Discontinuous Galerkin Methods for Hamilton-Jacobi Equations · DISCONTINUOUS GALERKIN METHODS FOR HAMILTON-JACOBI EQUATIONS Introduction of discontinuous Galerkin method How does

Discontinuous Galerkin Methods forHamilton-Jacobi Equations

Chi-Wang Shu

Division of Applied Mathematics

Brown University

DISCONTINUOUS GALERKIN METHODS FOR HAMILTON-JACOBI EQUATIONS

Outline

• Introduction of discontinuous Galerkin method for conservation laws

• Discontinuous Galerkin method for Hamilton-Jacobi equations of Huand Shu, reinterpreted by Li and Shu

• Discontinuous Galerkin method for Hamilton-Jacobi equations ofCheng and Shu

• Fast sweeping method for steady state solutions

Division of Applied Mathematics, Brown University


Introduction of discontinuous Galerkin method

How does the method work — an example

To solve a hyperbolic conservation law:

ut + f(u)x = 0 (1)

Multiplying with a test function v, integrate over a cell Ij = [xj− 12

, xj+ 12

],

and integrate by parts:∫

Ij

utvdx−∫

Ij

f(u)vxdx+ f(uj+ 12

)vj+ 12

− f(uj− 12

)vj− 12

= 0



Now assume both the solution u and the test function v come from a finite

dimensional approximation space Vh, which is usually taken as the space

of piecewise polynomials of degree up to k:

Vh ={

v : v|Ij ∈ P k(Ij), j = 1, · · · , N}

However, the boundary terms f(uj+ 12

), vj+ 12

etc. are not well defined

when u and v are in this space, as they are discontinuous at the cell

interfaces.



From the conservation and stability (upwinding) considerations, we take

• A single valued monotone numerical flux to replace f(uj+ 12

):

f̂j+ 12

= f̂(u−j+ 1

2

, u+j+ 1

2

)

where f̂(u, u) = f(u) (consistency); f̂(↑, ↓) (monotonicity) and f̂ isLipschitz continuous with respect to both arguments.

• Values from inside Ij for the test function v

v−j+ 1

2

, v+j− 1

2



Hence the DG scheme is: find u ∈ Vh such that∫

Ij

utvdx−∫

Ij

f(u)vxdx+ f̂j+ 12

v−j+ 1

2

− f̂j− 12

v+j− 1

2

= 0 (2)

for all v ∈ Vh.



Time discretization could be by the TVD Runge-Kutta method (Shu and

Osher, JCP 1988). For the semi-discrete scheme:

du

dt= L(u)

where L(u) is a discretization of the spatial operator, the third order TVD

Runge-Kutta is simply:

u(1) = un + ∆tL(un)

u(2) =3

4un +

1

4u(1) +

1

4∆tL(u(1))

un+1 =1

3un +

2

3u(2) +

2

3∆tL(u(2))



Advantages of the DG method:

• Easy handling of complicated geometry and boundary conditions(common to all finite element methods). Allowing hanging nodes in the

mesh;

• Compact. Communication only with immediate neighbors, regardlessof the order of the scheme;

• Explicit. Because of the discontinuous basis, the mass matrix is localto the cell, resulting in explicit time stepping (no systems to solve);

• Parallel efficiency. Achieves 99% parallel efficiency for static mesh andover 80% parallel efficiency for dynamic load balancing with adaptive

meshes (Flaherty et al.);



• Provable cell entropy inequality and L2 stability, for arbitrary scalarequations in any spatial dimension and any triangulation, for any order

of accuracy, without limiters;

• At least (k + 12)-th order accurate, and often (k + 1)-th order accurate

for smooth solutions when piecewise polynomials of degree k are

used, regardless of the structure of the meshes.

• Easy h-p adaptivity.

• Stable and convergent DG methods are now available for manynonlinear PDEs containing higher derivatives: convection diffusion

equations, KdV equations, ...



Three examples

We show three examples to demonstrate the excellent performance of the

DG method.

The first example is the linear convection equation

ut + ux = 0, or ut + ux + uy = 0,

on the domain (0, 2π) × (0, T ) or (0, 2π)2 × (0, T ) with thecharacteristic function of the interval (π

2, 3π

2) or the square (π

2, 3π

2)2 as

initial condition and periodic boundary conditions.



