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Uniquenessfor reactiondiffusionmodel

Stability fordiffusionmodel

Finitedifferences

Numericalanalysis

Finitedifferences forconvectiondiffusionmodel

Part III

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Tony Shardlow

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Numericalanalysis


Outline

1 Uniqueness for reaction diffusion model

2 Stability for diffusion model

3 Finite differences

4 Numerical analysis

5 Finite differences for convection diffusion model

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Numericalanalysis


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Finitedifferences

Numericalanalysis


A model reaction diffusionproblem

Model reaction diffusion problem

For a constant r 0,

u + ru = f for 0 < x < 1

u(0) = , u(1) = .

This is a two point Boundary Value Problem (BVP).

It is a simple model of a system with reaction and diffusion in

equilibrium.

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Numericalanalysis


Example soln 1

u = 1 for 0 < x < 1, (1.1)

u(0) = 0; u(1) = 0.

The solution is u(x) =1

2(x x2) .

0 0.5 10

1/8

x

temperature

Problem 1.1

0 0.5 10

1/(w^2)

Problem 1.2

defle

ction

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Numericalanalysis


Example soln 2

u + w2u = 1 for 0 < x < 1 (1.2)

u(0) = 0; u(1) = 0.

The solution is

u(x) =1

w2

(exp(wx) + exp(w(1 x)))

w2(1 + exp(w)).

0 0.5 10

1/8

temperature

Problem 1.1

0 0.5 10

1/(w^2)

Problem 1.2

deflection

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Uniqueness for reaction diffusionmodel

u + ru = f for 0 < x < 1

u(0) = , u(1) = .

Definition (well posed)

A boundary value problem is well posed ifexistence a solution exists

uniqueness the solution is unique

stability the solution depends continuously on the data.

We know solutions exists (using SoV) .

We now look at uniqueness .

HOMEWORK

You can now try Problem 17/120

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Stability for

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Finitedifferences

Numericalanalysis


Uniqueness for reaction diffusionmodel

u + ru = f for 0 < x < 1

u(0) = , u(1) = .

Definition (well posed)

A boundary value problem is well posed ifexistence a solution exists

uniqueness the solution is unique

stability the solution depends continuously on the data.

We know solutions exists (using SoV) .

We now look at uniqueness .

HOMEWORK

You can now try Problem 18/120

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Numericalanalysis


Theorem (uniqueness of soln)

There is at most one solution u to

u + ru = f, u(0) = , u(1) = ,

where r 0.

Proof.Suppose that u, v are solns.As the PDE is linear,

the difference w(x) = u(x) v(x) satisfies

w + r w = 0, w(0) = 0, w(1) = 0.

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Proof ctd.

10

(w)2 dx 0

+r1

0w2 dx 0

= 0,

with r 0.

The first term implies that w(x) = 0 for x (0, 1).Hence w is constant in (0, 1).

As w(0) = 0 and is constant, we conclude that w 0.As w = u v, we see u v and the two solutions are thesame.

We have proved uniqueness of solutions.

HOMEWORK

You can now try Problem 2 and 3

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Numericalanalysis


Proof ctd.

10

(w)2 dx 0

+r1

0w2 dx 0

= 0,

with r 0.




HOMEWORK


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Numericalanalysis


Proof ctd.

10

(w)2 dx 0

+r1

0w2 dx 0

= 0,

with r 0.




HOMEWORK


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Numericalanalysis


Proof ctd.

10

(w)2 dx 0

+r1

0w2 dx 0

= 0,

with r 0.




HOMEWORK


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Numericalanalysis


Proof ctd.

10

(w)2 dx 0

+r1

0w2 dx 0

= 0,

with r 0.




HOMEWORK


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Numericalanalysis


Outline






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Stability for

diffusionmodel

Finitedifferences

Numericalanalysis


Diffusion problem

The third condition for a well posed problem is stability or

continuous dependence on the problem data.For simplicity, we investigate stability for the diffusion problem(case r = 0 of the reaction diffusion model).

