1 Spring 2003 Prof. Tim Warburton MA557/MA578/CS557 Lecture 21

2 Recap So far we have derived the advection-(diffusion) equation in a one- dimensional domain. We created a numerical scheme based on piecewise polynomial approximations. Boundary conditions between sub-intervals are imposed by the use of fluxes. We proved stability and consistency for these schemes. But…. we do not live in a one-dimensional world So – on the march to our three-dimensional world we are going to investigate PDE’s with two spatial-dimensions.

3 Two-Dimensional Polynomial Space For the following we will use a two-dimensional space which is spanned by the polynomials of total order not greater than p: Later we will choose an orthonormal basis to discretize this space.

4 Two-Dimensional Advection Recall in one-dimensions we chose an arbitrary section of a pipe. We monitored the flux of fluid into and out of the ends of the pipe. The conservation equation in 1D we derived was: Now we start with a 2D domain and consider any simply connected sub-region:

5 Our Domain

6 Our Sub-region

7 Outward Normal n n

8 Conservation Suppose the fluid has a mean velocity everywhere. This time is a two-vector. The flux of fluid through the boundary is the integral of the concentration C being advected normal to the surface of w: n=outward pointing normal

9 Conservation Law Now we know how much of the concentrate is being advected through the boundary of our sub-region. The conservation law is now: We apply the divergence theorem (beware this relies on smoothness arguments):

10 In Component Form Assuming ubar is constant this becomes:

11 Finalize For all sub-regions: Assuming enough continuity we obtain:

12 Straight To DG The pde is: We substitute the DG scheme (using Lax-Friedrichs fluxes):

13 +/- Notation For the purpose of the numerical scheme notation: –The + notation implies the limit of C from outside the sub-region –The - notation implies the limit of C from inside the sub-region C-C- C+C+

14 Sample Domain Partition 1- Suppose we have a domain: Which we partition into 5 triangles: We note the unique edges, and arbitrarily label their edge limits with +/

15 Stability

16 Suppose We Partition

17 Continuing Where we have gathered like terms (note we ignored the boundary terms)

18 Finally -- Stability Where we have gathered like terms (note we dropped the boundary terms)

19 Now The Details Recall in 1D we were restricted to breaking the interval [a,b] into segments. In 2D there are many more choices. Two obvious shapes for sub-regions are the triangle and quadrilateral. There is plenty of software available for mesh generation using triangles. Building quadrilateral meshes is slightly tougher for a complex domain.

20 Mesh Generation For simplicity we will use the Triangle mesh generator. This has been built into WinUSEMe – review the previous notes. For now assume that the task of creating a mesh has been done – and – the mesh is held in a file. Note that for the DG we do not need to much information about how the element are connected – it is mostly sufficient to know which edge connects to which edge. We also need to know which faces lie on a domain boundary.

21 Reference Triangle The following will be our basic triangles: All straight sided triangles are the image of this triangle under the map: s r (-1,-1) (1,-1) (-1,1)

22 Reference Triangle The following will be our basic triangles: All straight sided triangles are the image of this triangle under the map: s r (-1,-1) (1,-1) (-1,1)

23 Orthonormal Basis for the Triangle Fortunately an orthonormal basis for the triangle has been discovered (and rediscovered multiple times). First we need to know some details about the Jacobi polynomials. These polynomials are parameterized by two reals: and their integer orders n,m such that they satisfy the orthogonality relationship (for integer alpha,beta): Basic identity:

24 Rodrigues’ Formula The Jacobi polynomials can be generated in a similar way to the Legendre polynomials: Note the Legendre polynomials are a special case of the Jacobi polynomials:

25 Recurrence Relation See:

26 Orthonormal Basis for the Triangle The following basis is due to Koornwinder (later revived by Proriol, Dubiner, Owens,….) Part of HW7 is to show that this is an orthogonal basis in the sense that:

27 Homework 7 Q1) Prove that the Koornwinder-Proriol-Dubiner-Owens… basis is orthogonal on the triangle: T={-1<=r,s;r+s<=0}: (HINT: under the mapping (r,s)->(a,b) the image of the triangle T is the square Q={-1<=a,b<=1} and the Jacobian is (1-b)/2 ) Q2) Determine an orthonormal basis for the triangle by normalizing the basis functions of the KPDO.. basis Q3) Construct a p’th order Vandermonde for the equally spaced points on the triangle (take half of the square of (p+1)x(p+1) points) using the orthonormal KPDO basis. In Matlab calculate the condition number of this Vandermonde basis as a function of p.

28 HW7 cont Q3cont) Use the lower left triangle of points:

29 HW7 cont Q3cont) The Vandermonde matrix V will be have (p+1)(p+2)/2 rows and columns. Q4) Extra credit: repeat 3 using half of the tensor product of the one-dimensional Chebychev nodes. For Q3 and (Q4) plot the polynomial order on the horizontal axis and cond(V) on the vertical axis

30 HW7 cont Q4cont) Use the lower left triangle of points:

31 Next Class We will create the derivative and surface matrices used in the DG scheme. We will transform the modal scheme into a nodal scheme. We will use a better set of nodes (improve the condition number of the Vandermonde matrix). Time permitting we will prove consistency. For more info on Jacobi polynomials: