By Lorenzi L., Lunardi A., Metafune G., Pallara D.

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Moreover f ∈ C α ([0, T ], X) by assumption, so we have only to show that t → etA (f (t) − f (0)) is α-H¨older continuous. 21) t σ α−1 dσ + M0 [f ]C α (t − s)α ≤ M1 [f ]C α s ≤ M1 + M0 (t − s)α [f ]C α . α Hence Au2 is α-H¨older continuous as well, and the estimate u C 1+α ([0,T ];X) + Au C α ([0,T ];X) ≤ c( f C α ([0,T ],X) + x X + Ax + f (0) DA (α,∞) ) follows, since u = Au + f and u = u1 + u2 . Let us now estimate [u (t)]DA (α,∞) . For 0 ≤ t ≤ T we have t Ae(t−s)A (f (s) − f (t))ds + etA (Ax + f (0)) + etA (f (t) − f (0)), u (t) = 0 so that for 0 < ξ ≤ 1 we deduce t ξ 1−α AeξA u (t) ≤ ξ 1−α A2 e(t+ξ−s)A (f (s) − f (t))ds 0 + ξ 1−α Ae(t+ξ)A (Ax + f (0)) + ξ 1−α Ae(t+ξ)A (f (t) − f (0)) 58 Chapter 4.

Hint. 3. Second method: determine explicitly the resolvent operator using the changes of variables x = et and x = −et ]. 7) where f is a given function in X, X = Lp (RN ), 1 ≤ p < +∞, or X = Cb (RN ). To get a representation formula for the solution, let us apply (just formally) the Fourier transform, denoting by u ˆ(t, ξ) the Fourier transform of u with respect to the space variable x. We get   u ˆt (t, ξ) = −|ξ|2 u ˆ(t, ξ), t > 0, ξ ∈ RN ,  u ˆ(0, ξ) = fˆ(ξ), ξ ∈ RN , 2 whose solution is u ˆ(t, ξ) = fˆ(ξ)e−|ξ| t .

For 0 ≤ t ≤ T we have t Ae(t−s)A (f (s) − f (t))ds + etA (Ax + f (0)) + etA (f (t) − f (0)), u (t) = 0 so that for 0 < ξ ≤ 1 we deduce t ξ 1−α AeξA u (t) ≤ ξ 1−α A2 e(t+ξ−s)A (f (s) − f (t))ds 0 + ξ 1−α Ae(t+ξ)A (Ax + f (0)) + ξ 1−α Ae(t+ξ)A (f (t) − f (0)) 58 Chapter 4. 22) 0 + M0 [Ax + f (0)]DA (α,∞) + M1 [f ]C α ξ 1−α (t + ξ)−1 tα ∞ ≤ M2 [f ]C α 0 σ α (σ + 1)−2 dσ + M0 [Ax + f (0)]DA (α,∞) + M1 [f ]C α . Then, [u (t)]DA (α,∞) is bounded in [0, T ], and the proof is complete. 7 implies that the condition Ax + f (0) ∈ DA (α, ∞) is necessary in order that Au ∈ C α ([0, T ]; X).

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Analytic semigroups and reaction-diffusion problems by Lorenzi L., Lunardi A., Metafune G., Pallara D.


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