I have a velocity field, say
$$
\mathbb{v}=\left\langle\frac{-2xy}{(x^2+y^2)^2},\frac{x^2-y^2}{(x^2+y^2)^2}\right\rangle
$$
First, I have to find the complex velocity potential, then I have to calculate the flux across a straight line from $2i$ to $1$ in the direction right to left.
So for the complex velocity potential, since this velocity field is both irrotational and incompressible $\phi$ and $\psi$ exists. thus the complex velocity is
$$
\Psi(x,y)=\phi(x,y)+i\psi(x,y)
$$
where the complex velocity potential here is
$$
\Psi(x,y)=\frac{x-y}{x^2+y^2}-i\frac{x^2-y^2}{x^2+y^2}.
$$
I am not sure that this is correct.
Lastly, the flux of $$\mathbb{v}=\int_\gamma\big(V\cdot n)d\vec{s}=\int_\gamma Pdy-Qdx$$
but this does not seem right, as I have a variable of two unknowns but only a 1D argument/input? Maybe if I use greens theorem?
""""Edit"""""
So if I do this into a surface integral using greens theorem
$$\int Pdy-Qdx = \iint \Big(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\Big)dxdy$$
But how would I know which limits I can use for which variables and how would one know the limit for the other variable, lets say that I use $2i$ to $1$ for $x$ how would I know what to use for $y$?
Best Answer
Denote the velocity components as
$$u = \frac{-2xy}{(x^2+y^2)^2}, \,\,\, v = \frac{x^2-y^2}{(x^2+y^2)^2}$$
The complex potential is an analytic function $\Psi(z) = \phi(x,y) + i \psi(x,y)$ of the complex variable $z = x+iy$ where the real part $\phi$ is the velocity potential and the imaginary part $\psi$ is the streamfunction.
Applying the Cauchy-Riemann equations we obtain
$$u = \frac{\partial \phi}{\partial x} = \frac{\partial \psi}{\partial y},\,\,\, v= \frac{\partial \phi}{\partial y} = -\frac{\partial \psi}{\partial x} $$
We can solve for $\phi$ by integrating first $u$ with respect to $x$,
$$\phi = \int \frac{-2xy}{(x^2+y^2)^2} \, dx = \frac{y}{x^2+y^2} + C(y)$$
Taking the derivative with respect to y we obtain
$$v = \frac{x^2 - y^2}{(x^2 + y^2)^2} = \frac{\partial \phi}{\partial y} = \frac{\partial}{\partial y}\frac{y}{x^2+y^2} + C'(y) = \frac{x^2 - y^2}{(x^2 + y^2)^2} + C'(y)$$
This implies that $C'(y) = 0$ and $C(y)$ is a constant which can be set arbitrarily to zero.
Thus, $\phi(x,y) = y/(x^2+y^2)$ and in a similar way we can find $\psi(x,y) = x/(x^2+y^2)$. The complex potential is then given by
$$\Psi(z) = \frac{y}{x^2+y^2} + i \frac{x}{x^2+ y^2}$$
We can compute the flux as a contour integral of the complex velocity $u -iv$, where the contour is the line segment with endpoints $1$ and $2i$. Equivalently, we can use a line integral in the real plane over the segment $C= \{(\mu(t),\lambda(t)) = (1-t, 2t): \, 0 \leqslant t \leqslant 1\}.$
Using the latter approach, we have
$$\text{flux} = \int_C \mathbf{v} \cdot \mathbf{n}\, ds,$$
where
$$\mathbf{n} = \left\langle\frac{-\lambda'(t)}{\sqrt{\mu'(t)^2 + \lambda'(t)^2}}, \frac{\mu'(t)}{\sqrt{\mu'(t)^2 + \lambda'(t)^2}} \right\rangle = \langle -2/\sqrt{3}, -1/\sqrt{3}\rangle,$$
$\mathbf{v} = \langle u(\mu(t), \lambda(t)), v(\mu(t), \lambda(t)) \rangle$ and $ds = \sqrt{\mu'(t)^2 + \lambda'(t)^2} dt = \sqrt{3} dt$.
You should be able to finish from here.