DGM for Scalar Conservation Laws∗

Amiya Kumar PaniDepartment of Mathematics

IIT Bombay, Powai, Mumbai-400076Email:akp@math.iitb.ac.in

December 6, 2014

1 Introduction

In this series of lectures, we briefly discuss the model and theoretical aspects like weak solution, conservequantities, and a priori bounds needed for the error analysis of discontinuous Galekin method (DGM). Inthe later part, we look at DGM and the related error analysis and super-convergence results. Althoughmaterial are available in many papers, but we make an attempt to unify the error analysis.

2 Mathematical Model

Mathematical equations very often provide a language to formulate physical phenomena in terms of sometractable forms so that it is possible to study them, predict their behaviour etc. Say for example, Newtoninvented Calculus to describe accurately the motion of bodies, Maxwell equations were written down todescribe electrodynamics. Similarly, Schrodinger’s equation was developed to describe aspects of quantummechanics, Navier-Stokes equations1 were formulated to model the motion of an incompressible viscousflow and Black-Scholes equations helped to understand option derivatives in stock market.

In fact, a mathematical model is an equation or a set of equations whose solution describes thephysical phenomenon in an approximate manner. It is indeed a simplified description of a physical realitywhich is expressed in the language of mathematics. It involves observation or experiments, and pickingup important factors influencing the system, and then a description in terms of mathematical equations.Once a mathematical formulation is achieved, analysis and simulation do provide results which are againto be validated against the observations or experiments. This will complete the cycle

I−−−−−−F−−−−−−S−−−−− I

of Identification (to extract mathematical essence out of the physical phenomenon at hand), Formulation (toput the problem in a familiar setting about which we already know something (sometimes called modelling

∗Advanced Workshop on Hyperbolic PDEs: Theory, Numerics & Applications and Conference on Partial DifferentialEquations (HPTNA-2014 & CPDE-2014) held at THE LNM INSTITUTE OF INFORMATION TECHNOLOGY, JAIPURduring December 01-11, 2014.

1Although this system of equations is used for the last two centuries by Engineers and Scientists to model the motionof an incompressible fluid flow with moderate velocity, the global existence of a unique smooth solution in 3-dimensionsremains an unsolved problem. In 2D, it is completely resolved by Olfa Ladyzhenskaya from Russia and J. L. Lions with Prodifrom France separately at the sametime. In fact it is one of the seven (now is called six, as Poincare conjecture has beenfully solved by Grusa Perelman) millenium prize problem proposed by the Clay Mathematical Institute (see the URL site:http://www.claymath.org/millennium/ for more detailed information).

1

activity)), Solution process (which involves mathematical, statistical and computational tools- a majorpart of it we shall be providing in courses given by Mathematics Department), and finally, Interpretation(to provide an answer to the physical problem that we have started with). We shall be talking about themodels described by conservation laws in this note.

Many PDE models come naturally from basic conservation principles or laws. In fact, we can simplydescribe the conservation principle as:

the rate at which a quantity of interest changes in a given domain must be equal to the rate atwhich the quantity flows across the boundary of the domain plus the rate at which the quantityis created or destroyed within the domain.

For example: in population dynamics, the rate of change of population of certain animal species ina fixed geographical region must be equal to the rate at which animals migrate into the region minus therate at which they migrate out plus the birth rate minus the death rate. Here population means populationdensity. Similar situations can occur in many physical procesess which may not be our focus in this note.

To quantify such verbal statements, we require to put them in some mathematical form. Say forexample, assume that the state variable u = u(x, t) denotes the density2 of a given quantity of interest. Forsimplicity, we assume that any variation of quantity is restricted to one spatial domain, say a tube withcross sectional area say A, where the cross section is labeled by the spatial variable x and the quantitiesof interest vary only in the x direction and also in time. The amount of the quantity in a small section ofwidth dx is given by u(x, t)Adx, where A is the cross sectional area of the tube. Further, let q = q(x, t)denote the flux of the quantity at x at time t. The flux is defined as the amount of the quantity crossingthe section x at time t and its unit is given by amount per unit area per unit time. Therefore, the amountof the quantity that is crossing the section at x and at time t is Aq(x, t). By convention, the flux is positiveif the flow is from left to right and negative if it is from right to left. Finally, let f = f(x, t) be the givenrate at which the quantity is created or destroyed within the section at x and at time t. It is called a sourceterm if it is positive and called a sink if it is negative. Hence, the amount of quantity that is created (ordestroyed) in a small width dx per unit time is f(x, t)Adx. We can formulate the conservation law byconsidering a fixed, but arbitrary section say a ≤ x ≤ b of the tube by requiring that the rate of change ofthe total amount of the quantity in this section must be equal to the rate at which it flows at x = a minusthe rate at which it flows out at x = b plus the rate at which it is created within the section a ≤ x ≤ b. Inthe language of Mathematics, we write this conservation law as:

(2.1)d

dt

∫ b

au(x, t)Adx = Aq(a, t)−Aq(b, t) +

∫ b

af(x, t)Adx.

This is a fundamental conservation principle,which is indeed written in an integral form. However, if uand q are sufficiently smooth (u and q have continuous first derivatives : will be sufficient for time being),then we write:

d

dt

∫ b

au(x, t) dx =

∫ b

aut(x, t) dx,

and

q(a, t)− q(b, t) = −∫ b

aqx(x, t) dx.

On substituting in (2.1), we now obtain∫ b

a(ut(x, t) + qx(x, t)− f(x, t)) dx = 0.

2 Density measured as mass per unit volume or length.

2

Since the section a ≤ x ≤ b is arbitrary and the integrand is continuous, it therefore, follows that theintegrand must vanish identically leading to :

(2.2) ut(x, t) + qx(x, t) = f(x, t).

This is a differential form of the conservation principle which is valid in the domain of interest. Unfor-tunately, this is one equation with two unknowns: u and q. Thus to close the system, we need to havea relation between the flux variable q and the density u, which is known as the constitutive relation orequation of state representing a particular physical situation.

For example, a model where the flux is proportional to the density, that is,

q = αu,

where α is a constant is called an advection model. The situation like spread of pollutants from an industrialchimney mainly due to a strong wind (with negligible diffusion) with velocity say α and no other externalinput, that is, with f = 0 can be described by

(2.3) ut + αux = 0, x ∈ IR, t > 0

and initial concentration u(x, 0) = u0(x), x ∈ IR, where u denotes the concentration of the pollutants andthe wind velocity α > 0 means wind is blowing from left to right. An exactly similar model is also valid forwater pollution if the velocity of the flow of water is given by α. This model is also known in literature astransport model. By a solution u of the above first order PDE (as the highest derivative is of first order),we mean that u is continuously differentiable and it satisfies the PDE and the initial condition.

Our next question is ‘How do we solve it ?’ that is, how to get an analytical or exact expression foru ?

Let us use a change of variable, that is; ξ = x− αt and τ = t. Let us denote u in the new variablesby U(ξ, τ), that is, U(ξ, τ) = u(ξ + ατ, τ) or u(x, t) = U(x− αt, t). Then, using chain rule, we write :

ut = Uξξt + Uττt = −αUξ + Uτ ,

andux = Uξξx + Uττx = Uξ.

On substituting in the main PDE, we obtain

Uτ = 0.