Model diffusion problem

u = f for 0 < x < 1

u(0) = ; u(1) = .

To establish stability, we show that u(x)

depends continuously on boundary data ( and ).

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diffusionmodel

Finitedifferences

Numericalanalysis


Diffusion problem




u = f for 0 < x < 1

u(0) = ; u(1) = .



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Finitedifferences

Numericalanalysis

Finite

differences forconvectiondiffusionmodel

Diffusion problem




u = f for 0 < x < 1

u(0) = ; u(1) = .



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Finitedifferences

Numericalanalysis

Finite


Maximum principle

Our main tool is the maximum principle.

Lemma (maximum principle)

Suppose that f (x) < 0 for all x (0, 1).Ifu = f , then u(x) attains its maximum value at one of thetwo end points x = 0, 1.

Proof.

Suppose for a contradiction that (0, 1) is a local maximum.

From calculus, u() 0 and u() = 0.Hence u() 0.But we assumed u = f < 0 for all x (0, 1).Proved by contradiction.

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diffusionmodel

Finitedifferences

Numericalanalysis

Finite


Maximum principle

Our main tool is the maximum principle.

Lemma (maximum principle)

Suppose that f (x) < 0 for all x (0, 1).Ifu = f , then u(x) attains its maximum value at one of thetwo end points x = 0, 1.

Proof.

Suppose for a contradiction that (0, 1) is a local maximum.

From calculus, u() 0 and u() = 0.Hence u() 0.But we assumed u = f < 0 for all x (0, 1).Proved by contradiction.

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( )

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Finitedifferences

Numericalanalysis

Finite


Upper bound on soln u(x)

Lemma

Ifu(x) = 0 for all x (0, 1),then u(x) M for x [0, 1] and M = max{u(0), u(1)}.

Proof.

For > 0, letv(x) = u(x) + x

2.

Then v = u 2 = 2 < 0 for x (0, 1).

By the maximum principle, v(x) max{v(0), v(1)} .

Now u(x) = v(x) x2 v(x). Hence,

u(x) v(x) max{v(0), v(1)} = max{u(0), u(1) + }.

As this holds for any > 0, we are done

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U b d l ( )

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Finitedifferences

Numericalanalysis

Finite



Lemma


Proof.


2.

Then v = u 2 = 2 < 0 for x (0, 1).



u(x) v(x) max{v(0), v(1)} = max{u(0), u(1) + }.


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Finite



Lemma


Proof.


2.

Then v = u 2 = 2 < 0 for x (0, 1).



u(x) v(x) max{v(0), v(1)} = max{u(0), u(1) + }.


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Finite



Lemma


Proof.


2.

Then v = u 2 = 2 < 0 for x (0, 1).



u(x) v(x) max{v(0), v(1)} = max{u(0), u(1) + }.


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Numericalanalysis

Finite


Lower bound on soln u(x)

Lemma

Ifu(x) = 0 for all x (0, 1)then m u(x) for x [0, 1] and m = min{u(0), u(1)}.

Proof.

Define w(x) = u(x).Then w(x) = 0 for x (0, 1).By previous lemma,

w(x) max{w(0), w(1)}.

and using w = u

u(x) max{u(0), u(1)} = min{u(0), u(1)}

and so u(x) min{u(0), u(1)} 26/120

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Finite


Lower bound on soln u(x)

Lemma

Ifu(x) = 0 for all x (0, 1)then m u(x) for x [0, 1] and m = min{u(0), u(1)}.

Proof.

Define w(x) = u(x).Then w(x) = 0 for x (0, 1).By previous lemma,

w(x) max{w(0), w(1)}.

and using w = u

u(x) max{u(0), u(1)} = min{u(0), u(1)}

and so u(x) min{u(0), u(1)} 27/120

Summary

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Stability for

diffusionmodel

Finitedifferences

Numericalanalysis

Finite


Summary

Theorem

Let u be a smooth solution of

u = 0 for 0 < x < 1

u(0) = ; u(1) = ;

For all x (0, 1),

min{, } u(x) max{, }.

