1. Introduction The study of dynamic equations on time scales is now an important object of research and has been extensively studied in recent years (see [1–21]). This is due to the fact that the theory of time scales which was introduced by Stefan Hilger [22] can unify and extend the difference and differential calculus in a consistent way.
Because dynamic inequalities play an important role in the study of qualitative properties of solutions of dynamic equations on time scales, many authors have expounded on various classes of dynamic inequalities in recent years; see [23–41] and the references cited therein. For instance, in 2013, Sun and Hassan [32] investigated the nonlinear integral inequality on time scales(1)yt≤ft+gt∫t0tasys+bsyασs-csyβσsΔs,where y,f,g,a,b,c:Tκ→R+ are rd-continuous functions and α,β are positive constants such that 0<α<1<β.
Tian et al. [37] investigated the nonlinear integral inequality on time scales(2)ypt≤ft+gt∫t0tasyps+bsyqs+cs+∫t0smτyrτΔτ+h1syασs-h2syβσsΔs, t∈Tκ,where p≥q>0,p≥r>0,0<α<p<β,p,q,r,α and β are real constants and y,f,g,a,b,c,m,h1,h2:Tκ→R+ are rd-continuous functions.
We note that the inequalities (1) and (2) have been proved in the cases when 0<α<1<β and 0<α<p<β, respectively. So it would be interesting to find the explicit bound for u of (2) in the following two cases when 0<p<α<β or 0<β<α<p.
In the present paper, we study some new half-linear integral inequalities on time scales. Our results not only complement the results established in [37] in the sense that the results can be applied in cases when 0<p<α<β or 0<β<α<p, but also furnish a handy tool for the study of qualitative properties of solutions of some complicated dynamic equations.
2. Preliminaries In what follows, we always assume that R+=[0,∞), T is an arbitrary time scale. The following lemmas are useful in the proof of the main results of this paper.
Lemma 1. Let m>0,n>0,p>0,α>0, and β>0 be given; then for each x≥0,(3)mxα-nxβ≤mβ-αβ-pβ-pnα-pmα-p/α-βxpholds for the cases when 0<p<α<β or 0<β<α<p.
Proof. If x=0, then it is easy to see that the inequality (3) holds. So we only prove that the inequality (3) holds in the case of x>0. For the case 0<p<α<β, set F(x)=mxα-p-nxβ-p,x>0, where m>0 and n>0. Letting F′(x)=0, we get x0=(m(α-p)/n(β-p))1/(β-α). Since x∈(0,x0), F′(x)>0; x∈(x0,+∞), F′(x)<0, F attains its maximum at x0=(m(α-p)/n(β-p))1/(β-α) and Fmax=F(x0)=m(β-α)/β-p((β-p)n/(α-p)m)(α-p)/(α-β). Thus, (3) holds. For the case 0<β<α<p, by a similar argument with the case p<α<β, we can get that (3) holds. The proof is complete.
Lemma 2 (see [42]). Assuming that y≥0,p≥q≥0, and p≠0, then for any K>0,(4)yq/p≤qpKq-p/py+p-qpKq/p.
Lemma 3 (see [1, Theorem 1.117]). Suppose that for each ε>0 there exists a neighborhood U of t, independent of τ∈[t0,σ(t)], such that(5)wσt,τ-ws,τ-wtΔt,τσt-s≤εσt-s, s∈U,where w:T×Tκ→R+ is continuous at (t,t), t∈Tκ with t>t0, and wtΔ(t,·) are rd-continuous on [t0,σ(t)]. Then(6)ft≔∫t0twt,τΔτimplies(7)fΔt=∫t0twtΔt,τΔτ+wσt,t, t∈Tκ.
Lemma 4 (see [1, Theorem 6.1]). Suppose that y and f are rd-continuous functions and p∈R+. Then(8)yΔt≤ptyt+ft, for all t∈Timplies(9)yt≤yt0ept,t0+∫t0tept,στfτΔτ, for all t∈T.
3. Main Results In this section, we deal with some half-line inequalities on time scales. For convenience, we always assume that t≥t0,t∈Tκ.