It is an ordinary differential equation in the new transformed dependent variable U. Hence we write thesolution as

U(ξ, τ) = C(ξ),

where C is a function of ξ and independent of τ. At τ = t = 0, we have U(ξ, 0) = u0(ξ). Hence, U(ξ, τ) =u0(ξ) and in the original variable u(x, t) = u0(x − αt). It can be shown easily that u given by the aboveexpression is a solution of the 1st order PDE provided u0 is continuously differentiable. If α > 0, thesolution is a right travelling wave which preserves the initial profile. In the context of air pollution modelthat too if the wind velocity is strong (with negligible diffusion), one can feel the same effect in near byareas which are in the direction of the wind even after a later time. This model can be modified to includethe effect of diffusion so that a distance far away from the chimney one does not feel the strong impact ofthe pollutant-a case of Bhopal gas tragedy is an example. Why should we bother about such a model? Incase such a model with a simulator is available, then one can take a decision to evacuate people so thatthe damage can be minimal, and hence, it helps in the level of planning.

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Exercise. Solve the following first order PDE using the transformation technique:

ut + αux + au = f(x, t), x ∈ IR, t > 0,

with initial condition : u(x, 0) = u0(x), x ∈ IR, where α and a are constants and f is a given function.Using the transformation technique we now arrive at:

Uτ + aU = f(ξ + ατ, τ)

with U(ξ) = u0(ξ). Its solution can be written as:

U(ξ, τ) := u0(ξ)e−aτ +

∫ τ

0e−a(τ−s)f(ξ + ατ, τ) dξ.

In the original variable, we obtain the solution as

u(x, t) := u0(x− αt)e−at +

∫ t

0e−a(t−s)f(x, t) dx.

One may be curious to know ‘How do we choose the transformations?’ Note that the straight linesξ = x− αt ≡ constant in space-time domain are solutions of

dx

dt= α,

which is called characteristic equation and the straight lines ξ = C, where C is an arbitrary constant arecalled characteristics. Thus it is possible to get the transformation by solving the characteristic equation.

Now for a problem like:ut + a(x, t)ux = f(x, t, u),

we again form the characteristic equation :

dx

dt= a(x, t),

let φ(x, t) = C be its general solution. Thus we obtain the characteristic coordinates as :

ξ = φ(x, t), τ = t.

Again using chain rule, we obtain the following transformed equation in these new co-ordinates :

Uτ = F (ξ, τ, U),

where U = U(ξ, τ).Another Example. Solve : ut + 2tux = 0, x ∈ R, t > 0. As a = 2t, the characteristic equation is given

by :dx

dt= 2t. On solving: we obtain x− t2 = C. Thus set ξ = x− t2 and τ = t. Using chain rule we obtain

ut = Uξ(−2t) + Uτ , ux = Uξ

and therefore, we obtain Uτ = 0. Its general solution is now given by U = F (ξ), where F is an arbitraryfunction. In the original variable we obtain : u(x, t) = F (x− t2).

In order to model a simple diffusion process, like diffusion of a solute in a solvent, diffusion of thepollutant in air which is coming from a chimney in a very calm morning, and heat conduction throughmaterials etc., the flux q is given by

q = −αux + βu,

4

where the constant α is known as conductivity in heat transfer problem, or called diffusivity in diffusionof a solute in a solvent (in that case u is known as concentration) and β is a constant. Thus, we write theequation as diffusion equation :

ut − αuxx + βux = f.

This is known as convection-diffusion equation where βux represents convective or transport term. If timepermit, we shall discuss briefly in these lectures.

Trafic Flow. Now we consider a trafic flow on a highway or in a tunnel under the following more realisticassumptions:

(i) There is only one lane and overtaking is not allowed. For example, a trafic in a tunnel is more closeto thisa assumption.

(ii) No car joins or goes away. This is possible for a section of road with out exit and entrance gates.

(iii) The average speed is not constant and depends on the density of cars ( that is number of cars perunit length).

We describe the flow by means of macroscopic variables such as : the density of cars say ρ, their averagespeed v and their flux q ( that is cars per unit time). The three variables are linked through q = vρ. Asfor our assumption (iii), v = v(ρ). Note that we expect the speed to decrease as the density increases.Therefore, we should have assuming that v is smooth:

v′(ρ) =dv

dρ≤ 0.

Using hypotheses, we can easily derive using conservation principle that ρ satisfies:

ρt + q(ρ)x = 0,

where q(ρ) = v(ρ)ρ. We need a relation for v = v(ρ) which is usually called constitutive relation and hasto be derived from experimental data. Observe that when ρ is small, it is more realistic to assume that theaverage speed v is more or less equal to the maximum speed vm, given by speed limit. When ρ increases,traffic slows down and stops at the maximum density ρm, say, from bumper to bumper traffic. Therefore,we resort to a simpler model which is consistent with the above reseaning, namely;

v(ρ) = vm

(1− ρ

ρm

).

Thus, q(ρ) becomes

q(ρ) = vm ρ(

1− ρ

ρm

).

Hence, the equation modelling the trafic flow now leads to

ρt + vm

(1− 2ρ

ρm

)ρx = 0, x ∈ IR, t > 0.(2.4)

To close the system, we need the following initial condition:

ρ(x, 0) = ρ0(x), x ∈ IR.(2.5)

The solution ρ can be written as :

ρ(x, t) = ρ0(x− q′(ρ0(x0))t),

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where (x0, 0) is the base point of the characterstic curve tracing back from the point (x, t).Initial Condition. When from bumper to bumper traffic is standing at red light, we consider that it isplaced at x = 0 while the road ahead is empty. Therefore, the initial density function ρ0 is given by ρm forx ≤ 0 and ahead of it, that is, for x > 0 ρ0 becomes zero. Now if at time t = 0 the traffic light turns green,traffic starts moving and we want to describe the evolution of the flow of traffic through the equations (2.4)- (2.5). We shall discuss this in details at a later stage.

Let us consider the case of traffic jam ahead. In such case, the initial density ρ0 becomes 1/8 ρm forx < 0, and ρm for x > 0 because the density is maximum for x > 0. The situation is again from bumperto bumper. Note that the cars on the left move with speed v = 7/8 vm so that it is expected that thecongestion is propagating back into the traffic.

We shall in the course of our lecture, discuss how to solve and interprete these results. One naturalquestion is Why does it work ?’ More precisely, we would like know the mathematical basis behind itand in a more natural way it also provides some theoretical results like existence and uniqueness.

To be more specific, we now consider nonlinear scalar conservation law as

ut + f(u)x = 0, (x, t) ∈ IR× (0,∞)(2.6)

with initial condition

u(x, 0) = u0(x), x ∈ IR,(2.7)

where f : IR −→ IR smooth nonlinear function. Then, we formally rewrite (2.6) in a non-conservative form

ut + f ′(u)ux = 0, (x, t) ∈ IR× (0,∞)(2.8)

Now the characteristic curves associated with (2.8) are given by

dx

dt(t) = f ′(u(x(t), t)), t > 0(2.9)

with initial condition x(0) = x0. Since f is smooth, an application of Picard’s theorem yields existence ofa unique solution x(t) locally, that is, for t in some interval (0, t∗). Since using chain rule and (2.8)

d

dtu(x(t), t) =

∂u

∂t(x(t), t) +

∂u

∂x(x(t), t)

dx

dt(t)

=(∂u∂t

+ f ′(u)∂u

∂x

)(x(t), t) = 0,

then, the solution u remains constant along such a curve. Note that from (2.9), the characteristic curveare given by

(2.10) x(t) = x0 + tf ′(u(x0)).

