Flux Conservative Problem

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Outline

Flux Conservative Problems

Claret FelicitasJoe Reyes

Institute of MathematicsUniversity of the Philippines - Diliman

2 April 2016Math 271.2

Claret Felicitas Joe Reyes Flux Conservative Problems

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Outline

Outline

1 Flux Conservative Equation

2 1st Order Flux Conservative EquationStability Analysis of Numerical SolutionsForward Time Centered Space (FTCS)Lax Method

3 Sources of Error


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Outline



3 Sources of Error


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Outline



3 Sources of Error


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Flux Conservative Equation1st Order Flux Conservative Equation

Sources of Error

Flux Conservative Equation

A large class of PDEs can be cast into the form of a ux

conservative equation:∂ U ∂ t

= −∂ f ∂ x

∂ U , ∂ U x , ∂ U xx , . . .


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Sources of Error

Flux Conservative Equation

Example: Flux Conservative Form for the Wave EquationWe consider the 1-D wave equation U tt = c 2U xx .If we let:

w = r s , where r = c ∂ U ∂ x and s = ∂ U

∂ t .

Then ∂ w ∂ t =∂ r ∂ t ∂ s ∂ t

= ∂ ∂ t c

∂ U ∂ x

∂ ∂ t

∂ U ∂ t

=c ∂ ∂ x

∂ U ∂ t

∂ 2U ∂ t 2

=c ∂ s ∂ x c ∂ r ∂ x

.

That is,∂ w ∂ t

= − ∂ ∂ x

0 − c − c 0

w = − ∂ ∂ x

f ( w ).


f

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Sources of Error

Stability Analysis of Numerical SolutionsForward Time Centered Space (FTCS)Lax Method

1st Order Flux Conservative Equation

∂ U ∂ t

= − c ∂ U ∂ x

(1)

We introduce the change of variable ξ = x − ct and:

∂ ∂ t

= ∂ξ ∂ t

∂ ∂ξ

= − c ∂ ∂ξ

∂ ∂ x

= ∂ξ ∂ x

∂ ∂ξ

= ∂ ∂ξ

Then (1) holds.So U (x , t ) = U (ξ ) = f (x − ct ) is the analytic general solution of (1).


Fl C i E i S bili A l i f N i l S l i

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Sources of Error


1st Order Flux Conservative Equation

We study the stability of different nite difference schemes insolving the 1st order Flux Conservative or 1-D Advection Equation:

U t = − cU x , x 0 ≤ x ≤ x 1, t 0 ≤ t ≤ T

We discretize the problem: ∆ x = x 1− x 0n+1 , ∆ t = T − t 0

m .And let the following:

x j = x 0 + j ∆ x , j = 0 , . . . , n + 1 ,

t k = t 0 + k ∆ t , k = 0 , . . . , m , and

U k j = U (x j , t k ).


Fl C ti Eq ti St bilit A l i f N i l S l ti

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Sources of Error


Forward Time Centered Space (FTCS)

Using forward difference in time t and central difference in space x:

∂ U k j ∂ t

=U k +1 j − U

k j

∆ t + O (∆ t )

∂ U k j ∂ x

=U k j +1 − U

k j − 1

2∆ x + O (∆ x 2)

Using FTCS Method on U t = − cU x , we get:

U k +1 j = U k

j − c ∆ t 2∆ x

[U k j +1 − U k

j − 1] (2)


Flux Conservative Equation Stability Analysis of Numerical Solutions

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Sources of Error


von Neumann Stability Analysis of FTCS Method

DenitionLet p be a real spatial wavenumber and ξ = ξ (p ) be a complexnumber that depends on p . Then a difference equation is unstableif |ξ (p )| > 1 for some p . ξ is called the amplication factor .



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Sources of Error



We assume that independent solutions (eigenmodes) of ( 2) (or anydifference equation) are of the form:

U k j = ξ k e ipj ∆ x (3)



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To nd ξ (p ) for FTCS Method, substitute ( 3) into (2):

ξ k +1 e ipj ∆ x = ξ e ipj ∆ x − c ∆ t 2∆ x

(ξ e ip ( j +1)∆ x − ξ e ip ( j − 1)∆ x )

= ξ e ipj ∆ x − c ∆ t 2∆ x (ξ e ipj ∆ x + ip ∆ x − ξ e ipj ∆ x − ip ∆ x )

= ξ k e ipj ∆ x 1 − c ∆ t 2∆ x

(e ip ∆ x − e − ip ∆ x )

⇒ ξ (p ) = 1 − ic ∆ t

∆ x sin(p ∆ x )⇒ | ξ (p )| ≥ 1 ∀p ⇒ FTCS Method is unconditionally unstable for solvingU t = − cU x .



