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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?

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    $\begingroup$ 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. $\endgroup$ – nicoguaro Mar 7 '17 at 13:29
  • $\begingroup$ Why do you exclude the possibility of a sharp cusp in your variables at the symmetry BC? $\endgroup$ – DanielRch Mar 7 '17 at 18:33
  • $\begingroup$ @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. $\endgroup$ – nicoguaro Mar 7 '17 at 19:47
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    $\begingroup$ @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. $\endgroup$ – Stefano M Mar 7 '17 at 22:11
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    $\begingroup$ @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. $\endgroup$ – Stefano M Mar 8 '17 at 18:39
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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.

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