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Alternative Development of Semigroups Consider the abstract initial value problem u v ÝtÞ + AuÝtÞ = 0, uÝ0Þ = u 0 where A denotes an unbounded but closed and densely defined linear operator on a Banach space X. We recall from the previous development that if we approximate this problem by uÝt + hÞ ? uÝtÞ B ? h AuÝt + hÞ, then ÝI + hAÞ uÝt + hÞ B uÝtÞ for h ³ 0, or uÝt + hÞ B ÝI + hAÞ ?1 uÝtÞ This leads to or uÝhÞ B ÝI + hAÞ ?1 u 0 uÝ2hÞ B ÝI + hAÞ ?1 uÝhÞ B ÝI + hAÞ ?2 u 0 _ uÝnhÞ B ÝI + hAÞ ?n u 0 , uÝtÞ B I + nt A ?n u 0 for t ³ 0. This suggests that the solution of the abstract IVP is given by uÝtÞ = Lim n¸K I + nt A ?n u 0 for t ³ 0, Ý1Þ and raises the question of whether this limit exists under appropriate conditions on A. Theorem (Hille-Yosida) In order that the limit in Ý1Þ exist it is sufficient that i) A is closed and densely defined ii) ÝVI + AÞ ?1 exists and is bounded for V ³ 0 iii) ÝVI + AÞ ?1 LÝXÞ ² 1 for V > 0. V Proof- It follows from iii) that for W ³ 0, ÝI + WAÞ ?1 Now for n = 1, 2, ... define LÝXÞ ²1 S n ÝtÞ = Ý2Þ I + nt A ?n . Then 1 ||S n ÝtÞ|| LÝXÞ ² 1 for n = 1, 2, ... t ³ 0. i.e., the family áS n ÝtÞâ is uniformly bounded in the operator norm on LÝXÞ. Now write u ? ÝI + WAÞ ?1 u = WÝI + WAÞ ?1 Au -u 5 D A . Then it follows from Ý2Þ that u ? ÝI + WAÞ ?1 u X ² W || Au|| X -u 5 D A ., and this implies that -u 5 D A , we have u ? ÝI + WAÞ ?1 u i.e., X ¸ 0 as W ¸ 0; ÝI + WAÞ ?1 u ¸ u in X as W ¸ 0 -u 5 D A . Since D A is dense in X and ÝI + WAÞ ?1 is uniformly bounded for W ³ 0, it follows that ÝI + WAÞ ?1 converges to I (in the norm topology on X) as W ¸ 0 and S n ÝtÞ converges to I (in the norm topology on X) as t ¸ 0. Now suppose for the moment that we can show 2 || S n ÝtÞu ? S m ÝtÞu|| X ² t Ý 1n + 2 1 m Þ||A 2 u|| X -u 5 D 2 = DÝA 2 Þ. Ý3Þ Then Ý3Þ implies áS n ÝtÞuâ is a Cauchy sequence in X for every u in D 2 . But D 2 = ÝVI + AÞ ?1 D A = ÝVI + AÞ ?2 X, and since D A is dense in X, it follows that D 2 is also dense in X. Then, áS n ÝtÞâ is a uniformly bounded family of bounded operators on X with áS n ÝtÞuâ converging strongly for all u in a dense subset of X. It follows that for all t ³ 0, S n ÝtÞ converges strongly to a bounded linear operator, SÝtÞ, as n ¸ K. It remains only to prove Ý3Þ. Write S n ÝtÞu ? S m ÝtÞu = lim P¸0 X t?P d ds P ÝS m Ýt ? sÞ S n ÝsÞuÞ ds t?P = lim P¸0 X ßS m Ýt ? sÞ S vn ÝsÞu ? S vm Ýt ? sÞ S n ÝsÞuà ds. P But S vn ÝtÞ = A I + nt A ?n?1 hence S m Ýt ? sÞ S vn ÝsÞu ? S vm Ýt ? sÞ S n ÝsÞu = ?sA = A I+ tm = ?m?1 I + ns A t ? s ? s A2 I + t ? s A m n m ?n ?sA ?A I+ tm ?m?1 I + ns A ?m I + ns A ?n?1 ?n?1 2 Now A commutes with its resolvent and the family áS n ÝtÞâ is uniformly bounded which leads to t ?s + s || S n ÝtÞu ? S m ÝtÞu|| X ² ||A 2 u|| X X 0 t m n 2 ² t 2 ds 1 + 1 ||A 2 u|| X m n -u 5 D 2 = DÝA 2 Þn Recall that in the previous approach we began with the assumption that the IVP had a unique solution, which we expressed as uÝtÞ = SÝtÞu 0 , and we then proceeded to prove that S(t) had certain properties. In particular, we showed that there was an operator B, called the generator, associated with S(t) and we showed that B was an extension of the operator -A from the IVP. Finally we proved the Hille-Yosida theorem which specified conditions on A sufficient to imply that -A was the generator of a semigroup. Here we have begun with A and showed that under certain conditions on A, the limit (1) must exist. We must now show that the operator S(t) obtained in this limit has properties that are useful for solving the IVP. Properties of S(t) Let A denote a linear operator on X satisfying the hypotheses of the theorem and let SÝtÞ = Lim n¸K I + nt A ?n . Then 1) SÝtÞ is a strongly continuous function of t ³ 0; i.e., -u 5 X ||SÝtÞu ? SÝt 0 Þu|| X ¸ 0 as t ¸ t 0 ³ 0. 