≥1 be a real number. Consider
(27)u0∈D(A1/2),u1∈H,μsatisfying(15)2withμ′∈L∞(0,T),f∈W1,1(0,T;D(A-1/2)).
Then there exists a unique solution u of Problem (26) in the class (25).We do not know if there is uniqueness of solutions for Theorem3, even when θ=0.In what follows we prove the above results.


## 3. Proof of Theorem1

Before proving the theorem, we make some considerations on the operatorAα. Recall hypothesis (13). By solving the spectral problem ((u,v))=λ(u,v), for all v∈V, we determine the eigenfunctions and eigenvalues, respectively, (wν) and (λν) of the operator A, that is,
(28)Awν=λνwν,ν=1,2,…,λν⟶∞,ν⟶∞.
Note that (wν) is a Hilbert basis of H (cf. Brezis [9] and Komornik [10]).Let beα∈ℝ, α≥0. Then the linear operator
(29)Aα:D(Aα)⟶H
is continuous, bijective, and
(30)Aαu=∑ν=1∞‍λνα(u,wν)wν,∀u∈D(Aα).
Also,
(31)(Aα)-1:H⟶D(Aα)
is given by
(32)(Aα)-1f=∑ν=1∞‍λν-α(f,wν)wν,∀f∈H.
These results can be found in Lions [1] and Medeiros and Milla Miranda [11].Introduce the adjoint operator(Aα)* of Aα, that is,
(33)(Aα)*:H⟶D(Aα)′〈(Aα)*f,z〉𝒳′×𝒳=(f,Aαz),∀z∈𝒳,
where 𝒳=D(Aα). Note that H is identified with H′. By the properties of Aα, we have that
(34)(Aα)*islinear,continuousandbijective.
Thus, the linear operator
(35)[(Aα)*]-1:D(Aα)′⟶H
is continuous and bijective.Proposition 7.
Letf∈H and g∈D(Aα)′. Then one has the following.(i)
∑ν=1∞‍λνα(f,wν)wνconverges inD(Aα)′,
and(36)(Aα)*f=∑ν=1∞‍λνα(f,wν)wν.(ii)
∑ν=1∞‍λν-α〈g,wν〉𝒳′×𝒳wνconverges inH,
and(37)[(Aα)*]-1g=∑ν=1∞‍λν-α〈g,wν〉𝒳′×𝒳wν,where 𝒳=D(Aα).Proof.
We prove(i). As f∈H, we have
(38)f=∑ν=1∞‍(f,wν)wν.
Consider z∈𝒳. Then noting that 〈wν,wμ〉𝒳′×𝒳=(wν,wμ), we obtain
(39)〈∑ν=mn‍λνα(f,wν)wν,z〉𝒳'×𝒳=〈∑ν=mn‍λνα(f,wν)wν,∑μ=1∞‍(z,wμ)wμ〉𝒳′×𝒳=∑ν=mn‍λνα(f,wν)(z,wν).
On the other hand, by (30) we derive
(40)(∑ν=mn‍(f,wν)wν,Aαz)=(∑ν=mn‍(f,wν)wν,∑μ=1∞‍λμα(z,wμ)wμ)=∑ν=mn‍λνα(f,wν)(z,wν).
The last two expressions give
(41)〈∑ν=mn‍λνα(f,wν)wν,z〉𝒳′×𝒳=(∑ν=mn‍(f,wν)wν,Aαz).
This and (38) provide that
(42)∑ν=mn‍λνα(f,wν)wν⟶0in𝒳′asm,n⟶∞.
So (i)1 is proved. Taking the limit in (41) and observing (33)2, we obtain (36).
We prove(ii). We have that there exists a unique f∈H such that
(43)(Aα)*f=g.
By (33)2 and (30)1, we have
(44)〈g,wμ〉𝒳′×𝒳=(∑ν=1∞‍(f,wν)wν,Aαwμ)=λμα(f,wμ).
Then
(45)∑ν=mn‍λν-α〈g,wν〉𝒳′×𝒳wν=∑ν=mn‍(f,wν)wν.
This implies that
(46)∑ν=mn‍λν-α〈g,wν〉𝒳′×𝒳wν⟶0inHasm,n⟶∞.
Thus (ii)2 is proved. By (43) and (45), we obtain
(47)∑ν=1∞‍λν-α〈g,wν〉𝒳′×𝒳wν=∑ν=1∞‍(f,wν)wν=f=[(Aα)*]-1g.
