Scalar Conservation Laws and Godunov's Scheme · GF NPY_]PQZ]LYLWd^T^...

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Unsteady Problems

Scalar Conservation Laws and Godunov’sScheme

K. Malakpoor

CASACenter for Analysis, Scientific Computing and Applications

Department of Mathematics and Computer Science

7-July-2005


Unsteady Problems

Conservation equations. Mass conservation

ρt +∇ · (ρv) = 0.

Two-dimensional Euler equation

ut + f1(u)x + f2(u)y = 0,

u =

ρ

ρν1ρν2e

, f1(u) =

ρν1

ρν21 + p

ρν1ν2ν1(e + p)

f2(u) =

ρν2

ρν1ν2ρν2

2 + pν2(e + p)

p = (γ − 1)ρε, e = ρε + 1/2ρ(ν2

1 + ν22).


Unsteady Problems

Conservation equations. Mass conservation

ρt +∇ · (ρv) = 0.

Two-dimensional Euler equation

ut + f1(u)x + f2(u)y = 0,

u =

ρ

ρν1ρν2e

, f1(u) =

ρν1

ρν21 + p

ρν1ν2ν1(e + p)

f2(u) =

ρν2

ρν1ν2ρν2

2 + pν2(e + p)

p = (γ − 1)ρε, e = ρε + 1/2ρ(ν2

1 + ν22).


Unsteady Problems

Finite Volume Schemes

ut +∇ · f (u) = 0 in Ω× [0, T ]

Discretization of space

Ci

x

t

1−ix ix 1+ix

1+nt

nt

Ω =⋃

Ωi , Ωj ∩ Ωk = ∅, forj 6= k

boundary ∂Ωj piecewise smooth.


Unsteady Problems

Finite VolumeIntegration over Ωj × [tn, tn+1]

|Ωj |un+1j = |Ωj |un

j −∫ tn+1

tn

∫∂Ωj

f (u(x , t)) · ndSdt

Finite Volume Scheme in One-DimensionIntegrating over [xi−1/2, xi+1/2]× [tn, tn+1]∫ tn+1

tn

∫ xi+1/2

xi−1/2

ut(x , t) +

∫ tn+1

tn

∫ xi+1/2

xi−1/2

f (u(x , t))x dxdt = 0.


Unsteady Problems

Finite VolumeIntegration over Ωj × [tn, tn+1]

|Ωj |un+1j = |Ωj |un

j −∫ tn+1

tn

∫∂Ωj

f (u(x , t)) · ndSdt

Finite Volume Scheme in One-DimensionIntegrating over [xi−1/2, xi+1/2]× [tn, tn+1]∫ tn+1

tn

∫ xi+1/2

xi−1/2

ut(x , t) +

∫ tn+1

tn

∫ xi+1/2

xi−1/2

f (u(x , t))x dxdt = 0.


Unsteady Problems

Finite Volume Scheme in Conservation formWe know that the weak solution satisfies in∫ xi+1/2

xi−1/2

[u(x , tn+1)− u(x , tn+1)] = −∫ tn+1

tn

[f (u(xi+1/2, t))dt − f (u(xi−1/2, t))dt

]

un+1i = un

i −∆t∆x

(gni+1/2 − gn

i−1/2)

with

uni approximates

1∆x

∫ xi+1/2

xi−1/2

u(x , tn)dx ,

gi+1/2 approximates1∆t

∫ tn+1

tnf (u(xi+1/2, t))dt ,


Unsteady Problems

Linear Advection Equation

ut + aux = 0, a ∈ R

Solution:

C : x = x(t) withdx(t)

dt= a, (characteristics)

A function u = u(x(t), t) satisfies

ddt

u(x(t), t) = ut + aux

u Solution ⇒ u = constant along C.


Unsteady Problems

Linear Advection Equation

ut + aux = 0, a ∈ R

Solution:

C : x = x(t) withdx(t)

dt= a, (characteristics)

A function u = u(x(t), t) satisfies

ddt

u(x(t), t) = ut + aux

u Solution ⇒ u = constant along C.


Unsteady Problems

Solution of Initial Value Problem

u(x , 0) = q(x) for all x ∈ R

u(x , t) = q(x − at) for all x , t .


Unsteady Problems

Upwind Scheme

x

t

1−ix ix 1+ix

1+nt

nt

a > 0 :un+1

i − uni

∆t+ a

uni − un

i−1

∆x= 0

stable for:∆t∆x

a < 1 (CFL-Condition)


Unsteady Problems

CIR-MethodCourant, Isaacson, Rees (1946)

un+1i = un

i − a∆t∆x

un

i − uni−1 for a > 0,

uni+1 − un

i for a < 0,

Reformulation

un+1i = un

i − a∆t

2∆x(un

i+1 − uni−1), central difference

+|a|∆t2∆x

(uni+1 − 2un

i + uni−1), dissipation


Unsteady Problems

CIR-MethodCourant, Isaacson, Rees (1946)

un+1i = un

i − a∆t∆x

un

i − uni−1 for a > 0,

uni+1 − un

i for a < 0,

Reformulation

un+1i = un

i − a∆t

2∆x(un

i+1 − uni−1), central difference

+|a|∆t2∆x

(uni+1 − 2un

i + uni−1), dissipation


Unsteady Problems

Reformulation of the CIR-Method, Godunov’s Ideaun piecewise constant

uni (x) = un

i for x ∈ [xi−1/2, xi+1/2].