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Sources of Error


Lax Method

Again, we are solving the Flux Conservative Equation: U t = − cU x .Instead of U k j in approximating U t , the Lax Method uses the

average for U k j = U k j +1 + U

k j − 1

2 :

∂ U k j ∂ t

=U k +1 j −

12 [U

k j +1 + U

k j − 1]

∆ t

Using centered difference again for U x , U t = − cU x becomes:

U k +1 j = 12

[U k j +1 + U k

j − 1] − c ∆ t 2∆ x

[U k j +1 − U k

j − 1] (4)



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q1st Order Flux Conservative Equation

Sources of Error

y yForward Time Centered Space (FTCS)Lax Method

von Neumann Stability Analysis of Lax Method

Substituting U k j = ξ k e ipj ∆ x into (4):

ξ k +1 e ipj ∆ x = 1

2[ξ k e ip ( j +1)∆ x + ξ k e ip ( j − 1)∆ x ]

− c ∆ t 2∆ x

[ξ k e ip ( j +1)∆ x − ξ k e ip ( j − 1)∆ x ]

= ξ k e ipj ∆ x 12

(e ip ∆ x + e − ip ∆ x ) − c ∆ t 2∆ x

(e ip ∆ x − e − ip ∆ x )

⇒ ξ = cos (p ∆ x ) − ic ∆ t ∆ x sin(p ∆ x )



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q1st Order Flux Conservative Equation

Sources of Error

y yForward Time Centered Space (FTCS)Lax Method

von Neumann Stability Analysis of Lax Method

The Lax Method is stable when |ξ |2 ≤ 1 :

|cos 2(p ∆ x ) + c 2∆ t 2

∆ x 2 sin

2(p ∆ x )| ≤ 1

⇒ | (1 − sin2(p ∆ x )) + c 2∆ t 2∆ x 2 sin

2(p ∆ x )| ≤ 1⇒ | 1 − (1 − c

2∆ t 2∆ x 2 )sin

2(p ∆ x )| ≤ 1⇒ 1 − c

2∆ t 2∆ x 2 ≥ 0

⇒ c 2∆ t

2

∆ x 2 ≤ 1⇒ ∆ t ≤ ∆ x c , (c > 0)



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1st Order Flux Conservative EquationSources of Error

Forward Time Centered Space (FTCS)Lax Method

Courant Condition

DenitionThe Courant Condition means Lax Method is stable when∆ t ≤ ∆ x c .


Flux Conservative Equationd l

Stability Analysis of Numerical Solutionsd d ( )

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Courant Condition: Stable SchemesGeometrically, if U k

j is computed from information at points j − 1

and j + 1 at time k ,the scheme is stable when U k +1 j does notrequire information from points outside [U k j − 1, U

k j +1 ].


Flux Conservative Equation1 t O d Fl C ti E ti

Stability Analysis of Numerical SolutionsF d Ti C t d S (FTCS)



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Courant Condition: Unstable SchemesSimilarly, unstable schemes arise when the time step ∆ t becomestoo large because U k +1 j requires information from points outside[U k j − 1, U

k j +1 ].



Stability Analysis of Numerical SolutionsForward Time Centered Space (FTCS)

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von Neumann Stability Analysis for Wave Equation

U tt = c 2U xx .

Let w =

r s =

cU x U t

.

∂ w ∂ t =

∂ r ∂ t ∂ s ∂ t

=c ∂ s ∂ x c ∂ r ∂ x

= c ∂ ∂ x r s .






