2.25 Advanced Fluid Mechanics, Solutions to Vector Calculus Review Problems (3.12)

MIT Department of Mechanical Engineering 2.25 Advanced Fluid Mechanics Panton 3.12 This problem is from “Incompressible Flow” by Ronald L. Panton Write the following formulas in Gibbs’s notation using the symbol \. Convert the expressions to Cartesian notation and prove that the equations are correct. div(φv) = φ div v + v . grad φ div(u × v) = v · curl u − u · curl v curl(u × v) = v · grad u − u · grad v + u div v − v div u Solution: a. div(φv) = φ divv + v · gradφ (1) Using Gibbs Notation we can rewrite equation 1 as: \ · (φ v) = φ \ · v + v · \φ (2) In order to write the equation in index notation, starting from left hand side we have: \ · (φv) = ( ∂ φvj ∂ ∂ φvj ei ) · (φ vj ej ) = (ei · ej ) = δij ∂xi ∂xi ∂xi (3) We know that δij is only non zero when i = j, therefore: ∂φvj ∂φvi δij = ∂xi ∂xi (4) where i is the summation variable. Then for the first term on the right hand side, following the same method as above: ∂ ∂vj ∂vj ∂vi φ \·v =φ ( e ) · vj e j = φ (e · e ) = φ δij = φ (5) ∂xi i j ∂xi ∂xi ∂xi i And for the second term: v · \φ = vi ei · ∂φ ∂φ ∂φ ∂φ e = vi e · e = vi δij = vi ∂xj i j ∂xj ∂xi ∂xj j (6) And thus equation 1 in index notation has a form of: ∂vi ∂φ ∂φvi =φ + vi ∂xi ∂xi ∂xi (7) Now in order to prove equation 1, we start from the left hand side of equation 7 and use the chain rule to open the derivative: ∂vi ∂φ ∂φvi =φ + vi (8) ∂xi ∂xi ∂xi Multiplying the last term by δii = 1: φ ∂φ ∂vi ∂φ ∂vi ∂φ ∂vi + vi δii = φ + vi (e · e ) = φ + vi ei · e ∂xi ∂xi ∂xi i i ∂xi ∂xi i ∂xi (9) By multiplying the last term with δii = 1 we are able to get to the dot product of two vector quantities. Also we know that in index notation: ∂vi =\·v (10) ∂xi ∂φ e = \φ ∂xi i (11) Therefore substituting back from equations 10 and 11 into 9 we get: \ · (φv) = and equation 1 is proved. ∂vi ∂φ ∂φvi =φ + vi ei · e = φ\ · v + v · \φ ∂xi ∂xi i ∂xi (12) b. div(u × v) = v · curlu − u · curlv (13) Equation 13 can be written in Gibbs notation as: \ · (u × v) = v · \ × u − u · \ × v (14) In index notation, the left hand side can be written as: \ · (u × v) = ( ∂uj vk ∂uj vk ∂ e ) · uj vk εjkl el = ( ) εjkl ei · el = εjkl δil ∂xi ∂xi ∂xi i (15) δil is only non zero when i = l. Thus: ∂uj vk ∂uj vk εjkl δil = εjki ∂xi ∂xi Therefore: \ · (u × v) = ∂uj vk εjki ∂xi (16) (17) As for the right hand side: v·\×u= ∂ v i ei · e × uk ek = ∂xj j ∂uk e · (ej × ek ) = vi ∂xj i ∂uk e · εjkl el = vi ∂xj i ∂uk εjkl δil = vi ∂xj ∂uk εjki vi ∂xj u·\×v = ∂ ui ei · e × vk ek = ∂xj j ∂vk (e × ek ) = ui ei · ∂xj j ∂vk εjkl ei · el = ui ∂xj ∂vk εjkl δil = ui ∂xj ∂vk εjki ui ∂xj (18) (19) Therefore, equation 13 in index notation is written as: ∂uk ∂vk ∂uj vk εjki = vi εjki − ui εjki ∂xi ∂xj ∂xj (20) In order to prove equation 13, we start from the left hand side and use the chain rule to open the derivatives: \ · (u × v) = ∂uj ∂vk ∂uj vk εjki = εjki vk + εjki uj ∂xi ∂xi ∂xi (21) Then multiplying the first term by δkk and the second term by δjj ( δkk = δjj = 1 does not add anything to the equation, however it helps in creating the dot product needed to prove the equations) : ∂uj ∂vk ∂uj ∂vk εjki vk + εjki uj = εjki vk δkk + εjki uj δjj ∂xi ∂xi ∂xi ∂xi (22) We