0 1 2 3 4 5 6x

-0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

1.1u

k=1, t=100π, solid line: exact solution;dashed line / squares: numerical solution

0 1 2 3 4 5 6x

-0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

1.1u

k=6, t=100π, solid line: exact solution;dashed line / squares: numerical solution

Figure 1: Transport equation: Comparison of the exact and the RKDG so-

lutions at T = 100π with second order (P 1, left) and seventh order (P 6,

right) RKDG methods. One dimensional results with 40 cells, exact solution

(solid line) and numerical solution (dashed line and symbols, one point per

cell)



0

0.2

0.4

0.6

0.8

1

1.2

u

0

1

2

3

4

5

6

x

0

1

2

3

4

5

6

y

P1

0

0.2

0.4

0.6

0.8

1

1.2

u

0

1

2

3

4

5

6

x

0

1

2

3

4

5

6

y

P6

Figure 2: Transport equation: Comparison of the exact and the RKDG so-

lutions at T = 100π with second order (P 1, left) and seventh order (P 6,

right) RKDG methods. Two dimensional results with 40 × 40 cells.



The second example is the double Mach reflection problem for the two

dimensional compressible Euler equations.



0.0 0.5 1.0 1.5 2.0 2.5 3.0

0.0

0.2

0.4

0.6

0.8

1.0

Rectangles P1, ∆ x = ∆ y = 1/240

0.0 0.5 1.0 1.5 2.0 2.5 3.0

0.0

0.2

0.4

0.6

0.8

1.0


Figure 3: Double Mach reflection. ∆x = ∆y = 1240

. Top: P 1; bottom:

P 2.Division of Applied Mathematics, Brown University


2.0 2.2 2.4 2.6 2.8

0.0

0.1

0.2

0.3

0.4

0.5


2.0 2.2 2.4 2.6 2.8

0.0

0.1

0.2

0.3

0.4


Figure 4: Double Mach reflection. Zoomed-in region. Top: P 2 with ∆x =

∆y = 1240

; bottom: P 1 with ∆x = ∆y = 1480

.Division of Applied Mathematics, Brown University


2.0 2.2 2.4 2.6 2.8

0.0

0.1

0.2

0.3

0.4

0.5


2.0 2.2 2.4 2.6 2.8

0.0

0.1

0.2

0.3

0.4

Rectangles P2, ∆ x = ∆ y = 1/480Rectangles P2, ∆ x = ∆ y = 1/480

Figure 5: Double Mach reflection. Zoomed-in region. P 2 elements. Top:

∆x = ∆y = 1240

; bottom: ∆x = ∆y = 1480

.Division of Applied Mathematics, Brown University


The third example is the flow past a forward-facing step problem for the

two dimensional compressible Euler equations. No special treatment is

performed near the corner singularity.



0.0 0.5 1.0 1.5 2.0 2.5 3.0

0.0

0.2

0.4

0.6

0.8

1.0


0.0 0.5 1.0 1.5 2.0 2.5 3.0

0.0

0.2

0.4

0.6

0.8

1.0


Figure 6: Forward facing step. Zoomed-in region. ∆x = ∆y = 1320

. Left:

P 1 elements; right: P 2 elements.Division of Applied Mathematics, Brown University


History of the DG method:

Here is a (very incomplete) history of the study of DG methods:

• 1973: First discontinuous Galerkin method for steady state linearscalar conservation laws (Reed and Hill).

• 1974: First error estimate (for tensor product mesh) of thediscontinuous Galerkin method of Reed and Hill (LeSaint and Raviart).

• 1986: Error estimates for discontinuous Galerkin method of Reed andHill (Johnson and Pitkäranta).

• 1989-1998: Runge-Kutta discontinuous Galerkin method for nonlinearconservation laws (Cockburn, Shu, ...).



• 1994: Proof of cell entropy inequality for discontinuous Galerkinmethod for nonlinear conservation laws in general multidimensional

triangulations (Jiang and Shu).

• 1997-1998: Discontinuous Galerkin method for convection diffusionproblems (Bassi and Rebay, Cockburn and Shu, Baumann and Oden,

...).

• 2002: Discontinuous Galerkin method for partial differential equationswith third or higher order spatial derivatives (KdV, biharmonic, ...) (Yan

and Shu, Xu and Shu, ...)



Collected works on the DG methods:

• Discontinuous Galerkin Methods: Theory, Computation andApplications, B. Cockburn, G. Karniadakis and C.-W. Shu, editors,

Lecture Notes in Computational Science and Engineering, volume 11,

Springer, 2000. (Proceedings of the first DG Conference)

• Journal of Scientific Computing, special issue on DG methods, 2005.

• Computer Methods in Applied Mechanics and Engineering, specialissue on DG methods, 2006.

• Journal of Scientific Computing, special issue on DG methods, toappear in 2009.



DG method for Hamilton-Jacobi equations of Hu and Shu (SISC 1999)

Reinterpreted by Li and Shu (Appl. Math. Let 2005)

Idea: convert HJ to conservation laws and then use DG

In one space dimension the HJ equation is

φt +H(φx) = 0, φ(x, 0) = φ0(x). (3)

This is a relatively easy case because (3) is equivalent to the conservation

law

ut +H(u)x = 0, u(x, 0) = u0(x) (4)

if we identify u = φx.



