自伴微分運算元 的英文怎麼說

中文拼音 [bànwéifēnyùnsuànyuán]
自伴微分運算元 英文
self-adjoint differential operator
  • : Ⅰ代詞(自己) self; oneself; one s own Ⅱ副詞(自然;當然) certainly; of course; naturally; willin...
  • : Ⅰ名詞(夥伴; 伴侶) companion; partner Ⅱ動詞(隨同; 配合) accompany
  • : 分Ⅰ名詞1. (成分) component 2. (職責和權利的限度) what is within one's duty or rights Ⅱ同 「份」Ⅲ動詞[書面語] (料想) judge
  • : Ⅰ動詞1 (物體位置不斷變化) move; revolve 2 (搬運; 運輸) carry; transport 3 (運用) use; wield...
  • : Ⅰ動詞1 (計算數目) calculate; reckon; compute; figure 2 (計算進去) include; count 3 (謀劃;計...
  • 運算 : [數學] operation; arithmetic; operating
  1. The resolvent operator of the 2n - order j - selfadjoint vector differential operator with one endpoint singularity

    向量的預解
  2. 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式生成的積性; 4
  3. On self - adjointness of the product of three second - order differential operators

    三個二階積的
  4. 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,討論了的本徵函數的正交性和完備性,介紹了尋求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,通過對近似方程中的特徵態的討論,引入適當的「導出態」 ,建立了在l _ 2空間的特徵態的完備性。
  5. 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積的性和一個s - l與一個四階積情形。
  6. 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

    關于積性,兩個正則或奇異的二階對稱式生成的積性已取得一些結果。
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