How to check if a function is a stream function? I am completely stuck on the following question:
Given the function $\psi=\frac{1}{2}({x^2-y^2})$, check whether it is a stream function for some $F=\langle y, x\rangle$.
Also show that the above $F$ has zero convergence and zero curl.
I know that a stream function is a scalar function of space and time such that its partial derivative with respect to any direction gives the velocity component at right angles.
But I am confused how to use this definition mathematically to compute if a function is a stream function or not. Can someone please help me and point me to the right direction?
 A: A streamfunction is defined with respect to the velocity field of an incompressible flow,  that is, a vector field $(x,y) \mapsto F(x,y)=\langle u(x,y),v(x,y) \rangle$ with Cartesian components u in the x-direction and v in the y-direction such that $$\tag{1}\nabla \cdot F =\frac{\partial u}{\partial x}+ \frac{\partial v}{\partial y}= 0$$
A (smooth) function $(x,y) \mapsto \psi(x,y)$ is a streamfunction for the velocity field $\langle u,v \rangle$  if and only if up to an arbitrary sign
$$\tag{2}u = -\frac{\partial \psi}{\partial y},\quad v = \frac{\partial \psi}{\partial x}$$
A velocity field given by (2) automatically satisfies the incompressibility condition (1) when $\psi$ is continuously differentiable since by Clairaut's theorem
$$\frac{\partial u }{\partial x} = - \frac{\partial^2 \psi}{\partial x \partial y}= -\frac{\partial^2 \psi}{\partial y \partial x} = - \frac{\partial v}{\partial y}$$
In this case we have $\psi(x,y) = \frac{1}{2}(x^2 - y^2)$ and
$$u(x,y) = -\frac{\partial \psi}{\partial y}= y, \quad v(x,y) =  \frac{\partial \psi}{\partial x}=x$$
We can easily confirm that  the divergence and curl of the velocity field $F = \langle y,x\rangle$ both vanish:
$$\nabla \cdot F = \frac{\partial u}{\partial x}+ \frac{\partial v}{\partial y}=\frac{\partial}{\partial x}(y) + \frac{\partial}{\partial y}(x)=0, \\ \nabla \times F = \left(\frac{\partial v}{\partial x} - \frac{\partial u}{\partial y}\right)\mathbf{e_z}= \left(\frac{\partial }{\partial x}(x) - \frac{\partial }{\partial y}(y)\right)\mathbf{e_z}= 0 \cdot \mathbf{e_z}$$
For more about streamfunctions, see this answer.