Assuming (3) is solved in the interval a ≤ x ≤ b and it is divided into thefollowing cells:

a = x 12

< x 32

< · · · < xN+ 12

= b, (5)

we denote

Ij = (xj− 12

, xj+ 12

), xj =1

2

(

xj− 12

+ xj+ 12

)

, (6)

hj = xj+ 12

− xj− 12

, h = maxjhj ,

and define the following approximation space

V kh ={

v : v|Ij ∈ P k(Ij), j = 1, ..., N}

. (7)

Here P k(Ij) is the set of all polynomials of degree at most k on the cell

Ij .



A k-th order discontinuous Galerkin scheme for the one dimensional

Hamilton-Jacobi equation (3) can then be defined as follows: find ϕ ∈ V kh ,such that∫

Ij

ϕxtv dx−∫

Ij

H(ϕx)vx dx+Ĥj+ 12

v−j+ 1

2

−Ĥj− 12

v+j− 1

2

= 0, j = 1, ..., N,

(8)

for all v ∈ V k−1h . Here

Ĥj+ 12

= Ĥ(

(ϕx)−

j+ 12

, (ϕx)+j+ 1

2

)

(9)

is a monotone flux, Ĥ(↑, ↓). Notice that the method described above isexactly the discontinuous Galerkin method for the conservation law

equation (4) satisfied by the derivative u = φx.



The DG scheme described above only determines ϕ for each element up

to a constant, since it is only a scheme for ϕx. The missing constant can

be obtained in one of the following two ways:

1. By requiring that∫

Ij

(ϕt +H(ϕx)) v dx = 0, j = 1, ..., N, (10)

for all v ∈ V 0h , that is,∫

Ij

(ϕt +H(ϕx)) dx = 0, j = 1, ..., N. (11)



2. By using (11) to update only one (or a few) elements, e.g., the

left-most element I1, then use

ϕ(xj, t) = ϕ(x1, t) +

∫ xj

x1

ϕx(x, t) dx (12)

to determine the missing constant for the cell Ij .



Both approaches are used in our numerical experiments. They perform

similarly for smooth problems, with the first approach giving slightly better

results. However, it is our numerical experience that, when there are

singularities in the derivatives, the first approach will often produce dents

and bumps when the integral path in time passes through the singularities

at some earlier time. The philosophy of using the second approach is that

one could update only a few elements whose time integral paths do not

cross derivative singularities.



Lemma 2.1.. The following L2 stability result for the derivative ϕx holds

for the discontinuous Galerkin method defined above, of any order of

accuracy k applied to any nonlinear Hamilton-Jacobi equation (3):

d

dt

∫ b

a

ϕ2x dx ≤ 0. (13)

�

For a finite interval [a, b], (13) trivially implies TVB (total variation

bounded) property for the numerical solution ϕ:

TV (ϕ) =

∫ b

a

|ϕx| dx ≤√b− a

√

∫ b

a

(

d

dxφ0(x)

)2

dx. (14)



This is a rather strong stability result, considering that it applies even if the

derivative of the solution φx develops discontinuities, no limiter has been

added to the numerical scheme, and the scheme can be of arbitrary high

order in accuracy. It also implies convergence of at least a subsequence of

the numerical solution ϕ when h→ 0. However, this stability result is notstrong enough to imply that the limit solution is the viscosity solution of (3).



Two Dimensional Case

We consider in this section the case of two spatial dimensions. The

algorithm in more spatial dimensions is similar. This time, the scalar

Hamilton-Jacobi equation

φt +H(φx, φy) = 0, φ(x, y, 0) = φ0(x, y) (15)

is in some sense equivalent to the following conservation law system

ut+H(u, v)x = 0, vt+H(u, v)y = 0, (u, v)(x, y, 0) = (u, v)0(x, y).

(16)

if we identify

(u, v) = (φx, φy). (17)

For example, a vanishing viscosity solution of (15) corresponds, via (17),

to a vanishing viscosity solution of (16), and vice versa.



However, (16) is not a strictly hyperbolic system, which may cause

problems in its numerical solution if we treat u and v as independent

variables. Instead, we would like to still use φ as our solution variable (a

polynomial) and take its derivatives as u and v.

Assuming we are solving (15) in the domain Ω, which has a triangulation

Th consisting of triangles or general polygons of maximum size (diameter)h, with the following approximation space

V kh ={

v : v|K ∈ P k(K), ∀K ∈ Th}

, (18)

where P k(K) is again the set of all polynomials of degree at most k on

the cell K .



