NM Ln3 4 PDE-2slides Page

I.Popescu:Numerical Methods I 1/25/2012

1

Hydroinformatics yModule 4: Numerical Methods I

Lecture 3 and 4: PDE, Hyperbolic PDE, Stability, Accuracy

1/25/2012 Numerical Methods 1

I.Popescu

5-Partial Differential Equations



2

PDEsparabolic

PDEs hyperbolic

eliptic

1n

x

n

t

)1,( nj


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


Other schemes– MoC ( 1-st and 2-nd order)– Preissmann scheme– Abbott-Ionescu scheme


3

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


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


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


4

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


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


xNeed to specify the solution at t=0

x=a x=b


5

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


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.


t

u

u

u

t

u

tt1 t2 t3 t4 t5


x


6


)()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.


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



7

(A)Hyperbolic PDEs


(A)Hyperbolic PDEs

Method of Characteristics



8

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


(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


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


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


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


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


)()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


)()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

1/25/2012 I.Popescu: Numerical Methods 22

adx

dtslope

1

t = t0

du = 0

x


12

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


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


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


)2(

)2(

)(

cuJ

cuJ

Ccudt

dx

Characteristic directions

A B

P

unique solution


A B


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Characteristic directions

P PP

Flow characteristics: Fr number

A BSub critical flow Super critical flow

A B


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)


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


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


P1 : t = t1

Initial conditionsC

D

A B


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Characteristics: Boundary conditions


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


x

t

C- = x ct C+ = x + ct

P(x,t)

A B


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(A)Hyperbolic PDEs

Finite difference method


FDM to hyperbolic PDE

Notations F- forward

B b d B- bacward T-time S –space

Schemes FTBS – explicit


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:


nj

nj

nj UCrCrUU 11

1

x

taCr n

j

1j0 j 1j0 j

Courant number


FTBS- boundary and initial conditions0

x

Ua

t

U

1 nj

nj

nj UCrCrUU 11

1

x

1n

t

t)1,( nj

Boundary


1j0 j 1j0 j

t),( njnconditions

Initial conditions


19


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”


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


1j0 j 1j0 j

t),( njn


20

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


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


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


1j0 j 1j0 j

tn

nj

nj

nj

nj UU

xUU 11

• two solutions for special BC

Other schemes

Lax-WendorfCranck Nicholson



22

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


• 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


Kinematic wave- pure convective



23



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


Example discretization


24

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


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)


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)


A0 B0 C0

A1 B1 C1

A2 B2 C2

. . . . . .

Ajj Bjj Cjj

0

1

2

.

. jj

D0

D1

D2

.

. Djj

=.all zeros

all zeros


25

Less equation than unknowns Use of suitable boundary conditions I d i ddi i l i bl


Introducing additional variables

Substitution of into the linear equations

Derivation of recurrence relations

jnjj

nj FE

111


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


xn

nx

f

x

ff

x

fnj

nj

n

1

with

j-1 j


26

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


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


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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