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Nonlinear Hyperbolic Systems of Conservation
Laws and Related Applications I
Mapundi K. Banda
Applied Mathematics Division - Mathematical Sciences, Stellenbosch University
[email protected]
Jul 23, 2013
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
1 / 84
All models are wrong, but some models are useful
George Box
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
2 / 84
Section 1: Mathematical Models - Evolution Equations
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CIMPA/MPE 2013
Jul 22 - Aug 3
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Modelling with Partial Differential Equations
Evolution equations to be studied are hyperbolic systems of
conservation laws:
First-order Hyperbolic Partial differential equation.
Interesting cases are usually nonlinear.
Play a prominent part in modelling flow and transport processes.
Discussion of the mathematical aspects.
Discussion of the ideas for designing numerical schemes.
Presentation of some case studies.
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
4 / 84
Why study Hyperbolic Conservation Laws
Special Difficulties:
shock formation (jump discontinuities).
Great deal is not yet known about the mathematical structure.
Provide efficient approximations of many complex models.
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
5 / 84
Reynold’s Transport Theorem
Consider an arbitrary physical quantity ρ (mass, energy, charge,
concentration of species of matter, etc).
ρ occupies Ω0 ⊂ Rn at time t = 0.
Let domain move with time:
consider a map
Φ : Ω0 × [0, T ] → Rn
define Ωt = {x = Φ(y , t) : y ∈ Ω0 }
Ω0 - reference configuration; t 7→ Φ(y , t) - trajectory of the particle (or
point) y ∈ Ω0
Using x and t is referred to as using the Eulerian description.
Using y and t is referred to as the Lagrangian description.
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
6 / 84
Reynold’s Transport Theorem
The velocity of the particle at point x = Φ(y , t) is given by:
u(x, t) = u(Φ(y , t), t) =
∂
Φ(y , t)
∂t
The total amount (mass) of the quantity ρ inside the domain Ωt is
given by
Z
Mt =
ρ(x, t) dx
Ωt
How does the total mass change with time - Reynolds Transport
theorem
Theorem
(Reynold’s Transport Theorem)
Z
Z
d
∂ρ
ρ(x, t) dx =
+ ∇ · (ρu) dx
dt Ωt
Ωt ∂t
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
7 / 84
Reynold’s Transport Theorem
d
dt
Z
Z
ρ(x, t) dx =
Ωt
Ωt
∂ρ
+ ∇ · (ρu) dx
∂t
Integral relation between change of total mass, Mt , and local changes
of density ρ and the transportation u in an arbitrary domain Ωt ⊂ Rn
To derive the partial differential equations from Reynolds Transport
Theorem use:
Lemma
(Variational Lemma) Let f ∈
L2 (Ω)
Then f ≡ 0 almost everywhere.
Mapundi K. Banda (Maties)
Z
with
fg dx = 0 for all g ∈ L2 (Ω).
Ω
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Jul 22 - Aug 3
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Application: Gas Flow in a pipe
Denote gas density by ρ and velocity in x-direction by u(x, t)
Domain Ωt = [x1 (t), x2 (t)] of moving gas with
d
xi (t) = u(xi (t), t)
dt
Total amount of gas
Z
x2 (t)
Mt =
ρ(x, t) dx
x1 (t)
The Reynolds Transport Theorem yields:
Z
d
∂ρ ∂(ρu)
Mt =
+
dx
dt
∂x
Ωt ∂t
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
9 / 84
Application: Gas Flow in a pipe
Axiom of Conservation of Mass,
Z
0=
Ωt
d
Mt = 0:
dt
∂ρ
∂
+
(ρu) dx
∂t
∂x
for any arbitrary domain Ωt of gas.
Apply the Variational Lemma to get the Continuity Equation
∂ρ
∂
+
(ρu) = 0
∂t
∂x
(ρu)(x, t) is the mass flux of gas through the cross-section of the pipe
at point x and t.
To get a closed relation between ρ and u, we need a constitutive
relation between ρ and u or additional conservation laws.
If u is known, the equation is referred to as an advection equation.
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
10 / 84
Mass Conservation Axiom
The change of Mass is given by the difference of mass fluxes at the
boundaries.
Consider the continuity equation on an interval [x1 , x2 ] during time
[t1 , t2 ]:
∂ρ
∂
+
(ρu) = 0
∂t
∂x
Rewrite the flux as f (ρ) = ρu, assuming u is given
Integrating with respect to space:
Z x2
∂ρ
dx = f (ρ(x1 , t)) − f (ρ(x2 , t))
x1 ∂t
Integrating with respect to time and space:
Z x2
Z x2
Z t2
Z
ρ(x, t2 ) dx −
ρ(x, t1 ) dx =
f (x1 , t) dt −
x1
Mapundi K. Banda (Maties)
x1
t1
CIMPA/MPE 2013
t2
f (x2 , t) dt
t1
Jul 22 - Aug 3
11 / 84
Application: Traffic Flow
Consider a highway section with car density ρ, cars are a conserved
quantity:
∂
∂ρ
+
(ρu) = 0
∂t
∂x
To close the equation, need a relation between ρ and u:
ρ u = u(ρ) = umax 1 −
ρmax
i.e. the denser the traffic, the slower the cars will move.