We propose a discontinuous Galerkin method for (15) as follows: find

ϕ ∈ V kh , such that∫

K

ϕxtv dxdy −∫

K

H(ϕx, ϕy)vx dxdy +∑

e∈∂K

∫

e

Ĥ1,e,K v dΓ = 0

(19)

and∫

K

ϕytv dxdy −∫

K

H(ϕx, ϕy)vy dxdy +∑

e∈∂K

∫

e

Ĥ2,e,K v dΓ = 0

(20)

for all v ∈ V k−1h and all K ∈ Th, in a least square sense. Here thenumerical flux is

Ĥi,e,K = Ĥi,e,K(

(∇ϕ)int(K), (∇ϕ)ext(K))

, i = 1, 2 (21)



where the superscript int(K) implies that the value is taken from within

the element K , and the superscript ext(K) implies that the value is taken

from outside the element K and within the neighboring element K ′

sharing the edge e with K . The flux (21) satisfies the following properties:

1. Ĥi,e,K is Lipschitz continuous with respect to all its arguments;

2. Consistency:

Ĥi,e,K (∇φ,∇φ) = H(∇φ)ni,where n = (n1, n2) is the unit outward normal to the edge e of the

element K ;

3. Conservation:

Ĥi,e,K(

(∇ϕ)int(K), (∇ϕ)ext(K))

= −Ĥi,e,K′(

(∇ϕ)int(K′), (∇ϕ)ext(K′))

.



where K ∩K ′ = e.We will again mainly use the simple (local) Lax-Friedrichs flux

Ĥ1,e,K((u, v)−, (u, v)+) = H

(

u− + u+

2,v− + v+

2

)

n1−1

2α(u+−u−)

(22)

and

Ĥ2,e,K((u, v)−, (u, v)+) = H

(

u− + u+

2,v− + v+

2

)

n2−1

2β(v+−v−)

(23)

where α = maxu,v |∂H(u,v)∂u | and β = maxu,v |∂H(u,v)∂v

|, with themaximum being taken over the relevant (local) range.

Notice that (19)-(20) are exactly the discontinuous Galerkin method for the

conservation law system (16) satisfied by the derivatives (u, v) = ∇φ.



For a rectangular mesh and for k = 1 this recovers the (local)

Lax-Friedrichs monotone scheme for (15) if we identify

uij =φi+1,j−φi−1,j

2∆xand vij =

φi,j+1−φi,j−12∆y

.

Of course, (19)-(20) have more equations than the number of degrees of

freedoms (an over-determined system) for k > 1, thus a least square

solution is needed. In practice, the least square procedure is performed as

follows: we first evolve (16) for one time step (one inner stage for high

order Runge-Kutta methods), using the discontinuous Galerkin method

(19)-(20) with u = ϕx and v = ϕy ; then ϕ at the next time level (stage) is

obtained (up to a constant) by least square:

||(ϕx−u)2+(ϕy−v)2||L1(K) = minψ∈Pk(K)

||(ψx−u)2+(ψy−v)2||L1(K).



Important note: In Li and Shu (Appl. Math. Let. 2005), the DG space for

(u, v) is chosen to contain only locally curl-free polynomial pairs. This

yields the same algorithm mathematically as described above, however it

has the following two advantages:

• The least square procedure is no longer needed, as now the numberof the degree of freedom is equal to the number of equations

• The computational cost is reduced, because the locally curl-freespace is smaller than the original unconstrained space



This determines ϕ for each element up to a constant, since it is only a

scheme for ∇ϕ. The missing constant can again be obtained in one of thefollowing two ways:

1. By requiring that∫

K

(ϕt +H(ϕx, ϕy)) v dxdy = 0, (24)

for all v ∈ V 0h and for all K ∈ Th, that is,∫

K

(ϕt +H(ϕx, ϕy)) dxdy = 0, ∀K ∈ Th ; (25)



2. By using (25) to update only one (or a few) elements, e.g., the corner

element(s), then use

ϕ(B, t) = ϕ(A, t) +

∫ B

A

(ϕx dx+ ϕy dy) (26)

to determine the missing constant. The path should be taken to avoid

crossing a derivative discontinuity, if possible.

We remark that the procedure discussed above is easily implemented in

any triangulation, e.g., for both rectangles and triangles.



DG method for Hamilton-Jacobi equations of Cheng and Shu (JCP 2007)

Hamilton-Jacobi equation:

ϕt +H(ϕx1, . . . , ϕxd , x1, . . . , xd) = 0, ϕ(x, 0) = ϕ0(x)

• We only consider linear or convex H as a function of ϕ in this lecture.

• Generic solution are continuous but may admit discontinuousderivatives.

• We are only interested in the viscosity solution, which is the uniquepractically relevant solution and satisfies the entropy condition.



