Fluid Mechanics, 6th Edition Solution Manual

Fluid Mechanics, 6th Ed. € € € € € € € € € € Kundu, Cohen, and Dowling Exercise 2.1. For three spatial dimensions, rewrite the following expressions in index notation and evaluate or simplify them using the values or parameters given, and the definitions of δij and εijk wherever possible. In b) through e), x is the position vector, with components xi. a) b ⋅ c where b = (1, 4, 17) and c = (–4, –3, 1) b) (u ⋅ ∇ ) x where u a vector with components ui. c) ∇φ , where φ = h ⋅ x and h is a constant vector with components hi. d) ∇ × u, where u = Ω × x and Ω is a constant vector with components Ωi. “1 2 3& $ $ e) C€ ⋅ x , where C = #0 1 2′ $0 0 1$ % ( Solution 2.1. a) b ⋅ c = bic i = 1(−4) + 4(−3) + 17(1) = −4 −12 + 17 = +1 +x . + % ∂ ( % ∂ ( % ∂ (.- 1 0 € ∂ b) (u ⋅ ∇ ) x = u j x i = -u1′ * + u2 ‘ * + u3 ‘ *0 x 2 ∂x j ∂x1 ) ∂x 2 ) ∂x 3 )/- 0 & & & , € -,x 3 0/ ) # ∂x & # ∂x & # ∂x & , + u1% 1 ( + u2 % 1 ( + u3 % 1 (. $ ∂x 2 ‘ $ ∂x 3 ‘. )u ⋅1+ u ⋅ 0 + u ⋅ 0, + $ ∂x1 ‘ ) u1 , 2 3 # ∂x 2 & # ∂x 2 &. + 1 + # ∂x 2 & . + . = +u1% ( + u2 % ( + u3 % (. = +u1 ⋅ 0 + u2 ⋅1+ u3 ⋅ 0. = u jδij = +u2 . = ui $ ∂x 2 ‘ $ ∂x 3 ‘. + $ ∂x1 ‘ + . +*u3 .# & # & # ∂x 3 &. *u1 ⋅ 0 + u2 ⋅ 0 + u3 ⋅1∂x 3 + ∂x 3 +u1%$ ∂x (‘ + u2 %$ ∂x (‘ + u3 %$ ∂x (‘. * 1 2 3 ∂φ ∂ ∂x i c) ∇φ = = ( hi x i ) = hi = hiδij = h j = h ∂x j ∂x j ∂x j € × u = ∇ × (Ω × x ) = ε ∂ (ε Ω x ) = ε ε Ω δ = (δ δ − δ δ )Ω δ = (δ δ − δ δ )Ω d) ∇ ijk klm l m ijk klm l jm il jm im jl l jm il jj ij jl l ∂x j = ( 3δil − δil )Ωl = 2δil Ωl = 2Ωl = 2Ω Here, the following identities have been used: εijkεklm = δilδ jm − δimδ jl , δijδ jk = δik , δ jj = 3 , and δij Ω j = Ωi #1 2 3’# x1 ‘ # x1 + 2x 2 + 3x 3 ‘ € % %% % % % e) C ⋅ x = Cij x j = $0 1 2($ x 2 ( = $€ x 2 + 2x 3 ( € € %0 0 1%% x % % % x3 & )& 3 ) & ) Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling Exercise 2.2. Starting from (2.1) and (2.3), prove (2.7). Solution 2.2. The two representations for the position vector are: x = x1e1 + x 2e 2 + x 3e 3 , or x = x1″e1″ + x 2″ e”2 + x 3” e”3 . Develop the dot product of x with e1 from each representation, e1 ⋅ x = e1 ⋅ ( x1e1 + x 2e 2 + x 3e 3 ) = x1e1 ⋅ e1 + x 2e1 ⋅ e 2 + x 3e1 ⋅ e 3 = x1 ⋅1+ x 2 ⋅ 0 + x 3 ⋅ 0 = x1 , and e1 ⋅ x = e1 ⋅ ( x1#e1# + x #2e#2 +€x #3e#3 ) = x1#e1 ⋅ e1# + x #2e1 ⋅ e#2 + x #3e1 ⋅ e#3 = x #iC1i , € set these equal to find: x1 = x “iC1i , € where Cij = e i ⋅ e#j is a 3 × 3 matrix of direction cosines. In an entirely parallel fashion, forming € the dot product of x with e2, and x with e2 produces: € x 2 = x “iC2i and x 3 = x “iC3i . Thus, for any component xj, where j = 1, 2, or 3, we have: € x j = x “iC ji , which is (2.7). € € € Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling Exercise 2.3. For two three-dimensional vectors with Cartesian components ai and bi, prove the Cauchy-Schwartz inequality: (aibi)2 ≤ (ai)2(bi)2. Solution 2.3. Expand the left side term, (ai bi )2 = (a1b1 + a2 b2 + a3b3 )2 = a12 b12 + a22 b22 + a32 b32 + 2a1b1a2 b2 + 2a1b1a3b3 + 2a2 b2 a3b3 , then expand the right side term, (ai )2 (bi )2 = (a12 + a22 + a32 )(b12 + b22 + b32 ) = a12 b12 + a22 b22 + a32 b32 + (a12 b22 + a22 b12 ) + (a12 b32 + a32 b12 ) + (a32 b22 + a22 b32 ). Subtract the left side term from the right side term to find: (ai )2 (bi )2 − (ai bi )2 = (a12 b22 − 2a1b1a2 b2 + a22 b12 ) + (a12 b32 − 2a1b1a3b3 + a32 b12 ) + (a32 b22 − 2a2 b2 a3b3 + a22 b32 ) 2 = (a1b2 − a2 b1 )2 + (a1b3 − a3b1 )2 + (a3b2 − a2 b3 )2 = a × b . Thus, the difference (ai )2 (bi )2 − (ai bi )2 is greater than zero unless a = (const.)b then the difference is zero. Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling Exercise 2.4. For two three-dimensional vectors with Cartesian components ai and bi, prove the triangle inequality: a + b ≥ a + b . Solution 2.4. To avoid square roots, square both side of the equation; this operation does not change the equation’s meaning. The left side becomes: 2 2 2 (a + b) = a +2 a b + b , and the right side becomes: 2 2 2 a + b = (a + b)⋅ (a + b) = a ⋅ a + 2a ⋅ b + b ⋅ b = a + 2a ⋅ b + b . So, 2 2 ( a + b ) − a + b = 2 a b − 2a ⋅ b . Thus, to prove the triangle equality, the right side of this last equation must be greater than or equal to zero. This requires: a b ≥ a ⋅ b or using index notation: ai2 bi2 ≥ ai bi , which can be squared to find: ai2 bi2 ≥ (ai bi )2 , and this is the Cauchy-Schwartz inequality proved in Exercise 2.3. Thus, the triangle equality is proved. Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling Exercise 2.5. Using Cartesian coordinates where the position vector is x = (x1, x2, x3) and the fluid velocity is u = (u1, u2, u3), write out the three components of the vector: (u ⋅ ∇)u = ui ∂u j ∂x i . ( ) Solution 2.5. € € + % ∂u ( % ∂u ( % ∂u ( / – u1’ 1 * + u2 ‘ 1 * + u3 ‘ 1 * & ∂x 2 ) & ∂x 3 ) – & ∂x1 ) % ∂u ( % ∂u ( % ∂u ( % ∂ u ( – % ∂u ( % ∂u ( % ∂u (a) (u ⋅ ∇ )u = ui ‘ j * = u1’ j * + u2 ‘ j * + u3 ‘ j * = , u1’ 2 * + u2 ‘ 2 * + u3 ‘ 2 *0 & ∂x i ) & ∂x1 ) & ∂x 2 ) & ∂x 3 ) – & ∂x1 ) & ∂x 2 ) & ∂x 3 )% ( % ( % ∂u3 (- ∂u3 ∂ u3 u + u + u ‘ * ‘ * *- 1 2 3’ & ∂x 2 ) & ∂x 3 )1 . & ∂x1 ) ) # ∂u & # ∂u & # ∂u & + u% ( + v% ( + w% ( + $ ∂z ‘ + + $ ∂x ‘ $ ∂y ‘ + # ∂v & # ∂v & # ∂v & + = * u% ( + v% ( + w% ( . $ ∂z ‘ + + $ ∂x ‘ $ ∂y ‘ # & + # ∂w & # ∂w &+ ∂w + u%$ (‘ + v% ( + w%$ (‘+ ∂z / $ ∂y ‘ , ∂x ( ) The vector in this exercise, (u ⋅ ∇ )u = ui ∂u j ∂x i , is an important one in fluid mechanics. As described in Ch. 3, it is the nonlinear advective acceleration. € € Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling Exercise 2.6. Convert ∇ × ∇ρ to indicial notation and show that it is zero in Cartesian coordinates for any twice-differentiable scalar function ρ. Solution 2.6. Start with the definitions of the cross product and the gradient. ∂ ∂ 2ρ ∇ × ( ∇ρ ) = εijk (∇ρ )k = εijk ∂xj ∂ x j∂ xk Write out the vector component by component recalling that εijk = 0 if any two indices are equal. Here the “i” index is the free index. ! 2 2 # ε123 ∂ ρ + ε132 ∂ ρ ∂ x2∂ x3 ∂ x3∂ x2 # # 2 2 # ∂ ρ ∂ ρ ∂ 2ρ εijk = ” ε 213 + ε 231 ∂ x j∂ xk # ∂ x1∂ x3 ∂ x3∂ x1 # 2 ∂ ρ ∂ 2ρ # ε312 + ε321 ∂ x1∂ x2 ∂ x2∂ x1 # $ % ! 2 2 # # ∂ ρ – ∂ ρ # # ∂ x2∂ x3 ∂ x3∂ x2 # # # # ∂ 2ρ ∂ 2ρ = – + & ” # # ∂ x1∂ x3 ∂ x3∂ x1 # # ∂ 2ρ ∂ 2ρ # # − # ‘ # $ ∂ x1∂ x2 ∂ x2∂ x1 % # # # # &=0, # # # # ‘ where the middle equality follows from the definition of εijk (2.18), and the final equality follows ∂ 2ρ ∂ 2ρ when ρ is twice differentiable so that . = ∂x j∂x k ∂x k∂x j € Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling Exercise 2.7. Using indicial notation, show that a × (b × c) = (a ⋅ c)b − (a ⋅ b)c. [Hint: Call d ≡ b × c. Then (a × d)m = εpqmapdq = εpqmapεijqbicj. Using (2.19), show that (a × d)m = (a ⋅ c)bm − (a ⋅ b)cm.] Solution 2.7. Using the hint and the definition of εijk produces: (a × d)m = εpqmapdq = εpqmapεijqbicj = εpqmεijq bicjap = –εijqεqpm bicjap. Now use the identity (2.19) for the product of epsilons: (a × d)m = – (δipδjm – δimδpj) bicjap = – bpcmap + bmcpap. Each term in the final expression involves a sum over “p”, and this is a dot product; therefore (a × d)m = – (a ⋅ b)cm + bm(a ⋅ c). Thus, for any component m = 1, 2, or 3, a × (b × c) = − (a ⋅ b)c + (a ⋅ c)b = (a ⋅ c)b − (a ⋅ b)c. Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling Exercise 2.8. Show that the condition for the vectors a, b, and c to be coplanar is εijkaibjck = 0. Solution 2.8. The vector b × c is perpendicular to b and c. Thus, a will be coplanar with b and c if it too is perpendicular to b × c. The condition for a to be perpendicular with b × c is: a ⋅ (b × c) = 0. In index notation, this is aiεijkbjck = 0 = εijkaibjck. Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling Exercise 2.9. Prove the following relationships: δijδij = 3, εpqrεpqr = 6, and εpqiεpqj = 2δij. Solution 2.9. (i) δijδij = δii = δ11 + δ22 + δ33 = 1 + 1 + 1 = 3. For the second two, the identity (2.19) is useful. (ii) εpqrεpqr = εpqrεrpq = δppδqq – δpqδpq = 3(3) – δpp = 9 – 3 = 6. (iii) εpqiεpqj = εipqεpqj = – εipqεqpj = – (δipδpj – δijδpp) = – δij + 3δij = 2δij. Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling Exercise 2.10. Show that C⋅CT = CT⋅C = δ, where C is the direction cosine matrix and δ is the matrix of the Kronecker delta. Any matrix obeying such a relationship is called an orthogonal matrix because it represents transformation of one set of orthogonal axes into another. Solution 2.10. To show that C⋅CT = CT⋅C = δ, where C is the direction cosine matrix and δ is the matrix of the Kronecker delta. Start from (2.5) and (2.7), which are x “j = x iCij and x j = x “iC ji , respectively, and change the index “i” into “m” on (2.5): x “j = x m Cmj . Substitute this into (2.7) to find: x j = x “iC ji = ( x m Cmi )C ji = CmiC ji x m . € € However, we also have xj = δjmxm, so € δ jm x m = CmiC ji x m → δ jm = CmiC ji , which can be written:€ δ jm = CmiCijT = C⋅CT, and taking the transpose € of the thisT produces: T (δ jm ) = δmj = (CmiCijT ) = CmiT Cij = CT⋅C. € € Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling Exercise 2.11. Show that for a second-order tensor A, the following quantities are invariant under the rotation of axes: I1 = Aii , I 2 = A11 A12 A21 A22 + A22 A23 A32 A33 + A11 A13 A31 A33 , and I3 = det(Aij). [Hint: Use the result of Exercise 2.8 and the transformation rule (2.12) to show that Iʹ′1 = Aʹ′ii = Aii.= I1. Then show that AijAji and AijAjkAki are also invariants. In fact, all contracted scalars of the I 2 = 12 “# I12 − Aij A ji $% , form AijAjk ⋅⋅⋅ Ami are invariants. Finally, verify that I 3 = 13 “# Aij A jk Aki − I1 Aij A ji + I 2 Aii $% . Because the right-hand sides are invariant, so are I2 and I3.] Solution 2.11. First prove I1 is invariant by using the second order tensor transformation rule (2.12): ” = Cim C jn Aij . Amn Replace Cjn by CnjT and set n = m, ” = CimCnjT Aij → Amm ” = Cim CmjT Aij . Amn T Use the result of Exercise 2.8, € δij = Cim Cmj = , to find: ” = δij Aij = Aii . I1 = Amm € Thus, the first invariant € is does not depend on a rotation of the coordinate axes. Now consider€whether or not AmnAnm is invariant under a rotation of the coordinate axes. Start with a double application of (2.12): € T ” Anm ” = Cim C jn Aij C pn Cqm A pq = C jn CnpT Cim Cmq Amn Aij A pq . From the result of Exercise 2.8, the factors in parentheses in the last equality are Kronecker delta functions, so ” Anm ” = δ jpδiq Aij A pq = Aij A ji . Amn € Thus, the matrix contraction AmnAnm does not depend on a rotation of the coordinate axes. The manipulations for AmnAnpApm are a straightforward extension of the prior efforts for Aii and AijAji. € ” Anp ” A”pm = Cim C jn Aij Cqn Crp Aqr Csp Ctm Ast = C jn CnqT Crp C psT CimCmtT Aij Aqr Ast . Amn Again, the factors in parentheses are Kronecker delta functions, so ” Anp ” A”pm = δ jqδrsδit Aij Aqr Ast = Aiq Aqs Asi , Amn which implies that the matrix contraction AijAjkAki does not depend on a rotation of the coordinate € axes. Now, for the second invariant, verify the given identity, starting from the given definition € for I2. A A12 A22 A23 A11 A13 I2 = 11 + + A21 A22 A32 A33 A31 A33 = A11 A22 − A12 A21 + A22 A33 − A23 A32 + A11 A33 − A13 A31 = A11 A22 + A22 A33 + A11 A33 − ( A12 A21 + A23 A32 + A13 A31 ) ( ( € € € € € ) ( )( )( )( ) ( )( ) )( )( ( ) = 12 A112 + 12 A222 + 12 A332 + A11 A22 + A22 A33 + A11 A33 − A12 A21 + A23 A32 + A13 A31 + 12 A112 + 12 A222 + 12 A332 = 1 2 2 [ A11 + A22 + A33 ] − (2A12 A21 + 2A23 A32 + 2A13 A31 + A + A + A ) 1 2 2 11 2 22 2 33 ) Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling = 12 I12 − 12 ( A11 A11 + A12 A21 + A13 A31 + A12 A21 + A22 A22 + A23 A32 + A13 A31 + A23 A32 + A33 A33 ) = 12 I12 − 12 ( Aij A ji ) = 12 (I12 − Aij A ji ) Thus, since I2 only depends on I1 and AijAji, it is invariant under a rotation of the coordinate axes because I1 and AijAji are invariant under a rotation of the coordinate axes. The manipulations for the third invariant are a tedious but not remarkable. Start from the given definition for I3, and group like terms. € € I3 = det ( Aij ) = A11 (A22 A33 − A23 A32 ) − A12 (A21 A33 − A23 A31) + A13 (A21 A32 − A22 A31 ) = A11 A22 A33 + A12 A23 A31 + A13 A32 A21 − ( A11 A23 A32 + A22 A13 A31 + A33 A12 A21 ) € € € € € € € € (a) Now work from the given identity. The triple matrix product AijAjkAki has twenty-seven terms: A113 + A11 A12 A21 + A11 A13 A31 + A12 A21 A11 + A12 A22 A21 + A12 A23 A31 + A13 A31 A11 + A13 A32 A21 + A13 A33 A31 + 3 € A21 A11 A12 + A21 A12 A22 + A21 A13 A32 + A22 A21 A12 + A22 + A22 A23 A32 + A23 A31 A12 + A23 A32 A22 + A23 A33 A32 + A31 A11 A13 + A31 A12 A23 + A31 A13 A33 + A32 A21 A13 + A32 A22 A23 + A32 A23 A33 + A33 A31 A13 + A33 A32 A23 + A333 These can be grouped as follows: Aij A jk Aki = 3(A12 A23 A31 + A13 A32 A21 ) + A11(A112 + 3A12 A21 + 3A13 A31) + 2 2 (b) A22 (3A21 A12 + A22 + 3A23 A32 ) + A33 (3A31 A13 + 3A32 A23 + A33 ) The remaining terms of the given identity are: −I1 Aij A ji + I2 Aii = I1(I2 – Aij A ji ) = I1 (I2 + 2I2 − I12 ) = 3I1I2 – I13 , where€the result for I2 has been used. Evaluating the first of these two terms leads to: 3I1I2 = 3(A11 + A22 + A33 )(A11 A22 − A12 A21 + A22 A33 − A23 A32 + A11 A33 − A13 A31 ) = 3(A11 + A22 + A33 )(A11 A22 + A22 A33 + A11 A33 ) − 3(A11 + A22 + A33 )(A12 A21 + A23 A32 + A13 A31 ) . € to (b) produces: Adding this Aij A jk Aki + 3I1I2 = 3(A12 A23 A31 + A13 A32 A21 ) + 3(A11 + A22 + A33 )(A11 A22 + A22 A33 + A11 A33 ) + 2 2 A11 (A112 − 3A23 A32 ) + A22 (A22 − 3A13 A31 ) + A33 (A33 − 3A12 A21 ) = 3(A12 A23 A31 + A13 A32 A21 − A11 A23 A32 − A22 A13 A31 − A33 A12 A21 ) + 3 3 (c) 3(A11 + A22 + A33 )(A11 A22 + A22 A33 + A11 A33 ) + A113 + A22 + A33 The last term of the given identity is: €3 3 3 2 2 2 2 2 2 I13 = A€ 11 + A22 + A33 + 3(A11 A22 + A11 A33 + A22 A11 + A22 A33 + A33 A11 + A33 A22 ) + 6A11 A22 A33 3 3 = A113 + A22 + A33 + 3(A11 + A22 + A33 )(A11 A22 + A11 A33 + A22 A33 ) – 3A11 A22 A33 € Subtracting this from (c) produces: Aij A jk Aki + 3I1I2 − I13 = 3(A12 A23 A31 + A13 A32 A21 − A11 A23 A32 − A22 A13 A31 − A33 A12 A21 + A11 A22 A33 ) = 3I3 . This verifies that the given identity for I3 is correct. Thus, since I3 only depends on I1, I2, and AijAjkAki, it is invariant under a rotation of the coordinate axes because these quantities are € invariant under a rotation of the coordinate axes as shown above. € Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling Exercise 2.12. If u and v are vectors, show that the products uiυj obey the transformation rule (2.12), and therefore represent a second-order tensor. Solution 2.12. Start by applying the vector transformation rule (2.5 or 2.6) to the components of u and v separately, u”m = Cim ui , and v “n = C jn v j . The product of these two equations produces: u”m v “n = Cim C jn uiv j , which is the same as (2.12) tensors. € for second order € € Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling Exercise 2.13. Show that δij is an isotropic tensor. That is, show that δʹ′ij = δij under rotation of the coordinate system. [Hint: Use the transformation rule (2.12) and the results of Exercise 2.10.] Solution 2.13. Apply (2.12) to δij, ” = CimC jnδij = CimCin = CmiT Cin = δmn . δmn where the final equality follows from the result of Exercise 2.10. Thus, the Kronecker delta is invariant under coordinate rotations. € Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling Exercise 2.14. If u and v are arbitrary vectors resolved in three-dimensional Cartesian 2 coordinates, use the definition of vector magnitude, a = a ⋅ a , and the Pythagorean theorem to show that u⋅v = 0 when u and v are perpendicular. € Solution 2.14. Consider the magnitude of the sum u + v, 2 u + v = (u1 + v1 ) 2 + (u2 + v 2 ) 2 + (u3 + v 3 ) 2 = u12 + u22 + u32 + v12 + v 22 + v 32 + 2u1v1 + 2u2v 2 + 2u3v 3 2 2 = u + v + 2u ⋅ v , which can be rewritten: 2 2 2 u + v − u − v = 2u ⋅ v . € €When u and v are perpendicular, the Pythagorean theorem requires the left side to be zero. Thus, u ⋅ v = 0. € € Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling Exercise 2.15. If u and v are vectors with magnitudes u and υ, use the finding of