Unsteady Problems

1.Solve the initial value problemut + aux = 0, u(x , 0) = un(x), for x ∈ Run(x) = un

i for x ∈ [xi−1/2, xi+1/2]

2. Average the exact solution

uni :=

1∆x

∫ xi+1/2

xi−1/2

u(x ,∆t)dx

CFL-condition:∆t∆x

a ≤ 1


Unsteady Problems

1.Solve the initial value problemut + aux = 0, u(x , 0) = un(x), for x ∈ Run(x) = un

i for x ∈ [xi−1/2, xi+1/2]

2. Average the exact solution

uni :=

1∆x

∫ xi+1/2

xi−1/2

u(x ,∆t)dx

CFL-condition:∆t∆x

a ≤ 1


Unsteady Problems

Reformulation of the CIR-MethodIntegration over [xi−1/2, xi+1/2]× [tn, tn+1]∫ tn+1

tn

∫ xi+1/2

xi−1/2

u(x , t) dxdt +

∫ tn+1

tn

∫ xi+1/2

xi−1/2

au(x , t) dxdt = 0

∆xun+1i −∆xun

i = ∆tgni+1/2 −∆tgn

i−1/2

with

uni approximates

1∆x

∫ xi+1/2

xi−1/2

u(x , tn)dx ,


∫ tn+1

tnau(xi+1/2, t)dt ,

numerical flux: gi+1/2 = g(ui , ui+1).


Unsteady Problems

Reformulation of the CIR-MethodIntegration over [xi−1/2, xi+1/2]× [tn, tn+1]∫ tn+1

tn

∫ xi+1/2

xi−1/2

u(x , t) dxdt +

∫ tn+1

tn

∫ xi+1/2

xi−1/2

au(x , t) dxdt = 0

∆xun+1i −∆xun

i = ∆tgni+1/2 −∆tgn

i−1/2

with

uni approximates

1∆x

∫ xi+1/2

xi−1/2

u(x , tn)dx ,


∫ tn+1

tnau(xi+1/2, t)dt ,

numerical flux: gi+1/2 = g(ui , ui+1).


Unsteady Problems

Scalar Conservation Laws

ut + f (u)x = 0,

Quasilinear form:

ut + a(u)ux = 0

Example: Burger’s equation, f (u) =12

u2

ut + uux = 0

Characteristics: C : x = x(t) withdf (u)

du= a(u)

u = constant along C.


Unsteady Problems


ut + f (u)x = 0,

Quasilinear form:

ut + a(u)ux = 0


u2

ut + uux = 0


du= a(u)



Unsteady Problems


ut + f (u)x = 0,

Quasilinear form:

ut + a(u)ux = 0


u2

ut + uux = 0


du= a(u)



Unsteady Problems


ut + f (u)x = 0,

Quasilinear form:

ut + a(u)ux = 0


u2

ut + uux = 0


du= a(u)



Unsteady Problems

Riemann ProblemExample 1:

ut + uux = 0, u(x , 0) =

1, x < 0,

−1, x > 0.

t

1−=ru1=lu

1−=t

x1=

t

x

evawkcohs


Unsteady Problems

Riemann ProblemExample 2:

ut + uux = 0, u(x , 0) =

−1, x < 0,

1, x > 0.

x

t

1=ru1−=lu

evawnoitcaferar


Unsteady Problems

Solution of the Riemann problem

Case f ′′(u) > 0,

ul > ur : Shock wave u(x , t) =

ul , x/t < s,

ur , x/t > s.with

s =f (ur )− f (ul)

ur − ul.

ul < ur : rarefaction wave

u(x , t) =

ul , x/t < a(ul),

ur , x/t > a(ur ),

a−1(x/t), else


Unsteady Problems

Solution of the Riemann problem

Case f ′′(u) > 0,

ul > ur : Shock wave u(x , t) =

ul , x/t < s,

ur , x/t > s.with

s =f (ur )− f (ul)

ur − ul.

ul < ur : rarefaction wave

u(x , t) =

ul , x/t < a(ul),

ur , x/t > a(ur ),

a−1(x/t), else


Unsteady Problems

Godunov’s Method:

un+1i = un

i −∆t∆x

(gni+1/2 − gn

i−1/2)

with the numerical flux

g(ul , ur ) =

f (ul), for ul > ur , s > 0,

f (ur ), for ul > ur , s < 0,

f (ul), for ul < ur , a(ul) > 0,

f (ur ), for ul < ur , a(ur ) < 0,

a−1(0), else.

with ul := ui and ur := ui+1


Unsteady Problems

Higher Order, Generalized Godunov’s Scheme

To compute the numerical flux gi+1/2, one has to solve theRiemann problem ut + f (u)x = 0, t ≥ tn,

u(x , tn) =

un

i + (x − xi)sni , x < xi+1/2,

uni+1 + (x − xi+1)sn

i+1, x > xi+1/2,

where sni and sn

i+1 called slopes, are constants.