Available numerical methods on structured meshes:

• Essentially non-oscillatory (ENO) or weighted ENO (WENO) finitedifference schemes (Osher and Shu SINUM 1991; Jiang and Peng

SISC 2000) on structured meshes:

– very simple

– highly accurate

– non-oscillatory

– very popular in applications



• WENO finite difference schemes on unstructured meshes (Zhang andShu SISC 2003):

– flexible in geometry and adaptive meshes

– accurate

– non-oscillatory

– rather complicated in coding



• DG schemes of Hu and Shu (SISC 1999)– First convert a HJ equation ϕt+H(ϕx) = 0 to a conservation law

ut +H(u)x = 0 (27)

by defining u = ϕx, then apply the usual DG scheme to (27) plus

evolving the cell average of ϕ.

– In multi-dimensions, a scalar HJ equation is converted to a system,

which is not strongly hyperbolic at u = 0. Additional cost and

complexity related to systems.

– Reinterpretation by Li and Shu (Appl. Math. Letters 2005): using

locally curl-free DG basis to save cost and to avoid the least square

procedure to reconstruct ϕ from u.



A new DG scheme (Cheng and Shu, JCP 2007): works directly on the HJ

equation without first converting it to a hyperbolic system for the

derivatives.

Motivation: look at a simple linear HJ equation

ϕt + a(x)ϕx = 0

with a′(x) ≥ 0. It can be rewritten as a conservation law with a sourceterm

ϕt + (a(x)ϕ)x = a′(x)ϕ (28)



The usual DG method can be applied to (28) as finding ϕh ∈ Vh such that∫

Ij

∂tϕh(x, t)vh(x)dx−∫

Ij

a(x)ϕh(x, t)∂xvh(x)dx

+a(xj+ 12

)ϕh(x−

j+ 12

, t)vh(x−

j+ 12

) − a(xj− 12

)ϕh(x−

j− 12

, t)vh(x+j− 1

2

)

=

∫

Ij

a′(x)ϕh(x, t)vh(x)dx (29)

holds for all test functions vh(x) ∈ Vh. The scheme (29) can be rewrittenas

∫

Ij

(∂tϕh(x, t) + a(x)∂xϕh(x, t)) vh(x)dx

+a(xj− 12

)(

ϕh(x+j− 1

2

, t) − ϕh(x−j− 12

, t))

vh(x+j− 1

2

) = 0



This motivates the following definition of our DG scheme for general HJ

equations: Find ϕh ∈ Vh, such that∫

Ij

(∂tϕh(x, t) +H(∂xϕh(x, t), x))vh(x)dx

+1

2

(

minx∈I

j+12

H1(∂xϕh, xj+ 12

) −∣

∣

∣

∣

∣

minx∈I

j+12

H1(∂xϕh, xj+ 12

)

∣

∣

∣

∣

∣

)

[ϕh]j+ 12

(vh)−

j+ 12

+1

2

(

maxx∈I

j− 12

H1(∂xϕh, xj− 12

) +

∣

∣

∣

∣

∣

maxx∈I

j− 12

H1(∂xϕh, xj− 12

)

∣

∣

∣

∣

∣

)

[ϕh]j− 12

(vh)+j− 1

2

= 0, j = 1, . . . , N

holds for any vh ∈ Vh.



Remarks about the scheme:

• We need ∂xϕh on the cells Ij− 12

and Ij+ 12

, which include the points

xj− 12

and xj+ 12

respectively, where the numerical solution ϕh(x, t) is

discontinuous. We use a L2 type projection to reconstruct a

continuous approximation of ϕh(x, t).

• The purpose of taking the maximum and minimum is to obtain betterstability by adding more viscosity, while still maintaining accuracy.

• The scheme is Roe type which may generate entropy violatingsolutions. We perform an entropy correction in sonic expanding cells

by simply switching to the Hu-Shu DG scheme in those cells.



We have the following stability and accuracy results for our scheme

applied to linear HJ equation

ϕt + a(x)ϕx = 0

Proposition 1. Suppose there is a constant β such that the derivative of

a(x) satisfies ax(x) < β for x ∈ [a, b], then we have the following L2stability for our scheme:

||ϕh(t)||L2 ≤ eβt/2||ϕh(0)||L2



Proposition 2. If a(x) and the solution ϕ are smooth and the scheme

with the finite element space consisting of piecewise polynomials of

degree ≤ k is used, then we have the following optimal L2 error estimate

||ϕh(t) − ϕ(t)||L2 ≤ Chk+1



Numerical results

Example 1. We solve the one dimensional problem

ϕt + sin(x)ϕx = 0

ϕ(x, 0) = sin(x)

ϕ(0, t) = ϕ(2π, t)

We clearly observe (k + 1)-th order of accuracy for P k polynomials.



Table 1: Errors and numerical orders of accuracy for Example 1 when us-

ing P 0 polynomials and Runge-Kutta third order time discretization on a

uniform mesh of N cells. Final time t = 1. CFL = 0.9.

