# Help needed to find the flow potential function

1. Nov 30, 2007

### Hells_Kitchen

Here is the catch:

We are given a 2-D velocity flow incompressible, irrotational of the form:

--> ^ ^ ^ ^
V = u i + v j = [4y -x(1+x)] i + y(2x+1) j

and we are asked to find the flow potential which obeys the Laplace Eq. for 2-D incompressible, irrotational flow: dΦ = u dx + v dy

in other words:

∂Φ
---- = u
∂x

∂Φ
---- = v
∂y

I integrated the first one and then the second one and compared the two functions and combined the terms, but at the end the Φ does not satisfy the first equation only the second one.

Another technique, I integrated the first function with respect to x and Φ is expressed as 4xy - x^2/2 - x^3/3 + f(y) = Φ (x,y)

now I differentiate with respect to y and equate it to v:

4x - f'(y) = y(2x+1) which solves to f(y) = xy^2 + y^2/2 -4xy + C

Plug it in the above expression and get:

Φ (x,y) = xy^2 + y^2/2 - x^2/2 - x^3/3 + C

now the first partial diff. eq is not satisfied but the first is.

Can someone explain what is wrong here?

2. Nov 30, 2007

### HallsofIvy

Staff Emeritus
What reason do you have to thinks such a potential exists?

If there exist $\phi$ such that $d\phi = udx+ vdy$ (as long as u and v are differentiable) then it must be true that the mixed derivatives are equal:
$$\frac{\partial^2 \phi}{\partial x \partial y}= \frac{\partial u}{\partial y}= \frac{\partial v}{\partial x}= \frac{\partial^2 \phi}{\partial y\partial x}$$
Here it is clear that that is not true: $(4y -x(1+x))_y= 4 \ne (y(2x+1))_x= 2y$. This is not an "exact differential" and there is no "flow potential".