This concludes the proof of the proposition.Motived by (37), we equip the space D(Aα)′ with the scalar product
(48)(g,h)D(Aα)′=([(Aα)*]-1g,[(Aα)*]-1h)=(∑ν=1∞‍λν-α〈g,wν〉𝒳′×𝒳wν,∑μ=1∞‍λμ-α〈h,wμ〉𝒳′×𝒳wμ),
where 𝒳=D(Aα). This scalar product on 𝒳' yields a norm
(49)∥g∥D(Aα)′=∥[(Aα)*]-1g∥=|∑ν=1∞‍λν-α〈g,wν〉𝒳′×𝒳wν|
which is equivalent to the usual norm of D(Aα)′.By similarity between expressions (30) and (36) and between (32) and (37), respectively, we introduce the notations
(50)(Aα)*=Aα,[(Aα)*]-1=(Aα)-1.
Also we use the notation
(51)(Aα)-1=A-α=D(Aα)′.
With these considerations and expressions (29) and (33), we obtain
(52)D(Aα)⟶AαH⟶AαD(A-α)
and by expressions (35) and (31),
(53)D(A-α)⟶A-αH⟶A-αD(Aα).
Also by (37) and (48), (49), respectively, we find
(54)A-αg=∑ν=1∞‍λν-α〈g,wν〉D(A-α)×D(Aα)wν,∀g∈D(A-α),(55)(g,h)D(A-α)=(A-αg,A-αh),∥g∥D(A-α)=|A-αg|.Proposition 8.
Considerβ, γ∈ℝ. Then the linear operator
(56)Aγ:D(Aβ)⟶D(Aγ-β)′
defined by
(57)〈Aγy,z〉D(Aγ-β)′×D(Aγ-β)=(Aβy,Aγ-βz)
is continuous.Proof.
We obtain(58)|〈Aγy,z〉𝒴′×𝒴|≤|Aβy||Aγ-βz|=∥y∥D(Aβ)∥z∥y,
where 𝒴=D(Aγ-β). Then
(59)∥Aγy∥𝒴′≤∥y∥D(Aβ),∀y∈D(Aβ)
which proves the proposition.Proof of Theorem1.
We use the Galerkin method (cf. Lions [12] and Vicente and Frota [13]). Thus consider an approximate solution um of Problem (14); that is,
(60)um(t)=∑j=1m‍gjm(t)wj;(61)(um′′,wj)+μ(Aum,wj)+|A-θ/2um|2p(um,wj)=〈f,wj〉D(Aγ0)′×D(Aγ0),j=1,2,…,m;um(0)=um0=∑j=1m‍〈u0,wj〉E′×Ewj;um′(0)=um1=∑j=1m‍〈u1,wj〉G′×Gwj,
where E=D(A-(1-θ)/2) and G=D(Aθ/2).Remark 9.
Note that(62)um0⟶u0inD(A(1-θ)/2),|A(1-θ)/2um0|≤|A(1-θ)/2u0|,um1⟶u1inD(A-θ/2),|A-θ/2um1|≤|A-θ/2u1|.Remark 10.
Observe that ifγ≤0 then
(63)〈z,wj〉D(A-γ)×D(Aγ)=(z,wj),j=1,2,…Multiply both sides of (61)1 by gjm′(t)λj-θ and add from j=1 up to j=m. We obtain
(64)(um′′,A-θum′)+μ(Aum,A-θum′)+|A-θ/2um|2p(um,A-θum′)=〈f,A-θum′〉W′×W,
where
(65)W=D(Aγ0),γ0=1+θ2.
Then,
(66)12ddt|A-θ/2um′|2+12ddt(μ|A(1-θ)/2um|2)+12(p+1)ddt|A-θ/2um|2(p+1)=12μ′|A(1-θ)/2um|2+〈f,A-θum′〉W′×W.
By Proposition 8 with γ=0 and β=-γ0 and noting that γ0-θ=(1-θ)/2, we obtain
(67)〈f,A-θwj〉W′×W=(A-γ0f,Aγ0-θwj)=(A-γ0f,A(1-θ)/2wj).
Substituting this equality into (66) and integrating on [0,t], we obtain
(68)12|A-θ/2um′(t)|2+12μ(t)|A(1-θ)/2um(t)|2+12(p+1)|A-θ/2um(t)|2(p+1)=12∫0t‍μ′(s)|A(1-θ)/2um(s)|2ds+∫0t‍(A-γ0f(s),A(1-θ)/2um′(s))ds+12|A-θ/2um1|2+12μ(0)|A(1-θ)/2um0|2+12(p+1)|A-θ/2um0|2(p+1).