N L1 error order L2 error order L∞ error order

40 0.49E-01 0.62E-01 0.29E+00

80 0.25E-01 0.95 0.32E-01 0.93 0.16E+00 0.86

160 0.13E-01 0.97 0.17E-01 0.96 0.83E-01 0.96

320 0.65E-02 0.98 0.84E-02 0.98 0.42E-01 0.99

640 0.33E-02 0.99 0.42E-02 0.99 0.21E-01 1.00







40 0.12E-02 0.25E-02 0.15E-01

80 0.31E-03 1.96 0.68E-03 1.90 0.43E-02 1.81

160 0.78E-04 1.97 0.18E-03 1.94 0.11E-02 1.92

320 0.20E-04 1.98 0.46E-04 1.97 0.29E-03 1.96

640 0.50E-05 1.99 0.12E-04 1.98 0.74E-04 1.98







40 0.48E-04 0.10E-03 0.52E-03

80 0.60E-05 2.99 0.14E-04 2.88 0.88E-04 2.58

160 0.75E-06 3.00 0.18E-05 2.90 0.14E-04 2.70

320 0.94E-07 2.99 0.24E-06 2.93 0.20E-05 2.78

640 0.12E-07 2.99 0.31E-07 2.95 0.27E-06 2.85







40 0.21E-05 0.51E-05 0.29E-04

80 0.14E-06 3.96 0.35E-06 3.88 0.22E-05 3.75

160 0.87E-08 3.97 0.23E-07 3.93 0.16E-06 3.78

320 0.55E-09 3.97 0.15E-08 3.96 0.10E-07 3.91

640 0.35E-10 3.98 0.94E-10 3.98 0.68E-09 3.95



Example 2.

We solve the two dimensional linear Hamilton-Jacobi equation with

variable coefficients

ϕt − yϕx + xϕy = 0.The computational domain is [−1, 1]2. The initial condition is given by

ϕ0(x, y) =

0 0.3 ≤ r0.3 − r 0.1 < r < 0.30.2 r ≤ 0.1

where r =√

(x− 0.4)2 + (y − 0.4)2. We also impose periodicboundary condition on the domain. This is a solid body rotation around the

origin.



For this problem, the derivatives of ϕ are not continuous. Therefore, we do

not expect to obtain (k + 1)-th order of accuracy for P k polynomials, see

the following table for the result of P 2 elements.





uniform mesh of N ×N cells. Final time t = 1. CFL = 0.1.

N ×N L1 error order L2 error order L∞ error order20 × 20 0.41E-03 0.13E-02 0.11E-0140 × 40 0.14E-03 1.58 0.55E-03 1.26 0.65E-02 0.8280 × 80 0.47E-04 1.54 0.24E-03 1.22 0.36E-02 0.84

160 × 160 0.15E-04 1.62 0.10E-03 1.23 0.21E-02 0.81



At t = 2π, i.e. the period of rotation, we take a snapshot at the line x = y

in Figure 7. We can see that a higher order scheme can yield better

results for this nonsmooth initial condition.



x

ϕ

-0.5 0 0.5

0

0.05

0.1

0.15

0.2

xϕ

-0.5 0 0.5

0

0.05

0.1

0.15

0.2

Figure 7: Example 2. 80× 80 uniform mesh. t = 2π. Solid line: the exactsolution; Rectangles: the numerical solution. One dimensional cut of 45◦

with the x axis. Left: P 1 polynomial; Right: P 2 polynomial.



Example 3.

We solve the same two dimensional linear Hamilton-Jacobi equation with


ϕt − yϕx + xϕy = 0as in the previous example but with a different initial condition as

ϕ0(x, y) = exp

(

−(x− 0.4)2 + (y − 0.4)22σ2

)

We take σ = 0.05 such that at the domain boundary, ϕ is very small,

hence imposing a periodic boundary condition will lead to small

non-smoothness errors. We then observe the desired order of accuracy in

the following table.





uniform mesh of N ×N cells. Final time t = 1. CFL = 0.1.

N ×N L1 error order L2 error order L∞ error order20 × 20 0.14E-02 0.10E-01 0.28E+0040 × 40 0.15E-03 3.21 0.15E-02 2.81 0.53E-01 2.4180 × 80 0.11E-04 3.82 0.11E-03 3.73 0.58E-02 3.19

160 × 160 0.11E-05 3.30 0.12E-04 3.26 0.90E-03 2.69



Example 4.

We solve the two dimensional linear Hamilton-Jacobi equation with


ϕt + f(x, y, t)ϕx + g(x, y, t)ϕy = 0

with

f(x, y, t) = sin2(πx) sin(2πy) cos

(

t

Tπ

)

,

g(x, y, t) = − sin2(πy) sin(2πx) cos(

t

Tπ

)

where T is the period of deformation.



