By Unknown Author

Hardbound.

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1 Well-Posedness for the Cauchy Problem with Fast Diﬀusion 19 In the second step we take θ0 ∈ L2 (Ω) = D(Bε ), f ∈ L2 (0, T ; V ). Since W 1,1 ([0, T ]; V ) is dense in L2 (0, T ; V ) and D(Bε ) = V is dense in L (Ω) we can take the sequences (fn )n≥1 ⊂ W 1,1 ([0, T ]; V ) and (θ0n )n≥1 ⊂ D(Bε ) such that 2 fn → f strongly in L2 (0, T ; V ), θ0n → θ0 strongly in L2 (Ω) as n → ∞. e. 58) for any t ∈ [0, T ]. We stress that ε is ﬁxed. 60) 2 dt + K T + 2ε, due to the strong convergence θ0n → θ0 and fn → f as n → ∞.

On Q. Proof. 27) corresponding to the same data f and θ0 . 27) written for y ∗ and y ∗ , multiply the diﬀerence scalarly in V by u(y ∗ − y ∗ )(t), and integrate over (0, t). 108) where A0 ψ = u(y ∗ − y ∗ ). 7) we obtain 1 u(y ∗ − y∗ )(t) 2 ≤ ρ 2 t 0 Ω 2 V t +ρ 0 Ω u(y ∗ − y ∗ )2 dxdτ u(y ∗ − y ∗ )2 dxdτ + 1 (N MK ku )2 2ρ t 0 u(y ∗ − y ∗ )(τ ) 2 V dτ. 2 Therefore, by Gronwall lemma (see [29]), u(y ∗ − y∗ )(t) V ≤ 0 and we deduce that uy ∗ (t) = uy∗ (t) for any t ∈ [0, T ]. e. on the set Qu where u(x) > 0.

1 Well-Posedness for the Cauchy Problem with Fast Diﬀusion 13 The operator Bε : D(Bε ) ⊂ V → V is single-valued, has the domain v uε v ∈ L2 (Ω); βε∗ D(Bε ) := ∈V and is deﬁned by Bε v, ψ := V ,V Ω v uε ∇βε∗ − K0 x, v uε · ∇ψdx, for any ψ ∈ V. 41) v uε In fact we note that Bε v = Aε and v ∈ D(Bε ) is equivalent to uvε ∈ D(Aε ). Also, it is easily seen that D(Bε ) = V. Indeed, if v ∈ D(Bε ) it follows that v ∗ uε ∈ V by the fact that the inverse of βε is Lipschitz, and from here we get 1,∞ that v ∈ V, since uε ∈ W (Ω).