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Periodicity and symmetry boundaries conditions 

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June 19, 2014, 12:17 
Periodicity and symmetry boundaries conditions

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Luis Delgado
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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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Follow this thread
http://www.cfdonline.com/Forums/mai...onditions.html Look at Dr Denaro's explanation (reply #10)
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Best wishes, Somdeb Bandopadhyay 

June 19, 2014, 15:15 

#3 
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cfdnewbie
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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 
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Yon Han Chong
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Periodic will have flow coming out from one side and go into the other match side.
Symmetry will mirror what's at that boundary. 

June 19, 2014, 20:37 

#5 
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Luis Delgado
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Thanks to everyone, I think I understand


June 20, 2014, 08:10 

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Adrien
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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(Lh) v(L+h)=v(Lh) 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 

July 20, 2014, 14:51 

#7 
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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(NI1) 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 
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Jonas T. Holdeman, Jr.
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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(NI1) .] 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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Quote:
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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[QUOTE=Jonas Holdeman;502375]leflix says:
Quote:
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(NI1) 

August 7, 2014, 22:39 

#11  
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Jonas T. Holdeman, Jr.
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[QUOTE=leflix;504933]
Quote:
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 
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Filippo Maria Denaro
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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)=((i1)*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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Jonas T. Holdeman, Jr.
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Quote:
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 degreesoffreedom 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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Filippo Maria Denaro
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Quote:
I think that it's the same procedure, indees i worked on unstructured triangular mesh similarily. As example, assume a 1D nonuniform 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. 

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boundaries condition, periodicity, symmetry 
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