$u– y^{3}/3 + 2x-x^{2}y, v=xy^{2} -2y - x^{3}/3$

i. whether the flow is possible

ii. obtain an expression for stream function

iii. obtain an expression for potential function

u=$\frac{y^3}{3}+2x-x^2 y \ \ \ \ \ \ \ v=xy^2-2y-\frac{x^3}{3}$

i)Whether the flow is possible

For this condition,

$\frac{∂u}{∂x}+\frac{∂v}{∂y}$=0

$\frac{∂u}{∂x}$=2-2.x.y,$\frac{∂v}{∂y}$=2.x.y-2

Which satisfy the condition hence flow is possible.

ii)To determine stream function,

u=$\frac{∂Ψ}{∂y}$

$\frac{y^3}{3}+2x-x^2 y=\frac{∂Ψ}{∂y}$

Integrating both the sides,

Ψ=$\frac{y^4}{12}$+$2xy-\frac{(x^2 y^2)}{2}+f(x)$

Differentiating the above equation w.r.t. x,

$\frac{∂Ψ}{∂x}=2y-\frac{(2xy^2)}{2}+f^{'} (x)$

but,$\frac{∂Ψ}{∂x}$=-v …