Define flux as
ρ f (ρ) = ρumax 1 −
ρmax
In general, the equation
∂
∂ρ
+
f (ρ) = 0;
or
∂t ρ + ∂x f (ρ) = 0
∂t
∂x
is the prototype of a hyperbolic scalar conservation law in one space
dimension
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
12 / 84
Application: Euler Equations of Inviscid Gas
Dynamics
Model the flow of a gas with density ρ, velocity v , energy E and
pressure p need additional conservation laws
ρt + (ρv )x = 0;
2
conservation of mass,
(ρv )t + (ρv + p) = 0;
conservation of momentum,
Et + (v (E + p))x = 0;
conservation of energy.
Need constitutive relations, where T is the absolute temperature of
the gas:
E=
ρv 2
2
Mapundi K. Banda (Maties)
p = ρT ;
ideal gas law,
+ cv ρT
kinetic + thermal (internal) energy.
CIMPA/MPE 2013
Jul 22 - Aug 3
13 / 84
Application: Euler Equations of Inviscid Gas
Dynamics
Model the flow of a gas with density ρ, velocity v , energy E and
pressure p need additional conservation laws
ρt + (ρv )x = 0;
2
conservation of mass,
(ρv )t + (ρv + p) = 0;
conservation of momentum,
Et + (v (E + p))x = 0;
conservation of energy.
Obtain 3 equations for 3 variables and p depends on ρ, v , E with
 


ρ
ρv
u = ρv  and F (u) =  ρv 2 + p(u) 
E
v (E + p(u))
giving a hyperbolic system of conservation laws in one space
dimension:
∂t u + ∂x F (u) = 0
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
14 / 84
Application: Euler Equations of Inviscid Gas
Dynamics
Assume temperature, T , is constant then
p = a2 ρ
where a is the speed of sound in the gas.
The flow is referred to as isothermal flow:
ρt + (ρv )x = 0;
2
2
(ρv )t + (ρv + a ρ) = 0;
Mapundi K. Banda (Maties)
conservation of mass,
conservation of momentum,
CIMPA/MPE 2013
Jul 22 - Aug 3
15 / 84
Equations of Internal Energy and Entropy
With some manipulation of the viscous heat conducting fluid dynamic
equations, one obtains:
ρ
de
= −p(∇ · v ) + εv + ∇ · (k∇T ) + qH
dt
where:
e is the internal energy.
p(∇ · v ) is the reversible work of the pressure forces (vanishes in
incompressible flows).
Other terms are considered as heat additions:
εv dissipation term - acting as an irreversible heat source
FC = −k∇T is flux due to heat conduction (Fourier’s Law).
qH are heat sources.
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
16 / 84
Equations of Internal Energy and Entropy
Introduce entropy per unit mass s of the fluid through the
thermodynamic relation:
T ds = de + p d
1
dp
= dh −
ρ
ρ
where h is the enthalpy.
Thus separation of reversible and irreversible heat additions:
T ds = dq + dq 0
where
dq is a reversible heat addition.
dq 0 is an irreversible heat addition.
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
17 / 84
Equations of Internal Energy and Entropy
Second Principle of Thermodynamics: dq 0 ≥ 0, i.e. in adiabatic flow
(dq = 0) entropy always increases.
We obtain:
ρT
ds
= εv + ∇ · (k∇T ) + qH
dt
where:
Last two terms are reversible heat addition by conduction and other
sources.
For qH = 0 and k = 0 the non-negative dissipation term εv is a
non-reversible heat source.
Entropy equation of the flow, equation is important but not
independent from energy equation
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
18 / 84
Equations of Internal Energy and Entropy
Either the entropy equation or the energy equation used with the
mass and momentum equation but not both.
Entropy is not ”conserved” in the sense of derived conservation
equations
ρT
ds
= εv + ∇ · (k∇T ) + qH
dt
Thus mathematical expression of second principle of thermodynamics
for an adiabatic flow without heat conduction or heat sources:
ds
= εv
dt
∂s
ρT
+ v · ∇s = εv
∂t
ρT
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
19 / 84
Physical Entropy condition
Since εv ≥ 0, viscous dissipation:
ρT
∂s
+ v · ∇s ≥ 0
∂t
Any solution of Euler equations with a physical meaning satisfies the
entropy condition above.
Hence entropy used as an additional condition, where necessary, to
exclude non-physical solutions for uniqueness.
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
20 / 84
Some remarks on the Euler Equations
Non-viscous, non-heat-conducting fluid, flow in the limit of vanishing
viscosity
Admits both continuous and discontinuous solutions:
Entropy variations: continuous flow variations entropy equation:
ρT
∂s
+ v · ∇s = 0
∂t
i.e. entropy constant along flow path
In absence of discontinuities - Euler equation describes isentropic flows.
Value of entropy can vary from one flow path to another.
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
21 / 84
Application: Cooling of Coffee
Consider a pot (domain P ⊂ R3 ) filled with coffee of constant density
ρ and with temperature θ.
Thermal energy contained in Ω ⊂ P is given by:
Z
ρcp θ(x, t) dx
Et =
Ω
Assume coffee is at rest u(x, t) ≡ 0. Then Reynolds Transport
Theorem yields:
Z
d
∂θ
Et =
ρcp
dx
dt
∂t
Ω
In general, change of thermal energy is a sum of heat flux through ∂Ω
and energy produced in Ω, here assumed to be zero.
