## anonymous 4 years ago V = u i + v j + w k and ∇= ∂╱∂x i + ∂╱∂y j + ∂╱∂z k that: (V∙∇)V = (1╱2)∇(V∙V)−V X (∇XV)

1. amistre64

this looks scary, what is it?

2. anonymous

i've been at this for 5 hours now

3. anonymous

i just have to prove the left side equals the right side

4. anonymous

u took calc 3 right?

5. amistre64

took, taking, taught meself, and all that stuff, yeah

6. anonymous

yea..figured since u helped me the other day you might be able to hand this

7. amistre64

can we clean it up some with a b c? V = ui + vj + wk ∇= ai + bj + ck ------------------ = au i +bv j +cw k

8. anonymous

so a b and c are the partials correct?

9. amistre64

yes (au +bv +cw) <u,v,k> is the left side then

10. anonymous

right

11. anonymous

so then it comes out to (au^2 + bv^2+ cw^2

12. amistre64

not quite, that dot is a scalar, not a vector so its applied to all the vector parts to stretch or shrink it to fit V.V = |V|^2 |V|^2 = u^2+v^2+w^2

13. anonymous

gotcha...distribute it

14. amistre64

edit:$\frac{1}{2}*(V.V)=\frac{ u^2+v^2+w^2}{2}$

15. amistre64

this is scaled to our gradient for that first part on the right

16. anonymous

isn't it 1/2 the gradient times V dot V?

17. amistre64

grad x V is: a b c u v w x = bw-vc y = -(aw-uc) z = av-bu and to cross that with V again we get: u v w bw-vc uc-aw av-bu x = v(av-bu) - w(uc-aw) y = -(u(av-bu)-w(bw-vc)) z = u(uc-aw) - v(bw-vc)

18. amistre64

yes, and V.V is just: u^2 + v^2 + w^2

19. amistre64

u v w u v w --------------- u^2 + v^2 + w^2

20. amistre64

i think i got all the parts ready :)

21. amistre64

$(au+bv+cw)(u\bar i+v\bar j+w\bar k)=...$ $...\frac{u^2+v^2+w^2}{2}(a\bar i+b\bar j+c\bar k)-((av^2-buv -wuc+aw^2)\bar i$$+(-uav+bu^2+bw^2-wvc)\bar j+ (u^2c-auw - vbw + v^2c)\bar k$ this look about right so far? :)

22. anonymous

its hard to say...b/c i been doing it partials and u v and w

23. anonymous

but...the setup looks correct from what i can decifer

24. amistre64

a b c just stand in for partials to make it cleam up

25. anonymous

then yes...it appears to be correct...it's only the right hand side of the equation right?

26. amistre64

the first line is the lhs; the rest is a bit lengthy and had to split it up some

27. amistre64

if we multiply it all by 2 we get rid of the fraction

28. anonymous

oh okay..yea

29. amistre64

$<(2au^2+2buv+2cuw),(2auv+2bv^2+2cwv),(2auw+2bvw+2cw^2)>$ is our LHS written out as its vector component stuff

30. amistre64

lets try to work up the RHS to match it

31. anonymous

alright...i'm gonna try and re copy this for my homework in long hand without the a b and c lol

32. amistre64

$<au^2,bv^2,cw^2>-etc$ $<(au^2-2av^2+2buv +2wuc-2aw^2),$$(bv^2+2uav-2bu^2-2bw^2+2wvc),$$(cw^2-2u^2c+2auw +2vbw -2v^2c)>$ with any luck is how the RHS is looking after combining it all together

33. anonymous

what is wuc?