Theorem 5. Assume that x,f,a,b,c,l,m,n:Tκ→R+, g,h,k:Tκ→(0,∞) are rd-continuous functions, (l/g)Δ(t)≥0, μ(t)F(t)<1 for t∈Tκ, θ:Tκ→Tκ is continuous function satisfying θ(t)≤t and θ(t0)=t0 for t∈Tκ, p,q,r,α, and β are constants satisfying (i) 0≤q≤p,0≤r≤p,p<α<β or (ii) 0≤q≤p,0≤r≤p,0<β<α<p.
Suppose that x satisfies(10)xpt≤ft+gt∫t0tasxps+bsxqs+cs+ms∫θt0θsnξxrξΔξΔs+lt∫t0thsxασs-ksxβσsΔs, t∈Tκ,then(11)xt≤ft+gt∫t0teG⊕Lt,σsAsΔs1/p, for any K1>0, K2>0, t∈Tκ,where(12)Ft≔lgΔt∫t0tHsgσsΔs+lσtgσtHtgσt,(13)Gt≔Ft1-μtFt,(14)Lt≔atgt+qpK1q-p/pbtgt+rpK2r-p/pmt∫θt0θtnξgξΔξ,(15)Mt≔atft+qpK1q-p/pbtft+p-qpK1q/pbt+ct+mt∫θt0θtnξrpK2r-p/pfξ+p-rpK2r/pΔξ+lgΔt∫t0tHsfσsΔs+lσtgσtHtfσt,(16)Ht≔htβ-αβ-pβ-pktα-phtα-p/α-β,(17)At≔1+μtGtMt.
Proof. From Lemma 1 and (10), we get that(18)xpt≤ft+gt∫t0tasxps+bsxqs+cs+ms∫θt0θsnξxrξΔξΔs+lt∫t0thsxασs-ksxβσsΔs≤ft+gt∫t0tasxps+bsxqs+cs+ms∫θt0θsnξxrξΔξΔs+lt∫t0tHsxpσsΔs, t∈Tκ,where H(t) is defined as in (16). Denote(19)zt=∫t0tasxps+bsxqs+cs+ms∫θt0θsnξxrξΔξΔs+ltgt∫t0tHsxpσsΔs, t∈Tκ.From the assumptions on x,a,b,c,m,n,θ,g,l,H, (18) and (19), we obtain that z is nondecreasing and(20)xt≤ft+gtzt1/p, t∈Tκ.Combining (19) and (20), we have that(21)zΔt=atxpt+btxqt+ct+mt∫θt0θtnξxrξΔξ+lgΔt∫t0tHsxpσsΔs+lσtgσtHtxpσt≤atft+gtzt+btft+gtztq/p+ct+mt∫θt0θtnξfξ+gξzξr/pΔξ+lgΔt∫t0tHsfσs+gσszσsΔs+lσtgσtHtfσt+gσtzσt, t∈Tκ.Applying Lemma 2 on the right side of (21), we get that(22)zΔt≤atft+gtzt+btqpK1q-p/pft+gtzt+p-qpK1q/p+ct+mt∫θt0θtnξrpK2r-p/pfξ+gξzξ+p-rpK2r/pΔξ+lgΔt∫t0tHsfσs+gσszσsΔs+lσtgσtHtfσt+gσtzσt≤atft+atgtzt+qpK1q-p/pbtft+gtzt+p-qpK1q/pbt+ct+rpK2r-p/pztmt∫θt0θtnξgξΔξ+mt∫θt0θtnξrpK2r-p/pfξ+p-rpK2r/pΔξ+zσtlgΔt∫t0tHsgσsΔs+lgΔt∫t0tHsfσsΔs+zσtlσtgσtHtgσt+lσtgσtHtfσt=Ltzt+Ftzσt+Mt,where F(t),L(t), and M(t) are defined as in (12), (14), and (15). From (13), we get (23)Ft=Gt1+μtGt, and by (22), we obtain (24)zΔt≤Ltzt+Gt1+μtGtzσt+Mt=Ltzt+Gt1+μtGtzt+μtzΔt+Mt, t∈Tκ. Hence(25)11+μtGtzΔt≤Gt1+μtGt+Ltzt+Mt, t∈Tκ, equivalently(26)zΔt≤Gt+1+μtGtLtzt+1+μtGtMt=G⊕Ltzt+1+μtGtMt=G⊕Ltzt+At, t∈Tκ,where A(t) is defined as in (17). Note that z is rd-continuous and G⊕L∈R+; from Lemma 4 and (26), we have that(27)zt≤∫t0teG⊕Lt,σsAsΔs, t∈Tκ.This and (20) imply the desired inequality (11). This completes the proof.