Also, thermal energy flows from hot to cold i.e. heat flux f :
f = −κ∇θ
where κ > 0 is heat conductivity.
Mapundi K. Banda (Maties)
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Application: Cooling of Coffee
Hence
Z
Z
Z
∂θ
dx = −
fn ds =
κn∇θ ds;
∂t
Ω
∂Ω
∂Ω
where n denotes the outer normal of the domain Ω considering the
heat flow out of domain.
Z
Z
ρcp
Apply Gauss divergence Theorem
Z
∂θ
dx =
ρcp
∂t
Ω
∇ · f dx:
fn ds =
∂Ω
Ω
Z
∇(κ∇θ) dx
∀Ω ⊂ P
Ω
Assume ρ, cp and κ to be constant, apply variational lemma:
∂θ
= α∇2 θ,
∂t
κ
> 0 is thermal conductivity.
ρcp
This is the heat equation or the diffusion equation.
where α =
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
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Application: Cooling of Coffee
The heat equation or diffusion equation models molecular diffusion
(Fick’s Law)!
To get a well-posed problem, specify initial and boundary conditions.
For flux due to advection at velocity u, then the equation takes the
form:
∂
∂θ
+
(θu) = α∇2 θ
∂t
∂x
The equation is referred to as the convection-diffusion or
advection-diffusion equation.
If inside the pot there is an extra source of heat, e.g. a heating
element, then a source term could be added:
∂θ
∂
+
(θu) = α∇2 θ + S
∂t
∂x
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
24 / 84
Solution of the Advection Equation
Consider the continuity equation with u(x, t) ≡ a, known then the
problem reduces to:
∂ρ
∂ρ
+a
= 0.
∂t
∂x
Given the initial condition
ρ(x, 0) = ρ0 (x)
−∞<x <∞
then the solution takes the form:
ρ(x, t) = ρ0 (x − at),
t>0
Note: initial profile simply moves downstream with velocity a, its
shape unchanged.
The solution above is true even if ρ0 (x) is not smooth!
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
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25 / 84
Example: The Burgers Equation
Consider the scalar parabolic equation
∂u
∂2u
∂u
+u
−ν 2 =0
∂t
∂x
∂x
Introduced by Burgers as simplest differential model for fluid flow (viscous) Burgers equation.
Though simple, it is regarded as model for decaying turbulence
Burgers studied the limit equation as ν → 0
∂u
∂ u2 +
=0
∂t
∂x 2
inviscid Burgers Equation
Inviscid Burgers Equation (or Burgers Equation) also studied in wave
theory to depict distortion of waveform in simple waves
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
26 / 84
Example: Solution of Viscous Burgers Equation
Consider the scalar parabolic equation
∂u
∂u
∂2u
+u
−ν 2 =0
∂t
∂x
∂x
The Cauchy problem of viscous Burgers Equation has an explicit
solution:
Apply Cole-Hopf transform:
u = −2ν
φx
φ
to transform the Burgers Equation to
∂φ
∂2φ
=ν 2
∂t
∂x
Use explicit expressions presented above for an explicit expression of
the solution.
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
27 / 84
Solution of the diffusion equation
Consider the diffusion equation:
∂θ
− ∇2 θ = f
∂t
where θ = θ(x, t), t ∈ R+ and x ∈ Ω ⊂ Rn
Initial conditions: differential equation first order in time:
θ(x, t) = θ0 (x),
in Ω
Boundary conditions:
Dirichlet conditions: θ(x, t) = θb (x) for x ∈ ∂Ω.
Neumann conditions: −(n · ∇)θ(x, t) = g (x) for x ∈ ∂Ω - heat flux.
Robin conditions: −(n · ∇)θ(x, t) − βθ(x, t) = g (x) for x ∈ ∂Ω and
β>0
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
28 / 84
Separation of variables
Consider an initial boundary value problem:
∂θ
∂2θ
=
∂t
∂x 2
on Q = R+ × (0, 1) with initial condition:
θ(x, 0) = θ0 (x),
x ∈ (0, 1)
and boundary data
θ(0, t) = 0
θ(1, t) = 0
for t ∈ R+
Seek a solution of the form:
θ(x, t) = T (t) · X (x)
Mapundi K. Banda (Maties)
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Separation of variables
Obtain
θ(x, t) =
∞
X
an e −n
2 π2 t
sin(nπx)
n=1
and the coefficients are given (using initial conditions) as the Fourier
coefficients of θ0 (x):
θ0 (x) =
∞
X
an sin(nπx)
n=1
Note: The solution is infinitely often differentiable - the smoothing
property of the diffusion equation.
The smoothing property also holds for non-continuous initial data and
inconsistent boundary data.
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
30 / 84
The Diffusion Kernel
Seek the fundamental solution - also called heat kernel:
∂ω
− κ∇2 ω = δx δt
∂t
in Rn × R+
Using Fourier Transforms:
1
2
ω(x, t) = √
e −x /(4κt)
n
( 4πκt)
ω(x, t) > 0 for all t > 0,
x ∈ Rn
Hence for f ≡ 0:
1
θ(x, t) = ω(·, t) ? θ0 = √
( 4πκt)n
Z
e −(x−y )
2 /(4κt)
θ0 (y ) dy
Rn
where ? is a convolution.