34. amistre64

as long as components wise we zero out, it should be proofed then au^2- 2au2 -2av^2+2buv-2buv +2wuc - 2cuw-2aw^2 = 0 wu is V parts, and c is a partial

35. anonymous

nevermind lol....partial of w u over z

36. amistre64

-au^2 -2av^2 -2aw^2 = 0 might need to rechk the math in all this, but this is the basic concept that im thinking of

37. anonymous

yea i'm getting lost

38. anonymous

trying to type it out with the partials

39. amistre64

V = <u,v,w> g = <a,b,c> (V∙g)V = (1╱2)g(V∙V)−V X (gXV) (V∙g) = au+bv+cw (au+bv+cw)V = < au^2+bvu+cwu, auv+bv^2+cwv, auw+bvw+cw^2> is our LHS ............................................ (1╱2)g(V∙V) (V∙V) = v^2+u^2+w^2 (1╱2)(v^2+u^2+w^2) = v^2/2 + u^2/2 + w^2/2 (1╱2)(V∙V)g = av^2/2 + au^2/2 + aw^2/2 , bv^2/2 + bu^2/2 + bw^2/2 , cv^2/2 +cu^2/2 + cw^2/2 .......................................................... − V X (gXV) gxV: a b c v u w ----- x = bw-cu y = cv-aw z = au-bv -V x that is bw-cu cv-aw au-bv -v -u -w --------------------- x = -wcv+aw^2 + au^2 -bvu y = -vau+bv^2 +bw^2-cuw z = -ubw+cu^2 + cv^2-awv ugh lol -------------------------- to get rid of the /2 *2 it all < 2au^2+2bvu+2cwu, 2auv+2bv^2+2cwv, 2auw+2bvw+2cw^2> = = <av^2 + au^2 + aw^2 , bv^2 + bu^2 + bw^2 , cv^2 +cu^2 + cw^2> + <-2wcv+2aw^2 +2au^2 -2bvu, -2vau+2bv^2 +2bw^2 -2cuw, -2ubw +2cu^2 + 2cv^2 -2awv >

40. amistre64

now compare component parts: 2au^2+2bvu+2cwu = av^2 + au^2 + aw^2-2wcv+2aw^2 +2au^2 -2bvu ? au^2 +2cwu = av^2 + aw^2-2wcv+2aw^2 +2au^2 ? u(au +2cw) = a(v^2 + w^2+2w^2) -2wcv +2au^2 ? yeah, where do you get these problems at anyhoos?

41. anonymous

lol...meteorology major

42. anonymous

sorry i'm trying to copy it all down lol

43. anonymous

its hard to see the difference between u and v when their close tother

44. amistre64

the wolf gives me this for VxgxV lets see if its does good with the rest too $<-a u^2+b u v+c v w-a w^2, a u v-b v^2+c u w-b w^2, -c u^2-c v^2+b u w+a v w>$

45. anonymous

alright i'm all caught up writing lol

46. anonymous

yea...from what i got..this does not equal each other lol

47. anonymous

3 terms per component on the left hand side to 5 terms per component in the rhs

48. amistre64

$<\frac a2u^2+\frac a2v^2+\frac a2w^2 , \frac b2u^2+\frac b2v^2+\frac b2w^2, \frac c2u^2+\frac c2v^2+\frac c2w^2>$$<a u^2-b u v-c v w+a w^2, -a u v+b v^2-c u w+b w^2, c u^2+c v^2-b u w-a v w>$ $av^2+buv+cwv=( \frac{au^2+ av^2+ aw^2+2au^2-2b u v-2c v w+2a w^2}{2})$ $2av^2+2buv+2cwv= au^2+ av^2+ aw^2+2au^2-2b u v-2c v w+2a w^2$ $av^2+4buv+4cwv= au^2+ aw^2+2au^2+2a w^2$ yeah, is it spose to proof correct? or is this just some sort of manuvuer on their part?

49. anonymous

the left side should equal the right side exactly

50. amistre64

try with some random easy numbers maybe, or functions lol just to test

51. amistre64

V = x,x^2,x^3 g = 1,2x,3x^2 at x=1

52. amistre64

V = 1,1,1 g = 1,2,3

53. amistre64

(v.g)v = (1+2+3)<1,1,1> = <6,6,6> lol, nice 1/2 g = (1/2, 1 , 3/2)(3) = <3/2, 3 , 9/2> <6,6,6> = <3/2, 3 , 9/2> - stuff <12,12,12> = <3, 6 , 9> - 2*stuff -3 -6 -9 ---------------- <9,6,3> = -2*stuff <-18,-12,-6> = stuff

54. amistre64

stuff = vxgxv 1,1,1 1,2,3 ----- x=1 y=-2 z=1 1,-2,1 1, 1, 1 ------- x=-3 y= 0 z= 3 <-18,-9,-3> = <-3,0,3> ??