If we let g(t)=l(t), m(t)=1, and θ(t)=t in Theorem 5, then we obtain the following corollary.
Corollary 6. Assume that x,a,b,c,f,g,n,h,k,p,q,r,α,β and H are defined the same as in Theorem 5 and μ(t)F(t)<1 for t∈Tκ. Suppose that u satisfies(28)xpt≤ft+gt∫t0tasxps+bsxqs+cs+∫t0snξxrξΔξ+hsxασs-ksxβσsΔs, t∈Tκ, then(29)xt≤ft+gt∫t0teG⊕Lt,σsAsΔs1/p, for any K1>0, K2>0, t∈Tκ, where(30)Ft≔Htgσt,Gt≔Ft1-μtFt,(31)Lt≔atgt+qpK1q-p/pbtgt+rpK2r-p/p∫t0tnξgξΔξ,(32)Mt≔atft+qpK1q-p/pbtft+p-qpK1q/pbt+ct+∫t0tnξrpK2r-p/pfξ+p-rpK2r/pΔξ+Htfσt,(33)At≔1+μtGtMt.
Remark 7. Corollary 6 extends Theorem 2.1 in [37] to the cases p<α<β and 0<β<α<p.
Remark 8. If l(t)=0,c(t)=0, and m(t)=0, then Theorem 5 reduces to Theorem 3.2 in [28]. If l(t)=0,a(t)=0, and m(t)=0, then Theorem 5 reduces to Theorem 2.4 in [29].
Theorem 9. Assume that x,a,b,c,f,g,l,m,n,θ,h,k,p,q, r,α,β,H, and L are defined the same as in Theorem 5, and μ(t)F~(t)<1 for t∈Tκ. Let w(t,s) and v(t,s) be defined as in Lemma 3 such that wtΔ(t,s)≥0 and vtΔ(t,s)≥0 for t≥s and (5) holds. Suppose that x satisfies(34)xpt≤ft+gt∫t0twt,sasxps+bsxqs+cs+ms∫θt0θsnξxrξΔξΔs+lt∫t0tvt,shsxασs-ksxβσsΔs, t∈Tκ,then(35)xt≤ft+gt∫t0teG~⊕L~t,σsA~sΔs1/p, for any K1>0, K2>0, t∈Tκ,where(36)F~t≔lgΔt∫t0tvt,sHsgσsΔs+lσtgσt∫t0tvtΔt,sHsgσsΔs+vσt,tHtgσt,(37)G~t≔F~t1-μtF~t,(38)L~t≔∫t0twtΔt,sLsΔs+wσt,tLt,(39)M~t≔∫t0twtΔt,sasfs+bsqpK1q-p/pfs+p-qpK1q/p+cs+ms∫θt0θsnξrpK2r-p/pfξ+p-rpK2r/pΔξΔs+wσt,tatft+btqpK1q-p/pft+p-qpK1q/p+ct+mt∫θt0θtnξrpK2r-p/pfξ+p-rpK2r/pΔξ+lgΔt∫t0tvt,sHsfσsΔs(40)+lσtgσt∫t0tvtΔt,sHsfσsΔs+vσt,tHtfσt,(41)A~t≔1+μtG~tM~t,and L(t) is defined as in (14).
