By E. A. Wallis Budge
Данный 2-х томный словарь - одна из многочисленных работ знаменитого английского египтолога Эрнеста Альфреда Уоллиса Баджа - главного хранителя отдела египетских и ассирийских древностей Британского музея с 1894 года по 1924 г. Он участвовал в археологических раскопках в Египте, Судане и Месопотамии, выполнил множество переводов с древних языков.
Read or Download An Egyptian Hieroglyphic Dictionary With an index of English words, king list and geographical list with indexes, list of hieroglyphic characters, Coptic and Semitic alphabets, etc PDF
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Additional resources for An Egyptian Hieroglyphic Dictionary With an index of English words, king list and geographical list with indexes, list of hieroglyphic characters, Coptic and Semitic alphabets, etc
Example text
Let us see now that ξ can be extended to (1, D). 1) 1 ¯ d¯ + δ) with v(1, ξ(d), d) = 1}. = sup{δ > 0; ξ can be extended to (d, If d¯+ 1 < D, by Lemma 8 and Theorem 11 we have that v(·, 1, d¯+ 1 ) − 1 oscillates. Let τ be the k-th zero of v(·, 1, d¯+ 1 ) − 1. Arguing as in the previous paragraph we see that there exists η > 0 and a continuous function ξ1 : (d¯+ 1 − η, d¯ + 1 + η) → (0, ∞) such that v(ξ1 (d), 1, d) − 1 = 0 and ξ1 (d¯+ 1 ) = τ . By the non-degeneracy of the zeroes, ξ1 (d) is the k-th zero of v(·, 1, d) − 1.
It follows from Theorem 9 that σ(1, d1 ) = ∞. 12) on initial data we obtain that limd→D ξj (d) = ∞, j = 2, 3, . . Case −1 < q < 1. We split the proof in two subcases: −1 < q ≤ 0 and 0 < q < 1. ´ A. CASTRO AND V. PADRON 50 Subcase −1 < q ≤ 0. There exist positive constants m1 , m2 such that m1 ≤ f (v)/(v − 1) ≤ m2 for any v ∈ (0, D]. Thus by the Sturm comparison theorem there exists of cj such that 0 < tj (d) ≤ cj for d ∈ (1, D), and j = 1, 2, . . Subcase 0 < q < 1. 5 and Theorem 1 we know that tj (d) is uniformly bounded on any compact subset of [1, D).
When k → ∞ we obtain that r 1 ¯ ¯ ) − f (τ ) dτ, k = 0, 1, . . 6). Therefore, by uniqueness H ¯ ≡ K. 11) ¯ ≤ K(r), r ∈ [0, w(0)]. 11). Hence there exists t ∈ (0, M ) such that v(t) = w(t). Suppose now that v(a) = w(a) = ρ > 1 for some a > 0, and v(t) > w(t) for 0 ≤ t < a. By uniqueness of solutions to initial values problems v (a) < w (a). Since ρ > 1 and (f (v)/(v − 1)) ≤ 0 on (1, ∞) (see lemma 22 below), it follows that (w(t) − 1)f (v(t)) − (v(t) − 1)f (w(t)) ≤ 0, 0 ≤ t ≤ a. 1), this inequality can be written as [t(−(w(t) − 1)v (t) + (v(t) − 1)w (t))] ≤ 0, 0 ≤ t ≤ a.