55. anonymous

so the answer above is correct as it stands?

56. amistre64

when i plug in some random function for V and grad it to g; this is what I get; so I dont see how it is spose to be true as a proof according to what youve listed

57. anonymous

it would just be weird b/c i have 5 of these to prove...and all 4 have matched perfectly

58. amistre64

<9,6,3> = -2*stuff /-2 -------- <-9/2, -3, -3/2> = stuff but still; thats not a match

59. anonymous

alright...i gotta work out some stuff with wolf then lol

60. amistre64

good luck :)

61. anonymous

u still around?

62. amistre64

im around

63. anonymous

i just emailed my professor..this is what i got back

64. anonymous

yep. they are supposed to match. Come by my office tomorrow with what you have done, and I can look at see where you are going wrong. Every year students think the 1/2 is an error, but it is correct......usually people go wrong when taking the derivative of u^2.

65. anonymous

now...i'm really confused b/c i don't see where i'm taking any derivatives

66. amistre64

hmmm, ...

67. amistre64

are v, u, amd w function of x?

68. amistre64

69. anonymous

i mean...exactly the way the question is written...is what i have

70. anonymous

del operator?

71. anonymous

is what she calls it

72. amistre64

so grad is the operator on V and not a specific vector ....

73. amistre64

del(V.V) is not what I had in mind at the begining

74. amistre64

but then what is V.del? del of what?

75. amistre64

or is V.del, just V' ?

76. amistre64

$\frac{d}{dx}u+\frac{d}{dx}v+\frac{d}{dx}w=<u',v',w'>$

77. amistre64

well, with the right parts that is :)

78. anonymous

well del is just = to partial/partial x etc.

79. anonymous

(V∙∇)V = (1╱2)∇(V∙V)−V X (∇XV) let's do it term by term....... Letting V=(u(x,y,z),v(x,y,z),w(x,y,z)) and start with the term 12∇(V⋅V). V⋅V=u2+v2+w2 and this is a scalar quantity. Then ∇(u2+v2+w2) is a vector quantity, and as above, is equal to: [∂∂x(u2+v2+w2)i+∂∂y(u2+v2+w2)j+∂∂z(u2+v2+w2)k] in calculating each component of this vector, we will use subscripts to indicate partial differentiation wrt the subscripted variable. We have [(2uux+2vvx+2wwx)i+(2uuy+2vvy+2wwy)j+(2uuz+2vvz+2wwz)k] So multiplying by the 1/2 will remove all these 2s meaning: 12∇(V⋅V)=[(uux+vvx+wwx)i+(uuy+vvy+wwy)j+(uuz+vvz+wwz)k] Next we want to deal with the term: V×(∇×V) First we calculate ∇×V=(wy−vz)i+(uz−wx)j+(vx−uy)k Next we calculate V×[(wy−vz)i+(uz−wx)j+(vx−uy)k] =[v(vx−uy)−w(uz−wx)]i+[w(wy−vz)−u(vx−uy)]j +[u(uz−wx)−v(wy−vz)]k and this equals V×(∇×V). Now we do the subtraction. In the i component we get: [uux+vvx+wwx−v(vx−uy)+w(uz−wx)]i which is equal to [uux+vuy+wuz]i Looking good? Let's do the j and k component now: [uuy+vvy+wwy−w(wy−vz)+u(vx−uy)]j which is equal to [uvx+vvy+wvz]j and at long last, the k component... [uuz+vvz+wwz−u(uz−wx)+v(wy−vz)]k which is equal to [uwx+vwy+wwz]k THEREFORE we have shown that the right hand side reduces to the vector: [uux+vuy+wuz]i+[uvx+vvy+wvz]j+[uwx+vwy+wwz]k Now let's stop here and think for a bit. V⋅∇=u∂∂x+v∂∂y+w∂∂z and this is an operator which acts on vectors to the right of it, if you like, so we get: (V⋅∇)V=[u∂∂xu+v∂∂yu+w∂∂zu]i+[u∂∂xv+v∂∂yv+w∂∂zv]j +[u∂∂xw+v∂∂yw+w∂∂zw]k Notice that this is exactly what we just proved, therefore the identity has been verified.