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
31 / 84
Solution of Diffusion Equations 1: Half-plane
Consider a 1D problem
θt = κθxx ,
0 < x < ∞,
θ(x, 0) = θ0 (x),
θ(0, t) = 0,
0<t<∞
for t = 0
for x = 0
The solution takes the form:
θ(x, t) =
1
√
( 4πκt)
Z h
i
2
2
×
e −(x−y ) /(4κt) − e −(x+y ) /(4κt) θ0 (y ) dy
R
Mapundi K. Banda (Maties)
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Solution of Diffusion Equations 1: Half-plane with
Neumann Boundary
Consider a 1D problem
θt = κθxx ,
0 < x < ∞,
θ(x, 0) = θ0 (x),
θx (0, t) = 0,
0<t<∞
for t = 0
for x = 0
The solution takes the form:
θ(x, t) =
1
√
( 4πκt)
Z h
i
2
2
×
e −(x−y ) /(4κt) + e −(x+y ) /(4κt) θ0 (y ) dy
R
Difference with the previous problem is a sign change.
Mapundi K. Banda (Maties)
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The Diffusion Kernel
The solution operator is strongly regularizing.
The solution of the diffusion equation has infinite speed of
propagation
Energy estimates: the homogeneous diffusion equation does not
conserve energy.
The solution of the non-homogeneous equation takes the form:
θ(x, t) = ω(·, t) ? θ0 + ω(·, t) ? f
Z
Z t
1
p
ω(x − y , t)θ0 (y ) dy +
=
n
0 ( 4πκ(t − τ ))
Rn
Z
2
×
e −(x−y ) /(4κ(t−τ )) f (y , τ ) dydτ
Rn
One can prove that the solution of the non-homogeneous diffusion
equation with Dirichlet boundary conditions is unique.
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
34 / 84
Cauchy and Riemann Problems: Introduction
Burgers possesses features of a scalar convex equation
∂u
∂
+
f (u) = 0
∂t
∂x
where f : R → R is a convex smooth function
The Cauchy Problem: Equation (1) with
(1)
u(x, 0) = u0 (x)
The Cauchy problem for the Burgers Equation admits discontinuous
weak solutions even for a smooth initial function u0
The Riemann Problem: Equation (1) with
(
uL
for x < 0
u(x, 0) =
uR for x ≥ 0
The solution for the Riemann Problem is either a shock propagating
or a rarefaction wave
Mapundi K. Banda (Maties)
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Summary
Derived the conservation laws based on the Reynolds Transport
Theorem.
Presented some special cases: linear advection, Burgers equation,
Euler equations of gas dynamics
Presented solution of the advection equation
Presented solutions of the diffusion equations based on the separation
of variables as well as the Green’s function.
Presented solutions of the advection-diffusion equation
Introduced some common concepts in the area of conservation laws
Mapundi K. Banda (Maties)
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Section 2: Mathematical Analysis - Hyperbolic Conservation Laws
Mapundi K. Banda (Maties)
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Preamble: Remarks on Hyperbolic Systems
Conservation Laws due to Lax
Systems of conservation laws in a single space variable is a
well-studied subject
Euler equations of compressible flow, are an important example of a
hyperbolic system of conservation laws
In phenomena governed by nonlinear hyperbolic equations signals
propagate with finite speed (recall the linear wave equation)
Singularities propagate along characteristics, BUT these arise
spontaneously, leading to formation of shocks
Mapundi K. Banda (Maties)
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38 / 84
Preamble: Remarks on Hyperbolic Systems
Conservation Laws due to Lax
Time is not reversible as for linear equations thus future and past are
different, as they are in real life
There is loss of information as time moves forward, which can be
interpreted as an increase in entropy
Basic existence theory of solutions of hyperbolic conservation laws in
single space variables is due to Jim Glimm
It is a scandal of mathematical physics that, apart from isolated
results, no comparable theory exists for more space variables
Mapundi K. Banda (Maties)
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39 / 84
Hyperbolic Conservation Laws
In general, seek a weak solution u : RD × [0, T ] → RK to the
Cauchy problem:
∂t u +
D
X
∂xj Fj (u) = 0;
u(x, 0) = u0 (x)
j=1
with Fj (u) : RK → RK , is a system of K conservation laws in RD with
flux functions Fj
Consider ∂t u +
D
X
Aj (u)∂xj u = 0;
u(x, 0) = u0 (x),
j=1
the system is (strictly) hyperbolic iff the matrix
D
X
ξj Aj (u) has only
j=1
real (pairwise disjoint) e-values (char. speeds) for all ξ ∈ RD .
Mapundi K. Banda (Maties)
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Examples of Hyperbolic Conservation Laws
Linear Equation:
Compressible Euler:
∂t u + ∂x u = 0;
∂t ρ + ∂x m = 0;
∂t m + ∂x (ρu 2 + p) = 0;
Burgers Equation:
∂t u + ∂x
u2 2
∂t E + ∂x (u(E + p)) = 0;
ρ
p = (γ − 1)(E − u 2 ).
2
= 0;
Incompressible Euler:
∇ · u = 0;
∂t u + ∇ · u ⊗ u + ∇p = 0.
Mapundi K. Banda (Maties)
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Hyperbolic Conservation Laws
For K = D = 1 i.e. one-dimensional scalar conservation laws, the
equation is always strictly hyperbolic
For D = 1 one needs to investigate the matrix
A(u) =
∂F
j
∂uk
(u)
1≤j,k≤K


F1
 F2 
 
with F =  . 
