Symmetric boundary condition

Question

I am solving a flow physics problem, in which I encounter a symmetric boundary. So I set the boundary conditions to be $$\frac{\partial u}{\partial r} = \frac{\partial v}{\partial r}=\frac{\partial w}{\partial r}=\frac{\partial T}{\partial r} =\frac{\partial \rho}{\partial r}$$ where $u,v,w$ are the 3 components of velocity and $T, \rho$ are temperature and density of the fluid respectively. $r$ is the direction perpendicular to the plane of symmetry. $v$ is the velocity in the $r$ direction. Since there will no flow across plane of symmetry, we also get another boundary condition $$v=0$$

My question is : Should I take $v=0$ and $\frac{\partial v}{\partial r}=0$ together for $v$ at the boundary or anyone one those will be sufficient?

For symmetric boundary conditions you get the normal derivative equals to zero. The principal variable equaling zero is normally used for anti-symmetric boundary conditions. — nicoguaro, Mar 07 '17 at 13:29
Why do you exclude the possibility of a sharp cusp in your variables at the symmetry BC? — DanielRch, Mar 07 '17 at 18:33
@DanielRch, that is how is normally done. If you want a sharp cusp, I suppose that you can't impose as a BC... or I have not seen it before, at least. — nicoguaro, Mar 07 '17 at 19:47
@nicoguaro: your remark is true for scalars and polar vector components parallel to a symmetry plane (or the tangent plane of a symmetry smooth surface.) Perpendicular components are reflected (change sign), so actually the "symmetry" becomes an "anti-symmetry" and $v=0$ holds. The opposite is true for axial vector components. — Stefano M, Mar 07 '17 at 22:11
@StefanoM, thanks for the clarification. Does this apply for pseudotensors in general? — nicoguaro, Mar 08 '17 at 04:27
@nicoguaro: also for (proper)-tensors some components are preserved, and some components change sign across a symmetry plane (eg. shear rate.) However while the distinction between proper- and pseudo-vectors in $\mathbb{R}^3$ is very important and their chirality properties easy to grasp (we all now that a right-hand becomes a left-one if reflected in a mirror) the distinction between proper and pseudo tensor is much more formal... And sorry I'm not able to give a concise description of symmetry methods in this more general setting. — Stefano M, Mar 08 '17 at 18:39

score 8 · Accepted Answer · answered Mar 07 '17 at 22:03

For the sake of simplicity let me slightly change your notation. Let $u, v, w$ be the components in the $x, y, z$ directions of a true (polar) vector, and $y=0$ the symmetry plane. For a true vector, symmetry conditions across the $xz$ plane are: \begin{eqnarray} u(x,y,z) &= u(x, -y, z) \\ -v(x,y,z) &= v(x, -y, z) \\ w(x,y,z) &= w(x,-y,z) \end{eqnarray} which imply \begin{eqnarray} -\frac{\partial}{\partial y}u(x,y,z) &= \frac{\partial}{\partial y} u(x, -y, z) \\ \frac{\partial}{\partial y}v(x,y,z) &= \frac{\partial}{\partial y}v(x, -y, z) \\ -\frac{\partial}{\partial y}w(x,y,z) &= \frac{\partial}{\partial y}w(x,-y,z) \end{eqnarray} Under a continuity hypothesis up to the first partial derivative, you obtain \begin{equation} v(x,0,z) = \frac{\partial}{\partial y}u(x,0,z) = \frac{\partial}{\partial y}w(x,y,z) = 0 \end{equation}

With reference to your problem you should not impose $\frac{\partial v}{\partial r} = 0$ across the symmetry plane, since velocity is a polar vector.

Symmetric boundary condition

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