# Periodicity and symmetry boundaries conditions

 Register Blogs Members List Search Today's Posts Mark Forums Read June 19, 2014, 12:17 Periodicity and symmetry boundaries conditions #1 New Member   Luis Delgado Join Date: Oct 2013 Posts: 19 Rep Power: 9 Hi, I have a simple question: What is the difference between periodicity boundary condition and symmetry boundary condition. Thanks. Luis   June 19, 2014, 13:28 #2
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Look at Dr Denaro's explanation (reply #10)

Quote:
 Originally Posted by FMDenaro periodicity: given a period L, the solution implies u(L +/- h) = u(+/- h) simmetry: given a plane symmetry z=L the solution implies u(L+h)=u(L-h)
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Best wishes,   June 19, 2014, 15:15 #3 Senior Member   cfdnewbie Join Date: Mar 2010 Posts: 557 Rep Power: 16 adding to this: symmetry is a local property, periodicity refers to the continuation from the "other end" of the domain. Just looking at a simple sine wave should answer your question. regards, cfdnwebie   June 19, 2014, 15:23 #4 Member   Yon Han Chong Join Date: Jun 2012 Posts: 77 Rep Power: 10 Periodic will have flow coming out from one side and go into the other match side. Symmetry will mirror what's at that boundary. leflix likes this.   June 19, 2014, 20:37 #5 New Member   Luis Delgado Join Date: Oct 2013 Posts: 19 Rep Power: 9 Thanks to everyone, I think I understand    June 20, 2014, 08:10 #6
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Rep Power: 8 I don't completely agree with the answer in the referred post by FMDenaro. As you can see in the enclosed picture, in my understanding a BC that has mirror symmetry with respect to a vertical plane means
u(L+h)=-u(L-h)
v(L+h)=v(L-h)
The advantage of this BC is mainly to cancel the normal velocity component to reproduce the effect of a solid wall (sometimes called "rigid lid BC").

Periodicity is even more straightforward (see picture).

Attached Images BC.jpg (36.5 KB, 29 views)   July 20, 2014, 14:51 #7 Senior Member   Join Date: Aug 2011 Posts: 271 Rep Power: 12 lets say the domain is 1D from indice I=1 to NI ,1 and NI are on the left and right boundaries respcetively and have to be precribed from boundary condition Periodic BC implies FI(NI)=FI(2) and FI(1)=FI(NI-1) for symmetry BC you have to set Neuman BC on the variable FI so dFI/dx=0 for 2D flows you have to add for the velocity: V_norm=0 d(V_para)/dn=0 V_norm is the normal componnent of the velocity fied to the boundary V_para is the parallel componnent of the velocity field to the boundary d/dn is the normal derivative to the boundary.   July 20, 2014, 17:05 #8 Senior Member   Jonas T. Holdeman, Jr. Join Date: Mar 2009 Location: Knoxville, Tennessee Posts: 126 Rep Power: 14 leflix says: [QUOTE=leflix;502365]lets say the domain is 1D from indice I=1 to NI ,1 and NI are on the left and right boundaries respcetively and have to be precribed from boundary condition Periodic BC implies FI(NI)=FI(2) and FI(1)=FI(NI-1) .] This assumes that the left and right sides overlap by one cell. By periodic BC we mean that the values on the left and right boundaries are identical, i.e. that FI(NI)=FI(1), and if derivatives are continuous, that d/dx(FI(NI))^- = d/dx(FI(1))^+ , where ^-/+ means taking the limit from the left or right. This is clear in the finite element method, but since ghost cells are sometimes used in finite difference methods, overlap may be necessary. In this case maybe something like FI(NI+1)=FI(2) and FI(NI)=FI(1) may be necessary, where FI(NI+1) is a ghost cell. Last edited by Jonas Holdeman; July 20, 2014 at 17:10. Reason: remove redundancy   July 21, 2014, 05:15 #9
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Filippo Maria Denaro
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 Originally Posted by lex78700 I don't completely agree with the answer in the referred post by FMDenaro. As you can see in the enclosed picture, in my understanding a BC that has mirror symmetry with respect to a vertical plane means u(L+h)=-u(L-h) v(L+h)=v(L-h) The advantage of this BC is mainly to cancel the normal velocity component to reproduce the effect of a solid wall (sometimes called "rigid lid BC"). Periodicity is even more straightforward (see picture). Adrien

yes, you are right for simmetry condition for a vector field, I just wrote condition for a scalar function    August 7, 2014, 18:50 #10
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Rep Power: 12 [QUOTE=Jonas Holdeman;502375]leflix says:
Quote:
 Originally Posted by leflix lets say the domain is 1D from indice I=1 to NI ,1 and NI are on the left and right boundaries respcetively and have to be precribed from boundary condition Periodic BC implies FI(NI)=FI(2) and FI(1)=FI(NI-1) .] This assumes that the left and right sides overlap by one cell. By periodic BC we mean that the values on the left and right boundaries are identical, i.e. that FI(NI)=FI(1),
I disagree with what you said Jonas. Actually keep in mind that points NI and 1 are located on the boundaries (I never spoke of ghost cells it is not necessary).
So FI(NI) cannot be equal to FI(1) since the both are unknown and must be set by the boundary condition. That's why we have to write FI(NI)=FI(2) and FI(1)=FI(NI-1)   August 7, 2014, 22:39 #11
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Quote:
 Originally Posted by Jonas Holdeman leflix says: I disagree with what you said Jonas. Actually keep in mind that points NI and 1 are located on the boundaries (I never spoke of ghost cells it is not necessary). So FI(NI) cannot be equal to FI(1) since the both are unknown and must be set by the boundary condition. That's why we have to write FI(NI)=FI(2) and FI(1)=FI(NI-1)
It seems we have reached an impasse. Maybe we should be sure we are talking about the same thing. Consider this example of a periodic mesh. Imagine a piece of paper wrapped around a cylinder, or a piece of paper with edges glued to make a cylinder. Now draw a continuous rectangular mesh on the paper cylinder, and start numbering the nodes on a circumferential line of nodes, 1, 2, 3, ... . When you get back to node 1, then you give it the alternate number NI. Node 1 and node NI are just different names for the same node, NI is an alias for 1. Now if you compute some (single-valued) field on this mesh, you must have FI(1)=FI(NI) because you have the same one function at the same node. And if the field has continuous first derivatives, you must have grad FI(1)=grad FI(NI) as well. If you now cut the paper along the line of nodes (parallel to the cylinder axis) to lay the paper flat, that must not change anything. You still must have FI(1)=FI(NI). And, of course, FI is unknown at all of the internal nodes.