 .. 
FK
Mapundi K. Banda (Maties)
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Example: The p-system
Model one-dimensional isentropic gas dynamics in Lagrangian
coordinates:
∂u
∂v
−
= 0;
∂t
∂x
∂u
∂
+
p(v ) = 0;
∂t
∂x
v is specific volume, u is the velocity, and the pressure p(v )
For a polytropic isentropic ideal gas: p(v ) = Av −γ for some constant
A = A(s) > 0 (depending on entropy), γ > 1
Mapundi K. Banda (Maties)
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Example: The p-system
∂v
∂u
−
= 0;
∂t
∂x
∂u
∂
+
p(v ) = 0;
∂t
∂x
System is of the form:
∂w
∂
+
f(w) = 0;
The conservative form
∂t
∂x
∂w
∂w
+ A(w)
= 0;
The quasilinear form
∂t
∂x
where A(w ) is the Jacobian matrix with two real distinct eigenvalues
p
p
λ1 = − (−p 0 (v )) < λ2 = (−p 0 (v ))
i.e. p 0 (v ) < 0.
Hence system is strictly hyperbolic provided p 0 (v ) < 0
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
44 / 84
Example: The p-system
∂v
∂u
−
= 0;
∂t
∂x
∂
∂u
+
p(v ) = 0;
∂t
∂x
The p-system is the simplest nontrivial example of a nonlinear system
of conservation laws.
Any nonlinear wave equation
∂2g
∂ ∂g −
σ( ) = 0
∂t 2
∂x
∂x
can be re-written in the form of the p-system:
Set
u=
Mapundi K. Banda (Maties)
∂g
,
∂t
v=
∂g
,
∂x
CIMPA/MPE 2013
p(v ) = −σ(v )
Jul 22 - Aug 3
45 / 84
Example: Isothermal Flow
For isothermal flow:
u=
ρ
ρv
A(u) =
F (u) =
0
)2
− (ρv
ρ2
+ a2
!
ρv
(ρv )2
2
ρ +a ρ
!
1
2(ρv )
ρ
The matrix has the eigenvalues
λ1 = v + a;
λ2 = v − a
where a is the speed of sound
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
46 / 84
Mathematical Challenges
Burgers Equation:
2
∂t u + ∂x u2 = 0;
u(x, 0) = 0.5 + sin(x),
x ∈ [0, 2π].
t
t
t
t
t
t
1.4
1.2
1
=
=
=
=
=
=
0.0
0.5
1.0
1.5
2.0
2.5
0.8
u
0.6
0.4
0.2
0
−0.2
−0.4
0
1
2
3
4
Dimensionless Distance
5
6
Discontinuous solutions do not satisfy PDE in the classical sense at
all points.
Need a weaker definition of solution of PDE.
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
47 / 84
Example: Euler Equations of Gas Dynamics.
∂t ρ + ∂x m = 0;
∂t m + ∂x (ρu 2 + p) = 0;
∂t E + ∂x (u(E + p)) = 0;
ρ
p = (γ − 1)(E − u 2 ).
2
Riemann problems.
uL , x < 0;
u(x, 0) =
uR , x > 0.
Mapundi K. Banda (Maties)
+ transparent boundary conditions
CIMPA/MPE 2013
Jul 22 - Aug 3
48 / 84
Example: Sod’s Riemann initial data.
N = 400,T = 0.1644
N = 400,T = 0.1644
1
1
Exact
Upw
JX
RCWENO
0.9
0.8
0.8
0.7
Pressure
Density
0.7
0.6
0.5
0.4
0.6
0.5
0.4
0.3
0.3
0.2
0.2
0.1
0.1
0
Exact
Upw
JX
RCWENO
0.9
0.2
0.4
0.6
0.8
1
0
0
x
0.2
0.4
0.6
0.8
1
x
Three waves: Rarefaction, Contact Discontinuity, Shock wave
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
49 / 84
Example: Euler Equations of Gas Dynamics in Two
dimensions.
∂t ρ + ∂x m + ∂y n = 0;
∂t m + ∂x (ρu 2 + p) + ∂y (ρuv ) = 0;
∂t n + ∂x (ρuv ) + ∂y (ρv 2 + p) = 0;
∂t E + ∂x (u(E + p)) + ∂y (v (E + p)) = 0;
ρ
p = (γ − 1)(E − (u 2 + v 2 )).
2
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
50 / 84
Double Mach Reflection
Domain Ω = [0, 4] × [0, 1].
Reflecting wall in bottom interval: [1/6, 4].
Mach 10 shock at x = 1/6, y = 0, 600 angle with x-axis.
Impose exact post-shock condition at bottom boundary [0, 1/6].
Reflective boundary for the rest of the boundary.
Top boundary exact motion of a Mach 10 shock is used.
Results at t = 0.2.