HOMEWORK

You can now try Problem 4

We now use this to show stability with respect to changes inthe boundary data , .

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Stability for

diffusionmodel

Finitedifferences

Numericalanalysis

Finite


Summary

Theorem

Let u be a smooth solution of

u = 0 for 0 < x < 1

u(0) = ; u(1) = ;

For all x (0, 1),

min{, } u(x) max{, }.

HOMEWORK


We now use this to show stability with respect to changes inthe boundary data , .

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Stability for

diffusionmodel

Finitedifferences

Numericalanalysis

Finite


boundary data

Theorem

Suppose that

u = f for 0 < x < 1

u(0) = ; u(1) = .

(D)

u = f for 0 < x < 1

u(0) = + 0; u(1) = + 1.

(D)

Then

supx(0,1)

|u(x) u(x)| max{|0|, |1|}.

Small changes (0, 1) to boundary data

cause small changes to the solution u

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Stability for

diffusionmodel

Finitedifferences

Numericalanalysis

Finite


Proof

Let e = u u. Subtracting (D) from (D) gives

e = 0 for 0 < x < 1

e(0) = 0; e(1) = 1.

As homogeneous, previous theorem implies the stability bound:

min{0, 1} e(x) max{0, 1}

This implies|e(x)| max{|0|, |1|}.

HOMEWORK


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Stability for

diffusionmodel

Finitedifferences

Numericalanalysis

Finite


Proof


e = 0 for 0 < x < 1

e(0) = 0; e(1) = 1.


min{0, 1} e(x) max{0, 1}


HOMEWORK


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diffusionmodel

Finitedifferences

Numericalanalysis

Finite


Proof


e = 0 for 0 < x < 1

e(0) = 0; e(1) = 1.


min{0, 1} e(x) max{0, 1}


HOMEWORK


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Outline

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Numericalanalysis

Finite


Outline






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Scientific computing

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diffusionmodel

Finitedifferences

Numericalanalysis

Finite



Scientific computing involves constructing numerical

solution techniques for mathematical models and usingcomputers to analyse and solve models that arise inscience and engineering.

Without computing, we could not find or even approximatesolutions to most mathematical models and PDEs.

Numerical analysis is the mathematical theoryunderpinning the techniques used in computationalscience. It aims to show existing algorithms are efficientand accurate and develop better algorithms.

We introduce one numerical solution technique for PDEs,known as finite differences.

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diffusionmodel

Finitedifferences

Numericalanalysis

Finite








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diffusionmodel

Finitedifferences

Numericalanalysis

Finite


Sc e t c co put g






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Numericalanalysis

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p g






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Finite difference approximation

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diffusionmodel

Finitedifferences

Numericalanalysis

Finite


pp

The idea of the finite difference method is to approximate

u(x) by uj at x = xj for xj on a grid

Definition (grid)

A uniform grid on [0, 1] is defined by

xj = jh, j = 0, . . . , n, h = 1/n.

0 = x0 x1 x2 x3 xn = 1

h =1

n

h h

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Finitedifferences

Numericalanalysis

Finite


grid

Definition (centered differencing)

To approximate derivatives u

(xj) at a grid point x = xj, write

u(xj) 1

hu(xj) ,

where u(x) = u(x + h/2) u(x h/2).

xj h

2

xj +h

2

u(xj h)

u(xj)

u(xj +h)

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Error

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Finitedifferences

Numericalanalysis

Finite


Theorem (Taylors theorem)

u(x + h) = u(x) + hu(x) +1

2h2u(x) + +

1

n!hnu(n)()

for some (x, x + h).

Then

u(x + h) u(x) = hu(x) +1

2h2u()

= hu

(x) + O(h2

) ,

where we use the notation O(h2) for short. It means anyquantity bounded by Kh2 for some constant K.

