Does this velocity field have a potential?

Question

Let $\Psi=x-\frac{x^3y}{2}$ be the stream function. Then, by definition: $$ v_x=-\frac{\partial \Psi}{\partial y}, \, v_y=-\frac{\partial \Psi}{\partial x} $$ To determine whether this field have a velocity potential: $$ -\frac{\partial \Phi}{\partial x}=v_x, \, -\frac{\partial \Phi}{\partial y}=v_y $$ Solution: \begin{align*} &\Phi (x,y)= -\frac{x^4}{8}+f(y) \\ &\Phi (x,y)=\frac{3x^2y^2}{4}-y+g(x) \end{align*} Does the fact that $x,y$ are not separated ensures that $\Phi(x,y)$ does not exist and the fluid is viscid?

Answer 1

Here is what I think is a somewhat easier and more clear way to see that there is no velocity potential:

Given

$\Psi = x - \dfrac{x^3 y}{2}, \tag 1$

with

$v_x = - \dfrac{\partial \Psi}{\partial y} = \dfrac{x^3}{2} \tag 2$

and

$v_y = -\dfrac{\partial \Psi}{\partial x} = \dfrac{3x^2y}{2} - 1, \tag 3$

setting

$\vec v = (v_x, v_y) = \left (\dfrac{x^3}{2}, \dfrac{3x^2y}{2} - 1 \right ), \tag 4$

we know that $\vec v$ has a velocity potential, that is, $\vec v$ is a gradient vector field, if and only if

$\nabla \times \vec v = \dfrac{\partial v_y}{\partial x} - \dfrac{\partial v_x}{\partial y} = 0; \tag 5$

however, we have

$\nabla \times \vec v = \dfrac{\partial v_y}{\partial x} - \dfrac{\partial v_x}{\partial y} = 3xy \ne 0; \tag 6$

therefore, $v$ is not a gradient field; there is no velocity potential for $\vec v$.

Having written this, I can see no serious flaws in our OP Jevaut's approach which is based upon solving for $\Phi(x, y)$ in two different ways, one for each component of $\vec v$; the two solutions so obtained are clearly not consistent, hence the conclusion that $\Phi(x, y)$ does not exist appears valid on the basis of Jevaut's demonstration.

In fact, Jevaut's approach is really a special case of the general result I invoked. See my linked citing for more details.

Such a fluid is thus viscid.

Answer 2

In a slight generalisation of Robert Lewis's answer, we can recall when fluids have stream functions and velocity potentials.

Since an incompressible fluid satisfies $\nabla.\vec{v}=0$, there exists a stream function $\Psi$ such that $v_x=\frac{\partial \Psi}{\partial y}$ and $v_y=-\frac{\partial \Psi}{\partial x}$ - note the slight difference to Jevaut's question. In this way, $$\nabla.\vec{u}=\frac{\partial v_x}{\partial x} + \frac{\partial v_y}{\partial y}=\frac{\partial^2 \Psi}{\partial x \partial y} - \frac{\partial^2 \Psi}{\partial y \partial x} = 0.$$

Now we can ask about the vorticity, $\vec{\omega} = \nabla \times \vec{v}$, of this 2D inviscid fluid. It is $$\vec{\omega} = \nabla^2\Psi \vec{k}. $$ Velocity potentials only exist when the flow is irrotational (has zero vorticity) everywhere. The requirement on the stream function is thus that $\nabla^2 \Psi\equiv 0$, which is not true here and so no stream function exists.

This is not the same as the fluid being viscous, since inviscid flows can have non-zero vorticity.