Burgers` equation Notes - University of Washington

Outline
Notes:
• Scalar nonlinear conservation laws
• Shocks and rarefaction waves
• Entropy conditions
• Finite volume methods
• Approximate Riemann solvers
• Lax-Wendroff Theorem
Reading: Chapter 11, 12
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
Burgers’ equation
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 11.4]
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 11.4]
Notes:
Quasi-linear form: ut + uux = 0
The solution is constant on characteristics so each value
advects at constant speed equal to the value...
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 11.4]
Burgers’ equation
Notes:
Equal-area rule:
The area “under” the curve is conserved with time,
We must insert a shock so the two areas cut off are equal.
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 11.4]
Riemann problem for Burgers’ equation
ut +
1 2
2u x
= 0,
Notes:
ut + uux = 0.
f (u) = 12 u2 ,
f 0 (u) = u.
Consider Riemann problem with states u` and ur .
For any u` , ur , there is a weak solution consisting of this
discontinuity propagating at speed given by the
Rankine-Hugoniot jump condition:
s=
− 21 u2`
1
= (u` + ur ).
ur − u`
2
1 2
2 ur
Note: Shock speed is average of characteristic speed on each
side.
This might not be the physically correct weak solution!
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 11.4]
Burgers’ equation
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 11.4]
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 11.4]
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 11.13 ]
Notes:
The solution is constant on characteristics so each value
advects at constant speed equal to the value...
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 11.4]
Weak solutions to Burgers’ equation
ut + 12 u2 x = 0,
Notes:
u` = 1, ur = 2
Characteristic speed: u
Rankine-Hugoniot speed: 21 (u` + ur ).
“Physically correct” rarefaction wave solution:
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 11.13 ]
Weak solutions to Burgers’ equation
ut + 12 u2 x = 0,
Notes:
u` = 1, ur = 2
Characteristic speed: u
Rankine-Hugoniot speed: 21 (u` + ur ).
Entropy violating weak solution:
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 11.13 ]
Weak solutions to Burgers’ equation
ut + 12 u2 x = 0,
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 11.13 ]
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 11.13 ]
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 11.13 ]
Notes:
u` = 1, ur = 2
Characteristic speed: u
Rankine-Hugoniot speed: 21 (u` + ur ).
Another Entropy violating weak solution:
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 11.13 ]
Vanishing viscosity solution
Notes:
We want q(x, t) to be the limit as → 0 of solution to
qt + f (q)x = qxx .
This selects a unique weak solution:
• Shock if f 0 (ql ) > f 0 (qr ),
• Rarefaction if f 0 (ql ) < f 0 (qr ).
Lax Entropy Condition:
A discontinuity propagating with speed s in the solution of a
convex scalar conservation law is admissible only if
f 0 (q` ) > s > f 0 (qr ), where s = (f (qr ) − f (q` ))/(qr − q` ).
Note: This means characteristics must approach shock from
both sides as t advances, not move away from shock!
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 11.13 ]
Riemann problem for scalar nonlinear problem
Notes:
qt + f (q)x = 0 with data
q(x, 0) =
if x < 0
if x ≥ 0
ql
qr
Piecewise constant with a single jump discontinuity.
For Burgers’ or traffic flow with quadratic flux, the Riemann
solution consists of:
• Shock wave if f 0 (ql ) > f 0 (qr ),
• Rarefaction wave if f 0 (ql ) < f 0 (qr ).
Five possible cases:
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 12.1]
Transonic rarefactions
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 12.1]
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 11.13 ]
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 11.13 ]
Notes:
Sonic point: us = 0 for Burgers’ since f 0 (0) = 0.
Consider Riemann problem data u` = −0.5 < 0 < ur = 1.5.
In this case wave should spread in both directions:
