Help needed to find the flow potential function

[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?

 

[HallsofIvy]

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".

 

[tacman]

There are a few possible issues with the approach you have taken. First, it is important to note that the flow potential function, Φ, is a scalar field while the given velocity flow, V, is a vector field. This means that the equations for Φ should be in terms of the scalar components of V, which in this case would be u and v.

In your first approach, you integrated the first equation and then the second equation separately, but this may not result in the correct flow potential function. Instead, you should integrate both equations simultaneously to ensure that the resulting function satisfies both equations.

In your second approach, you integrated with respect to x and then differentiated with respect to y, which is not a valid operation. In order to find Φ, you should integrate both equations with respect to their respective variables, x and y, and then combine the resulting functions to find the overall flow potential function.

Additionally, it is important to check for any potential errors in your calculations or algebraic manipulations as they could also result in an incorrect flow potential function. It may be helpful to double check your work or seek assistance from a tutor or classmate to ensure that your calculations are correct.

Overall, the key is to integrate both equations simultaneously and ensure that the resulting function satisfies both equations. If you are still having trouble, it may be helpful to seek further guidance from a professor or teaching assistant.