1D time dependent PDE - using orthogonal collocation

  • #1
mayzal
2
3
TL;DR Summary
solving PDE numerically using orthogonal collocation on finite elements method
I’m trying to learn how to apply orthogonal collocation on finite elements method (OCFEM) for PDEs and I’m having a trouble with the number of unknown and equations. Suppose I want to solve a PDE numerically using 2nd order Legendre polynomial in three elements (2 interior collocation points per element). Therefore we use:

##u^1(x,t)=\sum_{i=0}^2 A_i^1(t)P_i^1(x) = A_0^1(t)+A_1^1(t)P_1^1(x)+A_2^1(t)P_2^1(x)##
Where ##P_i^1(x)## is shifted Legendre polynomial of order i to the range of element 1 ##[0 , 1/3]##

##u^2(x,t)=\sum_{i=0}^2 A_i^2(t)P_i^2(x) = A_0^2(t)+A_1^2(t)P_1^2(x)+A_2^2(t)P_2^2(x)##
Where ##P_i^2(x)## is shifted Legendre polynomial of order i to the range of element 2 ##[1/3 , 2/3]##

##u^3(x,t)=\sum_{i=0}^2 A_i^3(t)P_i^3(x) = A_0^3(t)+A_1^3(t)P_1^3(x)+A_2^3(t)P_2^3(x)##
Where ##P_i^2(x)## is shifted Legendre polynomial of order i to the range of element 3 ##[2/3 , 1]##

For example ##u^1(x,t)## will be equal to ##A_0^1(t)+A_1^1(t)(6x-1)+A_2^1(t)(-\frac{1}{2}+\frac{3(6x-1)^2}{2})##

I will be substituting those formulation in the PDE at the interior collocation points but I will be getting more equations than the number of unknowns. The equations:

  • 6 equations at interior points (2 in each element)
  • 2 continuity equations (between element 1 and 2, in addition to element 2 and 3)
  • 2 boundary conditions
  • Total: 10 equations

Unknowns:
  • ##A_0^1(t), A_1^1(t), A_2^1(t)##
  • ##A_0^2(t), A_1^2(t), A_2^2(t)##
  • ##A_0^3(t), A_1^3(t), A_2^3(t)##
  • Total: 9 unknowns

In some references, in addition to the continuity equations they are equalizing the derivatives as well which is going to produce even more equations.

Can somebody point out what I'm doing wrong.

Attached is a file describing my attempt at learning this method on one PDE.
 

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  • #2
Having two internal collocation points in each element gives four collocation points in each element:

Element 1: 0, 1/9, 2/9, 1/3
Element 2: 1/3, 4/9, 5/9, 2/3
Element 3: 2/3, 7/9, 8/9, 1

That requires a 3rd order polynomial in each element.

Thus you start with 12 unknowns. You have 10 distinct collocation points, but at each interior element boundary point you get two equations: one from collocating the PDE so that [itex]u[/itex] is known there, and one from requiring that the expansions on the adjoining elements both take this value.

Thus you could update [itex]A_1^1[/itex], [itex]A_2^1[/itex] and [itex]A_3^1[/itex] using the PDE and [itex]A_0^1[/itex] by enforcing the boundary condition at the left. In element 3, you could update [itex]
A_1^3[/itex], [itex]A_2^3[/itex] and [itex]A_3^3[/itex] using the PDE and [itex]A_0^3[/itex] by enforcing the boundary condition at the right. Then in element 2 you can update [itex]A_2^2[/itex] and [itex]A_3^2[/itex] using the PDE and [itex]A_0^2[/itex] and [itex]A_1^2[/itex] from the matching conditions with elements 1 and 3. This gives eight unknowns obtained by enforcing the PDE and four obtained by enforcing boundary or continuity conditions,
 
  • #3
Thank you very much for your reply. I think using your approach I will be able to solve it. But I have one concern:

For example if I use up to third degree legendre polynomial in the first element in the period ##[0,1/3]## I get :

##u^1(x,t)=A_0^1(t)+A_1^1(t)(6x-1)+A_2^1(t)(-\frac{1}{2}+\frac{3(6x-1)^2}{2})+A_3^1(t)(\frac{5(6x-1)^3}{2}-9x+\frac{3}{2})##

which will provide 4 unknowns ##A_0^1(t), A_0^2(t), A_0^3(t), A_0^4(t)##. My concern is that since I will be using a third order polynomial, I will get three roots in this interval. For third degree shifted legendre polynomial to the period ##[0,1/3]## I will get the roots ##[0.1666666667, 0.0375672218, 0.2957661116]##. So does the process involve omitting one of roots and assign the rest as interior collocation points?

I'm sorry if I am not understanding this correctly.
 
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