Download Analytic Inequalities and Their Applications in PDEs by Yuming Qin PDF

By Yuming Qin

This e-book offers a couple of analytic inequalities and their functions in partial differential equations. those comprise crucial inequalities, differential inequalities and distinction inequalities, which play a vital position in setting up (uniform) bounds, international life, large-time habit, decay charges and blow-up of ideas to numerous sessions of evolutionary differential equations. Summarizing effects from an unlimited variety of literature resources resembling released papers, preprints and books, it categorizes inequalities by way of their diverse properties.

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Since 1/p + 1/q = 1, H¨older’s inequality yields t u(t) ≤ a(t) + (t − s)β−1 F (s)ω(u(s))ds 0 t = a(t) + (t − s)−α es F (s)e−s ω(u(s))ds 0 t ≤ a(t) + 1/p t (t − s)−ap eps ds 0 1/q F q (s)e−qs ω q (u(s))ds . 72), we have t t (t − s)−αp eps ds = ept 0 τ −αp e−pτ dτ = 0 ≤ ept p1−αp Γ(1 − αp). 73) 36 Chapter 1. Integral Inequalities Obviously, 1 − αp = condition (q), we get 1 (z+1)2 > 0 and so Γ(1 − αp) ∈ R. 61). 76) 0 where v(t) = e−t u(t) q , φ(t) = 2q−1 aq (t). 76). 79) d d Ω(V (t)) ≤ Ω(φ(t)) + 2q−1 Kzq F q (t)R(t).

0 Then J(t) satisfies J (t) ≤ Ct−α t−1/2 + εt−1 J(t) which implies d −ε t J(t) dt ≤ Ct−α−1/2−ε . 4) over (0, t) gives us t t−ε J(t) ≤ C s−α−1/2−ε ds. 5) 0 Here we used the assumption that lim t−ε J(t) = 0. t→0+ Assume that α + ε < 1/2. 3 yields t s−α−1/2−ε ds ≤ Ct−α+1/2−ε . 3). 1. 3) still holds. 1. 2 (The Kawashima–Nakao–Ono Inequality [423]). 7) 0 with some constants k0 , k1 > 0, α, β, γ ≥ 0 and 0 ≤ μ < 1.

4. The inequalities of Henry’s type 29 ∗ Proof. Let E := F := R and k(t, s) := B(t − s)−α for (t, s) ∈ JΔ . Then that −β ˙ Let a(t) := At and observe that k ∈ K(E, α) and ||k||(α),T = B for T ∈ J. 28) implies a ∈ K(E, β) and ||a||(β),T = A for T ∈ J. 23) that b := a + k ∗ u − u ∈ K(E, β). 2 implies that u = (a − b) + ω ∗ (a − b). 38) and that k ≥ 0 implies ω ≥ 0. Thus u ≤ a + ω ∗ a, ˙ that is, for almost all t ∈ J, t u(t) ≤ At−β + A ω(t − τ )τ −β dτ. 1, we have for all t > 0, ε > 0, ω(t) ≤ c(α, ε)Bt−α e(1+ε)μ(α,B)t .

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