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

- anonymous

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

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- amistre64

this looks scary, what is it?

- anonymous

i've been at this for 5 hours now

- anonymous

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

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## More answers

- anonymous

u took calc 3 right?

- amistre64

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

- anonymous

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

- amistre64

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

- anonymous

so a b and c are the partials correct?

- amistre64

yes
(au +bv +cw)

__is the left side then__- anonymous

right

- anonymous

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

- 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

- anonymous

gotcha...distribute it

- amistre64

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

- amistre64

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

- anonymous

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

- 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)

- amistre64

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

- amistre64

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

- amistre64

i think i got all the parts ready :)

- 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? :)

- anonymous

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

- anonymous

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

- amistre64

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

- anonymous

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

- amistre64

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

- amistre64

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

- anonymous

oh okay..yea

- amistre64

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

- amistre64

lets try to work up the RHS to match it

- anonymous

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

- amistre64

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

- anonymous

what is wuc?

- 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

- anonymous

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

- 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

- anonymous

yea i'm getting lost

- anonymous

trying to type it out with the partials

- 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?

- anonymous

lol...meteorology major

- anonymous

sorry i'm trying to copy it all down lol

- anonymous

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

- amistre64

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

- anonymous

alright i'm all caught up writing lol

- anonymous

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

- anonymous

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

- amistre64

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

- anonymous

the left side should equal the right side exactly

- amistre64

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

- amistre64

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

- amistre64

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

- 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

- 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> ??

- anonymous

so the answer above is correct as it stands?

- 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

- anonymous

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

- amistre64

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

- anonymous

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

- amistre64

good luck :)

- anonymous

u still around?

- amistre64

im around

- anonymous

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

- 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.

- anonymous

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

- amistre64

hmmm, ...

- amistre64

are v, u, amd w function of x?

- amistre64

gradV = V' right?

- anonymous

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

- anonymous

del operator?

- anonymous

is what she calls it

- amistre64

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

- amistre64

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

- amistre64

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

- amistre64

or is V.del, just V' ?

- amistre64

\[\frac{d}{dx}u+\frac{d}{dx}v+\frac{d}{dx}w=

__\]__- amistre64

well, with the right parts that is :)

- anonymous

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

- 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.

- 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)

- 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

- anonymous

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

- amistre64

yeah, james is smart that way

- anonymous

well the other guy if u scroll down solved it

- anonymous

but i can' t understand it

- amistre64

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

- amistre64

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

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

- amistre64

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

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