Green's Second Identity Prove Green's Second Identity forscalar-valued functions u and v defined on a region D:(uv²v – vv²u) dV = || (uvv – vVu) •n dS.(Hint: Reverse the roles of u and v in Green's First Identity.)

To prove: Green's second identity: ∭Du∇2v -v∇2u dV=∫Su∇v -v∇u·ndS

Answered: Green's Second Identity Prove Green's…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question: Let V be a vector space of all real-valued continuous functions defined on the interval [0, 1]. Define a linear functional L on V as follows: L(f) = f xf(x) dx Where f belongs to V. For a given function ƒ(x) = x² in V, find the value of L(f).
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Definition 28. Let V be a vector space over F (here F= R, C). Then the function (•₁•): V × V → F is said to be an inner product on V if the following conditions are satisfied: (x, x) is real and (x,x) ≥ 0 for all (x,x) = 0 if and only if x € 0₂. (i) (ii) (iii) (x, ay) = a(x, y) for all x, y V (iv) (x, y+z) = (x, y) + (x, z) for all (v) (x, y) = (y,x) for all x, y V. (a) (.,.): R¹ × R¹ → R is defined as xεν. (x, y) = x¹y, and for all x,y,z € V. Example: Verify that each of the following mappings define a real inner product. a EF. for all x, y ER".
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Green's Second Identity Prove Green's Second Identity for scalar-valued functions u and v defined on a region D: (uv²v – vv²u) dV = || (uvv – vVu) •n dS. (Hint: Reverse the roles of u and v in Green's First Identity.)