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Chemical Engineering Hand Written_Notes_Part_50

Chemical Engineering Hand Written_Notes_Part_50 - 102 3...

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102 3. LINEAR ALGEBRAIC EQUATIONS AND RELATED NUMERICAL SCHEMES (6.36) x ( k +1) = x ( k ) + 4 x ( k ) Step x ( k +1) predicts a function change (6.37) F ( k ) = F ( k +1) F ( k ) We impose the following two conditions to obtain estimate of J ( k +1) . (1) In the direction perpendicular to 4 x ( k ) , our knowledge about F is maintained by new Jacobian estimate J ( k +1) . This means for a vector, say r , if [ x ( k ) ] T r = 0 , then (6.38) J ( k ) r = J ( k +1) r In other words, both J ( k ) and J ( k +1) will predict some change in direc- tion perpendicular to x ( k ) . (2) J ( k +1) predicts for 4 x ( k ) , the same F ( k ) in linear expansion, i.e., (6.39) F ( k +1) = F ( k ) J ( k +1) 4 x ( k ) or (6.40) J ( k +1) 4 x ( k ) = F ( k ) Now, for vector r perpendicular to 4 x ( k ) , we have (6.41) J ( k +1) r = J ( k ) r + y ( k ) [ z ( k ) ] T r As (6.42) J ( k +1) r = J ( k ) r We have (6.43) y ( k ) [ z ( k ) ] T r = 0 Since 4 x ( k ) is perpendicular to r , we can choose z ( k ) = 4 x ( k ) . Substituting this choice of z ( k ) in equation (6.34) and post multiplying equation (6.34) by 4 x ( k ) , we get (6.44) J ( k +1) 4 x ( k ) = J ( k ) 4 x ( k ) + y ( k ) [ 4 x ( k ) ] T 4 x ( k ) Using equation (3), we have (6.45) F ( k ) = J ( k ) 4 x ( k ) + y ( k ) [ 4 x ( k ) ] T 4 x ( k ) which yields (6.46) y ( k ) =
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