know that εjki = −εikj . Thus: ∂uj εjki vk δkk + ∂xi ∂uj εijk vk δkk − ∂xi ∂vk εjki uj δjj = ∂xi ∂vk εikj uj δjj = ∂xi ∂uj ∂vk εjki vk ek · ek − εikj uj ej · ej = ∂xi ∂xi ∂uj ∂vk εjki ek · vk ek − εikj ej · uj ej ∂xi ∂xi (23) From the definition of curl in index notation we know: \×u= ∂uj εijk ek ∂xi (24) \×v = ∂vk εikj ej ∂xi (25) and Therefore, ∂uj ∂vk εjki ek · vk ek − εikj ej · uj ej = ∂xi ∂xi (\ × u) · v − (\ × v) · u (26) (27) Since dot product is commutative, we can rewrite equation 27 as: (\ × u) · v − (\ × v) · u = v · (\ × u) − u · (\ × v) (28) And thus equation 13 is proved. c. curl(u × v) = v · grad u − u · grad v + u · div v − v · div u (29) Equation 29 in Gibbs notation is presented as: \ × (u × v) = v · \ u − u · \ v + u \ · v − v \ · u (30) For the index notation, starting from the left hand side of equation 29: \ × (u × v) = ∂ e × (uj ej × vk ek ) = ∂xi i ∂uj vk ei × (ej × ek ) = ∂xi ∂uj vk ei × (εjkl el ) = ∂xi ∂uj vk εjkl ei × el = ∂xi ∂uj vk εjkl εilm em ∂xi (31) For the first term on the right hand side we have: v·\ u= ∂uj vi ei · ( e e )= ∂xk j k ∂uj vi ∂ e · (ej ek ) ∂xk i (32) Using the identity c · (a b) = (a · b)c: ∂uj e · (ej ek ) = ∂xk i ∂uj vi ∂ (e · e ) e = ∂xk i j k ∂uj vi ∂ δij ek = ∂xk ∂ui vi ∂ e ∂xk k vi ∂ (33) Same with the rest of the terms: v·\ u= ∂vj ui ei · ( e e )= ∂xk j k ∂vj ui e · (ej ek ) = ∂xk i ∂vj ui (e · e ) e = ∂xk i j k ∂vi ui e ∂xk k u \·v = ∂ u i ei ( e · v k ek ) = ∂xj j ∂vk ui e (e · e ) = ∂xj i j k ∂vk ui e δjk = ∂xj i ∂vj ui e ∂xj i v \·u= ∂ vi ei ( e · v k ek ) = ∂xj j ∂uk vi e (e · e ) = ∂xj i j k ∂uk vi e δjk = ∂xj i ∂uj vi e ∂xj i (34) (35) (36) Thus equation 29 in index notation is given as : ∂uj vk ∂ui ∂vi ∂vj ∂uj εjkl εilm em = vi ∂ e − ui e + ui e − vi e ∂xk k ∂xk k ∂xj i ∂xj i ∂xi (37) In order to prove this equation, starting from the left hand side (equation 31), we use the identity shown in equation 3.3.5 in Panton to change the alternating tensor into the Kronecker delta. Also, using the fact that εilm = εlmi and εjkl = εljk we have: ∂uj vk εjkl εilm em = ∂xi ∂uj vk εljk εlmi em = ∂xi ∂uj vk (δmj δik − δmk δij ) em = ∂xi ∂vk ∂uj (uj + vk )(δmj δik − δmk δij ) em = ∂xi ∂xi ∂vk ∂vk ∂uj ∂uj (uj δmj δik − uj δmk δij + vk δmj δik − vk δmk δij ) em ∂xi ∂xi ∂xi ∂xi (38) The first term in equation 38 is only non zero when m = j and i = k. uj ∂vk δmj δik em = ∂xi ∂vi uj e = ∂xi j ∂vi ui ej = ∂xi u \·v (39) The second term is only non zero when m = k and i = j: ∂vk δmk δij em = ∂xi ∂vk −ui e δii = ∂xi k ∂vk −ui e (e · e ) = ∂xi k i i ∂vk −ui e · (ek ei ) = ∂xi i ∂vk −ui ei · e e = ∂xi k i −u · \ v −uj (40) Here we used the identity c · (a b) = (a · b) c to change ek (ei · ei ) into the dot product of a vector and a tensor. The third term is only non zero when m = j and i = k: ∂uj δmj δik em = ∂xi ∂uj vi e δii = ∂xi j ∂uj vi e (e · e ) = ∂xi j i i ∂uj vi e · (ej ei ) = ∂xi i ∂uj vi ei · e e = ∂xi j i v·\ u vk (41) And the fourth term is only non zero when m = k and i = j: ∂uj δmk δij em = ∂xi ∂ui −vk e = ∂xi k ∂ui −vk ek = ∂xi −v \ · u −vk (42) Putting equations 39, 40, 41 and 42 back into equation 38 and equating it with equation 31, we get: \ × (u × v) = v · \ u − u · \ v + u \ · v − v \ · u (43) Therefore, equation 29 is proved. D