Thus,

u(x(t), t) = u0(x0)(2.11)

and the solution u is written asu(x, t) = u0(x− tf ′(u)).

If the nonlinear function f satisfies f ′′ > 0, For the non-viscous Burger’s equation in divergence form,the nonlinear function f(u) = 1/2u2 and exact solution u is written in implicit form as

u(x, t) = u0(x− tu(x, t)).

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3 Weak Formulation and A priori Bounds

This section deals with weak formulation, entropy condition, some conversed quanties and some a prioribounds.

For linear problem (2.3) discusses in Section 2, if the initial function u0 is not differentiable, thenthe corresponding solution will not qualify for a classical solution. Therefore, there is need for generalisingthe concept of solution.

To motivate this, multiply (2.6) by a test function φ ∈ C10 (IR× IR+) and integrate with respect space

and time domain to arrive at ∫ ∞0

∫IR

(∂u∂t

+∂f

∂x

)φ dx dt = 0,

where C10 (IR× IR+) is the space of all continuously differentiable functions having compact support. Apply

integration by part and use the fact that φ is zero outside the support to obtain

(3.1)

∫IR+×IR

(uφt + f(u)φx

)dx dt+

∫IRu0φ(x, 0) dx = 0 ∀φ ∈ C1

0 (IR× IR+).

We are now ready to define weak formulation.

Definition 3.1. A function u ∈ L∞(IR× IR+) is said to a weak solution of (??)-(??) if it satisfies

(3.2)

∫IR+×IR

(uφt + f(u)φx

)dx dt+

∫IRu0φ(x, 0) dx = 0 ∀φ ∈ C1

0 (IR× IR+).

Indeed, every weak solution u, which is in C10 (IR× IR+) is a classical solution of (2.6)- (2.7).

If u is discontinuous across a curve, say x(t), then the following Rankine-Hugnoit (R-H) condition

(3.3) f(u(x+(t), t))− f(u(x−(t), t)) = s(u(x+(t), t)− u(x−(t), t)

),

where s = ddtx(t) is speed of the discontinuity.

Unfortunately, weak solution may not be unique. Therefore, based on physical intuition variousentropy conditions imposed to single out physically more relevant solution and this may be discussedextensively in some lectures in this workshop. Hence, I refrain from describing it here in detail. But todescribe briefly, we need to impose an entropy condition. Let U(u) be a convex entropy function and F (u)be an associated entropy flux given by

F ′(u) = U ′(u)f ′(u).

In the region where u is smooth, then the following additional conservation law is satisfied:

(3.4) U ′(u)ut + U ′(u)f ′(u)ux = 0.

Essentially, we multiply (2.6) by U’(u) to obtain the above equation (3.4). Then this can be rewritten as

(3.5)∂U

∂t+∂F

∂x= 0.

The region where u is not smooth, the following entropy inequality is satisfied

(3.6)∂U

∂t+∂F

∂x≤ 0.

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4 DGM and Error Bounds

This Section focuses on DGM for the following conservation law: find u = u(x, t) such that

ut + f(u)x = 0, (x, t) ∈ (0, 1)× (0, T )(4.1)

with initial condition

u(x, 0) = u0(x), x ∈ (0, 1)(4.2)

and periodic boundary conditions. All our results are valid for x ∈ IR or any other boundary conditions.In this section, we discuss the Discontinuous Galerkin methods for the following scalar conservation law:Find u = u(x, t) such that

(4.3) ut + f(u)x = 0 x ∈ (0, 1), t ∈ (0, T )

with

(4.4) u(x, 0) = u0(x), x ∈ (0, 1)

and periodic boundary conditions. Note that, we have considered periodic boundary conditions for sim-plicity.For setting up the Galerkin approximation, we need weak formulation related to (4.3)-(4.4). Let ΠN ={xj+ 1

2}Nj=0 be a partition of (0, 1) and let Ij = (xj− 1

2, xj+ 1

2) with hj = xj+ 1

2− xj− 1

2, j = 1, · · · , N. Let

h = max1≤j≤N

hj . For weak formulation of (4.3)-(4.4), multiply (4.3) by a smooth function v and integrate

over Ij . After using integration by parts with respect to x, we arrive at for each j = 1, · · · , N

(4.5)

∫Ij

ut(x, t)v(x) dx−∫Ij

f(u(x, t))vx(x) dx+ f(u(xj+ 12, t))v(x−

j+ 12

)− f(u(xj− 12, t))v(x+

j− 12

) = 0

with

(4.6)

∫Ij

u(x, 0)v(x) dx =

∫Ij

u0(x)v(x) dx.

4.1 Discontinuous Galerkin Approximation.

Let Vh =

{vh ∈ L1(0, 1) : vh

∣∣∣Ij∈ P k(Ij), j = 1, · · · , N

}, where P k(I) denotes the space of polynomials

of degree ≤ k on I. From approximation theory, the following approximation property holds: For φ ∈Hk+1(Ij), there exists a positive constant Ck such that

(4.7) infχ∈Pk(Ij)

{‖φ− χ‖+ hj‖φx − χx‖} ≤ Ck|φ|Hk+1(Ij).

Further, for vh ∈ P k(Ij), the following inverse inequality holds:

(4.8) ‖vhx‖Ij ≤ Ch−1j ‖vh‖Ij , ‖vh‖L∞(Ij) ≤ Ch−1/2‖vh‖Ij .

As a consequence, there holds:

(4.9) ‖vh‖∂Ij :=(

(vh(x+j− 1

2

)2 + (vh(x−j+ 1

2

)2 ≤ Ch−1/2‖vh‖Ij .

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The DG approximation uh of u is to find uh(t) ∈ Vh such that∫Ij

uht(x, t)v(x)dx−∫Ij

f(uh(x, t))vhx(x)dx

+ f(uh)j+ 12(t)vh(x−

j+ 12

)− f(uh)j− 12(t))vh(x+

j− 12

) = 0 ∀vh ∈ P k(Ij)(4.10)

with

(4.11)

∫Ij

uh(x, 0)vh(x)dx =

∫Ij

u0(x)vh(x)dx. vh ∈ P k(Ij).

where the numerical flux f is defined as

(4.12) f(u)j+ 12(t) = f(u(x−

j+ 12

, t), u(x+j+ 1

2

, t)).

The numerical flux should be chosen suitably and that we shall discuss later. Note that the DG methodsgiven by (4.10)-(4.11) are conservative. Choose vh = 1 in (4.10) to obtain

(4.13)d

dt

∫Ij

uh(x, t) dx+ f(u)j+ 12(t)− f(u)j− 1

2(t) = 0.

Since we shall be looking forward to define monotone schemes or their variants. We now specify thatthe numerical flux f(a, b) satisfies the following properties, see [1]:

• locally Lipschitz condition.

• should be consistent with the flux f(u), that is, f(u, u) = f(u)

• should be a nondecreasing function of its first argument and a non increasing function of its secondargument.

Few examples of numerical fluxes satisfying the above properties are given as:

1. The Godunov flux:

fG(a, b) =

mina≤u≤b

f(u) if a ≤ b,

maxa≤u≤b

f(u) oherwise.

