min{γ, β}. We have the estimates
E(φ)≤KRnk+−l
k+−β,
F(φ)≤KRnk−−l
k−−β,
(37)
and
e
E(φ)≤KRnm+−l
m+−γ;
e
F(φ)≤KRnm−−l
m−−γ.
(38)
Finally, using (37)-(38) and the estimates (36), one
obtains
W≤K(Rζ1+Rζ2+Rζ3+Rζ4),(39)
Z≤K(Rλ1+Rλ2+Rλ3+Rλ4),(40)
where
ζ1=m+k+
m+k+−l2hnm+k+−l2
m+k+−β−lγ
k+i,
ζ2=m−k+
m−k+−l2hnm−k+−l2
m−k+−β−lγ
k+i,
ζ3=m+k−
m+k−−l2hnm+k−−l2
m+k−−β−lγ
k−i,
ζ4=m−k−
m−k−−l2hnm−k−−l2
m−k−−β−lγ
k−i,
and
λ1=k+m+
k+m+−l2hnm+k+−l2
m+k+−γ−lβ
m+i,
λ2=k−m+
k−m+−l2hnk−m+−l2
k−m+−γ−lβ
m+i,
λ3=k+m−
k+m−−l2hnk+m−−l2
k+m−−γ−lβ
m−i,
λ4=k−m−
k−m−−l2hnk−m−−l2
k−m−−γ−lβ
m−i.
Moreover, one has
ζ1= maxnζ1, ζ2, ζ3, ζ4o,
λ1= maxnλ1, λ2, λ3, λ4o.
(41)
1. If ζ1<0, the right hand side of (39) goes to
0, as Rgoes to innity, one can easily see that
Bm(.)(w) = ZΩ
|w(x)|m(x)dx = 0,
which implies that ∥w∥m(.)= 0 ie w= 0.
Then by 24 one obtains Y= 0 which yields
ϑ= 0. Hence (w, ϑ) = (0,0).
2. If λ1<0, the right hand side of (40) goes to
0, as Rgoes to innity, one obtains
Bm(.)(ϑ) = ZΩ
|ϑ(x)|k(x)dx = 0,
which implies that ∥ϑ∥p(.)= 0 ie ϑ= 0. Then
by 25 one obtains W= 0 which yields w= 0.
Hence (w, ϑ) = (0,0). Then, one can check
easily that under the condition 23, the only
weak solution to 2 is (w, ϑ) = (0,0). This
completes the proof of Theorem 2.2.
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WSEAS TRANSACTIONS on MATHEMATICS
DOI: 10.37394/23206.2024.23.52
Mohammed Yahiaoui, Ali Hakem