In this section we establish a local existence result for the solution to problem (1) under suitable conditions on m and p.

First, we give the following local existence result.

Proof.

Step
1. For v(x,t) given, we consider the local existence of the problem
(13)utt+A2u-γAut+autm-2ut=bvp-2v,x∈Ω, t∈0,T,u=∂u∂νA=0, x∈Γ, t∈(0,T),ux,0=u0x, ut(x,0)=u1(x), x∈Ω-.
We take sequences {u0μ},{u1μ}∈C0∞(Ω) to approximate u0 and u1, respectively, and take a sequence {vμ}∈C([0,T],C0∞(Ω)) to approximate v. Then we consider the problems
(14)uttμ+A2uμ-γAutμ+autμm-2utμ=bvμp-2vμ,x∈Ω, t∈0,T,uμ=∂uμ∂νA=0, x∈Γ, t∈(0,T),uμx,0=u0μ, utμ(x,0)=u1μ, x∈Ω-.
Using the same arguments as in [28], we get the existence of a sequence of unique solutions {(uμ,utμ)} of (14) satisfying
(15)uμ∈L∞0,T,H,utμ∈L∞0,T,W∩LmΩ×0,T,uttμ∈L∞0,T,L2Ω.

For R>0 large and T>0, we let ZT,R denote the set of all functions w which satisfy
(16)w∈C0,T,W,wt∈C([0,T],L2(Ω))∩Lm(Ω×(0,T)),w(x,0)=u0(x), wt(x,0)=u1(x), x∈Ω-,hwt∶=wtC([0,T],L2(Ω))2+AwC([0,T],L2(Ω))2 +∇gwgL2((0,T)×Ω)2+wtLm(Ω×(0,T))2≤R2.
It follows from the trace theorem that ZT,R is nonempty if R is sufficiently large and T is small. For example, if we let
(17)w(x,t)=u0xcost+u1xsint,
then w∈ZT,R for R suitable large and T small. In the present section, we always make this assumption.

Step
2. We proceed to show that the sequence {(uμ,utμ)} is Cauchy in ZT,R. For this aim, we set
(18)U∶=uμ-uν, V∶=vμ-vν,
and then U satisfies
(19)Utt+A2U-γAUt+autμm-2utμ-autνm-2utνUtt+=bvμp-2vμ-bvνp-2vν, x∈Ω, t∈(0,T),U=∂U∂νA=0, x∈Γ, t∈(0,T),Ux,0=U0x=u0μ-u0ν, Utx,0=U1x=u1μ-u1ν,x∈Ω-.
We multiply the equation of (19) by Ut and integrate over Ω×(0,t); then it follows that
(20)∫ΩUt2+AU2dx+2γ∫0t∫Ω∇gUsg2 dx ds +2a∫0t∫Ωusμm-2usμ-usνm-2usνusμ-usνdx ds =∫ΩU12+AU02dx +2b∫0t∫Ωvμp-2vμ-vνp-2vνUs dx ds.
To estimate from the above second term on the right side of (20), we use the inequality
(21)vμp-2vμ-vνp-2vν ≤C1vμ-vνvμp-2-vνp-2
for vμ,vν∈R and p≥2. Note that Ci (i≥1) denote positive constants depending only on p, m, γ, a, b, and Ω rather than the initial data (u0,u1) in this section. To estimate the second term on the left side of (20), we use the inequality
(22)wm-2w-w~m-2w~w-w~≥C2w-w~m
for w,w~∈R and m≥2 with w=usμ, w~=usν. Then (20)–(22) yield
(23)∫ΩUt2+AU2dx+2γ∫0t∫Ω∇gUsg2 dx ds +2aC2∫0t∫Ωusμ-usνm dx ds ≤∫ΩU12+AU02dx +2bC1∫0t∫Ωvμ-vνUsvμp-2-vνp-2dx ds.
From Hölder’s inequality and the Sobolev embedding theorem the last term of (23) takes the form
(24)I∶=2bC1∫0t∫Ωvμ-vνUsvμp-2-vνp-2dx ds≤2bC3∫0tUsV2n/(n-4)vμn(p-2)/2p-2+vνn(p-2)/2p-2ds≤2bC4∫0tUsAVAvμp-2+Avνp-2ds≤4bC4Rp-2∫0tUsAVds.
Using Young’s inequality, we have
(25)I≤2bC4Rp-2∫0tUs2+AV2ds≤2bC4Rp-2∫0tUs2ds+2bC4Rp-2Tmaxt∈0,TAV2.
Combining (23) and (25), we obtain, for any t∈[0,T],
(26)Ut2+AU2+2γ∇gUtgL2((0,T)×Ω)2 +2aC2UtLm((0,T)×Ω)m ≤U12+AU02+2bC4Rp-2Tmaxt∈0,TAV2 +2bC4Rp-2∫0tUs2 ds.
It follows from (26) and Gronwall’s inequality that, for any t∈[0,T],
(27)Ut2+AU2+2γ∇gUtgL2((0,t)×Ω)2 +2aC2UtLm((0,t)×Ω)m≤U12+AU02+2bC4Rp-2Tmaxt∈[0,T]AV2e2bC4Rp-2t.
Furthermore, we have
(28)maxt∈[0,T]Ut2+maxt∈0,TAU2+2γ∇gUtgL2((0,T)×Ω)2 +2aC2UtLm((0,T)×Ω)m≤U12+AU02+2bC4Rp-2Tmaxt∈[0,T]AV2e2bC4Rp-2T.
Since {vμ}, {u0μ}, and {u1μ} are Cauchy in C([0,T];W), W, and L2(Ω), respectively, we conclude that {uμ}, {utμ}, {|∇gutμ|g}, and {utμ} are Cauchy in C([0,T];W), C([0,T];L2(Ω)), L2((0,T)×Ω), and Lm((0,T)×Ω), respectively.