2. Engquist-Oscher flux:

fEO(a, b) =

∫ b

0min(f ′(s), 0) ds+

∫ a

0max(f ′(s), 0) ds+ f(0).

3. Lax-Friedrichs flux:

fLF (a, b) =1

2

[f(a) + f(b)− c(b− a)

]where

c = maxinf u0(x)≤s≤supu0(x)

|f ′(s)|.

4. Local Lax-Friedrichs flux

fLLF (a, b) =1

2

[f(a) + f(b)− c(b− a)

],

wherec = max

min(a,b)≤s≤max(a,b)|f ′(s)|.

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5. Roe flux with fix entropy

fR(a, b) =

f(a) if f ′(u) ≥ 0, x ∈ [min(a, b),max(a, b)],f(b) if f ′(u) ≤ 0, x ∈ [min(a, b),max(a, b)],fLLF , oherwise.

Remark 4.1. The Godunov flux fG produces smallest amount of artificial viscosity, while Local Lax-Friedrichs fLLF flux produces more artificial viscosity than fG, but their performance are quite similar innature. For complicated f, one may use the Lax-Friedrichs flux. It is observed that as we keep increasingthe degree k of the approximation, the choice of numerical flux does not have a significant influence on thequality of the approximations.

Remark 4.2. Note that for piecewise constant polynomials, that is, k = 0, we write (4.10)-(4.11) asfollows: For x ∈ Ij, set uh(x, t) = uj and for j = 1, · · · , N with (4.10)-(4.11) as

(4.14) ujt(t) + {f(uj(t), uj+1(t))− f(uj−1(t), uj(t))}/hj = 0

with

(4.15) uj(0) =1

hj

∫Ij

u0(x) dx.

Remark 4.3. Using Legendre polynomials Pl as local basis functions and exploiting the properties:

1. (Orthogonality Property).∫ 1

−1Pl(s)Pk(s)ds =

2

2l + 1δlk where δlk =

{1 l = k0 l 6= k

2. Pl(1) = 1, Pl(−1) = (−1)l

we obtain substituting uh(x, t) =∑k

l=0 uljφl(x), where φl(x) = Pl(2(x− xj)/∆xj) in (4.10)-(4.11) a system

of ODE’s

d

dtuh = Lh(uh) in (0, T )

uh(0) = u0h.(4.16)

Then, one can use ODE solver to compute the solution.

4.2 Cell Entropy inequality

For the monotone numerical flux, the following cell entropy condition holds.

Theorem 4.1. The semidiscrete solution uh of (4.10)-(4.11) satisfies

(4.17)d

dt

∫Ij

U(uh) dx+ Fj+1/2(t)− Fj−1/2(t) ≤ 0

for the entropy function U(uh) = 12u

2h with some consistent numerical entropy flux

Fj+1/2 = F (uh(x−j+1/2, t), uh(x+j−1/2, t))

given byFj+1/2 = −F (uh(x−j+1/2, t)) + fj+1/2uh(x−j+1/2, t),

where F (u) :=∫ u

0 f(s) ds and F ′(u) = f(u).

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Proof. Set vh = uh in (4.10) to find that∫Ij

uhtuh dx−∫Ij

f(uh)uhx dx+ fj+1/2(t)uh(x−j+1/2)− fj−1/2(t)uh(x+j−1/2) = 0,

and hence, we now rewrite it as∫Ij

dU

dt(uh) dx − F (uh(x−j+1/2, t)) + fj+1/2(t)uh(x−j+1/2)

+ F (uh(x+j−1/2, t))− fj−1/2(t)uh(x+

j−1/2) = 0.(4.18)

With

Θj−1/2 := −F (uh(x−j−1/2, t)) + fj−1/2(t)uh(x−j−1/2) + F (uh(x+j−1/2, t))− fj−1/2(t)uh(x+

j−1/2),

the equation (4.18) becomes

d

dt

∫Ij

U(uh) dx+ Fj+1/2 − Fj−1/2 + Θj−1/2 = 0.(4.19)

Note that using mean value property, it follows that for some u∗h in between u−h and u+h

Θ = −F (u−h ) + fu−h + F (u+h )− fu+

h

= (u+h − u

−h )(F ′(u∗h)− f)

= (u+h − u

−h )(f(u∗h)− f) ≥ 0,

where inequality comes from the monotone property of the numerical flux f . Hence, it completes the restof the proof.

Note that because of cell entropy,the semidiscrete approximation will converge to a physically relevantunique solution.

4.3 Convergence Analysis in Linear case

When f(u) = cu, all the numerical fluxes discussed above coincide and we write

f(a, b) = ca+ b

2− |c|

2(b− a).

For detailed convergence analysis, see [1] and references, therein.Below we discuss L2(0, 1) stability for the semidiscrete scheme (4.16) or (4.10)-(4.11).

Theorem 4.2. Let uh be a solution of (4.10)-(4.11). Then the scheme is stable with respect to L∞(L2)−norm in the following sense:

(4.20)1

2‖uh(T )‖2L2(0,1) + ΘT (uh) ≤ 1

2‖u0‖2L2(0,1)

where

(4.21) ΘT (uh) =|c|2

∫ T

0

N∑j=1

[uh(t)]2j+ 1

2

dt

and[uh(t)]j+ 1

2= uh(x+

j+ 12

, t)− uh(x−j+ 1

2

, t)

denotes the jump across xj+ 12.

11

Proof. Summing up (4.10) in j and then integrating (4.10) with f(u) = cu in time from 0 to T , we rewriteit as

(4.22) Ah(uh, vh) = 0 ∀vh ∈ Vh t ∈ (0, T ),

where

Ah(uh, vh) =

∫ T

0

∫ 1

0

∂uh(x, t)

∂tvh(x, t)dxdt

−∫ T

0

N∑j=1

∫Ij

f(uh(x, t))∂vh(x, t)

∂xdxdt

−∫ T

0

N∑j=1

f(uh)j+ 12(t)[vh(t)]j+ 1

2dt(4.23)

with f(uh) = cuh, set vh = uh to obtain

0 = Ah(uh, uh) =

∫ T

0

d

dt‖uh(t)‖2L2(0,1) dt−

∫ T

0

N∑j=1

∫Ij

cuh∂uh∂x

dx dt

−∫ T

0

N∑j=1

f(uh)j+ 12(t)[uh(t)]j+ 1

2dt.(4.24)

Since

−N∑j=1

∫Ij

cuh∂uh∂x

dx =c

2

N∑j=1

[u2h]j+ 1

2= c

N∑j=1

{uh[uh]}j+ 12,

where

uh(xj+ 12) =

1

2(uh(x−

j+ 12

) + uh(x+j+ 1

2

)),

and using the definition of f

−N∑j=1

f(uh)j+ 12[uh]j+ 1

2= −

N∑j=1

(cuh[uh]− |c|

2[uh]2

)j+ 1

2

,

we obtain from (4.24)

1

2‖uh(T )‖2 +

∫ T

0

N∑j=1

( |c|2

[uh]2)j+ 1

2

dt =1

2‖u0‖2L2(0,1).

This completes the rest of the proof.

Remark 4.4. From stability theorem, we observe that uh is controlled by the L2 norm of the initialcondition and this reflects the in-built dissipation mechanicsm of the DG method. Therefore, it is expectedthat the DG method is more accurate than the corresponding the standard Galerkins methods. Below, wediscuss the convergence analysis of the DG method.