Note that
(69)12∫0t‍μ′(s)|A(1-θ)/2um(s)|2ds≤2∫0t‍|μ′(s)|μ0[14μ(s)]|A(1-θ)/2um(s)|2ds.
Also
(70)∫0t‍(A-γ0f(s),A(1-θ)/2um′(s))ds=(A-γ0f(t),A(1-θ)/2um(t))-(A-γ0f(0),A(1-θ)/2um0)-∫0t‍(A-γ0f′(s),A(1-θ)/2um(s))ds,(71)|(A-γ0f(t),A(1-θ)/2um(t))|≤1μ0∥f∥L∞(0,T;D(A-γ0))2+14μ(t)|A(1-θ)/2um(t)|2,|∫0t‍(A-γ0f′(s),A(1-θ)/2um(s))ds|≤1μ0∫0T‍|A-γ0f′(s)|2ds+14∫0t‍|A-γ0f′(s)|[μ(s)|A(1-θ)/2um(s)|2ds].Substituting (69)–(71) into (68), we find
(72)12|A-θ/2um′(t)|2+14μ(t)|A(1-θ)/2um(t)|2+12(p+1)|A-θ/2um(t)|2(p+1)≤2∫0t‍|μ′(s)|μ0[14μ(s)|A(1-θ)/2um(s)|2ds]+M+|A-γ0f(0)||A(1-θ)/2u0|+N+14∫0t‍|A-γ0f′(s)|[μ(s)|A(1-θ)/2um(s)|2ds]+E(0),
where
(73)M=1μ0∥f∥L∞(0,T;D(A-γ0))2,N=1μ0∫0T‍|A-θ/2f′(s)|ds,E(0)=12|A-θ/2u1|2+12μ(0)|A(1-θ)/2u0|2+12(p+1)|A-θ/2u0|2(p+1).
Applying Gronwall inequality in (72), we deduce
(74)12|A-θ/2um′(t)|2+14μ(t)|A(1-θ)/2um(t)|2+12(p+1)|A-θ/2um(t)|2(p+1)≤(E(0)+M+|A-γ0f(0)||A(1-θ)/2u0|+N)×exp∫0T‍[(2/μ0)|μ′(t)|+|A-γ0f′(t)|]dt,hhhhhhhhhhhhhhhhhhhhhhhhhhhhhh∀t∈[0,T].With this inequality, we determine a subsequence of(um), still denoted by (um), and a function u such that
(75)um⟶uweakstarinL∞(0,T;D(A(1-θ)/2)),um′⟶u′weakstarinL∞(0,T;D(A-θ/2)).
By (13) we have that
(76)D(A(1-θ)/2)iscompactlyembeddinginD(A-θ/2)
(cf. [1] and [11]). Then convergences (75) and Aubin-Lions Theorem (cf. [14]) imply
(77)um⟶uinL∞(0,T;D(A-θ/2)).
Therefore,
(78)|A-θ/2um|2p⟶|A-θ/2u|2pinL∞(0,T).
This convergence and convergence (75)1 provide
(79)|A-θ/2um|2pum⟶|A-θ/2u|2puweakstarinL∞(0,T;D(A(1-θ)/2))
which implies
(80)|A-θ/2um|2pum⟶|A-θ/2u|2puweakstarinL∞(0,T;D(A-γ0))
since (1-θ)/2>-γ0,  γ0 defined in (65).In order to obtain an estimate for(um′′), we apply the method of projections to the approximate equation (61) (cf. Lions [12]). Thus, we consider the orthogonal projection
(81)Pm:H⟶Vm⊂H,Pmz=∑j=1m‍(z,wj)wj,
where Vm is the subspace generated by w1,w2,…,wm.By similar arguments employed to obtain (67) and by (54) or (30), we find
(82)〈f,wj〉D(A-γ0)×D(Aγ0)=(A-γ0f,Aγ0wj)=λjγ0(A-γ0f,wj).Multiply both members of (61)1 by λj-γ0wj and add from j=1 up to j=m. Then, applying to this result, expression (82), affirmation (54), or (30) and noting that A-γ0um′′, A-γ0(Aum),  |A-θ/2um|2pA-γ0um belong to Vm, we obtain
(83)A-γ0um′′+μA-γ0(Aum)+|A-θ/2um|2pA-γ0um=Pm(A-γ0f)
which gives
(84)|A-γ0um′′|≤μ|A(1-θ)/2um|+C|A-θ/2um|2p|A(1-θ)/2um|+|A-γ0f|.
