AMS 529: Finite Element Methods: Fundamentals, Applications, and New Trends Lecture 25: Introduction to Discontinuous Galerkin Methods Xiangmin Jiao SUNY Stony Brook Xiangmin Jiao Finite Element Methods 1 / 15
Outline 1 Overview of Discontinuous Galerkin 2 Discontinuous Galerkin for Hyperbolic Equations 3 Discontinuous Galerkin for Elliptic Equations Xiangmin Jiao Finite Element Methods 2 / 15
Discontinuous Basis Functions Discontinuous Galerkin uses discontinuous basis functions Continuous vs. discontinuous basis functions (Source: FEniCS Course). Xiangmin Jiao Finite Element Methods 3 / 15
Why Discontinuous Galerkin? High-order accuracy (higher than second order), which is desirable if 1 High precision is required Long-time integration High-dimensional problems are considered Memory restrictions become bottleneck Without continuity constraints, DG is easier with h- or p-adaptivity 2 Compact stencil (compared to finite volume methods) Better at solving hyperbolic problems (compared to finite elements) Multiphysics problems 1 1 J. S. Hesthaven, lecture notes on Discontinuous Galerkin Method. 2 FEniCS course on DG for elliptic problems Xiangmin Jiao Finite Element Methods 4 / 15
Outline 1 Overview of Discontinuous Galerkin 2 Discontinuous Galerkin for Hyperbolic Equations 3 Discontinuous Galerkin for Elliptic Equations Xiangmin Jiao Finite Element Methods 5 / 15
DG Scheme in 1-D Consider scalar problem u t + f (u) x = g(x), x [L, R] = Ω Form local residual x D k : R h (x, t) = uk h t + f h k x g(x) and require this to vanish locally in Galerkin sense D k R h (x, t)l k j (x) dx = 0 where l k j is restriction of test function ψ j over D k Xiangmin Jiao Finite Element Methods 6 / 15
DG Scheme in 1-D u h and ψ j is discontinuous across element boundaries Apply Gauss s theorem D k uh k t lk j fh k l k [ ] j x x glk j dx = fh k k+1 lk j x k Xiangmin Jiao Finite Element Methods 7 / 15
DG Scheme in 1-D Lack of uniqueness at interface is addressed by numerical flux f = f (u h, u+ h ) D k uh k t lk j fh k l k [ ] j x x glk j dx = f l k k+1 j x k Its corresponding variational form is then [ ] x R h (x, t)l k j (x) dx = (fh k f )l k k+1 j D k x k Choice of flux is important for accuracy and stability of DG! Flux scheme is adopted from finite-volume methods, such as Lax-Friedrich flux Riemann solvers Godunov fluxes, etc. Xiangmin Jiao Finite Element Methods 8 / 15
A Brief History of DG DG-FEM was first proposed by Reed/Hill in 1973 for advection-reaction equations First analysis was done by Lesaint and Raviart in 1974 to be O(h p ) in general and optimally O(h p+1 ) for special meshes with degree-p basis In 1986, Johnson showed sharp bound of O(h p+1/2 ) Extended to systems of conservation laws in 1990s by Cockburn and Shu It became popular in early 2000s Xiangmin Jiao Finite Element Methods 9 / 15
Outline 1 Overview of Discontinuous Galerkin 2 Discontinuous Galerkin for Hyperbolic Equations 3 Discontinuous Galerkin for Elliptic Equations Xiangmin Jiao Finite Element Methods 10 / 15
Derivation of a DG Formulation for Poisson Equation Consider Poisson s equation with homogeneous Dirichlet BCs u = f u = 0 in Ω on Ω Its variational form is Ω u v dx = Ω fv dx Assume we have mesh T on Ω with elements {T } Split left integral into sum over elements: u v dx = u v dx Ω T T T FEM applies integration by parts over Ω. DG applies it to elements: u v dx = u v dx nu v ds T T T Xiangmin Jiao Finite Element Methods 11 / 15
Derivation of a DG Cont d Then, T T T u v dx T T T nu v ds = Ω fv dx Each interior facet is shared by two elements T + and T Denote all interior facets by F i and all exterior faces by F e Redistribute integrals over element boundaries to integrals over F: nuv ds = ( n +u+ v + + n u v )ds+ nuv ds F i T T T F e Xiangmin Jiao Finite Element Methods 12 / 15
Derivation of a DG Formulation Cont d Let n + = n and n = n (i.e., geometry is continuous). Then, nuvds = ( u + v + u v ) nds + nuv ds F i T T T = u v ds + F i F e nu v ds F e where b denotes jump of vector field b, b (b + b ) n Jump identity: bv = b v + b v, where v = 1 2 (v + + v ), b = 1 2 (b+ + b ), and v = (v + v )n Applying jump identity to first term above, we have nuvds = u v + u v ds+ F i T T T F e nu v ds Xiangmin Jiao Finite Element Methods 13 / 15
Derivation of a DG Formulation Cont d Recall that T T T u v dx T T T nu v ds = Ω fv dx Weakly enforce following: Continuity of flux: u = 0 over all interior faces Continuity of solution: u = 0 over all interior faces Needs to stabilize by adding additional term Symmetric Interior Penalty (SIP/DG) formulation for Poisson equation: Find u V h = DG k (T ), such that u v dx T T T + F i + F e u v u v + S i (u, v) ds nu v + S e (u, v) ds = Ω fv dx where S i (u, v) = α h u v and S e(u, v) = γ h uv, with α, γ > 0, are introduced to stabilize solution Xiangmin Jiao Finite Element Methods 14 / 15
Example Problem Solves Poisson equation on unit square with source u(x, y) = f (x, y) f (x, y) = 100e 50((x 0.5)2 +(y 0.5) 2 ) and boundary conditions given by where u(x, y) = u 0 on x = 0 and x = 1 n u(x, y) = g on y = 0 and y = 1 using SIP/DG formulation Jupyter Notebook u 0 = x + 0.25 sin(2πx) g = (y 0.5) 2 Xiangmin Jiao Finite Element Methods 15 / 15