By James C. Robinson

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**Extra info for Infinite-Dimensional Dynamical Systems An Introduction to Dissipative Parabolic PDEs and the Theory of Global Attractors_SOLUTIONS**

**Example text**

In this example ω(0) = 0 and ω(x) = {|x| = 1} if x = 0. So (B) = {(0, 0)} ∪ {|x| = 1} (which is clearly not connected). 4 so that [ (B)] = (B) as claimed. We show that, for a bounded set X , ω1 (X ) = {y : S(tn )xn → y}, where tn → ∞ and xn ∈ X , is equal to ω2 (X ) = S(s)X . t≥0 s≥t If y ∈ ω1 (X ) then clearly y∈ S(s)X s≥t for all t ≥ 0 and hence in y ∈ ω2 (X ). So ω1 (X ) ⊂ ω2 (X ). Conversely, if y ∈ ω2 (X ) then for any t ≥ 0 y∈ S(s)X , s≥t and so there are sequences {τm(t) }, with τm(t) ≥ t, and {xm(t) } ∈ X with S(τm(t) )xm(t) → y.

2 10:52 Char Count= 0 Solutions to Exercises giving equality between box-counting dimension and fractal dimension in Rm . 3 log N (X, ) log N (X, n ) . ≤ lim sup − log − log n n→∞ That this inequality holds in the opposite sense is straightforward, and hence we obtain the desired equality. √ The sequence m = ( 2 log m)−1 , m ≥ 2, satisfies m+1 = m log m log 2 ≥ , log(m + 1) log 3 and so we can use the result of the previous exercise. Note that we have en ek − log n log k 2 = 1 1 2 + ≤ 2 2 (log n) (log k) (log n)2 for n > k, and so the first m − 1 elements from Hlog will belong to distinct balls of radius m .

J=1 Since u 2 = a(u, u) = Au, u = (A1/2 u, A1/2 u) = |A1/2 u|2 , this becomes |b(u, u, A2 u)| ≤ 3k|A3/2 u||Au| u ≤ as required. 23) from the previous exercise to write 1 2 d |Au|2 + ν|A3/2 u|2 dt ≤ |b(u, u, A2 u)| + f |A3/2 u| C ν ≤ |A3/2 u|2 + u 2 |Au|2 + 4 ν f ν 2 ν + |A3/2 u|2 , 4 so that d 2 f |Au|2 + ν|A3/2 u|2 ≤ dt ν 2 + C u 2 |Au|2 . 24). Integrating again with respect to s between t and t + 1 gives t+1 |Au(t + 1)|2 ≤ |Au(s)|2 ds + t 2M C + ν ν t+1 u(s) 2 |Au(s)|2 ds. 3 C 2M + ρV2 I A , ν ν an absorbing set in D(A).