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

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

Therefore, $2y-\frac{(2xy^2)}{2}+f^{'} (x)=-xy^2+2y+\frac{x^3}{3}$

$f^{'} (x)=\frac{x^3}{3}$

$f(x)=\frac{x^4}{12}$

Substituting f(x) in Ψ

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

iii) To determine potential function

$\frac{∂∅}{∂x}$=-u

$\frac{(∂∅)}{∂x}= \frac-{y}^{3}-2x+x^{2} y$

Integrating above expression

$∅=-\frac{(y^3 x)}{3}-\frac{(2x^2}{2}+\frac{(yx^3)}{3}+f(y)$

Differentiating w.r.t. y

$\frac{∂∅}{∂y}=-y^2 x+\frac{x^3}{3}+f^{'} (y)$

But $\frac{∂∅}{∂y}$=v

$-y^2 x+\frac{x^3}{3}+f^{'} (y)=xy^2-2y-\frac{x^3}{3}$

$f^{'} (y)=2xy^2-\frac{(2x^3)}{3}-2y$

$f(y)=\frac{(2xy^3)}{3}-\frac{2x^3 y}{3}-y^2$

$∅=\frac{-(y^3 x)}{3}-\frac{2x^2}{2}+\frac{yx^3}{3}+\frac{2xy^3}{3}-\frac{2x^3 y}{3}-y^2$

$∅=\frac{y^3 x}{3}-x^2-y^2-\frac{(yx^3)}{3}$

The above is potential function for given velocity component