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
51 / 84
Double Mach Reflection
∆x = ∆y = 1/120
1
0.8
0.6
0.4
0.2
0
0
0.5
1
1.5
2
2.5
3
3.5
4
3
3.5
4
∆x = ∆y = 1/240
1
0.8
0.6
0.4
0.2
0
0
0.5
Mapundi K. Banda (Maties)
1
1.5
2
CIMPA/MPE 2013
2.5
Jul 22 - Aug 3
52 / 84
Existence of Solution in scalar case
Definition (the Cauchy Problem(IVP))
For scalar conservation laws:
ut + f (u)x = 0;
u : R × R+ → R,
u(x, 0) = u0 (x),
x ∈R
f :R→R
Example:
Advection Equation:
a∈R
ut + aux = 0,
Burgers Equation:
ut +
Mapundi K. Banda (Maties)
u2 2
x
CIMPA/MPE 2013
=0
Jul 22 - Aug 3
53 / 84
The Linear Advection Equation
a∈R
ut + aux = 0,
The solution of the Cauchy problem is:
u(x, t) = u0 (x − at)
Note that initial data travels unchanged with velocity a.
The left side can be interpreted as a directional derivative with:
du
= 0,
dt
dx
=a
dt
The equation dx
dt = a, x(0) = x0 is the characteristic which travels
through x0 in the (x, t)-plane, i.e. t
x = x0 + at
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
54 / 84
The Linear Advection Equation
ut + aux = 0,
a∈R
Along the characteristics the solution is constant:
Suppose (x, t) is such that x = x0 + at holds
Then
u(x, t̄) = u0 (x − at̄) = u0 (x0 )
is constant
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
55 / 84
The General Scalar Equation
ut + f (u)x = 0
which for smooth f can be written in the form
ut + f 0 (u)ux = 0
f 0 (u) is referred to as the characteristic velocity - velocity with which
information propagates
To find characteristics solve:
x 0 (t) = f 0 (u(x(t), t)),
x(0) = x0
Along the solution x(t), u(x, t) is constant:
d
∂u(x(t), t)
∂
u(x(t), t) =
+
u(x(t), t)x 0 (t)
dt
∂t
∂x
= ut + f 0 (u)ux
= 0
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
56 / 84
The General Scalar Equation
ut + f (u)x = 0
which for smooth f can be written in the form
ut + f 0 (u)ux = 0
x 0 (t) is constant since f 0 (u(x(t), t) is constant.
Characteristics are straight lines given by:
x = x(t) = x0 + f 0 (u(x0 , 0))t = x0 + f 0 (u0 (x0 ))t
If the above can be solved for x0 for all (x, t), then the solution of the
conservation law takes the form:
u(x, t) = u(x0 , 0) = u0 (x0 )
where x0 is implicitly given as:
x = x0 + f 0 (u0 (x0 ))t
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
57 / 84
Example of Mathematical challenge
ut + f (u)x = 0
which for smooth f can be written in the form
ut + f 0 (u)ux = 0
with characteristics:
x = x(t) = x0 + f 0 (u(x0 , 0))t = x0 + f 0 (u0 (x0 ))t
Equation above cannot always be uniquely solved - the characteristics
can intersect after some time.
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
58 / 84
Example: Burgers Equation
Note: f (u) =
u2
2 .
Let the initial value be given by u0 (x) = −x.
Then the characteristics are
x = x(t) = x0 − x0 t
and this gives
t=
x
x0 − x
=1−
x0
x0
and also
u(x, t) = u0 (x0 ) = u0
x
1−t
.
The characteristics intersect at the point (x, t) = (0, 1).
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
59 / 84
Example: Burgers Equation
The equations for the characteristics are x(t) = x0 + u0 (x0 ) · t.
No unique solution for (x, t) anymore; the characteristics intersect.
t
x
x
Figure: Characteristics and shock development
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
60 / 84
Example: Burgers
x
x
Figure: Refraction of waves
One attains multivalued solution of the equation.
For example, the density of gas cannot be multivalued.
In this case the discontinuity propagates as a Shock !
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
61 / 84
Recall Example
Burgers Equation:
2
∂t u + ∂x u2 = 0;
u(x, 0) = 0.5 + sin(x),
x ∈ [0, 2π].
t
t
t
t
t
t
1.4
1.2
1
=
=
=
=
=
=
0.0
0.5
1.0
1.5
2.0
2.5
0.8
u
0.6
0.4
0.2
0
−0.2
−0.4
0
Mapundi K. Banda (Maties)
1
2
3
4
Dimensionless Distance
CIMPA/MPE 2013
5
6
Jul 22 - Aug 3
62 / 84
A case for weak solutions
This can now be mathematically accurately treated with the help of
weak (not differentiable) solutions.
Definition
A function u = u(x, t) is called weak solution of the Cauchy–Problem
ut + (f (u))x = 0,
if
Z Z
∞
Z
[uφt + f (u)φx ] dt dx +
R
u(x, 0) = u0 (x),
u0 (x) φ(x, 0) dx = 0,
∀φ ∈ C01 (R × R).
0
R
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
63 / 84
A case for weak solutions
If u is a smooth solution of the equation, then
Z Z ∞
[φut + φf (u)x ] dt dx = 0, ∀φ ∈ C01 (R × R).
R
0
Integration by parts gives
Z Z ∞
Z
∞ −
φt udt + φu dx+
R
0
0
∞
Z
−
0
R
∞ φx f (u)dx + φf (u)
= 0.
−∞
So smooth solutions are also weak.