First of all, we present a general prescription on the error analysis. Very often, a direct comparisionbetween the exact solution u and DG solution uh does not yield optimal order of convergence. Hence,there is a need of introducing an intermediate projection or auxiliary function in the discrete space, whichdepends on the exact solution, satisfying the following properties:

12

• it is easy derive from the knowledge of u and its comparision with u gives optimal estimates.

• its comparision with the DG solution uh yields improved error estimates.

Then the whole process shows optimal order of convergence for the error u− uh. Given such intermediateprojection, say uh ∈ Vh, split the error e := u− uh as

e := (u− uh) + (uh − uh) =: η + ξ.

Note that for vh(t) ∈ Vh, t ∈ (0, T )Ah(u, vh) = 0

This is called consistency condition. Hence, for vh(t) ∈ Vh, t ∈ (0, T ), we arrive at

(4.25) Ah(u− uh, vh) = 0.

Now with the help of uh, rewrite (4.25) as

(4.26) Ah(ξ, vh) = −Ah(η, vh).

Since

f(φ)j+1/2 = f(φ−j+1/2, φ+j+1/2) = cφj+1/2 −

|c|2

[φ]j+1/2,

where φ and [φ] denote the average value and jump, respectively, we write (4.26) as∫ T

0

N∑j=1

∫IJ

ξtvh dx dt−∫ T

0

N∑j=1

∫IJ

cξvhx dx dt−∫ T

0

N∑j=1

(cξj+1/2 −

|c|2

[ξ]j+1/2

)[vh]j+1/2 dt(4.27)

=

∫ T

0

N∑j=1

∫IJ

ηtvh dx dt−∫ T

0

N∑j=1

∫IJ

cηvhx dx dt−∫ T

0

N∑j=1

(cηj+1/2 −

|c|2

[η]j+1/2

)[vh]j+1/2 dt.

In subsequent error analysis, we choose uh appropriately. Note that due to presence of the derivative of vhin second term on the right hand side of (4.27), it may reduce the order of convergence. Therefore, in thenext theorem we choose uh appropriately to kill this term.

Theorem 4.3. (A priori L2 estimate) Assume that u0 ∈ Hk+1(0, 1). Then there exists a positive constantC independent of k, may depend on |c| and T such that

(4.28) ‖u(T )− uh(T )‖L2(0,T ) ≤ C hk+ 12 |u0|Hk+1(0,1).

Proof. We now introduce the following L2- projection. For w ∈ L2(0, 1), let Phw ∈ Vh defined by

(4.29)

∫ 1

0(Phw − w)vhdx = 0 ∀vh ∈ Vh

Note that for w ∈ Hk+1(0, 1), the following approximation property holds:

(4.30) ‖Phw − w‖L2(0,1) ≤ Chk+1‖w‖Hk+1(0,1).

Using Ph with uh = Phu, we now rewrite the error e := u− uh as

e := (u− Phu) + (Phu− uh) =: η + ξ.

13

Now using consistency condition (4.25) and vh = ξ, we obtain

(4.31) Ah(ξ, ξ) = −Ah(η, ξ).

For the term on the left hand side of (4.31), a use of the proof of Theorem 4.2 yields

(4.32) Ah(ξ, ξ) =1

2‖ξ(T )‖2L2(),1) + ΘT (ξ)− 1

2‖ξ(0)‖2L2(0,1).

With uh(0) = Phu0, it is observed that the last term of (4.32) vanishes as ξ(0) = 0. Substituting (4.32) in(4.31), we arrive at

(4.33)1

2‖ξ(T )‖2L2(0,1) + ΘT (ξ) = −Ah(η, ξ).

A use of the triangle inequality now shows

1

2‖e(T )‖2 + ΘT (ξ) ≤1

2‖ξ(T )‖2L2(0,1) + ΘT (ξ) +

1

2‖η(T )‖2 + ΘT (η)

=1

2‖η(T )‖2 + ΘT (η)−Ah(η, ξ).(4.34)

Thus to complete the estimate, we need to estimate the last two terms on the right hand side of (4.34).

Lemma 4.1. There is a positive constant C > 0 such that

(4.35) Ah(η, ξ) ≤ Chk+1T |u0|2Hk+1(0,1) +1

2ΘT (ξ).

Before proving Lemma 4.1, we need a couple of propositions.

Proposition 4.1. For w(t) ∈ L2(0, 1) and vh(t) ∈ Vh,there holds

(4.36) Ah(w − Phw, vh) = −∫ T

0

N∑j=1

f(w − Phw)j+ 12(t)[vh(t)]j+ 1

2dt.

Proof. From the definition of Ah(·, ·) in (4.23)

Ah(w − Phw, vh) =

∫ T

0

∫ 1

0(w − Phw)tvh(x, t) dx dt−

∫ T

0

N∑j=1

∫Ij

c(w − Phw)∂xvh dx dt

−∫ T

0

N∑j=1

f(w − Phw)j+ 12[vh(t)]j+ 1

2dt.(4.37)

Since c is a constant, by definition of L2 projection the first and second term on the right hand side of(4.37) becomes zero. Hence, it completes the rest of the proof.

Proposition 4.2. Let w ∈ Hk+1(Ij ∪ Ij+1). then, there is a positive constant Ck > 0 such that

|f(w − Phw)x(j +1

2)| ≤ Ckhk+ 1

2|c|2|w|Hk+1(Ij∪Ij+1).

14

Proof. Using the definition of the numerical flux f , note that

|f(w − Phw)x(j +1

2)| =|e

2

((Phwj+ 1

2−)+ + (Phw)+

j+ 12

)− |e|

2(Phw)+

j+ 12

− (Phw)−j+ 1

2

)− ew|

=|e− |e|2

(Phw+ − w)j+ 1

2+e+ |e|

2(Phw

− − w)j+ 12|

≤ |e|max{|(Phw+ − w)j+ 12|, |(Phw− − w)j+ 1

2|}

Using the inverse property (4.9),

|(Phw+ − w)j+ 12| ≤ Ch−

12 ‖Phw − w‖L2(Ij+1)

and|(Phw− − w)j+ 1

2| ≤ Ch−

12 ‖Phw − w‖L2(Ij)

along with approximation property of Ph, we complete the rest of the proof.

Proof of Lemma 4.1 Using proposition 4.1 with w = u and vh = ξ, it now follows from the definition ofAh(., .)

Ah(η, ξ) = −∫ T

0

N∑j=1

f(η)j+ 12[ξ]j+ 1

2dt.

An application of Young’s inequality yields

Ah(η, ξ) ≤∫ T

0

N∑j=1

1

|c||f(η)j+ 1

2(t)|2 dt+

∫ T

0

N∑j=1

|c|4

[ξ]2j+ 1

2

dt(4.38)

Use proposition 4.2 and definition of ΘT in (4.38) to obtain

Ah(η, ξ) ≤C2kh

2k+1 |c|4

∫ T

0

N∑j=1

|u|2Hk+1(Ij∪Ij+1) +1

2ΘT (ξ)

≤ C2k h

2k+1 |c|2T‖u‖2L∞(Hk+1(I)) +

1

2ΘT (ξ).

Using of regularity result|u(t)|Hk+1(I) ≤ |u0|Hk+1(I),

we complete the rest of the proof.Proof of theorem 4.3 In (4.34), substitute (4.35) to find that

‖ξ(T )‖2 + ΘT (ξ) ≤ Ch2k+1 |c|2T |u0|2Hk+1(0,1).

