# Download Nonlinear Numerical Analysis in Reproducing Kernel Space by Minggen Cui PDF

By Minggen Cui

Even supposing the applying of reproducing kernel has been explored in several fields long ago twenty to thirty years and the appropriate researches are energetic within the contemporary 5 years, there's nonetheless now not a publication at the software of reproducing kernel. This booklet makes an attempt to introduce to the readers engaged in mathematical program those options, particularly the developing concept of the reproducing kernel house that the authors initially created and steadily more advantageous. Reproducing kernel area is a different Hilbert area. the authors were engaged within the developing thought learn of the reproducing kernel area because 1980's, and labored out a sequence of particular structural equipment for reproducing kernel house and reproducing kernel capabilities.

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If S = {x1, x2, . } is a dense set of [0, 1], put Φij (x) = rxi (x)ej , Ψij (x) = (A∗Φij )(x), j = 1, 2, then {Ψij (x)j=1,2 }∞ i=1 is the complete system of W2 . Proof. Given u(x) ∈ W2 , if u(x), Ψij (x) W2 = 0 (i = 1, 2, . ; j = 1, 2), (Au)(x), Φij (x) W1 = 0 (i = 1, 2, . ; j = 1, 2). 7) Note that Au = A11u1 + A12 u2, A21u1 + A22u2 , Φij = Rxi ej . Hence (Au)(x), Φij(x) W1 = Aj1 u1 + Aj2 u2 , Rxi W21 = [Aj1 u1 + Aj2 u2](xi ), that is (Au)(xi) = 0 Since S is dense on [0, 1], we must have (Au)(x) = 0.

1. L is a bounded linear operator. 4) 32 Minggen Cui and Yingzhen Lin Proof. 1 Ly 2 L2 2 y (x) − G(x)y(x) dx = 0 1 0 0  1 [G(z)y(z)]2dx + 2  [y (x)]2dx + ≤  1 |y (x)||G(x)||y(x)|dx . 0 Note that |y(x)| = y(ξ), Rξ(x) ≤ M1 y oW 3 2 oW 3 2 , where M1 is a positive real number and G(x) is the bounded function, let |G(x)| ≤ M2 . This follows that 1 Ly 2 L2 [y (x)]2dx + M22 M12 y ≤ 0  1 + 2M2  2 oW 3 2 1/2 1 |y(x)|2dx |y (x)|2dx 0 0 1 [y (x)]2dx + M22 M12 y ≤ 2 oW 3 2 0 1 + 2M2 M1 y |y (x)|2dx oW 3 2 0 ≤ y 2 oW 3 2 + + 2M2 M1 y M22 M12 oW 3 2 y 2 oW 3 2 y 2 oW 3 2 ≤M y 2 oW 3 .

C   Fi (a, y)(dy)   n c n m x y ... a y (y − c)j Gj (x, c) j! x + y  (x − a)i   i!  +  n y H(x, y)(dy)m(dy)n. a c c (1) Since y ∂ m+n−2 f (x, y) = ∂xm−1 ∂y n−1 Fm−1 (a, y)dy + Gn−1 (x, c) c y x + H(x, y)dxdy, a hence ∂ m+n−2 ∂xm−1 ∂y n−1 c f (x, y) is completely continuous in D, and is true almost everywhere in D. So, f (x, y) ∈ (2) Note that fk (x, y) − f (x, y) m−1 d = i=0 c n−1 + j=0 ∂ m+n ∂xm ∂y n f (x, y) (m,n) W2 (D). 5) −→ k→∞ 0. D (m,n) Hence, function space W2 (D) is a Hilbert Space.