Solutions of a Class of Sixth Order Boundary Value Problems Using the Reproducing Kernel Space
نویسندگان
چکیده
and Applied Analysis 3 Since i ∞ i=1 is dense in [0, 1], (Lu)(x) = 0, which implies that u ≡ 0 from the existence of L. Using reproducing kernel property, it can be written as i (x) = i (y) , x (y)⟩ = ⟨(L ∗ i (x) , x (y)⟩ = i (y) , x (y)⟩
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تاریخ انتشار 2014