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Velocity Averaging and Mean Value Property

An explanation on why the naive averaging of velocity works for point vortex simulation on a discrete triangle mesh.

Point Vortex Dynamics

For our point vortex case, we have:

Δψ=ω    ψ=Δ1ω-\Delta \psi = \omega \implies \psi = - \Delta ^{-1}\omega

Where ω=k=1NΓkδk\omega = \sum_{k=1}^{N} \Gamma_{k}\delta_{k} where δk\delta_{k} is a delta function at the position of the kthk^{th} point vortex. And as Δ\Delta is a linear operator, we have:

ψ1=Δ1δk1,ψ2=Δ1δk2,\psi_{1} = - \Delta ^{-1}\delta_{k_{1}}, \quad \psi_{2} = - \Delta ^{-1}\delta_{k_{2}}, \quad\dots ψ=k=1Nψk\therefore \psi = \sum_{k=1}^{N} \psi_{k}

That means on a small circular patch around point vortex k1k_{1} we can write the global stream function as:

ψ=ψ1+k=2Nψk\psi = \psi_{1} + \sum_{k=2}^{N}\psi_{k}

ψ1\psi_{1} is singular and blows up at the center of point vortex. But h=k=2Nψkh = \sum_{k=2}^{N}\psi_{k} is a smooth harmonic scalar function on this small patch.

ψ=ψ1+h    h=ψψs(1)\therefore \psi = \psi_{1} + h \implies \boxed{h = \psi - \psi_{s}} \tag{1}

where ψs\psi_{s} is the singular stream function part of a point vortex inside the patch BrB_{r} centered at aa (position of the point vortex).

In the classical term, Eq. (2)(2) can be interpreted as hh is the actual stream function value obtained as a combination of Robin’s function and pairwise interaction of Green’s function (excluding the self interaction).

So for correct Advection, we want to get rid of the ψs\psi_{s} part and then have the velocity at the point vortex (point aa) as:

u(a)=Jh(a)\vec{u}(a) = J \nabla h(a)

For this we need the h(a)\nabla h(a).

Mean Value Property of the Harmonic Function

For a harmonic, function hh the value at a point x0x_{0} is given as:

h(x0)=1BrBr(x0)h(x)dxh(x_{0}) = \frac{1}{\lvert B_{r} \rvert }\int_{B_{r(x_{0})}}h(x)\,dx

where BrB_{r} is a ball of radius rr around the point x0x_{0}. Similarly this holds for the boundary of the ball as well.

h(x0)=1BrBr(x0)hdSh(x_{0}) = \frac{1}{\lvert \partial B_{r} \rvert}\int_{\partial B_{r(x_{0})}}h\,dS

In this case we are integrating over a sphere or a circle.

For 2D case, integral over the circle:

h(x0)=12π02πh(x0+r(cosθ,sinθ))dθh(x_{0}) = \frac{1}{2\pi}\int_{0}^{2\pi}h(x_{0} + r(\cos\theta, \sin\theta))\,d\theta

Consequently, we can take gradient of hh in terms of xx as:

h(x0)=12π02πh(x0+r(cosθ,sinθ))dθ\nabla h(x_{0}) = \frac{1}{2\pi}\int_{0}^{2\pi} \nabla h(x_{0} + r(\cos\theta, \sin\theta))\,d\theta

Equivalently,

h(a)=1BrBrh(x)dx(2)\boxed{\nabla h(a) = \frac{1}{\lvert \partial B_{r} \rvert }\oint_{\partial B_{r}}\nabla h(x)\,dx} \tag{2}

where BrB_{r} is a small circular patch around point aa with radius rr.

Velocity Averaging

Now, from Eq. (1)(1) and (2)(2):

h(a)=1BrBrh(x)dx\nabla h(a) = \frac{1}{\lvert \partial B_{r} \rvert }\oint_{\partial B_{r}}\nabla h(x)dx =1BrBr(ψψs)dx= \frac{1}{\lvert \partial B_{r} \rvert }\oint_{\partial B_{r}}\nabla(\psi - \psi_{s})\,dx h(a)=1BrBrψdx1BrBrψsdx(3)\nabla h(a)= \frac{1}{\lvert \partial B_{r} \rvert }\oint_{\partial B_{r}}\nabla\psi\,dx - \frac{1}{\lvert \partial B_{r} \rvert }\oint_{\partial B_{r}}\nabla\psi_{s}\,dx \tag{3}

One important thing about ψs\psi_{s} we know is that it’s radially symmetrical around a point aa. For a small enough patch, in continuum case it’s equal to ψs=Γ2πlogxa\psi_{s} = -\frac{\Gamma}{2\pi}\log \lvert x - a \rvert.

Given that BrB_{r} is a circular patch with center aa, ψs\psi_{s} is constant over Br\partial B_{r} i.e. it’s a contour line, and ψs(r)\psi_{s}(r) is constant for any value of rRr\leq R where RR is the radius of patch BrB_{r}. Therefore, magnitude of ψs\nabla \psi_{s} is constant over Br\partial B_{r} and direction is pointing radially outwards in the direction of normal n\vec{n}. Therefore,

Brψsdx=BrCndx=CBrndx=0\oint_{\partial B_{r}}\nabla \psi_{s}\,dx = \oint_{\partial B_{r}} C\vec{n}\,dx = C \oint_{\partial B_{r}}\vec{n}\,dx = 0 Brψsdx=0(4)\therefore \boxed{\oint_{\partial B_{r}} \nabla \psi_{s}\,dx = 0} \tag{4}

Now, Eq. (3)(3) becomes:

h(a)=1BrBrψdx\nabla h(a) = \frac{1}{\lvert \partial B_{r} \rvert }\oint_{\partial B_{r}}\nabla \psi\,dx u(a)=Jh(a)=1BrBrJψdx(5)\boxed{\vec{u}(a) = J\nabla h(a) = \frac{1}{\lvert \partial B_{r} \rvert }\oint_{\partial B_{r}} J\nabla \psi\,dx} \tag{5}

Written by Rudresh Veerkhare.