Now using triangle inequality , we complete the rest of the proof.

Remark 4.5. In the standard Galerkin approximation the order of convergence in L2- norm is k, where asin DGM , it is k + 1

2 for the smoothness on the initial data. However, if u0 ∈ Hk+2, then ‖(u− uh)(t)‖ =O(hk+1), which is proved below.

Theorem 4.4. Let u0 ∈ Hk+2(I). Then for T > 0

‖(u− uh)(t)‖L2(I) ≤ Chk+1|u0|Hk+2(0,1).

15

Note that the order of convergence in L2- norm is of order k+ 12 and the loss of 1

2 is due to proposition4.2 for establishing the numerical flux at the nodal points. One of the strategy is to choose appropriateauxiliary projection rather than L2-projection so that its comparison with DG solution yields an improvedresult.One such choice of the intermediate projection uh is to define Rh : L∞(0, 1) −→ Vh such that for w ∈L∞(0, T ) and for j = 1, · · · , N

(4.39) Rh(w)(xj,l)− w((xj,l)) = 0 l = 0, · · · , k

where xj,l’s are the Gauss-Randau quadrature points in Ij . Set

xj,k =

{xj+ 1

2if c < 0

xj− 12

if c > 0

Note that Rh satisfied for w ∈ P 2k−l(Ij)

(4.40)

∫Ij

(Rhw − w)(x)φ(x)dx = 0 ∀φ ∈ P l(Ij), l ≤ k

For l = k, Rh is the L2− projection.Below, we discuss the following approximation property for Rh.

Lemma 4.2. Let w ∈ Hk+2(Ij) and vh ∈ P k(Ij); Then

(4.41) |∫Ij

(Rhw − w)(x)vh(x)dx| ≤ Ckhk+1|w|HK+1(Ij)‖vh‖L2(Ij)

and

(4.42) |∫Ij

(Rhw − w)(x)∂xvh(x)dx| ≤ Ckhk+1|w|HK+2(Ij)‖vh‖L2(Ij)

Proof. For the first inequality (4.41), use (4.40) with l = k and then employ the approximation propertyof Rh to complete the estimate.for (4.42), apply (4.40) with l = k − 1, then use scaling argument to complete the proof.

Proof of theorem 4.4. Using Rh, split

e := u− uh = (u−Rhu) + (Rhu− uh) =: η + ξ.

Now

Ah(ξ, ξ) =1

2‖ξ(T )‖2L2(0,1) + ΘT (ξ)− 1

2‖ξ(0)‖2L2(0,1).

Observe that using consistency condition

Ah(ξ, ξ) = Ah(e− η, ξ) = −Ah(η, ξ),

and hence,

(4.43)1

2‖ξ(T )‖2L2(0,1) + ΘT (ξ) =

1

2‖ξ(0)‖2L2(0,1) −Ah(η, ξ).

16

To estimate the last term in (4.43), we observe that

Ah(η, ξ) =

∫ T

0

N∑j=1

∂tη(x, t)ξ(x, t) dx dt

−∫ T

0

N∑j=1

∫Ij

cη(x, t)∂xξ(x, t) dx dt

−∫ T

0

N∑j=1

f(η)j+ 12(t)[ξ(t)]j+ 1

2dt.(4.44)

For the last term of (4.44), using definition of the numerical flux, it follows that

f(η)j+ 12(t) =

c

2(Rhu

+ +Rhu−)− |c|

2(Rhu

+ −Rhu−)− cu

c− |c|2

(Rhu+ − u) +

c+ |c|2

(Rhu− − u)

= 0,

and therefore,

Ah(η, ξ) =

∫ T

0

N∑j=1

∂tηξ dx dt−∫ T

0

N∑j=1

∫Ij

eη∂xξ dx dt.

A use of Lemma 4.2 shows that

|Ah(η, ξ)| ≤ Ckhk+1|u0|Hk+2(I)

∫ T

0‖ξ‖L2(0,1) dt.

On substitution in (4.43) yields

‖ξ(T )‖2L2(0,1) + ΘT (ξ) ≤ ‖ξ(0)‖2 + Ckhk+1|u0|Hk+2(I)

∫ T

0‖ξ‖t2 ds

≤ ‖ξ(0)‖2 + CkT h2(k+1)‖u0‖2Hk+2(I) + Ck

∫ T

0‖ξ‖2L2(0,1) ds.

An application of Gronwall’s inequality 3 with triangle inequality yields the desired result.

4.4 Superconvergence Results

Subsequently efforts are on superconvergence results. To be more specific, let us present briefly the resultof Cheng and Shu [3].

Consider the linear conservation law with c = 1, that is,

ut + ux = 0 0 < x < 1, t > 0(4.45)

u(x, 0) = u0(x) 0 < x < 1(4.46)

3 If a nonnegative function φ satisfies

φ(T ) ≤ C1 + C2

∫ T

0

φ(t) dt,

then φ(T ) ≤ C1eC2T .

17

with both the periodic boundary condition u(0, t) = u(1, t) and the initial-boundary values problemu(0, t) = g(t). With Vh as above, that is,

Vh =

{vh ∈ L2(0, 1) : vh

∣∣∣Ij∈ P k(Ij), j = 1, · · · , N

},

the DG scheme with upwind flux is to seek uh(t) ∈ Vh such that∫Ij

uhtvhdx−∫Ij

uhvhxdx+ u−h v−h |j+ 1

2− u+

h v+h |j− 1

2= 0 ∀ vh ∈ Vh.(4.47)

For improving the estimates, we need to define appropriate intermediate projection of u onto Vh for k ≥ 1as

(4.48)

∫Ij

P−h uvhdx =

∫Ij

uvhdx ∀ vh ∈ P k−1(Ij)

with

(4.49) (P−h u)− = u− at xj+ 12.

The projection P−h u is well defined for a given u. Below, we recall some of its properties.

Lemma 4.3. There holds for z ∈ Hk+1(Ij)

(4.50) ‖P−h z − z‖L2(Ij) ≤ Ckhk+1|z|Hk+1(Ij)

Since u satisfies for uh ∈ Vh

(4.51)

∫Ij

utvhdx−∫Ij

uvhxdx+ u−v−h |j+ 12− u+v+

h |j− 12

= 0

Then from (4.47) and (4.51) with e = u− uh, we find that

(4.52)

∫Ij

etvhdx−∫Ij

evhxdx+ e−v−h |j+ 12− e+v+

h |j− 12

= 0 ∀ vh ∈ Vh.

Using projection P−h with uh = P−h u, we now split the error e as

e := (u− P−h u) + (P−h u− uh) =: η + ξ.

Since using Projection P−h e−j+ 1

2

= η−j+ 1

2

+ ξ−j+ 1

2

= ξ−j+ 1

2

and∫Ij

ηvhxdx = 0

then from (4.52), it follows that

(4.53)

∫Ij

ξtvhdx−∫Ij

ξvhxdx+ ξ−v−h |j+ 12− ξ+v+

h |j− 12

= −∫Ij

ηtvhdx ∀ vh ∈ Vh

Integrate by parts of the second term of (4.53) to find that

(4.54)

∫Ij

ξtvhdx+

∫Ij

ξxvhdx+ [ξ]v+h |j− 1

2= −

∫Ij

ηtvh dx,

where the jump [ξ] = ξ+ − ξ−.Below, we state the main theorem of this section.