Exercise 2.14 to show that u⋅v = uυcosθ where θ is the angle between u and v. Solution 2.15. Start with two arbitrary vectors (u and v), and view them so that the plane they define is coincident with the page and v is horizontal. Consider two additional vectors, βv and w, that are perpendicular (v⋅w = 0) and can be summed together to produce u: w + βv = u. u w θ v βv Compute the dot-product of u and v: u⋅v = (w + βv) ⋅v = w⋅v + βv⋅v = βυ2. where the final equality holds because v⋅w = 0. From the geometry of the figure: βv βυ u , or β = cosθ . cosθ ≡ = υ u u Insert this into the final equality for u⋅v to find: %u ( u ⋅ v = ‘ cos θ *υ 2 = uυ cosθ . &υ € ) € € Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling Exercise 2.16. Determine the components of the vector w in three-dimensional Cartesian coordinates when w is defined by: u⋅w = 0, v⋅w = 0, and w⋅w = u2υ2sin2θ, where u and v are known vectors with components ui and υi and magnitudes u and υ, respectively, and θ is the angle between u and v. Choose the sign(s) of the components of w so that w = e3 when u = e1 and v = e2. Solution 2.16. The effort here is primarily algebraic. Write the three constraints in component form: u⋅w = 0, or u1w1 + u2 w 2 + u3 w 3 = 0 , (1) v⋅w = 0, or υ1w1 + υ 2 w 2 + υ 3 w 3 = 0 , and (2) The third one requires a little more effort since the angle needs to be eliminated via a dot product: € w⋅w = u2υ2sin2θ = u2υ2(1 – cos2θ) = u2υ2 – (u⋅w)2 or € w12 + w 22 + w 32 = (u12 + u22 + u32 )(υ12 + υ 22 + υ 32 ) − (u1υ1 + u2υ 2 + u3υ 3 ) 2 , which leads to (3) w12 + w 22 + w 32 = (u1υ 2 − u2υ1 ) 2 + (u1υ 3 − u3υ1 ) 2 + (u2υ 3 − u3υ 2 ) 2 . Equation (1) implies: (4) w1 = −(w 2 u2 + w 3 u3 ) u1 € Combine (2) and (4) to eliminate w1, and solve the resulting equation for w2: € $ υ ‘ $ υ ‘ −υ1 (w 2 u2 + w 3 u3 ) u1 + υ 2 w 2 + υ 3 w 3 = 0 , or &− 1 u2 + υ 2 )w 2 + & − 1 u3 + υ 3 ) w 3 = 0 . % u1 ( % u1 ( € Thus: $υ ‘ $ υ ‘ $u υ − uυ ‘ (5) w 2 = +w 3 & 1 u3 − υ 3 ) &− 1 u2 + υ 2 ) = w 3 & 3 1 1 3 ) . € % u1 ( € % u1 ( % u1υ 2 − u2υ1 ( Combine (4) and (5) to find: ‘ w $$ υ u − υ 3 u1 ‘ w 3 $ υ1u3 u2 − υ 3 u1u2 + υ 2 u1u3 − υ1u2 u3 ‘ w1 = − 3 && 1 3 +) ) u2 + u3 ) = − & u1 %% υ 2 u1 − υ1u2 ( u1 % υ 2 u1 − υ1u2 ( ( € $ u2υ 3 − u3υ 2 ‘ w $ −υ u u + υ 2 u1u3 ‘ (6) =− 3& 3 1 2 ) = w3& ). u1 % υ 2 u1 − υ1u2 ( % u1υ 2 − u2υ1 ( Put € (5) and (6) into (3) and factor out w3 on the left side, then divide out the extensive common factor that (luckily) appears on the right and as the numerator inside the big parentheses. $€ (u2υ 3 − u3υ 2 ) 2 + (u3υ1 − u1υ 3 ) 2 + (u1υ 2 − u2υ1 ) 2 ‘ 2 2 2 w & ) = (u1υ 2 − u2υ1 ) + (u1υ 3 − u3υ1 ) + (u2υ 3 − u3υ 2 ) 2 (u1υ 2 − 2u$2υ1 ) 1 ‘ % ( , so w 3 = ±(u1υ 2 − u2υ1 ) . w3 & = 1 ) 2 % (u1υ 2 − u2υ1 ) ( If u = (1,0,0), and v = (0,1,0), then using the plus sign produces w3 = +1, so w 3 = +(u1υ 2 − u2υ1 ) . Cyclic permutation of the indices allows the other components of w to be determined: w1€= u2υ 3 − u3υ 2 , € w 2 = u3υ1 − u1υ 3 , € w 3 = u1υ 2 − u2υ1 . 2 3 € € € € Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling Exercise 2.17. If a is a positive constant and b is a constant vector, determine the divergence and the curl of u = ax/x3 and u = b×(x/x2) where x = x12 + x 22 + x 32 ≡ x i x i is the length of x. € € € € € € € € € € € Solution 2.17. Start with the divergence calculations, and use x = x12 + x 22 + x 32 to save writing. $ ‘ $ ‘ $ ax ‘ $ ∂ ∂ ∂ ‘ & x1€, x 2 , x 3 ) = a& ∂ , ∂ , ∂ ) ⋅ $& x1, x 2 , x 3 ‘) ∇ ⋅ & 3 ) = a& , , )⋅ % x ( % ∂x1 ∂x 2 ∂x 3 ( & [ x 2 + x 2 + x 2 ] 3 2 ) % ∂x1 ∂x 2 ∂x 3 ( % x 3 ( 2 3 % 1 ( € # ∂ # x1 & ∂ # x 2 & ∂ # x 3 && # 1 3 x1 & 1 3 x2 1 3 x3 = a% % 3 ( + 2×1 ) + 3 − 2x 2 ) + 3 − 2x 3 )( % 3(+ % 3 (( = a% 3 − 5 ( 5 ( 5 ( ‘ 2x x 2x x 2x $ ∂x1 $ x ‘ ∂x 2 $ x ‘ ∂x 3 $ x ” $ x # 3 3( x 2 + x 2 + x 2 ) & # 3 3& 1 2 3 ( = a% 3 − 3 ( = 0 . = a%% 3 − 5 ( x x ‘ $x ‘ $x Thus, the vector field ax/x3 is divergence free even though it points away from the origin everywhere. % b × x ( % ∂ ∂ ∂ ( % b2 x 3 − b3 x 2 ,b3 x1 − b1 x 3 ,b1 x 3 − b2 x1 ( ∇⋅’ 2 * =’ , , *⋅’ * & x ) & ∂x1 ∂x 2 ∂x 3 ) & x12 + x 22 + x 32 ) $ ∂ $ b x − b x ‘ ∂ $ b3 x1 − b1 x 3 ‘ ∂ $ b1 x 2 − b2 x1 ” =& & 2 3 2 3 2)+ & )+ & )) ( ∂x 2 % ( ∂x 3 % (( x x2 x2 % ∂x1 % # 2 & # 2 & # 2 & = (b2 x 3 − b3 x 2 )%− 4 (2×1 )( + (b3 x1 − b1 x 3 )% − 4 (2x 2 )( + (b1 x 2 − b2 x1 )%− 4 (2x 3 )( $ x ‘ $ x ‘ $ x ‘ 4 = − 4 (b2 x 3 x1 − b3 x 2 x1 + b3 x1 x 2 − b1 x 3 x 2 + b1 x 2 x 3 − b2 x1 x 3 ) = 0 . x This field is divergence free, too. The curl calculations produce: $ ax ‘ $ ∂ ∂ ∂ ‘ $ x1, x 2 , x 3 ‘ $ ∂x −3 ∂x −3 ∂x −3 ∂x −3 ∂x −3 ∂x −3 ‘ ∇ × & 3 ) = a& , , × = a x − x , x − x , x − x ) & ) & ) 2 1 3 2 1 % x ( % ∂x1 ∂x 2 ∂x 3 ( % x 3 ( % 3 ∂x 2 ∂x 3 ∂x 3 ∂x1 ∂x1 ∂x 2 ( # 3 x3 & 3 x2 3x 3 x3 3 x2 3 x1 = a% − 2x 2 ) + 2x 3 ),− 15 (2x 3 ) + 2×1 ),− 2×1 ) + 2x 2 )( = (0,0,0) 5 ( 5 ( 5 ( 5 ( 5 ( $ 2x ‘ 2x 2x 2x 2x 2x 3 Thus, thus the vector field ax/x is also irrotational. $ b × x ‘ $ ∂ ∂ ∂ ‘ $ b2 x 3 − b3 x 2 ,b3 x1 − b1 x 3 ,b1 x 2 − b2 x1 ‘ ∇ ×& 2 ) = & , , ) ×& ). % x ( % ∂x1 ∂x 2 ∂x 3 ( % x12 + x 22 + x 32 ( There are no obvious simplifications here. Therefore, compute the first component and obtain the others by cyclic permutation of the indices. $b × x’ ∂ $ b1 x 2 − b2 x1 ‘ ∂ $ b3 x1 − b1 x 3 ‘ ∇ ×& 2 ) = & )− & ) % x (1 ∂x 2 % ( ∂x 3 % ( x2 x2 # & # −2 & b −2 b = 12 + (b1 x 2 − b2 x1 )% 4 (2x 2 + 12 − (b3 x1 − b1 x 3 )% 4 (2x 3 $x ‘ $x ‘ x x 2 2 2 2b x − 4b1 x 2 + 4b2 x1 x 2 + 4b3 x1 x 3 − 4b1 x 3 2b 4 x = 1 = − 21 + 41 (b1 x1 + b2 x 2 + b3 x 3 ) 4 x x x This field is rotational. The other two components of its curl are: € € Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling $b × x’ $b × x’ 2b 4 x 2b 4 x ∇ × & 2 ) = − 22 + 42 (b1 x1 + b2 x 2 + b3 x 3 ) , ∇ × & 2 ) = − 23 + 43 (b1 x1 + b2 x 2 + b3 x 3 ) . % x (2 % x (3 x x x x € € Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling Exercise 2.18. Obtain the recipe for the gradient of a scalar function in cylindrical polar coordinates from the integral definition (2.32). € Solution 2.18. Start from the appropriate form of (2.32), ez 1 “z z ∇Ψ = lim ∫∫ ΨndA , where Ψ is a scalar function of e! V →0 V A “! position x. Here we choose a nearly rectangular volume “R eR V = (RΔϕ)(ΔR)(Δz) centered on the point x = (R, ϕ, z) with sides aligned perpendicular to the coordinate y directions. Here the e unit vector depends on ϕ so its ! direction is slightly different at ϕ ± Δϕ/2. For small Δϕ, R x this can be handled by keeping the linear term of a simple Taylor series: [eϕ ] ≅ eϕ ± (Δϕ 2)(∂eϕ ∂ϕ ) = eϕ  (Δϕ 2)e R . Considering the drawing ϕ ϕ ±Δϕ 2 and noting that n is an outward normal, there are six contributions to ndA: # $ ΔR & ΔR ‘ outside = % R + inside = −& R − (ΔϕΔze R , )ΔϕΔze R , $ % 2 ‘ 2 ( € % % Δϕ ( Δϕ ( e R *, eR * , close vertical side = ΔRΔz’−eϕ − more distant vertical side = ΔRΔz’eϕ − & ) & ) 2 2 top = RΔϕΔRe z , € €and bottom = −RΔϕΔRe z . Here all the unit vectors are evaluated at the center of the volume. Using a two term Taylor series approximation for Ψ on each of the six surfaces, and taking the six contributions in the same € € order, the integral definition becomes a sum of six terms representing ΨndA. € € 5.( 9 1 .( 1 ΔR ∂Ψ +( ΔR + ΔR ∂Ψ +( ΔR + -* R + -e R ΔϕΔz3 − 0* Ψ − -* R − -e R ΔϕΔz3 + 70* Ψ + 7 2 ∂R ,) 2 , 2 ∂R ,) 2 , 2 /) 2 7/) 7 7 77 . 1 . 1 ( 7( 1 Δϕ ∂Ψ +( Δϕ + Δϕ ∂Ψ +( Δϕ + ∇Ψ = lim −e − e ΔRΔz + Ψ + e − e ΔRΔz + 60* Ψ − : * * 3 0 3 – ϕ * – ϕ R R ΔR →0 RΔϕΔRΔz ) , ) , 2 ∂ϕ 2 2 ∂ϕ 2 ) , ) , / 2 / 2 7 7 Δϕ →0 Δz →0 7.( 7 1 .