Then estimates (74) and (80) provide
(85)(um′′)isboundedinL∞(0,T;D(A-γ0)).
Thus, there exists a subsequence of (um′′), still denoted by (um′′), such that
(86)um′′⟶u′′weakstarinL∞(0,T;D(A-γ0)).Expressions (75) and (86) tell us that u belongs to class (16). Convergences (75)1, (80), and (86) allow us to pass to limit in (83) and to obtain
(87)A-γ0u′′+A-γ0(μAu)+|A-θ/2u|2pA-γ0u=A-γ0f
which provides (17). Initial conditions (18) follow from convergences (75) and (86).


## 4. Proof of Theorem3

The idea is to apply a fixed point argument to the problem(88)uk′′+μAuk+kuk=fin(0,T),uk(0)=0,uk′(0)=u1,
where k∈ℝ,k≥0.We solve (88). Consider an approximate solution ukm of (88) given by
(89)ukm(t)=∑j=1m‍gjkm(t)wj,(90)(ukm′′,wj)+μ(Aukm,wj)+k(ukm,wj)=〈f,wj〉D(Aγ0)′×D(Aγ0),j=1,2,…,m;ukm(0)=0,ukm′(0)=um1.By similar arguments used to obtain (74), we derive
(91)12|A-θ/2ukm′(t)|2+14μ(t)|A(1-θ)/2ukm(t)|2+k2|A-θ/2ukm(t)|2≤Lexp∫0T‍a(t)dt=R,hhhhhhhhhhhhhhh∀t∈[0,T],∀k≥0,∀m,
where
(92)L=12|A-θ/2u1|2+M+N(MandNdefinedin(73)),a(t)=2μ0|μ′(t)|+|A-γ0f′(t)|.The preceding inequality gives(93)(ukm)isboundedinL∞(0,T;D(A(1-θ)/2)),hhhhhhhhhhhhhhhhhhhhhhhhhhh∀m,∀k≥0;(ukm′)isboundedinL∞(0,T;D(A-θ/2)),hhhhhhhhhhhhhhhhhhhhhhhhhh∀m,∀k≥0.
By the projection method, we obtain, as in (83),
(94)A-γ0ukm′′+μA-γ0(Aukm)+kA-γ0ukm=Pm(A-γ0f).
This and estimate (91) provide
(95)(ukm′′)isboundedinL∞(0,T;D(A-γ0)),∀m.Estimates (93) and (95) allow us to find a subsequence of (ukm), still denoted by (ukm), and a function uk such that, by passing to limit in (94), we obtain
(96)A-γ0uk′′+A-γ0μ(Auk)+kA-γ0uk=A-γ0finL∞(0,T;H).
This, initial conditions (90)2, and estimates (95) imply
(97)uk′′+μAuk+kuk=finL∞(0,T;D(A-γ0));uk(0)=0,uk′(0)=u1.By taking the lim inf in both side of (91), we obtain
(98)∥uk′∥L2(0,T;Z)2+∥uk∥L2(0,T;Y)2≤P,∀k≥0.AsY↪X↪Z and the embedding Y in X are compact (λ<(1-θ)/2), it follows from of Aubin-Lions Theorem (see Simon [14]) that
(99)∥uk∥C0([0,T];X)≤P1,∀k≥0.Define the map(100)ψ:[0,∞)⟶ℝ,ψ(k)=∥uk∥C0([0,T];X),
where uk is the solution of Problem (97). We will prove that ψ has a fixed point. Consider only the case f≠0. The case f=0 is outside of our attention. We will prove the following results.(I) One has
ψ(0)>0.In fact ifψ(0)=0, we have that u0=0 is a solution of (97) with k=0, but this a contradiction since f≠0.(II) One has
ψ is continuous on [0,∞).Letk0>0. Consider k>0. By (94) and (90)2 we obtain
(101)ukm′′+μAukm+kukm=Aγ0Pm(A-γ0f);ukm(0)=0,ukm′(0)=um1,uk0m′′+μAuk0m+k0uk0m=Aγ0Pm(A-γ0f),uk0m(0)=0,uk0m′(0)=um1.
Use the notation zm=ukm-uk0m. Then the preceding problems give
(102)zm′′+μAzm+k0zm=-(k-k0)ukm;zm(0)=0,zm′(0)=0.Taking the scalar product ofH of both sides of this equation with A-θzm′, we find
(103)12ddt|A-θ/2zm′|2+12ddt[μ|A(1-θ)/2zm|2]+k02ddt|A-θ/2zm|2=12μ′|A(1-θ)/2zm|2-(k-k0)(ukm,A-θzm′).