The initial condition is given by

ϕ0(x, y) =

0 0.3 ≤ r0.3 − r 0.1 < r < 0.30.2 r ≤ 0.1

where r =√

(x− 0.4)2 + (y − 0.4)2. This is a numerical test forincompressible flow. During the evolution, the initial data is severely

deformed, then it returns to the original shape after one period. At

t = 1.5, i.e. the period of rotation, we take a snapshot at the line x = y in

Figure 8. We can clearly observe that a higher order scheme yields better

results for this nonsmooth initial condition.



x

ϕ

-0.5 0 0.5

0

0.05

0.1

0.15

0.2

xϕ

-0.5 0 0.5

0

0.05

0.1

0.15

0.2

Figure 8: Example 4. 80×80 uniform mesh. t = 1.5. Solid line: the exactsolution; Rectangles: the numerical solution. One dimensional cut of 45◦

with the x axis. Left: P 1 polynomial; Right: P 2 polynomial.



Example 5.

We solve the model linear problem with a discontinuous coefficient

ϕt + sign(cos(x))ϕx = 0

ϕ(x, 0) = sin(x)

ϕ(0, t) = ϕ(2π, t)

For the viscosity solution, at x = π2

, there will be a shock forming in ϕx,

and at x = 3π2

, there is a rarefaction wave. For this example we need an

entropy correction.



x

ϕ

1 2 3 4 5 6

-1

-0.5

0

0.5

xϕ

0 1 2 3 4 5 6

-1

-0.5

0

0.5

Figure 9: Example 5. t = 1, CFL = 0.1, N = 80, using P 2 polynomi-

als. Solid line: the exact solution; Rectangles: the numerical solution. Left:

without entropy correction; Right: with entropy correction.



Example 6.

One dimensional Burgers’ equation

ϕt +ϕ2x2

= 0

ϕ(x, 0) = sin(x)

ϕ(0, t) = ϕ(2π, t)

When t = 0.5, the solution is still smooth, and the expected third order

accuracy is obtained for P 2 polynomials, see Table 7. After t = 1, a

shock will form in ϕx, our scheme can resolve the derivative singularity

sharply, see Figure 10.





uniform mesh of N cells. Final time t = 0.5. CFL = 0.1.


40 0.13E-04 0.22E-04 0.84E-04

80 0.17E-05 2.97 0.29E-05 2.93 0.12E-04 2.86

160 0.22E-06 2.98 0.37E-06 2.96 0.15E-05 2.92

320 0.27E-07 2.98 0.47E-07 2.97 0.20E-06 2.95

640 0.34E-08 2.99 0.59E-08 2.99 0.25E-07 2.97



x

ϕ

1 2 3 4 5 6-1

-0.5

0

0.5

Figure 10: Example 6. Numerical solution. Solid line: N = 500; Rectan-

gles: N = 40. Final time t = 1.5, CFL = 0.05, P 2 polynomials.



Example 7.

One dimensional Burgers’ equation with a nonsmooth initial condition

ϕt +ϕ2x2

= 0

ϕ(x, 0) =

π − x if 0 ≤ x ≤ πx− π elsewhere in[0, 2π],

ϕ(0, t) = ϕ(2π, t)

For the viscosity solution, the sharp corner at π will be smoothed out, and

a rarefaction wave will form in the derivative. For this example we also

need the entropy correction.



x

ϕ

1 2 3 4 5 6

-0.5

0

0.5

1

1.5

2

2.5

xϕ

1 2 3 4 5 6

-0.5

0

0.5

1

1.5

2

2.5

Figure 11: Example 7. t = 1, CFL = 0.05, N = 80, using P 2 poly-

nomials. Solid line: the exact solution; Rectangles: the numerical solution.

Left: without entropy correction; Right: with entropy correction.



Example 8.

Two dimensional Burgers’ equation.

ϕt +(ϕx + ϕy + 1)

2

2= 0

ϕ(x, y, 0) = − cos(x+ y)

with periodic boundary condition on the domain [0, 2π]2.

We use a uniform rectangular mesh. At t = 0.1, the solution is still

smooth. Numerical errors and order of accuracy are listed in Table 8,

demonstrating the expected order of accuracy. At t = 1, the solution is no

longer smooth. We plot the numerical solution in Figure 12. We observe

good resolution of the kinks in the solution.





uniform mesh of N ×N cells. Final time t = 0.1. CFL = 0.1.

N ×N L1 error order L2 error order L∞ error order10 × 10 0.30E-02 0.43E-02 0.35E-0120 × 20 0.38E-03 2.98 0.58E-03 2.90 0.56E-02 2.6440 × 40 0.48E-04 2.97 0.77E-04 2.91 0.80E-03 2.8180 × 80 0.66E-05 2.87 0.11E-04 2.83 0.14E-03 2.55



x

0

2

4

6

y

0

2

4

6

ϕ

-1.4

-1.2

-1

-0.8

-0.6

Figure 12: Example 8. Numerical solution when t = 1, CFL = 0.1,

40 × 40 uniform mesh, using P 2 polynomials.