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
64 / 84
The Riemann Problems
ut + f (u)x = 0
(
ul ,
u(x, 0) = u0 (x) =
ur ,
A weak solution is e.g. the shock wave
(
ul ,
u(x, t) =
ur ,
x <0
x >0
x < st
x > st,
where s is the shock speed.
ul
ul
ur
ur
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
65 / 84
The Riemann Problems
A weak solution is e.g. the shock wave
(
ul , x < st
u(x, t) =
ur , x > st,
where s is the shock speed.
s is given by the Rankine–Hugoniot conditions
s=
Mapundi K. Banda (Maties)
f (ul ) − f (ur )
[f ]
=:
.
ul − ur
[u]
CIMPA/MPE 2013
Jul 22 - Aug 3
66 / 84
Alternative 1: Integral Form
Integrate the differential equation:
Z t2 Z x2 n
o
∂t u(x, t) + ∂x F (u(x, t)) dx dt = 0;
t1
u(x, 0) = u0 (x)
x1
Giving the following integral forms:
1
Z t2
Z t2
Z x2
F (u(x2 , t))dt
F (u(x1 , t))dt −
u(x, t1 )dx +
u(x, t2 )dx =
t1
x1
t1
x1 Z
x2
d
u(x, t)dx = F (u(x1 , t)) − F (u(x2 , t)).
dt x1
Z
2
x2
More difficult to work with than the differential equation.
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
67 / 84
Alternative 2: Weak Formulation
Introduce the “weak” form of the PDE:
1
2
Z
0
Allows discontinuous solutions and easier to work with.
Fundamental in development and analysis of numerical methods.
∞Z ∞n
o
φ∂t u(x, t) + φ∂x F (u(x, t)) dx dt = 0;
u(x, 0) = u0 (x).
−∞
Integration gives:
Z ∞Z ∞n
Z
o
φt u(x, t) + φx F (u(x, t)) dxdt =
0
φ∈
−∞
1
C0 (R
∞
φ(x, 0)u(x, 0)dx
−∞
× R)
Integration over a bounded domain.
Problem: non-uniqueness of solutions with same initial data - seek
the physically relevant solution.
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
68 / 84
Example: Burgers Equation
Consider Burgers equation
f (u) =
u2
,
2
s=
1 ul2 − ur2
1
·
= (ul + ur ).
2 ul − ur
2
t
t
U
l
U
l
Ur
Ur
x
x
Figure: Characteristics for ul > ur (left) and ul < ur (right)
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
69 / 84
Example: Burgers Equation
If the characteristics run into the shock, then it is called stable
otherwise unstable
The mean value theorem:
f (ul ) − f (ur )
s=
= f 0 (ξ), ξ ∈ (ul , ur ) or (ur , ul ).
ul − ur
i.e. one has a stable shock (see figure 5.
t
U
l
Ur
x
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
70 / 84
Example: Burgers Equation
If f is convex, i.e. f 0 is monotone increasing, then
ul > ur ⇒ f 0 (ul ) > f 0 (ξ) > f 0 (ur ),
If on the other hand
ul < ur ⇒ f 0 (ul ) < f 0 (ξ) < f 0 (ur ),
then one has an unstable shock.
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
71 / 84
Example: Burgers Equation
Another weak solution for ul < ur !
Another solution is given by the so called rarefaction wave

x
0

ul ,

t ≤ f (ul )


u(x, t) = v ( xt ), f 0 (ul ) ≤ xt ≤ f 0 (ur )



u ,
x
0
r
t ≥ f (ur )
with f 0 (v (ξ)) = ξ.
ur
t=0
t=t
ul
f’l
f’r
x=0
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
72 / 84
Example: Burgers Equation
One can easily recalculate this, since
ut + f (u)x = 0
ξ= x
x
1
+ f 0 (v )v 0 (ξ) = 0
t2
t
x
=⇒ f 0 (v ) = = ξ.
t
=⇒t −v 0 (ξ)
u=v (ξ)
For the Burger’s equation (f 0 (u) = u, v (ξ) = ξ) then


ul , x ≤ ul t



u(x, t) = xt , ul ≤ xt ≤ ur ,



u , x ≥ u t
r
r
i.e. for ul < ur , we have found two weak solutions
Not a unique solution. But only one can be physically correct.
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
73 / 84
Alternative 3: Rankine-Hugoniot
Supplement PDE by additional “jump conditions” that are satisfied
across discontinuities.
Rankine-Hugoniot conditions give the shock speed, s:
s=
[F ]
F (ul ) − F (ur )
=
[u]
ul − ur
Problem: Consider:
Burgers Equation:
∂t u + ∂x
s1 =
u2 2
Multiply by 2u
∂t (u 2 ) + ∂x
= 0;
[ 12 u 2 ]
1
= (ul + ur ).
[u]
2
s2 =
[ 32 u 3 ]
[u 2 ]
2
u 3 = 0;
3
2 ur3 − ul3 =
3 ur2 − ul2
Same smooth solution but different discontinuous solutions!!!
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
74 / 84
Why all these Mathematical Challenges?
Nature has its own way of solving the same problems, physical
intuition necessary:
1
2
Equations are models of reality and some physical effects are ignored.
Example: Fluid flow always has viscous effects: strong near
discontinuity i.e. discontinuity := thin regions with very steep
gradients.
Thus, one needs further condition, in order to eliminate the non
physical weak solutions.
Such a simple additional assumption is the Entropy-condition
(Lax):
For convex f we get ul > ur , i.e. the shock is stable.
For ul < ur the shock solution disappears. In this case the rarefaction
wave is the correct physical solution.