We have
(104)(ukm,A-θzm′)=(A-γ0ukm,A(1-θ)/2zm′).
Integrating on [0,t] both sides of the last two expressions, we derive
(105)12|A-θ/2zm′(t)|2+12μ(t)|A(1-θ)/2zm(t)|2+k02|A-θ/2zm(t)|2=12∫0t‍μ′(s)|A(1-θ)/2zm(s)|2ds-(k-k0)∫0t‍(A-γ0ukm(s),A(1-θ)/2zm′(s))ds.
We obtain
(106)∫0t‍(A-γ0ukm(s),A(1-θ)/2zm′(s))ds=(A-γ0ukm(t),A(1-θ)/2zm(t))-∫0t‍(A-γ0ukm′(s),A(1-θ)/2zm(s))ds.
As (1-θ)/2>-γ0, we have
(107)|(A-γ0ukm(t),A(1-θ)/2zm(t))|≤C|A(1-θ)/2ukm(t)∥A(1-θ)/2zm(t)|≤C22μ0|A(1-θ)/2ukm(t)|+12μ(t)|A(1-θ)/2zm(t)|2.
Also,
(108)(A-γ0ukm′,A(1-θ)/2zm)=(A-θ/2ukm′,A-θ/2zm),(109)|∫0t‍(A-γ0ukm′(s),A(1-θ)/2zm(s))ds|≤∫0t‍|A-θ/2ukm′(s)∥A-θ/2zm(s)|ds≤∫0t‍C|A-θ/2ukm′(s)∥A(1-θ)/2zm(t)|ds≤C22μ0∫0T‍|A-θ/2ukm′(t)|2dt+12∫0t‍μ(s)|A(1-θ)/2zm(s)|2ds.Taking into account estimate (91) in (107) and (109), we find
(110)|(A-γ0ukm(t),A(1-θ)/2zm(t))|≤2C2Rμ02+12μ(t)|A(1-θ)/2zm(t)|2,|∫0t‍(A-γ0ukm′(s),A(1-θ)/2zm(s))ds|≤C2RTμ0+12∫0t‍μ(s)|A(1-θ)/2zm(s)|2ds.
Substituting the last two inequalities into (106), we obtain
(111)|k-k0|∫0t‍|(A-γ0ukm(s),A(1-θ)/2zm′(s))|ds≤2C2Rμ02|k-k0|+12|k-k0|μ(t)|A(1-θ)/2zm(t)|2+C2RTμ0|k-k0|+12|k-k0|∫0t‍μ(s)|A(1-θ)/2zm(s)|2ds.
Combining this inequality with (105), we derive
(112)12|A-θ/2zm′(t)|2+(1-|k-k0|)2μ(t)|A(1-θ)/2zm(t)|2+k02|A-θ/2zm(t)|2≤12μ0∫0t‍|μ′(s)|[μ(s)|A(1-θ)/2zm(s)|2]ds+[2C2Rμ02+C2μ0RT]|k-k0|+12|k-k0|∫0t‍μ(s)|A(1-θ)/2zm(s)|2ds.
Considering |k-k0|<1/2 and using the Gronwall inequality, this expression gives
(113)12|A-θ/2zm′(t)|2+14μ0|A(1-θ)/2zm(t)|2+k02|A-θ/2zm(t)|2≤R1|k-k0|,hhhhhhhhhhhhhhhhhhhhh∀t∈[0,T],
where the constant R1>0 is independent of m and k. Taking the lim inf in both sides of this inequality, we find
(114)12|A-θ/2(uk′(t)-uk0′(t))|2+μ04|A(1-θ)/2(uk(t)-uk0(t))|2≤R1|k-k0|,∀t∈[0,T].
By Simon [14] and noting that the embedding of Y in X is compact, we derive
(115)uk⟶uk0inC0([0,T];X),k⟶k0.
Thus.                       (116)∥uk∥C0([0,T];X)⟶∥uk0∥C0([0,T];X),k⟶k0
which proves the continuity of ψ at k0>0. In similar way we prove the continuity of ψ at k0=0.(III) One has
ψ(k)→0 as k→∞.Let(kl) be a sequence of positive numbers with kl→∞. It follows from (98) and the compactness of the embedding of Y in X that there exists a subsequence of (kl), still denoted by (kl), and a function χ such that
(117)ukl⟶χinC0([0,T];X).
