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I.Popescu:Numerical Methods I 1/25/2012
1
Hydroinformatics yModule 4: Numerical Methods I
Lecture 3 and 4: PDE, Hyperbolic PDE, Stability, Accuracy
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I.Popescu
5-Partial Differential Equations
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I.Popescu:Numerical Methods I 1/25/2012
2
PDEsparabolic
PDEs hyperbolic
eliptic
1n
x
n
t
)1,( nj
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eliptic1n
1j0 j 1j0 j
t),( nj
Solving PDEs
Method of characteristics Finite Difference method Explicit schemes Implicit schemes Upwind, forward/backward space/time
– FTCS– CTCS
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Other schemes– MoC ( 1-st and 2-nd order)– Preissmann scheme– Abbott-Ionescu scheme
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PDEs - Examples
The 1D advection-diffusion equation:
2
2
x
ud
x
ua
t
u
This PDE is first order in time and second order in space.
Simplify further (drop the second order diffusion or dissipation term):
0
x
ua
t
u
xxt
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This PDE is first order in time and first order in space It describes the time dependent shifting of the function u(x) along x with
a velocity a
Volunteer to solve this equation analytically?
Simple wave equation/ Advection equation
0
x
ua
t
u
Represents Kinematic wave
– u – water depth (m)– a –(celerity (m/s)
Transport of a polutant in a river
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Transport of a polutant in a river– u – concentration ( mg/l)– a –flow velocity (m/s)
Saint Venant equation – a system of two advection equations
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Necessary Information to Solve the IBVP
The Initial, Boundary Value Problem represented by the PDE
requires extra information in order to to be solvable.
What do we need?
0
x
ua
t
u
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What do we need?
IBVP Because of the hyperbolic nature of the PDE (solution travels
from right to left with increasing time), we need to supply:• Extent of solution domain• What is the solution at start of the solution process: u(x,0)• Boundary data: u(b,t), or u(a,t)
Final integration time• Final integration time.
t
we need to specify boundary ( e.g. inflow)data
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xNeed to specify the solution at t=0
x=a x=b
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Analytical solution of Hyperbolic PDEs 1
)()0(
0
fIC
x
ua
t
u Advection equation: • it describes the time dependent shiftingof the function u(x,0) along x with a velocity a
t
u
u
)()0,(: xfxuIC
x
u
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t
The solution at any time t > t0 can be described as a function of the state at time t0:
)0,(),( xutatxu This is a so-called initial value problem in which the state at any time t > t0 can be uniquelyfound when the state at time t = t0 is fully given.
Analytical solution of Hyperbolic PDEs 2
t
u
u
u
t
u
tt1 t2 t3 t4 t5
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x
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Analytical solution of Hyperbolic PDEs 3
)()0(
0
fIC
x
ua
t
u Advection equation: u
)()0,(: xfxuIC
)0,(),( atxutxu 1. This is a so-called initial value problem in which the state at any time t > t0 can be
uniquely found when the state at time t = t0 is fully given.
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2. The initial value problem is quite trivial, yet, as we will see below, this problemstands at the basis of numerical methods of hydrodynamics and is numerically surprisinglychallenging to solve!
Brief Summary
There is a checklist of conditions we will need to consider to obtain a unique solution of a PDE:solution of a PDE:
1) The PDE2) Boundary values (also known as boundary conditions)3) Initial values (if there is a time-like variable)4) Solution domain
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I.Popescu:Numerical Methods I 1/25/2012
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(A)Hyperbolic PDEs
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(A)Hyperbolic PDEs
Method of Characteristics
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The method of characteristics
The method of characteristics is a method which can be used to solve the initial value problem (IVP) for general first order PDEsproblem (IVP) for general first order PDEsby transforming them in ODEs.Consider the first order linear equation :
(5 1)0),(),(),(),(
txutxct
utxb
utxa
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(5.1)
With initial condition
tx
)()0,( xfxu
GoalChange coordinates from (x,t) to new coordinate
system (xo,s) so that PDE becomes ODE along certain curves in (x,t) plane.Such curves, along which the solution of the PDE
reduces to an ODE, are called the characteristic curves or just the characteristics. The new variable s will vary, and the new variable
x0 will be constant along the characteristics. The
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variable x0 will change along the initial curve in the x-t plane (along the line t=0). How do we find the characteristic curves?
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Finding characteristic curves
Notice that if we choose (5.2)
),(and),( txbd
dttxa
d
dx
Then
t
u
ds
dt
x
u
ds
dx
ds
du
)()(dsds
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And along the characteristic curves
(5.3) -ODE0)(),( sutxcds
du
Application steps Solve the two characteristic equations, (5.2). Find the constants of integration by setting x(0)=x0 (these will be
points along the t=0 axis in the x-t plane) and t(0)=0. We now have the transformation from (x,t) to (x0,s), x=x(x0,s) and t=t(x0,s)
xo x
t
Characteristic curve
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Solve the ODE (5.3) with initial condition u(0)=f(x0), where x0 are the initial points on the characteristic curves along the t=0 axis in the x-t plane.