HOMEWORK

You can now try Problem 7 and 8 41/120

Error

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Finitedifferences

Numericalanalysis

Finite



u(x + h) = u(x) + hu(x) +1

2h2u(x) + +

1

n!hnu(n)()


Then

u(x + h) u(x) = hu(x) +1

2h2u()

= hu

(x) + O(h2

) ,where we use the notation O(h2) for short. It means anyquantity bounded by Kh2 for some constant K.

HOMEWORK


Error

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Finitedifferences

Numericalanalysis

Finite



u(x + h) = u(x) + hu(x) +1

2h2u(x) + +

1

n!hnu(n)()


Then

u(x + h) u(x) = hu(x) +1

2h2u()

= hu

(x) + O(h2


HOMEWORK


Error

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Finitedifferences

Numericalanalysis

Finite



u(x + h) = u(x) + hu(x) +1

2h2u(x) + +

1

n!hnu(n)()


Then

u(x + h) u(x) = hu(x) +1

2h2u()

= hu

(x) + O(h2


HOMEWORK


Lemma

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Finitedifferences

Numericalanalysis

Finite


Lemma

The error in approximating u(xj) by u(xj)/h is

ej = u(xj) 1h

u(xj) = O(h2).

Proof.

By Taylors theorem,

u(x + h/2) =u(x) +h

2u(x) +

h2

8u(x) + O(h3)

u(x h/2) =u(x) h

2u(x) +

h2

8u(x) + O(h3)

so u(xj)

h=

(u(xj + h/2) u(xj h/2))

h

=hu(xj) + O(h

3)

h= u(xj) + O(h

2).

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Lemma

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Finitedifferences

Numericalanalysis

Finite


Lemma


ej = u(xj) 1h

u(xj) = O(h2).

Proof.

By Taylors theorem,

u(x + h/2) =u(x) +h

2u(x) +

h2

8u(x) + O(h3)

u(x h/2) =u(x) h

2u(x) +

h2

8u(x) + O(h3)

so u(xj)

h=

(u(xj + h/2) u(xj h/2))

h

=hu(xj) + O(h

3)

h= u(xj) + O(h

2).

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Lemma

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Finitedifferences

Numericalanalysis

Finite


Lemma


ej = u(xj) 1h

u(xj) = O(h2).

Proof.

By Taylors theorem,

u(x + h/2) =u(x) +h

2u(x) +

h2

8u(x) + O(h3)

u(x h/2) =u(x) h

2u(x) +

h2

8u(x) + O(h3)

so u(xj)

h=

(u(xj + h/2) u(xj h/2))

h

=hu(xj) + O(h

3)

h= u(xj) + O(h

2).

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Finitedifferences

Numericalanalysis

Finite


Definition (second centered difference)

To approximate the second derivative, write

1

h22u(xj) u

(xj)

where

2u(xj) =[u(xj)]

=u(xj +h

2) u(xj

h

2)

= u(xj + h) u(xj) (u(xj) u(xj h))

= u(xj + h) 2u(xj) + u(xj h) .

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Finitedifferences

Numericalanalysis

Finite


Definition (second centered difference)

To approximate the second derivative, write

1

h22u(xj) u

(xj)

where

2u(xj) =[u(xj)]

=u(xj +h

2) u(xj

h

2)

= u(xj + h) u(xj) (u(xj) u(xj h))

= u(xj + h) 2u(xj) + u(xj h) .

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Finite differences for modelreaction diffusion PDE

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Stability fordiffusion

model

Finitedifferences

Numericalanalysis

Finite


reaction diffusion PDE

Consider the reaction diffusion equation

u + ru = f, u(0) = u(1) = 0.

At the grid points xj = jh where h = 1/n, the PDE is

u(xj) + ru(xj) = f(xj) j = 1, 2, . . . , n 1.

Let fj = f(xj) and make approximations

u(xj) uj , u(xj)

2uj/h2.

Finite difference approximationuj is soln of

1

h22uj + ruj = fj j = 1, 2, . . . , n 1.

with boundary conditions u0 = un = 050/120


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model

Finitedifferences

Numericalanalysis

Finite


reaction diffusion PDE


u + ru = f, u(0) = u(1) = 0.


u(xj) + ru(xj) = f(xj) j = 1, 2, . . . , n 1.


u(xj) uj , u(xj)

2uj/h2.