This implies
(118)ukl⟶χinL2(0,T;Z).
By estimate (91), we obtain
(119)k2∥uk∥L2(0,T;Z)2≤R,∀k≥0.
Then
(120)ukl⟶0inL2(0,T;Z).
Convergences (118) and (120) provide
(121)χ=0.
Thus by (117) we find
(122)ukl⟶0inC0([0,T];X).
As the sequence (kl) was arbitrary and the limit of (ukl) is always the same, we conclude that
(123)uk⟶0inC0([0,T];X),k⟶∞.
Thus
(124)∥uk∥C0([0,T];X)⟶0,k⟶∞
which proves part (III).By (I)–(III), we deduce that there existsk>0,k∈ℝ such that
(125)ψ(k)=k.
Considering this k in (97), we obtain a solution u of (22) that satisfies all conditions of the theorem.The proof of Corollary5 follows by defining the map
(126)ψ:[0,∞)⟶ℝ,ψ(k)=∥uk∥Lp(0,T;X)p,
where uk is the solution of the problem
(127)uk′′+μAuk+kuk=finL∞(0,T;D(A-(1+θ)/2));u(0)=0,u′(0)=u1,
and applying similar arguments to those used in the proof of Theorem 3.


## 5. Proof of Theorem6

Letu and z be solutions of Problem (26) with u and z in class (25). Consider w=u-z. Then by (26) we have
(128)w′′+μAw+|u|2pw+(|u|2p-|z|2p)z=0inL∞(0,T;D(A-1/2)),w(0)=0,w′(0)=0.Fix0<s≤T. Consider 0≤t≤s. Introduce the function
(129)y(t)=∫st‍w(τ)dτ.
We have
(130)y∈C0([0,s];D(A1/2));y(s)=0,y′(t)=w(t)a.e.t∈(0,s);A1/2y(t)=∫st‍A1/2w(τ)dτ,A1/2y′(t)=A1/2w(t);y(t)=-∫0s‍w(τ)dτ+∫0t‍w(τ)dτ.
Apply the operator given by the first member of (128)1 to y(t). We obtain
(131)〈w′′(t),y(t)〉E′×E+μ(t)〈Aw(t),y(t)〉E′×E+|u(t)|2p(w(t),y(t))+(|u(t)|2p-|z(t)|2p)(z(t),y(t))=0,a.e.t∈(0,s),
where E=D(A1/2). By (130)2, we find
(132)〈w′′(t),y(t)〉E′×E=ddt〈w′(t),w(t)〉-12ddt|w(t)|2.
Also by (130)3,
(133)μ(t)〈Aw(t),y(t)〉E′×E=12ddt[μ(t)|A1/2y(t)|2]-12μ′(t)|A1/2y(t)|2.
Integrate (131) on [0,s] and use (128)1, (130), (132), and (133). We deduce
(134)12|w(s)|2+12μ(0)|∫0s‍A1/2w(t)dt|2=-12∫0s‍μ′(t)|A1/2y(t)|2dt+∫0s‍|u(t)|2p(w(t),y(t))dt+∫0s‍[|u(t)|2p-|z(t)|2p](z(t),y(t))dt.
We modify each term of the second member of (134). We have
(135)A1/2y(t)=∫0t‍A1/2w(τ)dτ-∫0s‍A1/2w(τ)dτ.
Then
(136)∫0s‍|μ′(t)∥A1/2y(t)|2dt≤2∫0s‍|μ′(t)||∫0t‍A1/2w(τ)dτ|2dt+2∫0s‍|μ′(t)||∫0s‍A1/2w(τ)dτ|2dt≤2∫0s‍|μ′(t)|μ0μ(0)|∫0t‍A1/2w(τ)dτ|2dt+2|∫0s‍A1/2w(τ)dτ|2∫0s‍|μ′(t)|dt.
First, we assume that μ′≠0. In this case, we choose s0=min⁡{μ(0)/8∥μ′∥L∞(0,T),1,T}>0. We have
(137)2|∫0s‍A1/2w(τ)dτ|2∫0s‍|μ′(t)|dt≤14μ(0)|∫0s‍A1/2w(τ)dτ|2,hhhhhhhhhhhhhhhhh0≤s≤s0.
Combining this last inequality with (136), we find
(138)∫0s‍|μ′(t)∥A1/2y(t)|2dt≤2∫0s‍|μ′(t)|μ0[μ(0)|∫0t‍A1/2w(τ)dτ|2]dt+14μ(0)|∫0s‍A1/2w(τ)dτ|2,0≤s≤s0.