We now have a solution u(x0,s) . Solve for s and x0 in terms of x and t (using the results of step 1) and substitute these values in u(x0,s) to get the solution to the original PDE as u(x,t)
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The MoC summary
0),(),(),(),(
txutxct
utxb
x
utxa
)()0,( xfxu
),(and),( txbds
dttxa
ds
dx
t
u
ds
dt
x
u
ds
dx
ds
du
For the selection :
and
)( td
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0)(),( sutxcds
du holds along the characteristic curves),(
),(
txb
txa
dt
dx
CasesDepending on values of a(x,t), b(x,t) and c(x,t)
there are 3 different PDEs: the constant coefficient advection equation,the constant coefficient advection equation, the variable coefficient advection equation,
inviscid Burgers' equation.
For all three examples, the initial conditions are specified as
)()0( f
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)()0,( xfxu
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Case 1: c(x,t)=0 and b(x,t)=1
0ds
du
),(),(
txatxadx
consttxu ),(
),( txadx
along
)()0,( xfxu
),(),(
txatxbdt
),(dt
)()0,( xfxu
0x
ua
t
u
Constant coefficient advection equation
tCharacteristic curve
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)()0,( xfxu I.C. Application steps:
dx/dt =a =>x=x0+at du/dt =0 =>u(t) =f(x0) u(t)=f(x-at) xo x
Characteristics of first-order PDEs
u = f(s ) along the characteristic direction u =constant
Characteristic t
t = t1
t = t2
t = t3lines= s1 s = s2 s = s3
d 0
t
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adx
dtslope
1
t = t0
du = 0
x
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Characteristics of first-order PDEs• Along the characteristic direction:
du = 0, u = constantu = f(s ) = constant
h l i i hNon-characteristic
•The solution remains the same along the characteristic direction•An observer moving with
s = constant sees no changes the form of u •The profile will change if the observer moves faster or slower than t t
t = t2
t = t3
Characteristic linelinet
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observer moves faster or slower than the characteristic line
t = t0
t = t1
x
Case 2: Variable coefficient advection equationApplication steps:
dx/dt =a =>x=G(a) du/dt =0 =>u(t) =f(x0)
U(t)
),(
0
txgax
ua
t
u
U(t)
)()0,(: xfxuIC
t
h
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xo x
Characteristic curve
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Case 2 Application: Saint Venant equations
Characteristic form of Saint Venant eq
0)2(
)()2( cu
cut
cu
0)2(
)()2(
x
cucu
t
cuxt
)(
dx
Ccudt
dx
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)2(
)2(
)(
cuJ
cuJ
Ccudt
dx
Characteristic directions
A B
P
unique solution
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A B
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Characteristic directions
P PP
Flow characteristics: Fr number
A BSub critical flow Super critical flow
A B
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Physical PlaneHyperbolic equation – propagation problem with no dissipation
t
C- = x ct C+ = x + ct
Domain of D d
Domain of Influence
P(x,t)
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xDependence
Initial conditions
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Characteristics - PropagationDomain of Dependence
P3 : t = t3
B d
P2 : t = t2
Boundary Conditions
Boundary Conditions
E
F
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P1 : t = t1
Initial conditionsC
D
A B
Characteristics - PropagationDomain of Influence
PB d
P2 : t = t2
P3 : t = t3Boundary Conditions
Boundary Conditions
E
F
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P1 : t = t1
Initial conditionsC
D
A B
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Characteristics: Boundary conditions
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SummaryThe conditions derived along characteristics enable us to
compute solutions from known conditions at an earlier point in time
Example If state of fluid in A is U=1m/s and h=5m, in B
u=1.2m/s and h=4.8 m, compute u and h in P
t
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x
t
C- = x ct C+ = x + ct
P(x,t)
A B
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(A)Hyperbolic PDEs
Finite difference method
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FDM to hyperbolic PDE
Notations F- forward
B b d B- bacward T-time S –space
Schemes FTBS – explicit
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FTBS explicit BTBS – implicit FTCS in n CTCS in n
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FDM-Hyperbolic PDE-Explicit schemes
FTBS
0
x
Ua
t
U
t
UU
t
Unj
nj
1
x
UU
x
Unj
nj
1
x
1n
t
t),( nj
)1,( nj
n
FT:
BS:
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nj
nj
nj UCrCrUU 11
1
x
taCr n
j
1j0 j 1j0 j
Courant number
FDM-Hyperbolic PDE-Explicit schemes
FTBS- boundary and initial conditions0
x
Ua
t
U
1 nj
nj
nj UCrCrUU 11
1
x
1n
t
t)1,( nj
Boundary
