Velocity and acceleration of a particle of a fluid

Woolyabyss
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Homework Statement


Fluid mech.png

Homework Equations


x = (x1,x2,x3) are the coordinates of the current configuration and X = (X1,X2,X3) are the coordinates of the reference configuration where x = y(X,t) where y is a deformation of X.

The Attempt at a Solution


(a)
taking the partial derivatives of xi with respect to t:
∂(x1)/∂t = αX2 + 2(α^2)X3*t
∂(x2)/∂t = 2(α^2)*X1*t + αX3
∂(x3)/∂t = αX1 +2(α^2)X2t

so v(X,t) = (αX2 + 2(α^2)X3*t , 2(α^2)*X1*t + αX3 , αX1 +2(α^2)X2t )
(1) v((d,d,d),0) = ( αd , αd , αd )

a(X,t) = ∂v/∂t = ( 2(α^2)X3, 2(α^2)*X1 , 2(α^2)X2 )

(2) a((d,d,d),0) = ( 2(α^2)d , 2(α^2)*d, 2(α^2)d )(b)
for this part we know that x = (d,d,d) , so we wan't to find the X that corresponds to this point so we can plug it into the functions v(X,t) and a(X,t).

rearranging the three equations of motion:
X1 = x1 - αX2t - (α^2)X3t^2 (a)
X2 = x1 - αX3t - (α^2)X1t^2 (b)
X3 = x1 - αX1t - (α^2)X2t^2 (c)

now, subbing (b) and (c) into (a) we have:

X1 = x1 - α( x1 - αX3t - (α^2)X1t^2 )t - (α^2)( x1 - αX1t - (α^2)X2t^2 )t^2

from the above it seems like like I won't be able to express X1 solely in terms of spatial coordinates because if I expand the second and third term I won't be able to get rid of X3 and X2.
Any help would be appreciated.
 
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You have 3 linear algebraic equations in three unknowns, X1, X2, and X3 corresponding to (d,d,d) @ t.
 
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Chestermiller said:
You have 3 linear algebraic equations in three unknowns, X1, X2, and X3 corresponding to (d,d,d) @ t.

Thanks for the reply,
looking at the system of equations:
d = X1 + αX2t + (α^2)X3t^2
d = X2 + αX3t + (α^2)X1t^2
d = X3 + αX1t + (α^2)X2t^2

if I begin subtracting the equations from one another to remove the d terms and then manipulating the resultant equations I will get:
Xi(terms involving α and t ) = 0 , for i =1,2,3
I'm not sure what I can say about this other than that either the Xi or () must be zero.
 
Woolyabyss said:
Thanks for the reply,
looking at the system of equations:
d = X1 + αX2t + (α^2)X3t^2
d = X2 + αX3t + (α^2)X1t^2
d = X3 + αX1t + (α^2)X2t^2

if I begin subtracting the equations from one another to remove the d terms and then manipulating the resultant equations I will get:
Xi(terms involving α and t ) = 0 , for i =1,2,3
I'm not sure what I can say about this other than that either the Xi or () must be zero.
Just use Gaussian elimination.
 
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Chestermiller said:
Just use Gaussian elimination.
Thanks again, using Gaussian elimination as you suggested I found X1,X2 and X3 all to be equal to d(1-αt)/(1-(α^3)(t^3)).
I suspect this to be the correct answer as its easy to see why when t approaches α^-1 this motion is unrealistic.
 
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