thus its natural to expect u to also be an element

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Unformatted text preview: xpect U to also be an element of C 2(0 1). Mathematically, we regard U as belonging to a nite-dimensional function space that is a subspace of C 2 (0 1). We express this condition by writing U 2 S N (0 1) C 2(0 1). (The restriction of these functions to the interval 0 < x < 1 will, henceforth, be understood and we will no longer write the (0 1).) With this interpretation, we'll call S N the trial space and regard the preselected functions j (x), j = 1 2 : : : N , as forming a basis for S N . Likewise, since v 2 L2, we'll regard V as belonging to another nite-dimensional ^ ^ function space S N called the test space. Thus, V 2 S N L2 and j (x), j = 1 2 : : : N , ^ provide a basis for S N . Now, replacing v and u in (1.2.2c) by their approximations V and U , we have (V L U] ; f) = 0 ^ 8V 2 S N : (1.2.4a) The residual r(x) := L U ] ; f (x) (1.2.4b) 1.2. Weighted Residual Methods 5 is apparent and clari es the name \method of weighted residuals." The vanishing of the inner product (1.2.4a) implies that the...
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