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
75 / 84
Unique Solutions - Vanishing Viscosity
Entropy condition I is reasonable for convex f , need a more general
solution:
1
Introduce a diffusive term in equation: Consider Linear Advection
Equation:
∂t u + a∂x u = 0;
2
u(x, 0) = u0 (x)
gives u(x, t) = u0 (x − at).
Advection-Diffusion Equation:
∂t u + a∂x u = Duxx ;
u(x, 0) = u0 (x)
F = F (u, ux ).
Parabolic, conservation law, always smooth solution t > 0 even if u is
discontinuous, for D 1 good approx. for linear advection equation.
3
Let D → 0 - “vanishing viscosity” method: useful for analysis, in
general, not optimal!
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
76 / 84
Unique Solutions - Entropy
Let η and ψ be two convex functions, for which we have for all
smooth solutions u of the conservation law:
η(u)t + ψ(u)x = 0,
i.e. η 0 (u)ut + ψ 0 (u)ux = 0.
For smooth solutions
ut + f 0 (u)ux = 0
or
η 0 (u)ut + η 0 (u)f 0 (u)ux = 0.
Giving
η 0 (u)f 0 (u) = ψ 0 (u)
and such functions (η, ψ) are called entropy–entropy flux pairs
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
77 / 84
Unique Solutions - Entropy
Sufficient for existence and uniqueness of weak solutions.
The function u(x, t) is the entropy solution if, for all convex entropy
functions, η(u), and corresponding entropy fluxes, ψ(u), the inequality
η(u)t + ψ(u)x ≤ 0,
is satisfied in the weak sense: i.e.
Z
Z Z∞
(φt η(u) + φx ψ(u)) dt dx + φ(x, 0)η(u)(x, 0) dx ≤ 0
R 0
(2)
R
∀φ ∈ C01 (R × R),
φ ≥ 0.
Definition
Let u be a weak solution of the conservation law. Moreover,suppose u
fulfills (2) for any entropy–entropy flux pair (η, ψ), then the function u is
called an entropy solution.
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
78 / 84
Unique Solutions - Entropy
For scalar conservation laws every convex function η leads to an
entropy:
i.e. there exist infinite many entropies!?
Theorem
Let f : R → R be a strict convex function and u is a piecewise smooth
solution of the conservation law ut + f (u)x = 0. If u satisfies the entropy
condition for one convex entropy η, then it will be fulfilled for every
entropy. Therefore u is an entropy solution.
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
79 / 84
Unique Solutions - Entropy
Appropriate function spaces.
Definition
Let u ∈ L∞ (Ω), Ω ⊂ Rn be open. Then the total variation of u is
defined by
Z
1
TV (u) = lim sup
|u(x + ε) − u(x)| dx.
ε→0 ε
Ω
The space of bounded variation is
BV (Ω) := {u ∈ L∞ (Ω) : TV (u) < ∞}.
If u 0 ∈ L1 (Ω) holds, then
Z
TV (u) =
|u 0 | dx.
Ω
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
80 / 84
Unique Solutions - Entropy
Theorem (Kruskov)
The scalar Cauchy-Problem
ut + (f (u))x = 0,
f ∈ C 1 (R)
u(x, 0) = u0 (x), u0 ∈ L∞ (R)
has a unique entropy solution u ∈ L∞ (R × R+ ), having the following
properties:
(i) ku(·, t)kL∞ ≤ ku0 (·)kL∞ ,
t ∈ R+
(ii) u0 ≥ v0 ⇒ u(·, t) ≥ v (·, t),
t ∈ R+
(iii) u0 ∈ BV (R) ⇒ u(·, t) ∈ BV (R) and TV (u(·, t)) ≤ TV (u0 )
Z
Z
(iv) u0 ∈ L1 (R) ⇒
u(x, t) dx = u0 (x) dx, t ∈ R+
(i)-(iv) are called
R
∞
L -stability,
Mapundi K. Banda (Maties)
R
monotonicity, TV-stability, conservativity.
CIMPA/MPE 2013
Jul 22 - Aug 3
81 / 84
Take Note!
The theorem can be extended to several dimensions x ∈ Rd , d > 1.
The theorem cannot be extended to the case of systems (n > 1).
Till now there is no general proposition proved for the system case.
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
82 / 84
References
E. Godlewski, P.-A. Raviart, Numerical Approximation of Hyperbolic
systems of Conservation Laws, Springer, Berlin, 1996
A. Klar, Numerical methods of partial differential equations II:
Hyperbolic conservation laws, Lecture Notes, Kaiserslautern.
C. Hirsch, Numerical Computation of Internal and External Flows I,
Wiley, New York, 2001.
C. Hirsch, Numerical Computation of Internal and External Flows II,
Wiley, New York, 2000.
B. Perthame, Transport Equations in Biology, Birkhäuser, Berlin,
2000.
P. D. Lax, Hyperbolic Partial Differential Equations, AMS, New York,
2006.
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
83 / 84
Acknowledgements
CIMPA, IMU, LMS-AMMSI, ICTP, AIMS
Prof. J. Banasiak, College of Mathematics, Computer Science and
Statistics, University of KwaZulu-Natal, Durban.
National Research Foundation, South Africa.
University of KwaZulu-Natal, Witwatersrand and Stellenbosch.
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
84 / 84