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Variable thickness shell axisymmetric ring complex variable equations and general solution


, May be the type (1.31) written as L () A z / V-12 (1 a.) Ru = o that, on the variables u and V, we get two simultaneous second order differential equations. Differential equations also rj D into ~ Step into a complex variable equation. Let, = cotlst. By equation (1.32) have h = h0 + port (+ sn) where h day and. Is a constant. Eq (1.35) into (1.32) can be obtained, so that S = U = now where the complex constants to be determined, then L (S) = L (U) + AL () j6 ( 1.33) If, as a constant, both (1.35) f】. 36) (1.37) (1.3 distressed Fiff ~ (ten) rU = as \and (1.33) into (1.3, and note that equation (1.36), we obtain L (S): = l0 (1 \1 A v) rA = day 2vh. A ~]: an AB may be the type (1.39) written as L (S) ÷ BS: F (1.41) can be obtained at A =: a 1 Standard d (1i) where: / 12 (: _) ~ the type (I.27) and (1.4) into (1.42) Quarry + (a 1 (k + sin) 2, etc.) + 【+ VC0S (.39) (1.43) (1.44) h (ten st ¨) dhd call-in a c +,] s: (. e,[link widoczny dla zalogowanych], s =, / and note the type (1.35), can be the type (1.46) written as) T = F Ⅱ where J (): sin2 (k ten sf) a ~ ± a hD + a (k + sr: mp) (1.47) (1.48 );(){+[ sin ~)], one chip, etc. ) (1.49 and F: F ... plastic inside ... ho + n (+ sinqp) and for that = where the boundary value, then the equation (1.4 can be transformed into the dZT +,[link widoczny dla zalogowanych], (p): F (B ) d82 ... ... ... where (1.52 )-___ l and let us solve the equation 1. homogeneous solution and equation (1.52) as for the mouth of the cycle, port) = (: a J ((1) F ) = sue (z ~ () Second, the differential equations (1.52). The corresponding homogeneous equation is + JIT = 0 is equal to the dual function, then the equation (2.1) the solution to (1.54) T = Σfe \Changes may be in the range of J 【0】 on the number developing into Fourier cosine series Jl = port o +2 Σacos2k ~ (2.3 'factor. can be a method to determine. For example, if the request: i) when 0 ≤ k ≤ 1 时 s hunger j + ≠ 0 (j: 1.2) (2.4) ii) the following inequality in the interval [., J] on the establishment of Shu everywhere o + port (+ sf jealous) ≠ 0 (2.5 ) the coefficient n used numerical integration method. characteristic index can use the following formula: black (-?-) zl (0) Emily () ∞ where + (+ +4'. --- 【police +3 Z-ao ... called ... * *) to be type A lI cotton by the Chinese (p + 2n) t + = 一 0: (= 0, ± 1 ± 2. ± 3,[link widoczny dla zalogowanych], ... ... what-.= 0} j1.2) to o is not an integer Therefore,[link widoczny dla zalogowanych], equation (2.1) of the two linearly independent solution is Tl = Σl \2.12) n = a.. 2. special solution Assuming F (mesh) can be in the interval [0, curse】 launched on the Fourier cosine series, and J and F write the form Qi tF '= Σ,[link widoczny dla zalogowanych],: e ... ,,:=, 2.14) = a oo tie set equation (1.52) the particular solution is.. rΣf: 3. (2.15) bar =- oo equation (2. 1a), (2.14) and (2.15) into (1.52) available.. ∞ ∞ a Σ (2) zt% 2. ¨-Σde2 \.. =- ∞ = a 。。。。= Σ,: e2 'month (2.16) bar = a oo Therefore, the coefficient of t: determined by the following equation:.. a (2) f: + Σ port t : A h =,: (2.17) =- ∞ (tadpole = 0, ± 1, ± 2, ± 3, ... ...)
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