Solved

# question in Diverge

Posted on 2012-09-09
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if u = x^2 y^2 + sin (t)

the diverge is

2xy^2 + 2x^2y - d(t)/dt cos (t)

is this correct ? Also , can it be considered incompressible flow ?
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Question by:c_hockland

LVL 26

Expert Comment

"is this correct ?" no
-
I am not sure what you mean by d(t)/dt cos(t)
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Author Comment

2xy^2 + 2x^2y -  cos (t)
0

LVL 26

Expert Comment

I have a hard time following your problem. What is the time doing in there?
The divergance operator operated on a vector field. I see no unit vectors in your equation
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LVL 31

Expert Comment

Divergence is a property of a vector field,

F(x,y) =(F1(x,y),F2(x,y))

then

Div F(x,y) = dF1(x,y)/dx  + dF2(x,y)/dy

The function you have shown is a scalar function, not a vector.
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Author Comment

i meant to write

x^2y^2i + sin(t) k
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LVL 26

Expert Comment

You are still in trouble, especially if you mean t to be time.

Strictly speaking, using standard notation where only the i component is a space function, your answer is
div (u) = 2 (y^2) x
0

LVL 31

Accepted Solution

## Divergence of a Vector field

If, where x1, x2, x3 are spatial coordinates,

F(x1,x2,x3) = f1(x1,x2,x3)i + f2(x1,x2,x3)j + f3(x1,x2,x3)k

then

div(F(x1,x2,x3) )= df1/dx1 + df2/dx2 + df3/dx3

Hence if

u(x,y,z,t) = x²y²i + 0j + sin(t)k

then,  (at some time t)

div( u(x,y,z,t) ) =2xy²

(remember your 3rd spatial coordinate is z not t)

## Incompressible Flow

For flow to be incompressible then the divergence of the flow field must be zero at all times t, ie

div( u(x,y,z,t) ) = 0

u(x,y,z,t) = x²y²i + 0j + sin(t)k

then at a given time t

div( u(x,y,z,t) ) = 2xy²

which is clearly not zero every where, ie your flow is compressible.  The sin(t)k  is just a rigid body motion term in the z axis, btw.
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