Finite difference approximation uj

is soln of

1

h22uj + ruj = fj j = 1, 2, . . . , n 1.



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model

Finitedifferences

Numericalanalysis

Finite


reaction diffusion PD


u + ru = f, u(0) = u(1) = 0.


u(xj) + ru(xj) = f(xj) j = 1, 2, . . . , n 1.


u(xj) uj , u(xj)

2uj/h2.

Finite difference approximation uj

is soln of

1

h22uj + ruj = fj j = 1, 2, . . . , n 1.


Finite differences for modelreaction diffusion problem

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model

Finitedifferences

Numericalanalysis


p

Definition (finite difference method)

The finite difference method is to find uj u(xj) such that

1

h22uj + ruj = fj j = 1, 2, . . . , n 1, ()

and u0 = un = 0.

As2uj = uj1 2uj + uj+1,

the finite difference method is

1

h2(uj1 2uj + uj+1) + r uj = fj j = 1, 2, . . . , n 1.

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Finite differences for modelreaction diffusion problem

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model

Finitedifferences

Numericalanalysis


Definition (finite difference method)

The finite difference method is to find uj u(xj) such that

1

h22uj + ruj = fj j = 1, 2, . . . , n 1, ()

and u0 = un = 0.

As2uj = uj1 2uj + uj+1,

the finite difference method is

1

h2(uj1 2uj + uj+1) + r uj = fj j = 1, 2, . . . , n 1.

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Linear system of eqns

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model

Finitedifferences

Numericalanalysis


1h2

(uj1 2uj + uj+1) + ruj = fj j = 1, 2, . . . , n 1.

and rearranging

1

h

2uj1 + (

2

h

2+ r)uj

1

h

2uj+1 = fj.

The BCs u(0) = u(1) = 0 give u0 = 0 and un = 0 . Writeas a system of linear equations:

( 2h2

+ r) 1h2

0

. . . . . . 1

h2( 2h2

+ r) 1h2

. . .. . .

0 1h2

( 2h2

+ r)

u1...

uj...

un1

=

f1...fj...

fn1

55/120

Example (r = 0, f(x) = 1)

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Part III

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model

Finitedifferences

Numericalanalysis


Take n = 6, i.e. h = 1/6. The system is

72 36 0 0 036 72 36 0 0

0 36 72 36 00 0 36 72 36

0 0 0 36 72

u1u2u3u4

u5

=

1111

1

Solve the system of equations (e.g., with MATLAB):

u1 = u5 = 5/72;

u2 = u4 = 1/9 = 8/72;u3 = 1/8 = 9/72.

Note that uj = u(xj) =12 (xj x

2j ). That is, the finite

difference solution is exact in this example.

56/120

Example (r = 0, f(x) = 1)

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Part III

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model

Finitedifferences

Numericalanalysis


Take n = 6, i.e. h = 1/6. The system is

72 36 0 0 036 72 36 0 0

0 36 72 36 00 0 36 72 36

0 0 0 36 72

u1u2u3u4

u5

=

1111

1

Solve the system of equations (e.g., with MATLAB):

u1 = u5 = 5/72;

u2 = u4 = 1/9 = 8/72;u3 = 1/8 = 9/72.

Note that uj = u(xj) =12 (xj x

2j ). That is, the finite

difference solution is exact in this example.

57/120

Example

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model

Finitedifferences

Numericalanalysis


u

+ 16u

= 1for

0 0 if the local error satisfies

|Tj| = O(hk) j = 1, 2, . . . , n 1.

Theorem (2nd order consistent)

The method is 2nd order consistent if the local truncation errorsatisfies

|Tj| = O(h2), j = 1, 2, . . . , n 1.

67/120

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model

Finitedifferences

Numericalanalysis

Finitedifferences for

convectiondiffusionmodel

Definition (kth order consistent)

A finite difference scheme is said to be kth order consistent fork > 0 if the local error satisfies

|Tj| = O(hk) j = 1, 2, . . . , n 1.