( 1 + + 70* Ψ + Δz ∂Ψ -e z RΔϕΔR3 − 0* Ψ − Δz ∂Ψ -e z RΔϕΔR3 + … 7 78/) 7; 2 ∂z , 2 ∂z , 2 /) 2 € € € Here the mean value theorem has been used and all listings of Ψ and its derivatives above are evaluated at the center of the volume. The largest terms inside the big {,}-brackets are proportional to ΔϕΔRΔz. The remaining higher order terms vanish when the limit is taken. /( Ψ 3 ( Ψ R ∂Ψ + R ∂Ψ + e R -ΔϕΔRΔz − *− e R − e R -ΔϕΔRΔz +1 1* e R + ) 2 2 ∂R , 2 ∂R , 1) 2 1 ( eϕ ∂Ψ e R + 1(eϕ ∂Ψ e R + 1 1 ∇Ψ = lim − Ψ-ΔϕΔRΔz + * − Ψ-ΔϕΔRΔz + 0* 4 ΔR →0 RΔϕΔRΔz ) 2 ∂ϕ 2 2 ∂ϕ 2 , ) , 1 1 Δϕ →0 Δz →0 1( R ∂Ψ + 1 ( R ∂Ψ + e z -ΔϕΔRΔz − *− e z -ΔϕΔRΔz + … 1* 1 ) 2 ∂z , 2) 2 ∂z , 5 ‘Ψ ∂Ψ 1 ∂Ψ Ψ ∂Ψ * ∂Ψ 1 ∂Ψ ∂Ψ ∇Ψ = ( e R + eR + eϕ − e R + ez + = eR + eϕ + ez ∂R R ∂ϕ R ∂z , ∂R R ∂ϕ ∂z )R Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling Exercise 2.19. Obtain the recipe for the divergence of a vector function in cylindrical polar coordinates from the integral definition (2.32). Solution 2.19. Start from the appropriate form of (2.32), ez 1 “z z ∇ ⋅ Q = lim ∫∫ n ⋅ QdA , where Q = (QR, Q , Qz) is a vector V →0 V A “! function of position x. Here we choose a nearly rectangular “R volume V = (RΔϕ)(ΔR)(Δz) centered on the point x = (R, ϕ, z) with sides aligned perpendicular to the coordinate directions. Here the e unit vector depends on ϕ so its ! direction is slightly different at ϕ ± Δϕ/2. Considering the x R drawing and noting that n is an outward normal, there are six contributions to ndA: # $ ΔR & ΔR ‘ outside = % R + (ΔϕΔze R , inside = −& R − )ΔϕΔze R , close vertical side = −ΔRΔz[eϕ ]ϕ −Δϕ 2 , $ % 2 ‘ 2 ( ϕ € ϕ more distant vertical side = ΔRΔz[eϕ ] , top = RΔϕΔRe z , and bottom = −RΔϕΔRe z . € otherwise specified. Using Here the unit vectors are evaluated at the center of the volume unless € € a two-term Taylor series approximation for the components of Q on each of the six surfaces, and taking the six contributions to n ⋅ QdA in€the same order, the integral € definition becomes: € 5.( 1 . 1 9 ( ΔR ∂QR +( ΔR + ΔR ∂QR +( ΔR + -* R + -ΔϕΔz3 − 0* QR − -* R − -ΔϕΔz3 +7 70* QR + 2 ∂R ,) 2 , 2 ∂R ,) 2 , 2 /) 2 7 7/) 77 1 .( 1 1 € 77.( Δϕ ∂Qϕ + Δϕ ∂Qϕ + ∇ ⋅ Q = lim −ΔRΔz + Q + ΔRΔz + 60* Qϕ − : )3 0* ϕ 3 -( ΔR →0 RΔϕΔRΔz 2 ∂ϕ 2 ∂ϕ ) , ) , / 2 / 2 7 7 Δϕ →0 Δz →0 7.( 7 1 .( 1 + + 70* Qz + Δz ∂Qz -RΔϕΔR3 − 0*Qz − Δz ∂Qz – RΔϕΔR3 + … 7 78/) 7; 2 ∂z , 2 ∂z , 2 /) 2 ϕ +Δϕ 2 € € € Here the mean value theorem has been used and all listings of the components of Q and their derivatives are evaluated at the center of the volume. The largest terms inside the big {,}brackets are proportional to ΔϕΔRΔz. The remaining higher order terms vanish when the limit is taken. 0(QR R ∂QR + 4 ( QR R ∂QR + Δ ϕ ΔRΔz − − − Δ ϕ ΔRΔz + 2* + 2 *) 2 2 ∂R -, 2 ∂R -, 2) 2 2 ( + ( + 2 2 1 1 ∂Qϕ 1 ∂Qϕ ∇ ⋅ Q = lim ΔϕΔRΔz + * ΔϕΔRΔz + 1−*− 5 ΔR →0 RΔϕΔRΔz )2 ∂ϕ , 2 ) 2 ∂ϕ , 2 Δϕ →0 Δz →0 2( R ∂Ψ + 2 ( R ∂Ψ + Δ ϕ ΔRΔz − − Δ ϕ ΔRΔz + … 2* 2 *) 2 ∂z -, 3) 2 ∂z , 6 &Q ∂Q 1 ∂Qϕ ∂Qz ) 1 ∂ 1 ∂Qϕ ∂Qz ∇⋅Q =’ R + R + + + *= (RQR ) + ∂R R ∂ϕ ∂z + R ∂ R R ∂ϕ ∂z (R e! eR y Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling Exercise 2.20. Obtain the recipe for the divergence of a vector function in spherical polar coordinates from the integral definition (2.32). z Solution 2.20. Start from the appropriate 1 form of (2.32), ∇ ⋅ Q = lim ∫∫ n ⋅ QdA , V →0 V A #r where Q = (Qr, Q , Q ) is a vector function of position x. Here we choose a nearly er #” rectangular volume V = (rΔ θ )(rsin θ Δ ϕ )(Δr) € ” centered on the point x = (r, θ, ϕ) with sides e” aligned perpendicular to the coordinate directions. Here the unit vectors depend on y θ and ϕ so directions are slightly different at e! rsin” ! θ ± Δθ/2, and ϕ ± Δϕ/2. Considering the x #! drawing and noting that n is an outward normal, there are six contributions to ndA: # Δr & # Δr & $ Δr ‘ $ Δr ‘ outside = % r + (Δθ% r + ( sin θΔϕ (e r ) , inside = & r − )Δθ& r − ) sin θΔϕ (−e r ) , $ % 2′ $ 2’ 2( % 2( bottom = [ r sin(θ + Δθ 2)ΔϕΔr](eθ )θ +Δθ 2 , top = [ r sin(θ − Δθ 2)ΔϕΔr](−eθ )θ −Δθ 2 , θ ϕ close vertical side = rΔθΔr(−eϕ ) , and more distant vertical side = rΔθΔr(+eϕ ) . ϕ −Δϕ 2 ϕ +Δϕ 2 €Here the unit vectors are evaluated at the center € of the volume unless otherwise specified. Using a two-term Taylor series approximation € for the corresponding components of Q on each of the € six surfaces produces: % € Δθ ∂Qθ ( $ € Δr ∂Qr ‘ % Δr ∂Qr ( outside: &Qr + , *(eθ ) )e r , inside: ‘Qr − *e r , bottom: ‘Qθ + & % & 2 ∂θ ) θ +Δθ 2 2 ∂r ( 2 ∂r ) & & Δθ ∂Qθ ) Δϕ ∂Qϕ ) top: (Qθ − +(eθ )θ −Δθ 2 , close vertical side : (Qϕ − +(−eϕ )ϕ −Δϕ 2 , and ‘ 2 ∂θ * 2 ∂ϕ * ‘ Δϕ ∂Qϕ ( € € % € more distant vertical side : ‘Qϕ + *(eϕ )ϕ +Δϕ 2 . 