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Lax Method

We will solve the wave equation using the Lax method:r t = cs x becomes

r k +1

j = 12[r

k j − 1 + r

k j +1 ] +

c ∆ t 2∆ x (s

k j +1 + s

k j − 1) (5)

s t = cr x becomes

s k +1 j = 1

2[s k j − 1 + s

k j +1 ] +

c ∆ t

2∆ x (r k j +1 + r

k j − 1) (6)

where r k j = r (x j , t k ) and s k

j = s (x j , t k )




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Lax Method

For von Neumann Stability Analysis assume eigen modes forr k j and s

k j are of the form

r k j s k j

= ξ k +1 e ipj ∆ x r 0 j s 0 j

⇒ solution is stable if |ξ | ≤ 1Equations (5) and (6) give

ξ − cos (p ∆ x ) − ic ∆ t ∆ x sin(p ∆ x )− ic ∆ t ∆ x sin(p ∆ x ) ξ − cos (p ∆ x )




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Lax Method

Remark

This has a solution only if determinant = 0.This gives ξ = cos (p ∆ x ) ± i c ∆ t ∆ x sin(p ∆ x ).This is stable if |ξ |2 ≤ 1 which gives same Courant condition∆ t ≤ ∆ x c .



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st O de u Co se vat ve quat oSources of Error

Phase Errors through dispersion

We will show the effects of phase errors by studying the numericalsolution of the 1-D advection equation using different time stepswhich lead to dispersion being absent or present.

The Fourier mode U (x , t ) = e i (px + ω t ) is an exact solution of U t = − cU x if ω and p satisfy the dispersion relation ω = − cp ,then U (x , t ) = e ip (x − ct ) = f (x − ct ) gives the exact solution of U t = − cU x .

This mode is completely undamped and the amplitude is constantfor the numerical solution using a time step which satises thedispersion relation.



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qSources of Error

Dispersion relation for the Lax Method

The dispersion relation is only satised if ∆ = ∆ x c .

Consider U k j = ξ k e ipj ∆ x

ξ = cos (p ∆ x ) − ic ∆ t

∆ x sin(p ∆ x )

= e − ip ∆ x + i (1 − c ∆ t ∆ x

)sin(p ∆ x )

If we let ∆t ≤ ∆ x c

⇒ ξ = e ip ∆ x and U k j

= ξ k e ipj ∆ x =e ip (− k ∆ x + j ∆ x ) . When we substitute x j = j ∆ x , t k = k ∆ t and thedispersion relation ∆x = c ∆ t then U k j = e

ip (− ck ∆ x + j ∆ x )

= e ip (x j − ct k ) = f (x j − ctk ).



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Sources of Error

Dispersion in the numerical solution of the 1-D advectionequation using the Lax Method

Use Lax Method to solve:

U t + U x = 0 , 0 ≤ x ≤ 2 ≈ ∞ , 0 ≤ t ≤ 1

initial conditions:

U (x , 0) = 1,0.2≤ x ≤ 0.40, otherwise = U 0(x )

boundary conditions:

U (0, t ) = U (2, t ) = 0





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Sources of Error


U (x , t ) = U 0(x − ct )U 0(x − t )(c = 1)We compare the above exact solution with the numerical solutionusing the Lax method with different time steps:

∆ t = ∆ x c → no dispersion matches analytic solution.∆ t = ∆ x 2c → dispersion present but pulse matches speed of wave.∆ t = 1.001∆ x c → courant condition not met : unstable!



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Sources of Error


Equation (4) gives U k +1 j = 12 [U

k j +1 + U

k j − 1] −

c ∆ t 2∆ x [U

k j +1 − U

k j − 1]

Let s = c∆ t ∆ x

⇒ U k +1 j

= 1

2(1 − s )U k

j +1 + U k

j − 1 +

1

2(1 + s )U k

j − 1

Solve for U k +1 j for 0 ≤ k ≤ m, 1 ≤ j ≤ 3 with boundaryconditions: U k 0 = 0, U

k 4 = 0. We have:



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Sources of Error


or U k +1 = A U k b

Dispersion means the initial pulse changes shape (unlike analyticalsolution) because wave components with different frequenciestravel at different speeds.



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Sources of Error


(a) ∆ t = ∆ x 2c where dispersion is present but the pulse matches theanalytical solution for the speed of the wave.



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(b) ∆ t = ∆ x c where no dispersion is present and numerical solutionmatches analytical solution exactly.



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(c) ∆ t = 1.001∆ x c where the Courant condition is not met andsolution is becoming unstable.



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Sources of Error

Error due to nonlinear terms

Example: Shock wave equation

U t + UU x = 0

nonlinear term causes wave prole to steepen resulting in ashock.

schemes stable for linear problems can become unstable.



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Aliasing error

Example:Aliasing error occurs when a short wavelength (λ 1) is notrepresented well by the mesh-spacing (∆x ), and may bemisinterpreted as a longer wavelength oscillation (λ 2).

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Flux Conservative Problem