Theorem (2nd order consistent)

The method is 2nd order consistent if the local truncation errorsatisfies

|Tj| = O(h2), j = 1, 2, . . . , n 1.

68/120

Proof.

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Part III

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model

Finitedifferences

Numericalanalysis



By Taylors theorem,

u(x h) =u(x) hu(x) + h22

u(x) h36

u(x) + O(h4)

u(x + h) =u(x) + hu(x) +h2

2u(x) +

h3

6u(x) + O(h4).

andTj = u

(xj) 1

h2(u(xj h) 2u(xj) + u(xj + h)).

We can add the expansions for u(x h) and u(x + h), to find

u(xj h) + u(xj + h) = 2u(xj) + h2u(xj) + O(h

4).

Rearranging givesTj = O(h

2).

69/120

Global error

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model

Finitedifferences

Numericalanalysis



HOMEWORK


We relate the global error ej at the grid point xj to the localtruncation error Tj

Tj =

1

h2 2

u(xj) + ru(xj) fj j = 1, 2, . . . , n 1,

0 = 1

h22uj + ruj fj j = 1, 2, . . . , n 1.

Subtracting these equations and letting ej = u(xj) uj gives

Tj = 1

h22ej + rej j = 1, 2, . . . , n 1.

We have a linear system of eqns for ej.70/120

Global error

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model

Finitedifferences

Numericalanalysis



HOMEWORK



Tj =

1

h2 2

u(xj) + ru(xj) fj j = 1, 2, . . . , n 1,

0 = 1

h22uj + ruj fj j = 1, 2, . . . , n 1.


Tj = 1

h22ej + rej j = 1, 2, . . . , n 1.

We have a linear system of eqns for ej.

71/120

Global error

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model

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Numericalanalysis



HOMEWORK



Tj =

1

h2 2

u(xj) + ru(xj) fj j = 1, 2, . . . , n 1,

0 = 1

h22uj + ruj fj j = 1, 2, . . . , n 1.


Tj = 1

h22ej + rej j = 1, 2, . . . , n 1.


72/120

Global error

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model

Finitedifferences

Numericalanalysis



HOMEWORK

You can now try Problem 9 and 10We relate the global error ej at the grid point xj to the localtruncation error Tj

Tj =

1

h2 2

u(xj) + ru(xj) fj j = 1, 2, . . . , n 1,

0 = 1

h22uj + ruj fj j = 1, 2, . . . , n 1.


Tj = 1

h22ej + rej j = 1, 2, . . . , n 1.


73/120

Stability theorem for finitedifferences

Theorem (Stability theorem)

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model

Finitedifferences

Numericalanalysis



Theorem (Stability theorem)

Suppose that {uj}nj=0 satisfy the tridiagonal system of

equations

auj1 + buj cuj+1 0 j = 1, 2, . . . , n 1.

Let a, b, c denote real coefficients with

a 0c 0

b a + c > 0

(S)

then

uj max{0, u0, un} for all j = 0, 1, 2, . . . , n.

74/120

Proof.

Suppose for a contradiction there exists u 0 for

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Numericalanalysis



Suppose for a contradiction, there exists uk 0 fork = 1, . . . , n 1 such that

uk =max{u0, u1, u2, . . . , un}

min{uk1, uk+1} 0, this is a contradiction.

75/120

Proof.

Suppose for a contradiction there exists uk 0 for

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Part III

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model

Finitedifferences

Numericalanalysis



Suppose for a contradiction, there exists uk 0 fork = 1, . . . , n 1 such that

uk =max{u0, u1, u2, . . . , un}

min{uk1, uk+1} 0, this is a contradiction.

76/120

Proof ctd.

W l d h i h

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model

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Numericalanalysis



We conclude that either

1

there is a maximum, which is negative (uk 0) or2 there is a maximum at the boundary uk max{u0, un}.

.