18

Theorem 4.5. Let u and uh be solution of (4.45)-(4.46) and (4.47), respectively. Further let uh(0) =P−h u0. Then, there holds

(4.55) ‖ξ(t)‖L2(0,1) = ‖(P−h u− uh)(t)‖L2(0,1) ≤ C1(t+ 1)hk+ 32 ,

and

(4.56) ‖e(t)‖L2(0,1) ≤ C1 t hk+ 3

2 + C2 hk+1,

where C1 = C1(‖u‖k+3) and C2 = C2(‖u‖k+1).

Before, we provide a proof of theorem 4.5, we state and prove some useful Lemmas.

Lemma 4.4. For φ ∈ C1 on Ij, let B−j (φ) and B+j (φ) be two functional given by

B−j (φ) =

∫Ij

φ(x)(x− xj− 1

2

hj)d

dx

(φ(x)(

x− xjh

))dx

and

B+j (φ) =

∫Ij

φ(x)(x− xj+ 1

2

hj)d

dx

(φ(x)(

x− xjh

))dx,

respectively. Then

(4.57) B−j (φ) =1

4hj

∫Ij

φ2(x)dx+1

4φ2(xj+ 1

2),

and

(4.58) B+j (φ) = − 1

4hj

∫Ij

φ2(x)dx− 1

4φ2(xj− 1

2).

Proof. It is sufficient to prove (4.57). From the definition of B−j (φ),

B−j (φ) =

∫Ij

φ(x)(x− xj− 1

2

hj)(φ′(x)(

x− xjh

) + φ(x)1

hj

)dx

=

∫Ij

(φφ′(x)

(x− xj)h2j

(x− xj− 12))dx+

∫Ij

φ2(x)(x− xj− 1

2

h2j

)dx

=

∫Ij

d

dx((φ)2

2)(x− xj)hj

(x− xj− 12)dx+

∫Ij

φ2(x)(x− xj− 1

2

h2j

)dx

Now integrate by parts yields

B−j (φ) =1

4φ2(xj+ 1

2)− 1

2

∫Ij

φ2(x)(2x− xj − xj− 1

2

h2j

)dx+

∫Ij

φ2(x)(x− xj− 1

2

h2j

) dx

=1

4hj

∫Ij

φ2(x) dx+1

4φ2(xj+ 1

2).

Hence, it completes the rest of the proof.

19

Lemma 4.5. Let wj be defined by

(4.59) wj(x) = (ξ − ξ(xj))hj

(x− xj), x ∈ Ij .

Then

(4.60) ‖w‖L2(0,1) =( N∑j=1

∫Ij

(wj(x))2dx)2≤ 4h‖et‖L2(0,1)

Proof. Note that wj ∈ P k−1(Ij). From the error equation (4.54)

(4.61)

∫Ij

etvhdx+

∫Ij

ξxvhdx+ [ξ]v+h |j− 1

2= 0

In case v+h (xj− 1

2) = 0 then (4.61) boils down to

(4.62)

∫Ij

etvhdx+

∫Ij

ξxvhdx = 0

Set vh = wj(x)(x−x

j− 12

)

hjin (4.62), then v+

h (xj− 12) = 0. Now using the definition of wj , note that

ξ = ξ(xj) + wj(x)(x− xj)hj

on Ij

and plugging in (4.62), it follows that∫Ij

etwj(x)(x− xj− 1

2)

hjdx+

∫Ij

(wj(x)

(x− xj)hj

)xwj(x)

(x− xj− 12)

hjdx = 0

In terms of B−j , we rewrite it as ∫Ij

etwj(x)(x− xj− 1

2)

hjdx+Bj−(wj) = 0

A use of Lemma 4.4 yields

(4.63)1

4hj

∫Ij

wj2(x)dx+

1

4wj

2(xj+ 12) = −

∫Ij

etwj(x)(x− xj− 1

2)

hjdx

Since the second term is nonnegative, then∫Ij

wj2(x) dx ≤− 4

∫Ij

etwj(x)(x− xj− 12)dx

≤4hj‖et‖L2(Ij)‖wj‖L2(Ij) ≤ 4h‖et‖L2(Ij)‖wj‖L2(Ij),

and hence, summing up from j = 1 to N , we arrive at

‖w‖L2(0,1) =( N∑j=1

∫Ij

(wj(x))2dx)2≤ 4h‖et‖L2(0,1).

This completes the rest of the proof.

20

We observe that et := ηt + ξt. Therefore to estimate ‖et‖L2(0,1), it is enough to estimate ξt. Nowdifferentiate (4.53) with respect to t to arrive at∫

Ij

ξttvhdx−∫Ij

ξtvhxdx+ ξ−t v−h |j+ 1

2− ξ−t v

+h |j+ 1

2= −

∫Ij

ηttvhdx

Setting vh = ξt and summing up over j, we find that∫Iξttξtdx−

∫Iξtξtxdx+

1

2

N∑j=1

[ξt]2j+ 1

2

= −∫Iηttξtdx

Here, the second term is zero using periodic boundary condition and third term is non-negative. Hence

1

2

d

dt(

∫I|ξt|2dx) ≤ (

∫I|ηtt|2dx)

12 (

∫I|ξt|2dx)

12

and

(4.64) ‖ξt‖L2(0,1) ≤ ‖ηtt‖L2(0,1) + ‖ξt(0)‖

Note that at t = 0, ξ(0) = 0 since uh(0) = P−h u0.Therefore from the error equation (4.53)∫

Ij

etvhdx = 0, that is,

N∑j=1

∫Ij

ξt(0)vhdx = −N∑j=1

∫Ij

ηt(0)vhdx.

Set vh = ξt(0) to arrive at

(4.65) ‖ξt(0)‖L2(I) ≤ ‖ηt(0)‖L2(I) ≤ Chk+1|ut(0)|Hk+1(I) ≤ Chk+1‖ut(0)‖Hk+2(I).

Using (4.65) in (4.64) and using approximation property (4.50) of P−h , we obtain

‖ξt(t)‖L2(0,1) ≤ Chk+1(‖utt‖H2k+1(I) + ‖u0‖H2k+2(I)) ≤ Chk+1‖u0‖H2k+3(I)

Lemma 4.6. If ‖u0‖ ∈ H2k + 3(I), then

(4.66) ‖(u− uht)(t)‖ = ‖et(t)‖ ≤ Chk+1‖u0‖H2k+3(I)

Moreover,

(4.67) ‖w(t)‖L(0,1) ≤ chk+1‖u0‖Hk+3(I).