80. amistre64

$V = u i + v j + w k$ $∇= \frac{∂}{∂x} i + \frac{∂}{∂y} j +\frac{∂}{∂z} k$ $(V∙∇)= \frac{∂u}{∂x} + \frac{∂v}{∂y} +\frac{∂w}{∂z}$ $(V∙∇)V= \left(\frac{∂u}{∂x}u + \frac{∂v}{∂y}u +\frac{∂w}{∂z}u\right)i+\left(\frac{∂u}{∂x}v + \frac{∂v}{∂y}v +\frac{∂w}{∂z}v\right)j+\left(\frac{∂u}{∂x}w + \frac{∂v}{∂y}w +\frac{∂w}{∂z}w\right)k$ ----------------------------------- $∇(V∙V)= \left(\frac{∂}{∂x}u^2i + \frac{∂}{∂y}v^2j +\frac{∂}{∂z}w^2k\right)$ $∇(V∙V)= \left(\frac{∂u}{∂x}2ui + \frac{∂v}{∂y}2vj +\frac{∂w}{∂z}2wk\right)$ $∇(V∙V)/2= \left(\frac{∂u}{∂x}ui + \frac{∂v}{∂y}vj +\frac{∂w}{∂z}wk\right)$ ------------------------------------ so far −V X (∇XV)

81. anonymous
82. amistre64

$-VxV=<0,0,0>$so i have to wonder about that part :) $(∇xV)=\frac{∂}{∂y}w-\frac{∂}{∂z}v,\ \frac{∂}{∂z}w-\frac{∂}{∂x}u,\ \frac{∂}{∂x}u - \frac{∂}{∂y} v$ $-Vx(∇xV)=(\frac{∂w}{∂y}-\frac{∂v}{∂z},,\ \frac{∂u}{∂x} - \frac{∂v}{∂y}$ $\ x=\frac{∂}{∂x}(wu)+\ \frac{∂}{∂x}(uv) - \frac{∂}{∂y}(v^2)-\frac{∂}{∂z}(w^2)$ $\ x=\frac{∂u}{∂x}w+\frac{∂w}{∂x}u+\ \frac{∂u}{∂x}v+\ \frac{∂v}{∂x}u - \frac{∂v}{∂y}2v-\frac{∂w}{∂z}2w$ if this is right, its a killer trying to keep track of it all

83. anonymous

yea...u see the link i just sent you.....a guy figured it all out...i can't make sense of it though

84. amistre64

yeah, james is smart that way

85. anonymous

well the other guy if u scroll down solved it

86. anonymous

but i can' t understand it

87. amistre64

ok, so we seem to be good up to that -v x del x v part; right?

88. amistre64

do you understand the notations?$du/dx=u_x$

89. amistre64

if so: $del(x)V=\begin{pmatrix}x&y&z\\d_x&d_y&d_z\\u&v&w\end{pmatrix}$ $x=(w_y-v_z)$ $y=(u_z-w_x)$ $z=(v_x-u_y)$ Vx(that stuff): \begin{pmatrix}x&y&z\\u&v&w\\(w_y-v_z)&(u_z-w_x)&(v_x-u_y) \end{pmatrix} $x=(vv_x-vu_y-wu_z+ww_x)$ $y=(uu_y-uv_x+ww_y-wv_z)$ $z=(vw_y-vv_z-uu_z+uw_x)$

90. amistre64

i can make it out, i just cant seem to get a good explanation on it tho