HOMEWORK


77/120

Proof ctd.

W l d th t ith

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model

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Numericalanalysis



We conclude that either

1

there is a maximum, which is negative (uk 0) or2 there is a maximum at the boundary uk max{u0, un}.

.

HOMEWORK


78/120

Stability+consisitency convergence

1 F PDE th i i i l ti

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Part III

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model

Finitedifferences

Numericalanalysis



1 For PDE, the maximum principle gave us continuousdependence on data and hence well posedness.

2 For numerical method, the discrete maximum principlegives us continuous dependence of global error ontruncation error.

3 This type of result for numerical methods is usually

referred to as stability and

Stability+consistency convergence

Recall, consistency means local truncation error Tj is

O(hk), some k > 1.convergence means global error u(xj) uj is small.

4 No proofs given, but Theorem given next.

79/120


1 For PDE the maximum principle gave us continuous

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Numericalanalysis











80/120



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model

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Numericalanalysis











81/120

P III



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model

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Numericalanalysis











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P t III

Theorem (Stability+constistency=convergence)

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model

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Numericalanalysis



(S y+ y g )

Suppose that

consistent: the finite difference method is kth order consistent.

stable the method has the form

auj1 + buj cuj+1 = fj

with a, c 0 and b a + c > 0.Then, the numerical approximation converges to the exact

solution,

ej = |uj u(xj)| = O(hk) as h 0.

83/120

Part III

Stability of finite differenceapprox

Recall the finite difference approximation

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Part III

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Finitedifferences

Numericalanalysis




1h2

uj1 + ( 2h2

+ r)uj 1h2

uj+1 = fj j = 1, 2, . . . , n 1,

To show that the approximation is stable,we apply the Stability Theorem and look at condition (S).

a 0 1h2

0

c 0 1

h2 0

b a + c 2

h2+ r

2

h2

The centred approximation method is stable as r 0.

84/120

Part III



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Part III

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Finitedifferences

Numericalanalysis




1h2

uj1 + ( 2h2

+ r)uj 1h2

uj+1 = fj j = 1, 2, . . . , n 1,


a 0 1h2

0

c 0 1

h2 0

b a + c 2

h2+ r

2

h2


85/120

Part III



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Part III

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Finitedifferences

Numericalanalysis




1h2

uj1 + ( 2h2

+ r)uj 1h2

uj+1 = fj j = 1, 2, . . . , n 1,


a 0 1h2

0

c 0 1

h2 0

b a + c 2

h2+ r

2

h2


86/120

Part III



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Part III

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Finitedifferences

Numericalanalysis



pp

1h2

uj1 + ( 2h2

+ r)uj 1h2

uj+1 = fj j = 1, 2, . . . , n 1,


a 0 1h2

0

c 0 1

h2 0

b a + c 2

h2+ r

2

h2


87/120

Part III

1 Proof of theorem not given See written notes

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Finitedifferences

Numericalanalysis



1 Proof of theorem not given. See written notes.

2 It is similar to the proof of stability for the diffusionproblem.

3 We conclude that the global error for the centereddifference approximation of the diffusion problem satisfies

global error = |uj u(xj)| = O(h2).

88/120

Part III

Outline


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Finitedifferences

Numericalanalysis







89/120

Part III

Convection diffusion problem

The problem

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Finitedifferences

Numericalanalysis



u + ru = f

is known as a reaction diffusion equation, as the term u

models diffusion and the term u models reaction.We now look at a convection diffusion equation, replacing uwith u.

Convection-diffusionu + wu = f for 0 < x < 1

u(0) = ; u(1) =

for some f : (0, 1) R

, a scalar w R

, and boundary values, R.The scalar w is known as the wind.

It controls the strength and direction of the convection.

90/120

Part III

Problem 1.3

u + wu = 0 for 0 < x < 1 (1.3)

(0) 1 (1) 0

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Finitedifferences

Numericalanalysis



u(0) = 1; u(1) = 0.