We introduce the notations
(139)∥u∥L∞(0,T;H)≤S1,∥z∥L∞(0,T;H)≤S2.
(Note that w∈C0([0,T];H) since w belongs to class (25)). We have
(140)|∫0s‍|u(t)|2p(w(t),y(t))dt|≤12S12p∫0s‍|w(t)|2dt+12S12p∫0s‍|y(t)|2dt.
By (130)4, we obtain
(141)|y(t)|2≤2[∫0s‍|w(τ)|dτ]2+2[∫0t‍|w(τ)|dτ]2≤4s∫0s‍|w(τ)|2dτ.
Therefore
(142)∫0s‍|y(t)|2dt≤4s2∫0s‍|w(t)|2dt.
Combining (140) and (142), we deduce
(143)|∫0s‍|u(t)|2p(w(t),y(t))dt|≤S12p∫0s‍12|w(t)|2dt+4S12ps2∫0s‍12|w(t)|2dt.The preceding inequality with0≤s≤s0 gives
(144)|∫0s|u(t)|2p(w(t),y(t))dt|≤5S12p∫0s12|w(t)|2dt.
On the other hand,
(145)|u(t)|2p-|z(t)|2p=(|u(t)|p+|z(t)|p)(|u(t)|p-|z(t)|p),||u(t)|p-|z(t)|p|≤p(S1+S2)p-1|w(t)|.
Hence,
(146)||u(t)|2p-|z(t)|2p|≤N|w(t)|,
where
(147)N=p(S1p+S2p)(S1+S2)p-1.
Thus estimate (146) gives
(148)|∫0s‍(|u(t)|2p-|z(t)|2p)(z(t),y(t))dt|≤NS2∫0s‍|w(t)||y(t)|dt≤NS2∫0s‍12|w(t)|2dt+12NS2∫0s‍|y(t)|2dt.
This expressions, (142), and 0≤s≤s0 provide
(149)|∫0s‍(|u(t)|2p-|z(t)|2p)(z(t),y(t))dt|≤5NS2∫0s‍12|w(t)|2dt.
Combining inequalities (138), (144), and (149) with inequality (134), we obtain
(150)12|w(s)|2+14μ(0)|∫0s‍A1/2w(τ)dτ|2≤2∫0s‍|μ′(t)|μ0[μ(0)|∫0t‍A1/2w(τ)dτ|2]dt+(5S12p+5NS2)∫0s‍12|w(t)|2dt,0≤s≤s0.
This implies
(151)w(s)=0,0≤s≤s0.We will prove thatw′(s0)=0. In fact as w belongs to class (25) we have that w′∈Cs0([0,s0];H); that is, (w′(t),ξ) is continuous on [0,s0] for all ξ∈H. Consider ξ∈D(A). Then by (128)1 we obtain
(152)〈w′′(t),ξ〉E′×E+〈Aw(t),ξ〉E′×E+|u(t)|2p(w(t),ξ)+(|u(t)|2p-|z(t)|2p)(z(t),ξ)=0,a.e.t∈(0,s0).
Integrating this equality on [0,s0] and using (151), we derive
(153)(w′(s0),ξ)+∫0s0‍(|u(t)|2p-|z(t)|2p)(z(t),ξ)dt=0.
This, (146), and (151) give
(154)|(w′(s0),ξ)|≤∫0s0‍|(|u(t)|2p-|z(t)|2p)∥z(t)∥ξ|dt≤∫01‍N|w(t)∥z(t)∥ξ|dt=0.
Therefore, (w′(s0),ξ)=0 for all ξ∈D(A). By density we obtain (w′(s0),ξ)=0 for all ξ∈H. Thus w′(s0)=0. Note that the constants S1,  S2, and N given, respectively, by (139) and (147) are independent of 0≤s≤s0.We apply similar arguments to the problem(155)w′′+Aw+|u|2pw+(|u|2p-|z|2p)z=0inL∞(s0,2s0;D(A-1/2)),w(s0)=0,w′(s0)=0
and we obtain w(s)=0 for s0≤s≤2s0. After a finite number of steps we proof that w(s)=0 for 0≤s≤T.Whenμ′=0, that is, μ(t)=μ0, for all t∈[0,T], expressions (144), (149), and similar arguments used to obtain the preceding result, allow us to deduce the uniqueness of solutions in this case.