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 11.13 ]
Transonic rarefactions
Notes:
Entropy-violating approximate Riemann solution:
1
s = (u` + ur ) = 0.5.
2
Wave goes only to right, no update to cell average on left.
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 11.13 ]
Transonic rarefactions
Notes:
If u` = −ur then Rankine-Hugoniot speed is 0:
Similar solution will be observed with Godunov’s method
if entropy-violating approximate Riemann solver used.
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 11.13 ]
Entropy-violating numerical solutions
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 11.13 ]
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 12.3]
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 15.3.2 ]
Notes:
Riemann problem for Burgers’ equation at t = 1
with u` = −1 and ur = 2:
Godunov with no entropy fix
Godunov with entropy fix
2.5
2.5
2
2
1.5
1.5
1
1
0.5
0.5
0
0
−0.5
−0.5
−1
−1.5
−3
−1
−2
−1
0
1
2
3
−1.5
−3
−2
High−resolution with no entropy fix
0
1
2
3
2
3
2.5
2
2
1.5
1.5
1
1
0.5
0.5
0
0
−0.5
−0.5
−1
−1.5
−3
−1
High−resolution with entropy fix
2.5
−1
−2
−1
0
1
2
3
R.J. LeVeque, University of Washington
−1.5
−3
−2
−1
0
1
IPDE 2011, July 1, 2011
[FVMHP Sec. 12.3]
Approximate Riemann solvers
Notes:
For nonlinear problems, computing the exact solution to each
Riemann problem may not be possible, or too expensive.
Often the nonlinear problem qt + f (q)x = 0 is approximated by
qt + Ai−1/2 qx = 0,
q` = Qi−1 ,
qr = Qi
for some choice of Ai−1/2 ≈ f 0 (q) based on data Qi−1 , Qi .
Solve linear system for αi−1/2 : Qi − Qi−1 =
P
p
p
p αi−1/2 ri−1/2 .
p
p
p
Waves Wi−1/2
= αi−1/2
ri−1/2
propagate with speeds spi−1/2 ,
p
ri−1/2
are eigenvectors of Ai−1/2 ,
p
si−1/2 are eigenvalues of Ai−1/2 .
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 15.3.2 ]
Approximate Riemann solvers
qt + Âi−1/2 qx = 0,
q` = Qi−1 ,
Notes:
qr = Qi
Often Âi−1/2 = f 0 (Qi−1/2 ) for some choice of Qi−1/2 .
In general Âi−1/2 = Â(q` , qr ).
Roe conditions for consistency and conservation:
• Â(q` , qr ) → f 0 (q ∗ ) as q` , qr → q ∗ ,
• Â diagonalizable with real eigenvalues,
• For conservation in wave-propagation form,
Âi−1/2 (Qi − Qi−1 ) = f (Qi ) − f (Qi−1 ).
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 15.3.2 ]
Approximate Riemann solvers
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 15.3.2 ]
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 12.2 ]
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 12.3]
Notes:
For a scalar problem, we can easily satisfy the Roe condition
Âi−1/2 (Qi − Qi−1 ) = f (Qi ) − f (Qi−1 ).
by choosing
Âi−1/2 =
f (Qi ) − f (Qi−1 )
.
Qi − Qi−1
1
Then ri−1/2
= 1 and s1i−1/2 = Âi−1/2 (scalar!).
Note: This is the Rankine-Hugoniot shock speed.
=⇒ shock waves are correct,
rarefactions replaced by entropy-violating shocks.
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 12.2 ]
Approximate Riemann solver
Notes:
∆t +
A ∆Qi−1/2 + A− ∆Qi+1/2 .
∆x
For scalar advection m = 1, only one wave.
Wi−1/2 = ∆Qi−1/2 = Qi − Qi−1 and si−1/2 = u,
Qn+1
= Qni −
i
A− ∆Qi−1/2 = s−
i−1/2 Wi−1/2 ,
A+ ∆Qi−1/2 = s+
i−1/2 Wi−1/2 .
For scalar nonlinear: Use same formulas with
Wi−1/2 = ∆Qi−1/2 and si−1/2 = ∆Fi−1/2 /∆Qi−1/2 .
Need to modify these by an entropy fix in the trans-sonic
rarefaction case.
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 12.3]
Entropy fix
Notes:
Qn+1
= Qni −
i
Revert to the formulas
∆t +
A ∆Qi−1/2 + A− ∆Qi+1/2 .
∆x
A− ∆Qi−1/2 = f (qs ) − f (Qi−1 )
A+ ∆Qi−1/2 = f (Qi ) − f (qs )
left-going fluctuation
right-going fluctuation
if f 0 (Qi−1 ) < 0 < f 0 (Qi ).
High-resolution method: still define wave W and speed s by
Wi−1/2 = Qi − Qi−1 ,
(f (Qi ) − f (Qi−1 ))/(Qi − Qi−1 )
si−1/2 =
f 0 (Qi )
R.J. LeVeque, University of Washington
if Qi−1 =
6 Qi
if Qi−1 = Qi .
IPDE 2011, July 1, 2011
[FVMHP Sec. 12.3]
Godunov flux for scalar problem
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 12.3]
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 12.1]
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 12.3]
Notes:
The Godunov flux function for the case f 00 (q) > 0 is