x

nu

+)1-i(u-iu

iu-)1i(u +

+iu


Unsteady Problems

Second order accuracy in space:

u(x , tn) = u(xi , tn) + (x − xi)ux(x , tn) +O((x − xi)2)

sni = ux(xi , tn) +O((x − xi)

2)

Taylor expansion in time:

u(x , t + ∆t/2) = u(x , t) +∆t2

ut(x , t) +O(∆t2), ut = −f (u)x

u(x , t +∆t2

) = u(x , t)− ∆t2

f (u(x , t))x +O(∆t2)

⇒ un+1i± = un

i± −∆t∆x

(f (uni+)− f (un

i−))


Unsteady Problems

Second order accuracy in space:

u(x , tn) = u(xi , tn) + (x − xi)ux(x , tn) +O((x − xi)2)

sni = ux(xi , tn) +O((x − xi)

2)

Taylor expansion in time:

u(x , t + ∆t/2) = u(x , t) +∆t2

ut(x , t) +O(∆t2), ut = −f (u)x

u(x , t +∆t2

) = u(x , t)− ∆t2

f (u(x , t))x +O(∆t2)

⇒ un+1i± = un

i± −∆t∆x

(f (uni+)− f (un

i−))


Unsteady Problems

MUSCL-ProcedureBoundary Values at tn

uni± = un

i ±∆x2

sni

tn → tn+1/2

un+1/2i± = un

± −∆t

2∆x(f (un

i+)− f (uni−))

FV-scheme:

un+1i = un

i −∆t∆x

(gn+1/2i+1/2 − gn+1/2

i−1/2 )

with gn+1/2i+1/2 = g(un+1/2

i+ , un+1/2(i+1)−).


Unsteady Problems


uni± = un

i ±∆x2

sni

tn → tn+1/2

un+1/2i± = un

± −∆t

2∆x(f (un

i+)− f (uni−))

FV-scheme:

un+1i = un

i −∆t∆x

(gn+1/2i+1/2 − gn+1/2

i−1/2 )

with gn+1/2i+1/2 = g(un+1/2

i+ , un+1/2(i+1)−).


Unsteady Problems


uni± = un

i ±∆x2

sni

tn → tn+1/2

un+1/2i± = un

± −∆t

2∆x(f (un

i+)− f (uni−))

FV-scheme:

un+1i = un

i −∆t∆x

(gn+1/2i+1/2 − gn+1/2

i−1/2 )

with gn+1/2i+1/2 = g(un+1/2

i+ , un+1/2(i+1)−).


Unsteady Problems

LimitersDefine ∆−un

i = uni − un

i−1, ∆+uni = un

i+1 − uni and

rni = ∆−un

i /∆+uni .

Define a coefficient function Φ(rni ) with

sni ∆x = Φ(rn

i )∆+uni

The coefficient Φ(rni ) is often referred to as the limiter.

A limiter has to satisfy

0 ≤Φ(rn

i )

rni

≤ 2 and 0 ≤ Φ(rni ) ≤ 2,∀i .


Unsteady Problems

Minmod limiter: Φ(r) = min (r , 1).Davis limiter: Φ(r) = min (2r , 1/2(1 + r), 2).

Harmonic limiter: Φ(r) =2r

1 + r

Jameson limiter: Φ(r) = 1/2(1 + r)[1−

∣∣∣∣ 1− r1 + |r |

∣∣∣∣q], q > 0.


Unsteady Problems

Generalized Godunov’s SchemeLet us consider a general transport equation ofconvection-diffusion-reaction type in the divergence form

∂u∂t

+∇ · f (u,∇u) = S(u).

A generalized Godunov scheme has the form:

ZΩα

(un+1 − un) dΩ +

Z tn+1

tndtZ

∂Ωα

f (Et u,∇Et u) · n dΓ =

Z tn+1

tndtZ

ΩαS(u)dΩ.

Ωα denotes an element of the mesh and Et is an approximationevolution operator over a time segment t ∈ [tn, tn+1].


Unsteady Problems

Numerical Scheme for Convection-Diffusion-Reaction Problems1

∂c∂t

+∂

∂x[−D

∂c∂x

+ vc] = S(c), v > 0, D > 0, x ∈ [0, L].

2

cn+1i −∆t/2S(cn+1

i ) = cni −

∆t∆x

[gni+1/2−gn

i−1/2]+∆t/2S(cni ).

wheregn

i+1/2 = vcn+1/2i+1/2 − Dsn

i ,

and sni is again the slope of the local linear recovery.


Unsteady Problems

Numerical Scheme for Convection-Diffusion-Reaction Problems1

∂c∂t

+∂

∂x[−D

∂c∂x

+ vc] = S(c), v > 0, D > 0, x ∈ [0, L].

2

cn+1i −∆t/2S(cn+1

i ) = cni −

∆t∆x

[gni+1/2−gn

i−1/2]+∆t/2S(cni ).

wheregn

i+1/2 = vcn+1/2i+1/2 − Dsn

i ,

and sni is again the slope of the local linear recovery.

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