By Knopp K.
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Example text
1. 8) and λn R for some subsequence. e. O . Proof. From now on, we will prove the lemma only for the sequence (un ), because the same arguments can be applied to (wn ). 6), that is, R |un |2 + R (1 + λn V )|un |2 ≤ R (1 + λn V )un u + R un u − R g(x, un )(u − un ). O. A. Corrˆea Using the fact that V (t)u(t) = 0 for all t ∈ R, it follows that R |un |2 + R (1 + λn V )|un |2 ≤ R un u + R un u − R g(x, un )(u − un ). 2, un → u in H 1 (R) for some subsequence. Hence, lim inf n→+∞ R (|un |2 + |un |2 ) = lim n→+∞ R (un u + un u) = R R (|u |2 + |u|2 ), (|u |2 + |u|2 ), and lim n→+∞ R g(x, un )(u − un ) = 0.
O . Proof. From now on, we will prove the lemma only for the sequence (un ), because the same arguments can be applied to (wn ). 6), that is, R |un |2 + R (1 + λn V )|un |2 ≤ R (1 + λn V )un u + R un u − R g(x, un )(u − un ). O. A. Corrˆea Using the fact that V (t)u(t) = 0 for all t ∈ R, it follows that R |un |2 + R (1 + λn V )|un |2 ≤ R un u + R un u − R g(x, un )(u − un ). 2, un → u in H 1 (R) for some subsequence. Hence, lim inf n→+∞ R (|un |2 + |un |2 ) = lim n→+∞ R (un u + un u) = R R (|u |2 + |u|2 ), (|u |2 + |u|2 ), and lim n→+∞ R g(x, un )(u − un ) = 0.
0 0 ··· λN −1 Thus, at z = (0, 0) we have N −1 D2 G(0)y, y E(w, y) dy RN −1 = i=1 RN −1 λi yi2 E(w, y) dy. ´ and E. Medeiros E. M. do O 20 By the definition of the mass moment of inertia we have that the moment of inertia about the yi -axis, i = 1, . . , N − 1, respectively the polar moment of inertia are given by N −1 Iyi = RN −1 yi2 E(w, y) dy, I0 = N −1 Iyi = i=1 RN −1 i=1 yi2 E(w, y) dy, respectively. Now, using the fact of E(w, y) is a symmetric function, we conclude that Iy1 = · · · = IyN −1 , which implies that I0 = (N − 1)Iy1 .