There is no special boundary condition, only a rule to preserve the original topology. You could have cut the paper along any of the lines of nodes in order to flatten the paper, and you must get the same result (but shifted on the flattened grid, depending on where you make the cut).

Now if you are using the finite difference method, your stencil must wrap around or involve "ghost cells" or such, but that is not my area of interest.   August 8, 2014, 05:37 #12 Senior Member   Filippo Maria Denaro Join Date: Jul 2010 Posts: 5,422 Rep Power: 58   well, I think that there is no real problem...periodicity means you have to fulfill a condition f(x0) = f(x0+L), where L is some periodicity lenght. In terms of index topology, for a mesh size h=L/N x(i)=((i-1)*h+x0 for i=1,..., N+1 therefore f(1)=f(N+1). This function values both appears as to be prescribed as a sort BC.s, but you just chose to write the equation in 1 or N+1 and solve. The resulting value is coupled by periodicity to the other one. For example, solve the equation in node i=1 and the value is set to f(N+1).   August 8, 2014, 10:43 #13
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 Originally Posted by FMDenaro well, I think that there is no real problem...periodicity means you have to fulfill a condition f(x0) = f(x0+L), where L is some periodicity lenght. In terms of index topology, for a mesh size h=L/N x(i)=((i-1)*h+x0 for i=1,..., N+1 therefore f(1)=f(N+1). This function values both appears as to be prescribed as a sort BC.s, but you just chose to write the equation in 1 or N+1 and solve. The resulting value is coupled by periodicity to the other one. For example, solve the equation in node i=1 and the value is set to f(N+1).
I think I understand your argument in the context of equally-spaced nodes in a finite difference context (you use a constant spacing h in your example). But I am thinking more generally in terms of an unequally-spaced FD or a finite element context. In this context, a mesh from 1, ... N does not provide enough information to solve the problem. What is missing for FD is the spacing (measured in a positive direction) between your node N and node 1, which would have to be a side condition, and which I suppose you could call a boundary condition.

In the finite element context we focus on elements in the space between the nodes where we define the field in terms of shape functions (or elements), and the nodes are simply a convenient place to store the data as degrees-of-freedom for the shape functions (well yes their coordinates do fix the shape as well). An element contributes to the nodal data on either side in 1D, and to all connected nodes in 2D. If the mesh runs from 1 to N, and N is not aliased to node 1, then we have left out the contributions from the element connecting these nodes in 1D, or whole rows or columns of elements in 2D. This problem is avoided by aliasing node 1 to node N, and F(1) to F(N) as I have suggested.   August 8, 2014, 11:54 #14
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 Originally Posted by Jonas Holdeman I think I understand your argument in the context of equally-spaced nodes in a finite difference context (you use a constant spacing h in your example). But I am thinking more generally in terms of an unequally-spaced FD or a finite element context. In this context, a mesh from 1, ... N does not provide enough information to solve the problem. What is missing for FD is the spacing (measured in a positive direction) between your node N and node 1, which would have to be a side condition, and which I suppose you could call a boundary condition. In the finite element context we focus on elements in the space between the nodes where we define the field in terms of shape functions (or elements), and the nodes are simply a convenient place to store the data as degrees-of-freedom for the shape functions (well yes their coordinates do fix the shape as well). An element contributes to the nodal data on either side in 1D, and to all connected nodes in 2D. If the mesh runs from 1 to N, and N is not aliased to node 1, then we have left out the contributions from the element connecting these nodes in 1D, or whole rows or columns of elements in 2D. This problem is avoided by aliasing node 1 to node N, and F(1) to F(N) as I have suggested.

I think that it's the same procedure, indees i worked on unstructured triangular mesh similarily.
As example, assume a 1D non-uniform mesh size going from x0 to x0+L, therefore x(2)-x(1)=h(1), x(3)-x(2)=h(2), .... x(N+1)-x(N)=h(N).

Now, you have hypotetically ghost nodes, x(0)=x(1)-h(N)=x(N)-L and x(N+2)=X(N+1)+h(1)=x(2)+L. This way, you can simply defines the shape functions respecting periodicity constraint.  Tags boundaries condition, periodicity, symmetry Thread Tools Search this Thread Show Printable Version Email this Page Search this Thread: Advanced Search Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is OffTrackbacks are Off Pingbacks are On Refbacks are On Forum Rules Similar Threads Thread Thread Starter Forum Replies Last Post Luis Delgado FLUENT 1 June 20, 2014 04:17 keeper CFX 5 April 3, 2014 05:19 lost.identity Main CFD Forum 9 August 30, 2012 12:16 OneDotThree ANSYS Meshing & Geometry 10 March 7, 2012 05:28 gman89 CFX 3 May 18, 2011 19:45

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