Proof of theorem 4.5 Choose vh = ξ in (4.54) and obtain

(4.68)

∫Ij

ξtξdx−∫Ij

ξxξdx+ ξ−ξ−|j+ 12− ξ−ξ+|j− 1

2= −

∫Ij

ηtξdx

For the second term

−∫Ij

ξxξdx = −1

2

∫ xj+1

2

xj− 1

2

(ξ2)xdx = −1

2

((ξ2)−

j+ 12

− (ξ2)+j− 1

2

)

21

On substitution in (4.68), we find that∫Ij

ξtξdx+ (ξ−)2|j+ 12

+1

2(ξ2)+

j− 12

− ξ−ξ+|j− 12

= −∫Ij

ηtξdx

and ∫Ij

ξtξdx+ Fj+ 12− Fj− 1

2+

1

2[ξ]2

j+ 12

= −∫Ij

ηtξdx

whereFj+ 1

2= −(ξ+

j+ 12

)2 + ξ−j+ 1

2

ξ+j+ 1

2

Summing up from j = 1 to j = N , we arrive at

1

2

d

dt

∫I|ξ(x, t)|2dx+

1

2

N∑j=1

[ξ]2j+ 1

2

= −∫Iηtξdx

Since ξ−12

= 0, the second term is nonnegative and hence

(4.69)d

dt‖ξ(t)‖2L2(0,1) ≤ 2|

∫Iηtξdx|

To estimate the term on the right hand side, we note that∫Iηtξdx =

N∑j=1

∫Ij

ηt

(ξ(xj) + wj(x)

(x− xj)h

)dx =

N∑j=1

∫Ij

ηtwj(x)(x− xj)

hdx

Therefore using (4.50), (4.60) and (4.66) we obtain

|∫Iηtξdx| ≤ ‖ηt‖L2(I)‖w‖L2(I) ≤ Chk+1‖u‖Hk+2hk+2(t+ 1)

≤ Ch2k+3(t+ 1)‖u‖2Hk+3(I)(4.70)

Substitute (4.70) in (4.69) and then integrate from 0 to t we find with ξ(0) = 0 that

‖ξ(t)‖2 ≤ Ch2k+3(t+ 1)

∫ t

0‖u‖2Hk+3(I)ds ≤ C(t+ 1)h2k+3‖u0‖Hk+3(I)

This completes the proof of (4.55). For (4.66) a use of triangle inequality with (4.50) completes the restof the proof.

Remark 4.6. Note that the comparison between the Galerkin approximation uh and the intermediateprojection P−h u yields a superconvergence results, that is

‖(P−h u− uh)(t)‖L2(0,1) = O(hk+ 32 ).

22

5 DGM for Nonlinear scalar Conservation laws.

Consider the following scalar nonlinear conservation laws: Find u = u(x, t), x ∈ (0, 1), t > 0 such that

(5.1) ut + f(u)x = g(x, t), x ∈ (0, 1), t > 0

with initial condition

(5.2) u(x, 0) = u0(x), x ∈ (0, 1)

and periodic boundary condition of period 1, that is, u(x+ 1, t) = u(x, t) ∀x ∈ (0, 1).The result can be extended to compactly supported boundary conditions.This section deals with superconvergence of semidiscrete DG method with an upwind numerical flux, seeMeng et al. [4].

For Galerkin method, let

Vh =

{vh ∈ L2(I) : vh

∣∣∣Ij∈ P k(Ij), j = 1, · · · , N

}Then DG formulation is to seek uh(t) ∈ Vh, t > 0 such that∫

Ij

uhtvh dx−∫Ij

f(uh)vhx dx+ fj+ 12v−h |j+ 1

2− fj− 1

2v+h |j− 1

2+

1

2[ξ]2

j+ 12

= −∫Ij

gvh dx ∀vh ∈ Vh, j = 1, · · · , N,(5.3)

where the numerical flux fj+ 12

is chosen as an upwind flux on each element boundary which includes the

Godunov flux, the Engquist-Osher flux and the Roe flux with an entropy. For our future analysis, weassume that

A1. f satisfies |f ′(u)| ≥ δ uniformly on the interval [m0,M0] with δ > 0.

A2. f ∈ C3 with derivatives bounded on [m0 − 1,M0 + 1].

Note that we can modify the flux f so that |f ′(u) ≥ 12δ| on [m0 − 1,M0 + 1].

Let

H1h(Ih) =

{v ∈ L2(I) : v

∣∣∣Ij∈ Hk(Ij), j = 1, · · · , N

}where Ih = ∪Nj=1Ij with Ij = (xj− 1

2, xj+ 1

2).

For our error analysis, we shall use the following two projections. Let P+−h : H1

h(I) −→ Vh be theprojections defined for φ by

(5.4)

∫Ij

(P+h φ− φ)vh dx = 0 ∀vh ∈ P k−1(Ij), (P+

h φ)+j− 1

2

= φ(x+j− 1

2

),

and

(5.5)

∫Ij

(P−h φ− φ)vhdx = 0 ∀vh ∈ P k−1(Ij), (P−h φ)−j− 1

2

= φ(x−j+ 1

2

).

Setting Qh = Ph or P±h , one can use standard scaling argument to derive the following approximationproperties:

(5.6) ‖φ−Qhφ‖L2(I) + h‖φx −Qhφx‖L2(I) + h12 ‖φ−Qhφ‖Γh

≤ Chk+1|φ|Hk+1(I),

23

where

(5.7) ‖φ‖2Γh=

N∑j=1

((φ+j+ 1

2

)2 + (φ−j+ 1

2

)2).

Moreover,

(5.8) ‖φ−Qhφ‖L∞(I) ≤ Chk+ 12 |φ|Hk+1(I).

Inverse property: For vh ∈ Vh, the following properties hold:

‖vhx‖ ≤ Ch−1‖vh‖, ‖vh‖∞ ≤ Ch−12 ‖vh‖

‖vh‖Γh≤ Ch−

12 ‖vh‖.(5.9)

5.1 Error Analysis.

Let e = u− uh. With the help of Qh, let us split it as

e := (u−Qhu) + (Qh − uh) =: η + ξ.

Here the projection Qh is defined according to the sign variation of f ′(u), that is,

Qh = P−h , if f ′(u(x, t)) > 0 ∀t ∈ [0, T ],

andQh = P+

h , provided f ′(u(x, t)) < 0 ∀t ∈ [0, T ].

Below, we state the main theorem of this section.

Theorem 5.1. Let u and uh be the solution of (5.1)-(5.2) and (5.3), respectively. Assume that u, ut, utt ∈Hk+1(I). Then under the assumptions A1−A2 with uh(0) = Qhu0, there holds for k ≥ 1

‖(Qhu− uh)(t)‖L2(0,1) ≤ Chk+ 32 ∀t ∈ (0, T ].

Moreover for t ∈ (0, T ]‖e(t)‖L2(I) = ‖(u− uh)(t)‖L2(I) ≤ Chk+1.

For more details, see, Meng et al. [4].

References

[1] B. Cockburn, Discontinuous Galerkin methods for convection-dominated problems, Lecture Notes inComputational Science and Engineering, Volume 9 (1999), pp 69-224.

[2] W. Cao, Z. Zhang and Q. Zou, Superconvergence of DGM for linear hyperbolic equations, SIAM J.Numer. Anal. 52 (2014), pp. 2555-2573.

[3] Y. Cheng and C-W. Shu, Superconvergence of DG and LDG schemes for linear hyperbolic andconvection-diffusion equations in one space dimension, SIAM J. Numer. Anal. 47 (2010), pp. 4044-4072.

[4] X.Meng, C-W. Shu, Q. Zhang and B.Wu, Superconvergence of DGM for scalar nonlinear conservationlaws in one space dimension, SIAM J. Numer. Anal. 50 (2012), pp. 2336-2356.

[5] Y. Yang and C-W. Shu, Analysis of optimal superconvergence of DGM for linear hyperbolic equations,SIAM J. Numer. Anal.50 (2012), pp. 3110-3133.

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