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1j0 j 1j0 j
t),( njnconditions
Initial conditions
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FDM-Hyperbolic PDE-Explicit schemes
FTBS- boundary and initial conditions0
x
Ua
t
U
nnn UCC UU 11 nj
nj
nj UCrCrUU 11
1
For a given point, a stencil is a fixed subset of nearest neighbors
A stencil code updates every point in a regular grid by “applying a stencil”
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FDM-Hyperbolic PDE-Implicit schemes
BTBS0
x
Ua
t
U
11 0)1( 111
nj
nj
nj UCrUCrCrU
x
taCr n
j
x
1n
t
t)1,( nj
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1j0 j 1j0 j
t),( njn
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FDM-Hyperbolic PDE-Implicit schemes
BTBS- boundary and initial conditions0
x
Ua
t
U
0)1( 11 nnn UCUCC U
x
1n
t
t)1,( nj
Boundary
0)1( 111
nj
nj
nj UCrUCrCrU
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1j0 j 1j0 j
t),( njnconditions
Initial conditions
FDM-Hyperbolic PDE – (Explicit)Schemes
FTCS0
x
Ua
t
U
x
1n
t
t),( nj
)1,( nj
n
FT: CS:t
UU nj
nj
1
x
UU nj
nj
211
02
111
x
UUa
t
UU nj
nj
nj
nj
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1j0 j 1j0 j n
jnj
nj
nj UUCrUU 11
1
2
1
• unconditionally unstable
• used to compute very first time steps
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FDM-Hyperbolic PDE-Upwind schemes Any BS scheme is also called upwind schemes, because
information comes from the upstream Similarly the CS schemes are called centered schemes The CTCS scheme
UU0
x
Ua
t
U
x
1n
t
t
022
1111
x
UUa
t
UU nj
nj
nj
nj
nnnn UUta
UU 11
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1j0 j 1j0 j
tn
nj
nj
nj
nj UU
xUU 11
• two solutions for special BC
Other schemes
Lax-WendorfCranck Nicholson
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FD Schemes for hyperbolic system of equations
Saint Venant equations: Unsteady, nearly horizontal flow
AQWhere:
0
q
2
2
ARC
QgQ
x
hgA
x
AQ
t
Q
t
A
x
Q
Q - discharge, m3 s-1
A - flow area, m2
q - lateral flow, m2s-1
h - depth above datum, mC - Chezy resistance coefficient, m1/2s-1
R - hydraulic radius, m- momentum distribution coefficient
Variables Conditions for solution
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• two independent (x, t)
• two dependent (Q, h)
• 2 point initial (Q, h)
• 1 point up/downstream
– h
– Q
– Q=f(h)
Fully dynamicDiffusive wave – no inertia
FD Schemes for hyperbolic system of equations
Kinematic wave- pure convective
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FD Schemes for hyperbolic system of equations
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Abbot – Ionescu scheme
Structured, cartesian grid Implicit scheme (Abbott-Ionescu)
C ti it ti h t d• Continuity equation - h centered
• Momentum equation - Q centered
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Example discretization
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Abbott Ionescu scheme
02
x
uugh
t
uu
t
hg 02
x
hugh
t
hu
t
uh
t
hh
t
h nj
nj
1
t
uu
t
uu
t
unj
nj
nj
nj 1
111
11
2
1
uuuuu
nj
nj
nj
nj1 11
11
11
t
uu
t
unj
nj
1
11
t
uu
t
uu
t
hnj
nj
nj
nj 2
12
1
2
1
uuuuh n
jnj
nj
nj1 2
112
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x
uu
x
uu
x
u jjjj
222
1 1111
xxx
h jjjj
222
1 22
Transformation into linear equations
jnjj
njj
njj DQChBQA 1111 1
111
1
(mass)
Abbot – Ionescu scheme
jnjj
njj
njj DhCQB1hA 111 1
111
1
jnjj
njj
njj DCBA 1111 1
111
1
Tri-diagonal matrix form of equation
A B C Dn+1 n
(momentum)
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A0 B0 C0
A1 B1 C1
A2 B2 C2
. . . . . .
Ajj Bjj Cjj
0
1
2
.
. jj
D0
D1
D2
.
. Djj
=.all zeros
all zeros
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Less equation than unknowns Use of suitable boundary conditions I d i ddi i l i bl
Abbot – Ionescu scheme
Introducing additional variables
Substitution of into the linear equations
Derivation of recurrence relations
jnjj
nj FE
111
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jjj
jjjj
jjj
jj
BEA
CADF
BEA
CE
1
1
4-points stencil (2 points in space, 2 levels in time)
t
x
f
Preismann scheme 0
x
fa
t
f
f(x,t) can be either u, either h
n+11-
1 nx
nn x
f
x
f
x
f
11
with
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xn
nx
f
x
ff
x
fnj
nj
n
1
with
j-1 j
I.Popescu:Numerical Methods I 1/25/2012
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4-points stencil (2 points in space, 2 levels in time)
Preismann scheme 0
x
fa
t
f
f(x,t) can be either u, either h
t 1
t
f
t
f
t
f tn+1
1-jt
f
1
jt
f
1 jj ttt
x
ff
t
fnj
nj
j
1
with
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xn
j-1 j
t
n+1
t
U
fffffnj
nj
nj
nj
1
11
1
1
FDM-Hyperbolic PDE-Preismann scheme
0
x
fa
t
f
x n
j-1 j
t
n+1
U
ttt
x
ff
x
ff
x
fnj
nj
nj
nj
1
11
1
1
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x n
j-1 j
x xxx
< 0.5 - fully unstable
= 0, the scheme is explicit, = 1, the scheme is fully implicit
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