自伴 的英文怎麼說

中文拼音 [zìbàn]
自伴 英文
self adjoint
  • 自 : Ⅰ代詞（自己） self; oneself; one s own Ⅱ副詞（自然;當然） certainly; of course; naturally; willin...
  • 伴 : Ⅰ名詞（夥伴; 伴侶） companion; partner Ⅱ動詞（隨同; 配合） accompany
  1. There are five natural sample zones of adiantum reniforme l. var. sinense y. x. lin investigated, and recorded 34 species. 10 dominant species were used to calculate the niche width of populations and the niche overlap interpopulations

    調查了5個自然樣地，記錄了34個伴生種，取10個優勢種進行種群生態位寬度與生態位重疊的計算。
  2. She has been slightly bothered by the lack of time her husband has for her, but she has led a peaceful and happy life in a rich neighborhood in gracemeria, a capital of republic of emmeria, until a sudden aerial bombing takes her away from her daughter

    她對丈夫缺少陪伴她的時間多少有些煩心，但是她依然在格瑞斯美瑞亞? ?伊美瑞亞共和國的一個富有社區里過著平和而開心的生活，直到突然的空襲把她和自己的女兒分開為止。
  3. The hurons stood aghast at this sudden visitation of death on one of their band.

    休倫人眼見自己一個同伴這樣突然喪命，不由得全都嚇呆了。
  4. Into the bargain and the greatest danger of all was who you got drunk with though, touching the much vexed question of stimulants, he relished a glass of choice old wine in season as both nourishing and blood - making and possessing aperient virtues notably a good burgundy which he was a staunch believer in still never beyond a certain point where he invariably drew the line as it simply led to trouble all round to say nothing of your being at the tender mercy of others practically

    說起來，最危險的一點是你跟什麼樣的夥伴一道喝得醉醺醺的。就拿這個非常令人困擾的酒精飲料來說吧，他本人總是按時津津有味地喝上一盅精選的陳葡萄酒，既滋補，又能造血，而且還是輕瀉劑尤其對優質勃艮第的靈效，他堅信不疑。然而他從來也不超過自己規定的酒量，否則確實會惹出無窮的麻煩，就只好乾脆聽任旁人的善心來擺布了。
  5. The resolvent operator of the 2n - order j - selfadjoint vector differential operator with one endpoint singularity

    自伴向量微分運算元的預解運算元
  6. On [ a, b ] the product l = l2 lt of operator l { and l2 is selfadjoint if and only if 4

    討論一類兩個四階對稱微分算式生成的積運算元的自伴性； 4
  7. On self - adjointness of the product of three second - order differential operators

    三個二階微分運算元積的自伴性
  8. And beginning with a perturbed nls equation, using a multi - scales perturbation expansion, we get the zero order and the first order equations, discuss the eigenstates of the operator in the equations, induct relevant " derivative states ", form the completeness of the bounded eigenstates of the associated operator in li space, and expand the corresponding parameters in the closure, get a series evolution equations of the coefficients in the expanded formulas, find the first order approximate solution by researching the evolution equations. this paper also gives the basis of this method - the completeness we have formed and the singular perturbation technique

    ） dinser方程的求解問題，討論了自伴運算元的本徵函數的正交性和完備性，介紹了尋求微分方程的近似解常用的攝動方法，並從帶有某種擾動項的nls方程出發，利用多重尺度的攝動方法得到了方程的零級近似方程和一級近似方程，通過對近似方程中運算元的特徵態的討論，引入適當的「導出態」 ，建立了運算元在l _ 2空間的特徵態的完備性。
  9. By resorting to the residue method, the asymptotic formulas for the eigenvalues and the expansion theorems of dirac eigenvalue problems are proved under the self - adjoint and non - self - adjoint boundary conditions

    本文用留數方法證明了自伴和非自伴的dirac運算元的特徵值估計和特徵展開定理。
  10. Multiplicative self - adjoint maps on a non - standard operator algebra which preserve spectrum

    一個非標準運算元代數上的保譜乘法自伴映射
  11. Functional inequalities for fractional powers of positive definite self - adjoint operators

    正定自伴運算元分數冪的泛函不等式
  12. Study of non - self - adjoint variational problem in low - frequency eddy current electromagnetic field

    低頻渦流電磁場非自伴變分問題的研究
  13. The abstract boundary value problem of non - selfadjoint and non - compact operator with reflective boundary condition

    具反射邊界條件的非自伴非緊的抽象邊值問題
  14. 5. on [ a, b ] the product l = l3l2l { of operator lltl2 and l3 is selfadjoint if and only if a and l2 is selfadjoint. 6

    討論三個正則的s - l運算元積的自伴性和一個s - l運算元與一個四階微分運算元積情形。
  15. For the non - self - adjoint dirac operators, there are plentiful content in the problems of eigenval ue expansion problems

    從所得結果來看，對于非自伴dirac運算元來說，特徵展開問題具有相當豐富的內容。
  16. For the expansion theorems of self - adjoint dirac operator, it is difficult to prove it by using the method of integral equation

    對于自伴dirac運算元的特徵展開定理的證明，用積分方程方法有一定的困難。
  17. About dirac eigenvalue problem with general two points " liner algebra, corresponding operator of which often is non - self - adjoint operator

    對於一般兩點線性（代數）邊界條件下的dirac特徵值問題，相應的運算元一般說是非自伴的。
  18. The condition under which the dirac operator is self - adjoint is discussed under the general linear boundary condition between the interval of two points. for the expansion theorem of non - self - adjoint dirac operator, it is unable to use the method of integral equation. but under the linear boundary condition and unlocal boundary condition, the eigenvalue expansion problems of non - self - adjoint operator can still be discussed by using the residue method

    對于非自伴dirac運算元的特徵展開定理已無法應用積分方程的方法，本文仍用留數方法對一個兩點非自伴邊界條件和一個非局部邊界條件下產生的非自伴運算元的特徵展開問題進行了討論，分別得到了它們的特徵展開定理。
  19. On [ 0, oo ) the product l = l2lt of operator z, j and l2 isselfadjointifandonly if where 4 5, c, ? are some ordinary matrix with rank ( a b ) = rank ( c d ) = 4. 2

    關于積運算元的自伴性，兩個正則或奇異的二階對稱微分算式生成的積運算元的自伴性已取得一些結果。
  20. Couples appreciate each other and enjoy mutual encouragement, especially in adversity, and some critical moments. we are more inspired by the energy and love from partners

    夫妻之間互相欣賞，互相激勵，尤其是在逆境和一些關鍵時刻，來自伴侶的激勵的能量遠比想象的要強的多。
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