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Unformatted text preview: residual is orthogonal in L2 to all functions V in
the test space S N .
Substituting (1.2.3) into (1.2.4a) and interchanging the sum and integral yields
j =1 dj ( j L U] ; f) = 0 8dj j = 1 2 : : : N: (1.2.5) Having selected the basis j , j = 1 2 : : : N , the requirement that (1.2.4a) be satis ed for
all V 2 S N implies that (1.2.5) be satis ed for all possible choices of dk , k = 1 2 : : : N .
This, in turn, implies that
( j L U] ; f) = 0 j = 1 2 : : : N: (1.2.6) Shortly, by example, we shall see that (1.2.6) represents a linear algebraic system for the
unknown coe cients ck , k = 1 2 : : : N .
One obvious choice is to select the test space S N to be the same as the trial space
and use the same basis for each thus, k (x) = k (x), k = 1 2 : : : N . This choice leads
to Galerkin's method
( j L u] ; f ) = 0 j = 1 2 ::: N (1.2.7) which, in a slightly di erent form, will be our \work horse." With j 2 C 2, j =
1 2 : : : N , the test space clearly has more continuity than necessary. Integrals like
(1.2.4) or (1.2....
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- Spring '14