Proof. From Lemma 1 and (34), we get that(42)xpt≤ft+gt∫t0twt,sasxps+bsxqs+cs+ms∫θt0θsnξxrξΔξΔs+lt∫t0tvt,shsxασs-ksxβσsΔs≤ft+gt∫t0twt,sasxps+bsxqs+cs+ms∫θt0θsnξxrξΔξΔs+lt∫t0tvt,sHsxpσsΔs, t∈Tκ,where H(t) is defined as in (16). Denote(43)zt=∫t0twt,sasxps+bsxqs+cs+ms∫θt0θsnξxrξΔξΔs+ltgt∫t0tvt,sHsxpσsΔs, t∈Tκ.From the assumptions on x,w,v,a,b,c,m,n,θ,g,l,H, (42) and (43), we obtain that z is nondecreasing and(44)xt≤ft+gtzt1/p, t∈Tκ.Combining (43) and (44), we have that(45)zΔt=∫t0twtΔt,sasxps+bsxqs+cs+ms∫θt0θsnξxrξΔξΔs+wσt,tatxpt+btxqt+ct+mt∫θt0θtnξxrξΔξ+lgΔt∫t0tvt,sHsxpσsΔs+lσtgσt∫t0tvtΔt,sHsxpσsΔs+vσt,tHtxpσt≤∫t0twtΔt,sasfs+gszs+bsfs+gszsq/p+cs+ms∫θt0θsnξfξ+gξzξr/pΔξΔs+wσt,tatft+gtzt+btft+gtztq/p+ct+mt∫θt0θtnξfξ+gξzξr/pΔξ+lgΔt∫t0tvt,sHsfσs+gσszσsΔs+lσtgσt∫t0tvtΔt,sHsfσs+gσszσsΔs+vσt,tHtfσt+gσtzσt, t∈Tκ.Applying Lemma 2 on the right side of (45), we have that (46)zΔt≤∫t0twtΔt,sasfs+gszs+bsqpK1q-p/pfs+gszs+p-qpK1q/p+cs+ms∫θt0θsnξrpK2r-p/pfξ+gξzξ+p-rpK2r/pΔξΔs+wσt,tatft+gtzt+btqpK1q-p/pft+gtzt+p-qpK1q/p+ct+mt∫θt0θtnξrpK2r-p/pfξ+gξzξ+p-rpK2r/pΔξ+lgΔt∫t0tvt,sHsfσs+gσszσsΔs+lσtgσt∫t0tvtΔt,sHsfσs+gσszσsΔs+vσt,tHtfσt+gσtzσt≤zt∫t0twtΔt,sasgs+bsqpK1q-p/pgs+ms∫θt0θsnξrpK2r-p/pgξΔξΔs+∫t0twtΔt,sasfs+bsqpK1q-p/pfs+p-qpK1q/p+cs+ms∫θt0θsnξrpK2r-p/pfξ+p-rpK2r/pΔξΔs+ztwσt,tatgt+btqpK1q-p/pgt+mt∫θt0θtnξrpK2r-p/pgξΔξ+wσt,tatft+btqpK1q-p/pft+p-qpK1q/p+ct+mt∫θt0θtnξrpK2q/pr-p/pfξ+p-rpK2r/pΔξ+zσtlgΔt∫t0tvt,sHsgσsΔs+lσtgσt∫t0tvtΔt,sHsgσsΔs+vσt,tHtgσt+lgΔt∫t0tvt,sHsfσsΔs+lσtgσt∫t0tvtΔt,sHsfσsΔs+vσt,tHtfσt≤∫t0twtΔt,sLsΔs+wσt,tLtzt+lgΔt∫t0tvt,sHsgσsΔs+lσtgσt∫t0tvtΔt,sHsgσsΔs+vσt,tHtgσtzσt+∫t0twtΔt,sasfs+bsqpK1q-p/pfs+p-qpK1q/p+cs+ms∫θt0θsnξrpK2r-p/pfξ+p-rpK2r/pΔξΔs+wσt,tatft+btqpK1q-p/pft+p-qpK1q/p+ct+mt∫θt0θtnξrpK2r-p/pfξ+p-rpK2r/pΔξ+lgΔt∫t0tvt,sHsfσsΔs+lσtgσt∫t0tvtΔt,sHsfσsΔs+vσt,tHtfσt=L~tzt+F~tzσt+M~t,where F~(t),L~(t), and M~(t) are defined as in (36), (38), and (40). By a similar argument with Theorem 5 in the remaining proof of theorem, one can prove (35). This completes the proof.