2 ∂ϕ ) & to n ⋅ QdA , the integral definition becomes: € € Collecting and summing the six contributions 1 ∇ ⋅ Q = lim × Δr →0 (rΔθ )(r sin θΔϕ )Δr € ΔΔθϕ →0 →0 € 2 2 70* 3 0* 3; Δr ∂Qr – * Δr Δr ∂Qr – * Δr 92,Qr + /Δθ, r + / sin θΔϕ5 − 2,Qr − /Δθ, r − / sin θΔϕ5 9 2 ∂r . + 2. 2 ∂r . + 2. 91+ 4 1+ 49 9 9 3 0* 39 9 0* * * Δθ ∂Qθ Δθ Δθ ∂Qθ Δθ 8+2, Qθ + /r sin,θ + /ΔϕΔr5 − 2, Qθ − / r sin,θ − /ΔϕΔr5< + + 2 ∂θ . 2 . 2 ∂θ . 2 . 4 1+ 49 9 1+ 9 0* 9 3 0* 3 ∂Q ∂Q 9−2, Qϕ − Δϕ ϕ /rΔθΔr5 + 2,Qϕ + Δϕ ϕ / rΔθΔr5 + … 9 2 ∂ϕ . 2 ∂ϕ . 9: 1+ 9= 4 1+ 4 € Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling The largest terms inside the big {,}-brackets are proportional to ΔθΔϕΔr. The remaining higher order terms vanish when the limit is taken. 0* ∂Q 4 2,r 2 r + 2rQr /Δθ sin θΔϕΔr 2 . 2+ ∂r 2 2 2 * 1 ∂Q ∇ ⋅ Q = lim × 1+,sin θ θ + cos θQθ /rΔθΔϕΔr5 Δr →0 (rΔθ )(r sin θΔϕ )Δr . ∂θ 2 + 2 Δθ →0 Δϕ →0 2 *∂Qϕ 2 2+, 2 /rΔθΔϕΔr + … 3 + ∂ϕ . 6 Cancel the common factors and take the limit, to find: .' ∂Q * ' * '∂Q * 1 1 ∂Q ∇⋅Q = × /)r 2 r + 2rQr , sin θ + )sin θ θ + cosθQθ ,r + ) ϕ ,r2 + ( + ( ∂ϕ + 3 (r)(r sin θ ) 0( ∂r ∂θ € &∂ ∂Q ) 1 ∂ = 2 × ' ( r 2Qr ) sin θ + r (sin θQθ ) + r ϕ * r sin θ ( ∂r ∂θ ∂ϕ + 1 ∂ 1 ∂ 1 ∂Qϕ € = 2 ( r 2Qr ) + (sin θQθ ) + r ∂r r sin θ ∂θ r sin θ ∂ϕ € € Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling Exercise 2.21. Use the vector integral theorems to prove that ∇ ⋅ (∇ × u) = 0 for any twicedifferentiable vector function u regardless of the coordinate system. Solution 2.21. Start with the divergence theorem for a vector € function Q that depends on the spatial coordinates, ∫∫∫ ∇ ⋅ QdV = ∫∫ n ⋅ QdA V A n t1 V nc1 nc2 t2 where the arbitrary closed volume V has surface A, and the A1 outward normal is n. For this exercise, let Q = ∇ × u so that A2 C ∫∫∫ ∇ ⋅ (∇ × u)dV = ∫∫ n ⋅ (∇ × u) dA . € V A Now split V into two sub-volumes V1 and V2, where the surface of V1 is A1 and the surface of V2 € surfaces, but A1+ A2 = A so: is A2. Here A1 and A2 are not closed ∫∫∫ ∇ ⋅ (∇ × u)dV = ∫∫ n ⋅ (∇ × u) dA + ∫∫ n ⋅ (∇ × u) dA . € V A1 A2 where n is the same as when the surfaces were joined. However, the bounding curve C for A1 and A2 is the same, so Stokes theorem produces: ∫∫∫ ∇ ⋅ (∇ × u) dV = ∫ u ⋅ t1 ds + ∫ u ⋅ t 2 ds . € V C C Here the tangent vectors t1 = nc1 × n and t 2 = nc 2 × n have opposite signs because nc1 and nc2, the normals to C that are tangent to surfaces A1 and A2, respectively, have opposite sign. Thus, the two terms on the right side of the last equation are equal and opposite, so (i) ∫∫∫ ∇ ⋅ (∇ × u)dV = 0 . € € V For an arbitrary closed volume of any size, shape, or location, this can only be true if ∇ ⋅ (∇ × u) = 0 . For example, if ∇ ⋅ (∇ × u) were nonzero at some location, then integration in small volume centered on this location would not be zero. Such a nonzero integral is not allowed € by (i); thus, ∇ ⋅ (∇ × u) must be zero everywhere because V is arbitrary. € € € Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling Exercise 2.22. Use Stokes’ theorem to prove that ∇ × (∇φ ) = 0 for any single-valued twicedifferentiable scalar φ regardless of the coordinate system. Solution 2.22. From (2.34) Stokes Theorem€is: ∫∫ (∇ × u) ⋅ ndA = ∫ u ⋅ tds . A € C Let u = ∇φ , and note that ∇φ ⋅ tds = (∂φ ∂s) ds = dφ because the t vector points along the contour C that has path increment ds. Therefore: (ii) (∇ × [∇φ ]) ⋅ ndA = ∫ ∇φ ⋅ tds = ∫ dφ = 0 , €∫∫ A C C € where the final equality holds for integration on a closed contour of a single-valued function φ. For an arbitrary surface A of any size, shape, orientation, or location, this can only be true if ∇ × ( ∇φ ) = 0 . For example, if ∇ × ( ∇φ ) = 0 were nonzero at some location, then an area € integration in a small region centered on this location would not be zero. Such a nonzero integral is not allowed by (ii); thus, ∇ × ( ∇φ ) = 0 must be zero everywhere because A is arbitrary.