Solution1 - AE 321 Solution to Practice Problems Chapter 1:...

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AE 321 – Solution to Practice Problems Chapter 1: Mathematical Preliminaries 1. The number of terms is given by the following expression n 3 , (1.1) where n is the number of free indices . Then, Number of free indices n Number of terms n 3 ij A 2 9 kl ij B , 4 81 ji ij C , 0 1 ijk D 3 27 qr pq B A 2 9 2. The number of equations is again obtained from the number of free indices by Equation (1.1). Number of free indices n Number of equations n 3 il kl jk ij D C B A = 2 9 0 , = j ij A 1 3 ijkl kl ij l ijk C B A = + , , 4 81
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3.1 ik jk ij C B A = Let’s first work with the dummy index and perform the summation ik k i k i k i C B A B A B A = + + 3 3 2 2 1 1 (1.2) In expression (1.2) we have 2 free indices, therefore it represents 9 equations that are obtained by expanding (1.2) for every combination of i and k : i.e. 1 = i & 1 = k , 1 = i & 2 = k , …. The 9 equations are given in the following table. 1 = k 2 = k 3 = k 1 = i 11 31 13 21 12 11 11 C B A B A B A = + + 12 32 13 22 12 12 11 C B A B A B A = + + 13 33 13 23 12 13 11 C B A B A B A = + + 2 = i 21 31 23 21 22 11 21 C B A B A B A = + + 22 32 23 22 22 12 21 C B A B A B A = + + 23 33 23 23 22 13 21 C B A B A B A = + + 3 = i 31 31 33 21 32 11 31 C B A B A B A = + + 32 32 33 22 32 12 31 C B A B A B A = + + 33 33 33 23 32 13 31 C B A B A B A = + + 3.2 k k i i C B A , = Once again, let’s start by working with the dummy index, i.e. k , and perform the summation A i = B i ,1 C 1 + B i ,2 C 2 + B i ,3 C 3 = ! B i ! x 1 C 1 + ! B i ! x 2 C 2 + ! B i ! x 3 C 3 (1.3) In this case we only have one free index; therefore expression (1.3) represents 3 equations:
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i = 1 ! A 1 = B 1,1 C 1 + B 1,2 C 2 + B 1,3 C 3 = " B 1 " x 1 C 1 + " B 1 " x 2 C 2 + " B 1 " x 3 C 3 i = 2 ! A 2 = B 2,1 C 1 + B 2,2 C 2 + B 2,3 C 3 = " B 2 " x 1 C 1 + " B 2 " x 2 C 2 + " B 2 " x 3 C 3 i = 3 ! A 3 = B 3,1 C 1 + B 3,2 C 2 + B 3, 3 C 3 = " B 3 " x 1 C 1 + " B 3 " x 2 C 2 + " B 3 " x 3 C 3 4.1 Notice that the first expression survives only if j i = . Therefore, ! ij ij = ii = 3 or alternatively ij ij = jj = 3 . 4.2 In the second expression we notice that jk ij becomes ij jk = ik " ij jk ki = ik ki = kk = 3 5(i) ( ) ( ) ( ) ( ) kq jr kr jq iqr ijk jq kr kq jr jr kq kr jq jr kq kr jq jq ir ki kq ir ji jr iq ki kr iq ji jr kq kr jq jq ki kq ji ir jr ki kr ji iq jr kq kr jq ii kr kq ki jr jq ji ir iq ii iqr ijk kr jr kq jq " # = $ # + + # # = # + + # # = # + # # # = = 3 3 3 ! " # ! " # ! " # ! " # 5(ii) Using the results from (i) and substituting q j =
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