Step
3. We now prove that the limit (u(x,t),ut(x,t)) is a weak solution of (13).

To this end, we multiply equation of (14) by w∈W and integrate over Ω; then we obtain that
(29)ddt∫Ωutμw dx+∫ΩAuμAw dx+γ∫Ω∇gutμ,∇gwg dx +a∫Ωutμm-2utμw dx=b∫Ωvμp-2vμw dx.
As μ→∞, the following hold:
(30)∫ΩAuμAw dx⟶∫ΩAuAw dx, in C([0,T]),∫Ωvμp-2vμw dx⟶∫Ωvp-2vw dx, in C([0,T]),∫Ω∇gutμ,∇gwg dx⟶∫Ω∇gut,∇gwg dx,in C0,T,∫Ωutμm-2utμw dx⟶∫Ωutm-2utw dx, in L1((0,T)).
Then it follows that
(31)∫Ωutw dx∶=limμ→∞∫Ωutμw dx
is an absolutely continuous function on [0,T]; thus for almost all t∈[0,T], (u(x,t),ut(x,t)) is a weak solution of problem (13). To prove uniqueness, we denote that uμ, uν are the corresponding solutions of problem (13) to vμ, vν, respectively. Then U=uμ-uν satisfies
(32)∫ΩUt2+AU2dx+2γ∫0t∫Ω∇gUsg2dx ds +2aC2∫0t∫Ωusμ-usνmdx ds≤2bC1∫0t∫Ωvμ-vνUsvμp-2-vνp-2dx ds.
This shows that U=0 for vμ=vν. The uniqueness follows.

Step
4. We denote by K the map which carries v∈ZT,R into u; that is,
(33)Kv=u,
where (u,ut) is the solution of problem (13). We establish a priori estimate below to show that K maps ZT,R into itself if R is sufficiently large and T is sufficiently small relative to R. We then equip ZT,R with the complete metric ρ defined by
(34)ρw,w-=supt∈0,Thwt-w-t
and show that K is strict contraction if T is sufficiently small. The contraction mapping principle thus implies that K has a unique fixed point which is obviously a solution to problem (1). For this purpose, we multiply the equation of (13) by 2ut and integrate over Ω×(0,T) to get that
(35)∫Ωut2+Au2dx+2γ∫0t∫Ω∇gusg2dx ds +2a∫0t∫Ωusm dx ds=∫Ωu12+Au02dx+2b∫0t∫Ωvp-2vus dx ds.
From Hölder’s inequality, it follows that, for t∈[0,T],
(36)∫Ωut2+Au2dx+2γ∫0t∫Ω∇gusg2dx ds +2a∫0t∫Ωusm dx ds≤∫Ωu12+Au02dx+2b∫0t∫Ωusvp-1dx ds≤∫Ωu12+Au02dx+2bC5∫0tusAvp-1ds≤∫Ωu12+Au02dx+2bC5maxt∈[0,T]Avp-1∫0tusds≤∫Ωu12+Au02dx+2bC5maxt∈0,TAvp-1 ×maxt∈[0,T]ut∫0tds≤∫Ωu12+Au02dx+2bC5TRp-1maxt∈[0,T]ut.
By choosing R large enough and then T sufficiently small, we obtain
(37)u∈ZT,R.
This shows that K maps ZT,R into itself.

Step
5. We verify that K is a contraction if T is sufficiently small. Let v,v-∈ZT,R, and set u=Kv and u-=Kv-. Clearly, U:=u-u- is the solution of the problem
(38)Utt+A2U-γAUt+autm-2ut-au-tm-2u-t =bvp-2v-bv-p-2v- x∈Ω, t∈(0,T),U=∂U∂νA=0, x∈Γ, t∈(0,T),U(x,0)=0, Ut(x,0)=0, x∈Ω-.
Multiplying the equation of (38) by 2Ut and integrating it over Ω, we have
(39)ddtUt2+AU2+2γ∫0t∫Ω∇gUsg2 dx ds + 2a∫0t∫Ωusm-2us-u-sm-2u-sus-u-sdx ds =2b∫ΩUtvp-2v-v-p-2v-dx.
Noticing U(x,0)=Ut(x,0)=0 and using the same arguments as (20)–(28), we get the estimate
(40)ρ(u,u-)≤2bC6RpTe2bC6Rp-2Tρ(v,v-),
where C6 is a positive constant independent of R and T. By choosing T so small that
(41)2bC6RpTe2bC6Rp-2T<1,
then the map K is a contraction. By the contraction mapping principle, the map K has a unique fixed point which is obviously a solution u=Kv∈ZR,T. It is clear that (u,ut) is the desired solution of problem (1), and the proof of Theorem 1 is completed.

Next we present two lemmas which will be used in the following sections.