Question

V = u i + v j + w k and ∇= ∂╱∂x i + ∂╱∂y j + ∂╱∂z k that:

(V∙∇)V = (1╱2)∇(V∙V)−V X (∇XV)

Answer 1

this looks scary, what is it?

Answer 2

i've been at this for 5 hours now

Answer 3

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

Answer 4

u took calc 3 right?

Answer 5

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

Answer 6

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

Answer 7

can we clean it up some with a b c?


V = ui + vj + wk
∇= ai + bj + ck
------------------
= au i +bv j +cw k

Answer 8

so a b and c are the partials correct?

Answer 9

yes

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

Answer 10

right

Answer 11

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

Answer 12

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

Answer 13

gotcha...distribute it

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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)

Answer 18

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

Answer 19

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

Answer 20

i think i got all the parts ready :)

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

oh okay..yea

Answer 29

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

Answer 30

lets try to work up the RHS to match it

Answer 31

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

Answer 32

\[<au^2,bv^2,cw^2>-etc\]
\[<(au^2-2av^2+2buv +2wuc-2aw^2),\]\[(bv^2+2uav-2bu^2-2bw^2+2wvc),\]\[(cw^2-2u^2c+2auw +2vbw -2v^2c)>\]

with any luck is how the RHS is looking after combining it all together

Answer 33

what is wuc?

Answer 34

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

Answer 35

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

Answer 36

-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

Answer 37

yea i'm getting lost

Answer 38

trying to type it out with the partials

Answer 39

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 >

Answer 40

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?

Answer 41

lol...meteorology major

Answer 42

sorry i'm trying to copy it all down lol

Answer 43

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

Answer 44

the wolf gives me this for VxgxV

lets see if its does good with the rest too
\[<-a u^2+b u v+c v w-a w^2, a u v-b v^2+c u w-b w^2, -c u^2-c v^2+b u w+a v w>\]

Answer 45

alright i'm all caught up writing lol

Answer 46

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

Answer 47

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

Answer 48

\[<\frac a2u^2+\frac a2v^2+\frac a2w^2 , \frac b2u^2+\frac b2v^2+\frac b2w^2, \frac c2u^2+\frac c2v^2+\frac c2w^2>\]\[<a u^2-b u v-c v w+a w^2, -a u v+b v^2-c u w+b w^2, c u^2+c v^2-b u w-a v w>\]

\[av^2+buv+cwv=( \frac{au^2+ av^2+ aw^2+2au^2-2b u v-2c v w+2a w^2}{2})\]
\[2av^2+2buv+2cwv= au^2+ av^2+ aw^2+2au^2-2b u v-2c v w+2a w^2\]
\[av^2+4buv+4cwv= au^2+ aw^2+2au^2+2a w^2\]
yeah, is it spose to proof correct? or is this just some sort of manuvuer on their part?

Answer 49

the left side should equal the right side exactly

Answer 50

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

Answer 51

V = x,x^2,x^3

g = 1,2x,3x^2

at x=1

Answer 52

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

Answer 53

(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

Answer 54

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> ??

Answer 55

so the answer above is correct as it stands?

Answer 56

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

Answer 57

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

Answer 58

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

Answer 59

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

Answer 60

good luck :)

Answer 61

u still around?

Answer 62

im around

Answer 63

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

Answer 64

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.

Answer 65

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

Answer 66

hmmm, ...

Answer 67

are v, u, amd w function of x?

Answer 68

gradV = V' right?

Answer 69

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

Answer 70

del operator?

Answer 71

is what she calls it

Answer 72

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

Answer 73

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

Answer 74

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

Answer 75

or is V.del, just V' ?

Answer 76

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

Answer 77

well, with the right parts that is :)

Answer 78

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

Answer 79

(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).

Vâ‹…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.

Answer 80

\[V = u i + v j + w k \]
\[∇= \frac{∂}{∂x} i + \frac{∂}{∂y} j +\frac{∂}{∂z} k\]

\[(V∙∇)= \frac{∂u}{∂x} + \frac{∂v}{∂y} +\frac{∂w}{∂z}\]

\[(V∙∇)V= \left(\frac{∂u}{∂x}u + \frac{∂v}{∂y}u +\frac{∂w}{∂z}u\right)i+\left(\frac{∂u}{∂x}v + \frac{∂v}{∂y}v +\frac{∂w}{∂z}v\right)j+\left(\frac{∂u}{∂x}w + \frac{∂v}{∂y}w +\frac{∂w}{∂z}w\right)k\]
-----------------------------------

\[∇(V∙V)= \left(\frac{∂}{∂x}u^2i + \frac{∂}{∂y}v^2j +\frac{∂}{∂z}w^2k\right)\]
\[∇(V∙V)= \left(\frac{∂u}{∂x}2ui + \frac{∂v}{∂y}2vj +\frac{∂w}{∂z}2wk\right)\]
\[∇(V∙V)/2= \left(\frac{∂u}{∂x}ui + \frac{∂v}{∂y}vj +\frac{∂w}{∂z}wk\right)\]
------------------------------------

so far


−V X (∇XV)

Answer 81

http://openstudy.com/users/sonofa_nh#/updates/4f258d94e4b0a2a9c266fb07

Answer 82

\[-VxV=<0,0,0>\]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

Answer 83

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

Answer 84

yeah, james is smart that way

Answer 85

well the other guy if u scroll down solved it

Answer 86

but i can' t understand it

Answer 87

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

Answer 88

do you understand the notations?\[du/dx=u_x\]

Answer 89

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}

\[x=(vv_x-vu_y-wu_z+ww_x)\]
\[y=(uu_y-uv_x+ww_y-wv_z)\]
\[z=(vw_y-vv_z-uu_z+uw_x)\]

Answer 90

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