## 6. Applications

LetΩ be an open bounded set of ℝn with boundary Γ of class C∞. Let A=-Δ be the self-adjoint operator defined by the triplet {H01(Ω),L2(Ω),((u,v))}, where ((u,v)) denotes the usual scalar product of H01(Ω). The norm of L2(Ω) is denoted by |u|.Lemma 11.
Letθ≥-1 be a real number. Then, D(A(1+θ)/2) is contained in H1+θ(Ω) and
(156)theembeddingofD(A(1+θ)/2)inH1+θ(Ω)iscontinuous.Proof.
First we prove (156) when (1+θ)/2=m, m a natural number. More precisely, we prove that if u∈D(Am), then u∈H2m(Ω) and
(157)∥u∥H2m(Ω)≤Cm|Amu|.
To prove (157), we use the method of mathematical induction. Consider m=1 and u∈D(A). Then, by the regularity of solutions of elliptic problems, we obtain u∈H2(Ω) and
(158)∥u∥H2(Ω)≤C1|Au|.
Assume that (157) holds for m=h. Consider u∈D(Ah+1). Then, Au∈D(Ah). By induction hypothesis it follows that Au∈H2h(Ω) and
(159)∥Au∥H2h(Ω)≤Ch|Ah(Au)|=|Ah+1u|.
Consider the problem
(160)Au=ginΩ,u=0on⁡Γ,
where g=Au. As g∈H2h(Ω), by the regularity of solutions of elliptic problems, we have u∈H2h+2(Ω) and
(161)∥u∥H2h+2(Ω)≤C*∥Au∥H2h(Ω).
Inequalities (159) and (161) provide
(162)∥u∥H2h+2(Ω)≤C*Ch|Ah+1u|=Ch+1|Ah+1u|.
This and (158) prove (157).
Next we use the interpolation of Hilbert spaces. Consider a natural numberm such that 2m>1+θ. By results of intermediate spaces, we have
(163)[H2m(Ω),L2(Ω)]γ=H(1-γ)2m(Ω),0≤γ≤1.
(cf. Lions and Magenes [8]). We have by (157) that the injections
(164)D(Am)⟶H2m(Ω),D(A0)=L2(Ω)⟶L2(Ω)
are continuous. Then, by interpolation of Hilbert spaces, we have that the injection
(165)[D(Am),L2(Ω)]γ⟶[H2m(Ω),L2(Ω)]γ,0≤γ≤1
is continuous. We choose γ0=1-(1+θ)/2m. Then, by (163), we obtain
(166)[H2m(Ω),L2(Ω)]γ0=H1+θ(Ω).
Also
(167)[D(Am),L2(Ω)]γ0=D(A(1+θ)/2).
These last two equalities and (165) give the lemma.Consider the operatorδ(x-x0), x0∈Ω. For θ>(n/2)-1; we have that
(168)H1+θ(Ω)↪C0(Ω¯).
This embedding and (156) imply that
(169)theembeddingofD(A(1+θ)/2)inC0(Ω¯)iscontinuous(θ>n2-1).Define(170)〈δ(x-x0),φ〉=φ(x0),φ∈D(A(1+θ)/2).
By (169) we have that δ(x-x0)∈D(A-(1+θ)/2). Thus, for
(171)f=vδ(x-x0)withv∈W1,1(0,T),δ(x-x0)∈D(A-(1+θ)/2)(θ>n2-1),
Theorems 1 and 3 give, respectively, solutions u of problems (14) and (22).Consider (171) for the particular case θ=0. Then Theorems 1 and 6 provide a unique solution u of the problem
(172)utt-μuxx+(∫ab‍|u|2dx)pu=vδ(x-x0)in(a,b)×(0,T)(p≥1);u(a,t)=u(b,t)=0,t∈(0,T);u(x,0)=u0(x),ut(x,0)=u1(x),x∈(a,b).
On the other hand, we have X=L2(Ω) if λ=0 (X defined in (21)). In this case -θ/2≤0<(1-θ)/2, that is, 0≤θ<1. The two restrictions θ>n/2-1 and 0≤θ<1 give, respectively, for n=1,2, and 3 the variations 0≤θ<1, 0<θ<1, and 1/2<θ<1. In all three cases, Corollary 5 gives a solution u of the problem
(173)u′′-μΔu+[∫0T‍(∫Ω‍|u|2dx)p/2dt]u=vδ(x-x0)inΩ×(0,T)(p≥1);u=0inΓ×(0,T);u(x,0)=u0(x),u′(x,0)=u1(x),x∈Ω,
where Ω is an open bounded set of ℝn, n=1,2,3.



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*Source: 305718-2014-03-24.xml*