If we let g(t)=l(t), m(t)=1, w(t,s)=v(t,s), and θ(t)=t in Theorem 9, then we obtain the following corollary.
Corollary 10. Assume that x,f,g,w,a,b,c,n,h,k,p,q,r,α,β, and H are defined the same as in Theorem 9 and μ(t)F~(t)<1 for t∈Tκ. Suppose that x satisfies(47)xpt≤ft+gt∫t0twt,sasxps+bsxqs+cs+∫t0snξxrξΔξ+hsxασs-ksxβσsΔs, t∈Tκ,then(48)xt≤ft+gt∫t0teG~⊕L~t,σsA~sΔs1/p, for any K1>0, K2>0, t∈Tκ, where(49)F~t≔∫t0twtΔt,sFsΔs+wσt,tFt,G~t≔F~t1-μtF~t,L~t≔∫t0twtΔt,sLsΔs+wσt,tLt,M~t≔∫t0twtΔt,sMsΔs+wσt,tMt,A~t=1+μtG~tM~t, and F(t), L(t), and M(t) are defined the same as in (30), (31), and (32).
Remark 11. Corollary 10 extends Theorem 2.2 in [37] to the cases p<α<β and 0<β<α<p.
Remark 12. If c(t)=0,m(t)=0, and l(t)=0, then Theorem 9 reduces to Theorem 3.8 in [28]. If m(t)=0 and l(t)=0, then Theorem 9 reduces to Theorem 2.11 in [29].
4. Applications In this section, we apply our results to study the boundedness of the solutions of a dynamic equation on time scales.
Example 13. Consider the following dynamic equation on time scales:(50)xptΔ=Pt,xt,xσt,∫θt0θtQs,xsΔs,(51)xt0=x0, t0, t∈Tκ, where p>0 and u0 are constants, θ:Tκ→Tκ is continuous function satisfying θ(t)≤t and θ(t0)=t0 for t∈Tκ, and P:Tκ×R×R×R→R and Q:Tκ×R→R are continuous functions.
Theorem 14. Suppose that the functions P and Q in (50) satisfy the conditions(52)Pt,u,v,w≤ct+atup+btuq+htvα-ktvβ+mtw, t∈Tκ, u,v,w∈R,(53)Qt,u≤ntur, t∈Tκ, u∈R,where a,b,c,m,n:Tκ→R+, and h,k:Tκ→(0,∞) are rd-continuous functions and q,r,α, and β are constants satisfying (i) 0≤q≤p,0≤r≤p,p<α<β or (ii) 0≤q≤p,0≤r≤p,0<β<α<p.
If x is a solution of (50) and μ(t)F(t)<1 for t∈Tκ, then(54)xt≤x0p+∫t0teG⊕Lt,σsAsΔs1/p, for any K1>0, K2>0, t∈Tκ,where(55)Ft≔htβ-αβ-pβ-pktα-phtα-p/α-β,(56)Gt≔Ft1-μtFt,(57)Lt≔at+qpK1q-p/pbt+rpK2r-p/pmt∫θt0θtnξΔξ,(58)Mt≔x0pat+qpK1q-p/px0pbt+p-qpK1q/pbt+ct+rpK2r-p/px0p+p-rpK2r/pmt∫θt0θtnξΔξ+x0pFt,(59)At≔1+μtGtMt.
Proof. The equivalent integral equation of (50) is denoted by(60)xpt=x0p+∫t0tPs,xs,xσs,∫θt0θsQξ,xξΔξΔs, t∈Tκ. Using the assumptions (52) and (53), we have (61)xpt≤x0p+∫t0tasxsp+bsxsq+cs+ms∫θt0θsnξxξrΔξ+hsxσsα-ksxσsβΔs, t∈Tκ.Then, a suitable application of Theorem 5 to (61) yields (54).