if Qi−1 > qs and s > 0
 f (Qi−1 )
n
f (Qi )
if Qi < qs and s < 0
Fi−1/2
=

f (qs )
if Qi−1 < qs < Qi .


f (q)
if Qi−1 ≤ Qi
 Q min
i−1 ≤q≤Qi
=

if Qi ≤ Qi−1 ,
max f (q)

Qi ≤q≤Qi−1
Here s =
f (Qi )−f (Qi−1 )
Qi −Qi−1
is the Rankine-Hugoniot shock speed.
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 12.1]
Entropy-violating numerical solutions
Notes:
Riemann problem for Burgers’ equation at t = 1
with u` = −1 and ur = 2:
Godunov with no entropy fix
Godunov with entropy fix
2.5
2.5
2
2
1.5
1.5
1
1
0.5
0.5
0
0
−0.5
−0.5
−1
−1.5
−3
−1
−2
−1
0
1
2
3
−1.5
−3
−2
High−resolution with no entropy fix
0
1
2
3
2
3
2.5
2
2
1.5
1.5
1
1
0.5
0.5
0
0
−0.5
−0.5
−1
−1.5
−3
−1
High−resolution with entropy fix
2.5
−1
−2
−1
0
1
2
3
R.J. LeVeque, University of Washington
−1.5
−3
−2
−1
0
IPDE 2011, July 1, 2011
1
[FVMHP Sec. 12.3]
Entropy (admissibility) conditions
Notes:
We generally require additional conditions on a weak solution
to a conservation law, to pick out the unique solution that is
physically relevant.
In gas dynamics: entropy is constant along particle paths for
smooth solutions, entropy can only increase as a particle goes
through a shock.
Entropy functions: Function of q that “behaves like” physical
entropy for the conservation law being studied.
NOTE: Mathematical entropy functions generally chosen to
decrease for admissible solutions,
increase for entropy-violating solutions.
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 11.4]
Entropy functions
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 11.4]
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 11.4]
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 11.4]
Notes:
A scalar-valued function η : lRm → lR is a convex function of q
if the Hessian matrix η 00 (q) with (i, j) element
00
ηij
(q) =
∂2η
∂q i ∂q j
is positive definite for all q, i.e., satisfies
v T η 00 (q)v > 0
for all q, v ∈ lRm .
Scalar case: reduces to η 00 (q) > 0.
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 11.4]
Entropy functions
Notes:
Entropy function: η : lRm → lR
Entropy flux: ψ : lRm → lR
chosen so that η(q) is convex and:
• η(q) is conserved wherever the solution is smooth,
η(q)t + ψ(q)x = 0.
• Entropy decreases across an admissible shock wave.
Weak form:
Z x2
Z
η(q(x, t2 )) dx ≤
x1
+
Z
x2
η(q(x, t1 )) dx
x1
t2
t1
ψ(q(x1 , t)) dt −
Z
t2
ψ(q(x2 , t)) dt
t1
with equality where solution is smooth.
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 11.4]
Entropy functions
Notes:
How to find η and ψ satisfying this?
η(q)t + ψ(q)x = 0
For smooth solutions gives
η 0 (q)qt + ψ 0 (q)qx = 0.
Since qt = −f 0 (q)qx this is satisfied provided
ψ 0 (q) = η 0 (q)f 0 (q)
Scalar: Can choose any convex η(q) and integrate.
Example: Burgers’ equation, f 0 (u) = u and take η(u) = u2 .
Then ψ 0 (u) = 2u2 =⇒
Entropy function: ψ(u) = 23 u3 .
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 11.4]
Weak solutions and entropy functions
The conservation laws
1 2
ut +
u
=0
2
x
u2
and
both have the same quasilinear form
IPDE 2011, July 1, 2011
[FVMHP Sec. 11.4]
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 11.4]
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 11.4]
Notes:
+
t
R.J. LeVeque, University of Washington
2 3
u
3
=0
x
ut + uux = 0
but have different weak solutions, different shock speeds!
Entropy function: η(u) = u2 .
A correct Burgers’ shock at speed s = 12 (u` + ur ) will have
total mass of η(u) decreasing.
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 11.4]
Entropy functions
Z
x2
x1
Notes:
η(q(x, t2 )) dx ≤
+
Z
x2
x1
Z t2
t1
η(q(x, t1 )) dx
ψ(q(x1 , t)) dt −
Z
t2
ψ(q(x2 , t)) dt
t1
comes from considering the vanishing viscosity solution:
qt + f (q )x = qxx
Multiply by η 0 (q ) to obtain:
η(q )t + ψ(q )x = η 0 (q )qxx
.
Manipulate further to get
η(q )t + ψ(q )x = η 0 (q )qx
R.J. LeVeque, University of Washington
x
− η 00 (q ) (qx )2 .
IPDE 2011, July 1, 2011
[FVMHP Sec. 11.4]