Example 9.

We solve the Eikonal equation given by

ϕt + |ϕx| = 0ϕ(x, 0) = sin(x)

ϕ(0, t) = ϕ(2π, t)

Figure 13 shows the space-time location where the entropy correction is

applied. We observe that the correction is mostly applied at a few cells

neighboring the boundary of the rarefaction wave. The number of cells in

which the correction is performed is relatively small compared to the total

number of cells.



x

t

1 2 3 4 5 60

2

4

6

8

Figure 13: Example 9. t = 10, CFL = 0.1, N = 160 uniform mesh,

using P 2 polynomials. ε = 10−10. Rectangular symbols mark the cells in

which the entropy correction is performed.Division of Applied Mathematics, Brown University


Example 10.

We solve the two dimensional Eikonal equation

ϕt +√

ϕ2x + ϕ2y = 1

First, we consider the case of the computational domain being

[0, 1]2 \ [0.4, 0.6]2. For the inner boundary along [0.4, 0.6]2, we imposethe boundary condition ϕ = 0. On the other hand, we impose free outflow

boundary conditions on the outer boundary. The initial condition is taken

as ϕ0(x, y) = max{|x− 0.5|, |y − 0.5|} − 0.1. The steady statesolution should give us a function that is equal to the distance of the point

to the inner boundary. We plot the numerical steady state solution in

Figure 14.



x

0

0.2

0.4

0.6

0.8

1

y

0

0.2

0.4

0.6

0.8

1

ϕ

0

0.1

0.2

0.3

0.4

0.5

x

y

0.2 0.4 0.6 0.8 1

0.2

0.4

0.6

0.8

Figure 14: Example 10. Steady state solution with 40 × 40 uniform mesh,using P 2 polynomials. Left: three dimensional plot; Right: contour plot.



Next, we consider this example with a point source condition; namely, we

take the inner boundary to be the center point (0.5, 0.5). In this case, we

would need ϕ in the center cell to be the L2 projection of the exact

distance function. For all other cells, the computation is the same as for

the previous case. The initial condition is taken as

ϕ0(x, y) = max{|x− 0.5|, |y − 0.5|}. We plot the numerical steadystate solution in Figure 15. We can see that in both cases we obtain very

good resolution to the viscosity solution.



x

0

0.2

0.4

0.6

0.8

y

0

0.2

0.4

0.6

0.8

ϕ

0

0.2

0.4

0.6

x

y

0.2 0.4 0.6 0.8 1

0.2

0.4

0.6

0.8

Figure 15: Example 10. Steady state solution with 39 × 39 uniform mesh,using P 2 polynomials. Left: three dimensional plot; Right: contour plot.



Fast sweeping method for steady state solutions

The DG method is quite local regardless of the order of accuracy, hence it

is well suited for the design of fast sweeping methods. Preliminary results

in Li, Shu, Zhang and Zhao (JCP 2008) for a P 1 fast sweeping DG

algorithm for the Eikonal equation based on the DG formulation of Cheng

and Shu are very promising. We have now obtained even better results,

with the number of iterations uniformly low (usually only 4 or at most 8

iterations) for all meshes.



References for the third lecture:

[1] Y. Cheng and C.-W. Shu, A discontinuous Galerkin finite element

method for directly solving the Hamilton-Jacobi equations, Journal of

Computational Physics, v223 (2007), pp.398-415.

[2] C. Hu and C.-W. Shu, A discontinuous Galerkin finite element method

for Hamilton-Jacobi equations, SIAM Journal on Scientific Computing, v21

(1999), pp.666-690.

[3] O. Lepsky, C. Hu and C.-W. Shu, Analysis of the discontinuous Galerkin

method for Hamilton-Jacobi equations, Applied Numerical Mathematics,

v33 (2000), pp.423-434.



[4] F. Li and C.-W. Shu, Reinterpretation and simplified implementation of a

discontinuous Galerkin method for Hamilton-Jacobi equations, Applied

Mathematics Letters, v18 (2005), pp.1204-1209.

[5] F. Li, C.-W. Shu, Y.-T. Zhang and H. Zhao, A second order

discontinuous Galerkin fast sweeping method for Eikonal equations,

Journal of Computational Physics, v227 (2008), pp.8191-8208.

[6] C.-W. Shu, High order numerical methods for time dependent

Hamilton-Jacobi equations, in Mathematics and Computation in Imaging

Science and Information Processing, S.S. Goh, A. Ron and Z. Shen,

Editors, Lecture Notes Series, Institute for Mathematical Sciences,

National University of Singapore, volume 11, World Scientific Press,

Singapore, 2007, pp.47-91.



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