# Analytic Inequalities and Their Applications in PDEs by Yuming Qin

By Yuming Qin

This e-book provides a couple of analytic inequalities and their purposes in partial differential equations. those contain vital inequalities, differential inequalities and distinction inequalities, which play a vital function in constructing (uniform) bounds, international life, large-time habit, decay charges and blow-up of options to numerous sessions of evolutionary differential equations. Summarizing effects from an unlimited variety of literature resources equivalent to released papers, preprints and books, it categorizes inequalities by way of their varied houses.

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Additional info for Analytic Inequalities and Their Applications in PDEs (Operator Theory: Advances and Applications)

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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.

Let b ≥ 0, β > 0, and let a(t) be a non-negative function locally integrable on [0, T ) for some T ≤ +∞. 50) 0 on this interval. Then for all 0 ≤ t < T , t u(t) ≤ a(t) + 0 +∞ (bΓ(β))n (t − s)nβ−1 a(s) ds. 6 ([355]). 5, let a(t) be a nondecreasing function on [0, T ). 52) +∞ zk k=0 Γ(kβ+1) . 6, which concerns a nonlinear integral inequality. 1) are special cases of this nonlinear one. First let us deﬁne a special class of nonlinear functions. 1. Let q > 0 be a real number and 0 < T ≤ +∞. We say that a function ω : R+ = [0, +∞) → R satisﬁes condition (q), if for all u ∈ R+ , t ∈ [0, T ), e−qt [ω(u)]q ≤ R(t)ω(e−qt uq ), where R(t) is a continuous non-negative function.

32) by means of Stirling’s formula, is taken from [941]. 4 with β = 0 and α ∈ (0, 1). 4 ([980]). Let v(t) ≥ 0 be continuous on [t0 , T ]. 40) t0 then there is a constant M > 0, independent of a, such that v(t) ≤ M a. 41) Proof. 40) and exploiting the identity t (t − s)−α−1 (s − τ )β−1 ds = (t − τ )α+β−1 0 Γ(α)Γ(β) , Γ(α + β) we obtain t v(t) ≤ a + b s (t − s)α−1 a + b t0 ≤a 1+b (T − t0 )α a (s − τ )α−1 v(τ )dτ ds t0 t t (t − s)α−1 (s − τ )α−1 ds v(τ )dτ + b2 t0 (T − t0 )α = a(1 + b) 1 + b a + b2 τ Γ2 (α) Γ(2α) t (t − τ )2α−1 v(τ )dτ, t0 which implies that n−1 v(t) ≤ a j=0 b(T − t0 )α a j + [bΓ(α)]n Γ(nα) t (t − τ )nα−1 v(τ )dτ.