Entropy functions
Notes:
Smooth solution to viscous equation satisfies
η(q )t + ψ(q )x = η 0 (q )qx
x
− η 00 (q ) (qx )2 .
Integrating over rectangle [x1 , x2 ] × [t1 , t2 ] gives
Z
x2
η(q (x, t2 )) dx =
x1
−
Z
+
Z
−
t2
t1
t2
t1
x2
η(q (x, t1 )) dx
x1
t1
t2
Z
Z
ψ(q (x2 , t)) dt −
Z
t2
t1
ψ(q (x1 , t)) dt
0 η (q (x2 , t)) qx (x2 , t) − η 0 (q (x1 , t)) qx (x1 , t) dt
Z
x2
x1
η 00 (q ) (qx )2 dx dt.
Let → 0 to get result:
Term on third line goes to 0,
Term of fourth line is always ≤ 0.
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 11.4]
Entropy functions
IPDE 2011, July 1, 2011
[FVMHP Sec. 11.4]
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 11.4]
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 12.10]
Notes:
Weak form of entropy condition:
Z ∞Z ∞
Z
φt η(q) + φx ψ(q) dx dt +
−∞
0
R.J. LeVeque, University of Washington
∞
−∞
φ(x, 0)η(q(x, 0)) dx ≥ 0
for all φ ∈ C01 (lR × lR) with φ(x, t) ≥ 0 for all x, t.
Informally we may write
η(q)t + ψ(q)x ≤ 0.
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 11.4]
Lax-Wendroff Theorem
Notes:
Suppose the method is conservative and consistent with
qt + f (q)x = 0,
Fi−1/2 = F(Qi−1 , Qi ) with F(q̄, q̄) = f (q̄)
and Lipschitz continuity of F.
If a sequence of discrete approximations converge to a function
q(x, t) as the grid is refined, then this function is a weak
solution of the conservation law.
Note:
Does not guarantee a sequence converges (need stability).
Two sequences might converge to different weak solutions.
Also need to satisfy an entropy condition.
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 12.10]
Sketch of proof of Lax-Wendroff Theorem
Notes:
Multiply the conservative numerical method
Qn+1
= Qni −
i
∆t n
n
(F
− Fi−1/2
)
∆x i+1/2
by Φni to obtain
Φni Qn+1
= Φni Qni −
i
∆t n n
n
Φ (F
− Fi−1/2
).
∆x i i+1/2
This is true for all values of i and n on each grid.
Now sum over all i and n ≥ 0 to obtain
∞
∞ X
X
n=0 i=−∞
Φni (Qn+1
− Qni ) = −
i
∞
∞
∆t X X n n
n
Φi (Fi+1/2 − Fi−1/2
).
∆x
n=0 i=−∞
Use summation by parts to transfer differences to Φ terms.
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 12.10]
Sketch of proof of Lax-Wendroff Theorem
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 12.10]
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 12.10]
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 12.11]
Notes:
Obtain analog of weak form of conservation law:
∆x∆t
"
∞ X
∞ n
X
Φi − Φn−1
i
Qni
∆t
n=1 i=−∞
n
∞ X
∞
∞
X
X
Φi+1 − Φni
n
+
Fi−1/2
= −∆x
Φ0i Q0i .
∆x
n=0 i=−∞
i=−∞
Consider on a sequence of grids with ∆x, ∆t → 0.
Show that any limiting function must satisfy weak form of
conservation law.
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 12.10]
Analog of Lax-Wendroff proof for entropy
Notes:
Show that the numerical flux function F leads to a
numerical entropy flux Ψ
such that the following discrete entropy inequality holds:
η(Qn+1
) ≤ η(Qni ) −
i
i
∆t h n
Ψi+1/2 − Ψni−1/2 .
∆x
Then multiply by test function Φni , sum and use summation by
parts to get discrete form of integral form of entropy condition.
=⇒ If numerical approximations converge to some function,
then the limiting function satisfies the entropy condition.
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 12.11]
Entropy consistency of Godunov’s method
Notes:
|
For Godunov’s method, F (Qi−1 , Qi ) = f (Q∨
i−1/2 )
|
where Q∨
i−1/2 is the constant value
along xi−1/2 in the Riemann solution.
|
Let Ψni−1/2 = ψ(Q∨
i−1/2 )
Discrete entropy inequality follows from Jensen’s inequality:
The value of η evaluated at the average value of q̃ n is less than
or equal to the average value of η(q̃ n ), i.e.,
!
Z xi+1/2
Z xi+1/2
1
1
n
q̃ (x, tn+1 ) dx ≤
η q̃ n (x, tn+1 ) dx.
η
∆x xi−1/2
∆x xi−1/2
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 12.11]
R.J. LeVeque, University of Washington
IPDE 2011, July 1, 2011
[FVMHP Sec. 12.11]

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