2) SÝ0Þ = I 3) ||SÝtÞ|| LÝXÞ ² 1. 4) SÝtÞ : D A ¸ D A and ASÝtÞu = SÝtÞAu -u 5 D A 5) S v ÝtÞu = ?ASÝtÞu = ?SÝtÞAu -u 5 D A 6) lim t¸0 SÝtÞ ? I u = ?Au t -u 5 D A To prove 1) we can write ||SÝtÞu ? SÝt 0 Þu|| X = ||SÝtÞu ? S m ÝtÞu + S m ÝtÞu ? S m Ýt 0 Þu + S m Ýt 0 Þu ? SÝt 0 Þu|| X and use (3) to write 3 2 || SÝtÞu ? S m ÝtÞu|| X ² t ||A 2 u|| X -u 5 D 2 = DÝA 2 Þ. 2m Since S m ÝtÞ is strongly continuous in t for every m, ||S m ÝtÞu ? S m Ýt 0 Þu|| X ¸ 0 as t ¸ t 0 ³ 0, and the result follows. The results 2) and 3) are proved similarly. To prove 4) note that since A commutes with its resolvent ÝI + WAÞ ?1 , it follows that for all u 5 D A A I + nt A A I + nt A and ?n?1 ?1 u = A S n ÝtÞ I + nt A u í A SÝtÞ Iu = ASÝtÞu ?n?1 ?n?1 u = I + nt A Au ?1 = S n ÝtÞ I + nt A Au í SÝtÞI Au = SÝtÞAu. S vn ÝtÞ = ? I + nt A This proves 4) and since S n ÝtÞu ? u = ? X t 0 I + nb A ?n?1 Au db ?n?1 A, we have -u 5 D A . Letting n tend to K we find t 7) SÝtÞu ? u = ? X SÝbÞAu db 0 -u 5 D A . Note that passing to the limit under the integral is valid since the convergence I + nb A ?n?1 Au í SÝbÞAu is uniform on any finite time interval. The result 7) is also of interest. To prove 6), use 7) to write t ÝSÝtÞ ? IÞu = ? X SÝbÞAu db -u 5 D A . 0 Now the MVT for integrals implies t X 0 SÝbÞAu db = SÝt 0 Þ t Au for 0 ² t 0 ² t, and SÝtÞ ? I u = ?SÝt 0 Þ Au t for 0 ² t 0 ² t. Now let t tend to zero and use 2) to get 6). 4 It is now possible to show that for all u 0 5 D A , the unique solution of is given by u v ÝtÞ + AuÝtÞ = 0, uÝ0Þ = u 0 uÝtÞ = SÝtÞu 0 . That uÝtÞ = SÝtÞu 0 solves the IVP follows from 5) and 2). To see that the solution is unique, let vÝtÞ satisfy v 5 CÝß0, Tà : D A Þ V C 1 ÝÝ0, T : D A Þ v v ÝtÞ + AvÝtÞ = 0, Let Then FÝsÞ = SÝt ? sÞvÝsÞ, vÝ0Þ = 0. 0 ² s ² t, t > 0 fixed. F v ÝsÞ = ?S v Ýt ? sÞvÝsÞ + SÝt ? sÞv v ÝsÞ = ASÝt ? sÞvÝsÞ ? SÝt ? sÞAvÝsÞ = 0. i.e., F is constant. But, FÝtÞ = SÝ0ÞvÝtÞ = vÝtÞ and FÝ0Þ = SÝtÞvÝ0Þ hence vÝtÞ = SÝtÞvÝ0Þ = 0 for all t ³ 0. Since the difference of any two solutions to the IVP must satisfy the IVP with a zero initial state, the uniqueness follows. Finally consider FÝsÞ = SÝt ? sÞvÝsÞ in the special case that vÝsÞ = SÝsÞu 0 . Then FÝsÞ = SÝt ? sÞSÝsÞu 0 and FÝ0Þ = SÝtÞu 0 . But F is constant even in this case so FÝsÞ = SÝt ? sÞSÝsÞu 0 = SÝtÞu 0 = FÝ0Þ. -u 0 5 D A . Since D A is dense in X we have .SÝt ? sÞSÝsÞ = SÝtÞ for 0 ² s ² t or 8) SÝb + sÞ = SÝbÞSÝsÞ for s, b ³ 0. Note that if we let JÝtÞu = e ?Vt SÝtÞu for V, t ³ 0 and u 5 X, then 5 J v ÝtÞu = ?Ve ?Vt SÝtÞu + e ?Vt S v ÝtÞu = ? e ?Vt SÝtÞÝVI + AÞu -u 5 D A i.e., J v ÝtÞu = ? JÝtÞÝVI + AÞu -u 5 D A and we see that JÝtÞ is a contraction semigroup generated by ÝVI + AÞ. Let v = ÝVI + AÞu. Then t JÝtÞu ? u = ? X JÝbÞÝVI + AÞu db -u 5 D A 0 and if we apply ÝVI + AÞ to both sides of this equation, then we get t JÝtÞv ? v = ?ÝVI + AÞ X JÝbÞv db -v 5 X. 0 Let t tend to infinity and use the fact that || JÝtÞu|| X = || e ?Vt SÝtÞu|| X ² e ?Vt ||u|| X to conclude that K u = X e ?Vb SÝbÞÝVI + AÞu db 0 -u 5 D A . Note that this implies ÝVI + AÞ is one to one. Moreover, K v = ÝVI + AÞ X e ?Vb SÝbÞv db 0 -v 5 X. implies that ÝVI + AÞ is onto and K ÝVI + AÞ ?1 v = X 0 e ?Vb SÝbÞv db -v 5 X, with ÝVI + AÞ ?1 v X ² 1 ||v|| X . -V > 0. V Together, these properties of S(t) show that the Hille Yosida conditions on A are not only sufficient for the generation of S(t), they are also necessary. 6