Science Mathematics

Nonlinear Numerical Analysis in Reproducing Kernel Space by Minggen Cui

By Minggen Cui

Even though the applying of reproducing kernel has been explored in numerous fields long ago twenty to thirty years and the correct researches are energetic within the contemporary 5 years, there's nonetheless now not a booklet at the software of reproducing kernel. This booklet makes an attempt to introduce to the readers engaged in mathematical program those strategies, specifically the developing idea of the reproducing kernel house that the authors initially created and steadily stronger. Reproducing kernel house is a different Hilbert house. the authors were engaged within the developing conception learn of the reproducing kernel house when you consider that 1980's, and labored out a chain of particular structural equipment for reproducing kernel house and reproducing kernel capabilities.

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Additional resources for Nonlinear Numerical Analysis in Reproducing Kernel Space

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3 1 −(x+t) + e−|x−t| . 7) x c(s, x)u(s) ds, 0 x < τ (x), λ(x) where λ(x) = max{x − τ (x), 0}. Now, Eq. 4) is rewritten with the transformation as follows: (Lu)(x) = F (x), 0 x < ∞, u(x) = h(x), −τ (x) < x < 0, where F (x) = f (x), τ (x) ≤ x < ∞, f (x) − b(x)h(x − τ (x)), 0 ≤ x < τ (x). 1. If a(x), b(x) are bounded on interval [0, ∞), c(x, s) is bounded on interval [0, ∞) × [0, ∞), then operator L : W22 [0, ∞) −→ W21[0, ∞) is a bounded linear operator. Proof. For u(x) ∈ W22 [0, ∞), note that x (Lu)(x) = u (x) + a(x)u(x) + w(x) + c(s, x)u(s) ds, λ(x) where b(x)u(x − τ (x)), τ (x) ≤ x < ∞, 0, 0 ≤ x < τ (x), w(x) = Lu ≤ W21 u (x) W21 + a(x)u(x) W21 + w(x) W21 2 W21 ≤ B2 u 2 W22 , x + c(s, x)u(s) ds λ(x) , W21 (1) a(x)u 2 W21 ≤ A2 u 2 W22 , w(x) ∞ u 2 W21 (u )2 + (u )2 dx ≤ u = 2 W22 , 0 where A, B are the upper bounds of |a(x)|, |b(x)|, respectively.

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. 2. Let f (x, y), g1(x)g2(y) ∈ W2 f (x, y), g1(x)g2(y) (m,n) W2 = (D) is a Reproducing Ker- (D), then f (x, y), g1(x) W2m , g2 (y) W n 2 .

2. Let f (x, y), g1(x)g2(y) ∈ W2 f (x, y), g1(x)g2(y) (m,n) W2 = (D) is a Reproducing Ker- (D), then f (x, y), g1(x) W2m , g2 (y) W n 2 . Proof. 1. Let f1 (x)f2 (y), g1(x)g2(y) ∈ W2 f1 (x)f2 (y), g1(x)g2(y) (m,n) W2 23 . (D), then = f1(x), g1(x) W2m f2(y)g2(y) W2n . 2. If f (x) ∈ W2m [a, b], then for n ∈ N, we have f (x) W2m = f (x) (m,n) W2 . 6) (m,n) Proof. 1, one gets f (x) ∈ W2 obtains immediately f (x) 2 (m,n) W2 = f (x), f (x) W2m = f (x) (D) and 2 W2m . 2, we can obtain the next theorem immediately.

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