The solution is

u(x) =exp(w) exp(wx)

exp(w) 1

0 0.50

1

x

temperature

Problem 1.3

w=0

w=5

w=20

0 0.5 10

1/8

Problem 1.4

w=20

w=5

w=0

91/120

Part III

Problem

u + wu = 1 for 0 < x < 1 (1.4)

(0) 0 (1) 0

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Finitedifferences

Numericalanalysis



u(0) = 0; u(1) = 0.

The solution is

u(x) =x

w

1 exp(wx)

w(exp(w) 1)

0 0.50

1

x

temperature

Problem 1.3

w=0

w=5

w=20

0 0.5 10

1/8

Problem 1.4

w=20

w=5

w=0

92/120

Part III

Centered finite differences

There are two derivatives to approximate.

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Finitedifferences

Numericalanalysis



u(xj)

u(xj) 2u(xj)

h2=

u(xj + h) 2u(xj) + u(xj h)

h2.

u

(xj)

We cannot use u(xj) = u(xj + h/2) u(xj h/2) as we canonly take values on the grid.We use the averaged centered difference:

u(xj) u(xj)

h= 1

2h

u(xj + h

2) + u(xj h

2)

=1

2h(u(xj + h) u(xj h)) .

93/120

Part III

Numerical approximation

The exact solution at the grid points u(xj) satisfies

8/3/2019 PDEs - Slides (3)

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Finitedifferences

Numericalanalysis



u(xj) + wu(xj) = fj j = 1, 2, . . . , n 1.

Replace derivatives by centered finite differences

Centered finite difference method

Find uj such that

1

h22uj +

w

huj = fj j = 1, 2, . . . , n 1.

That is,

1

h2(uj+1 2uj + uj1) +

w

2h(uj+1 uj1) = fj.

94/120

Part III

Linear system of equations

For j = 1, . . . , n 1,

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Finitedifferences

Numericalanalysis



1

h2 (uj+1 2uj + uj1) +w

2h (uj+1 uj1) = fj

1

h2

w

2h

a

uj1 +

2

h2

b

uj +

1

h2+

w

2h

c

uj+1 = fj

To take care of the boundary conditions The BC u(0) = gives u0 = and

a + bu1 cu2 = f0 bu1 cu2 = f0 + a.

The BC u(1) = gives un = and

aun2 + bun1 = fn1 + c.

95/120

Part III


For j = 1, . . . , n 1,

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Finitedifferences

Numericalanalysis


1

h2 (uj+1 2uj + uj1) +w

2h (uj+1 uj1) = fj

1

h2

w

2h

a

uj1 +

2

h2

b

uj +

1

h2+

w

2h

c

uj+1 = fj


a + bu1 cu2 = f0 bu1 cu2 = f0 + a.



96/120

Part III


For j = 1, . . . , n 1,

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Finitedifferences

Numericalanalysis


1

h2 (uj+1 2uj + uj1) +w

2h (uj+1 uj1) = fj

1

h2

w

2h

a

uj1 +

2

h2

b

uj +

1

h2+

w

2h

c

uj+1 = fj


a + bu1 cu2 = f0 bu1 cu2 = f0 + a.



97/120

Part III


Collect all the equations as a linear system.

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Finitedifferences

Numericalanalysis


b c 0. . .

. . .

a b c. . .

. . .

0 a b

u1...

uj...

un1

=

f1 + a...fj...

fn1 + c

.

Top row comes from left hand boundary condition and thebottom rows comes from the right boundary condition.

Also note the boundary condition affects the right hand sidevector in top/bottom entry.

In contrast to reaction diffusion equation,the matrix is non-symmetric as a = c.

98/120

Part III

Numerical analysis

HOMEWORK

Y t P bl 12

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Finitedifferences

Numericalanalysis



1 Is the approximation consistent? If u is smooth,

|Tj| Ch2, j = 1, 2, . . . , n 1.

In other words, Tj = O(h2).

Proof by Taylors theorem.

2 Is the approximation stable?

99/120

Part III

Numerical analysis

HOMEWORK

Y t P bl 12

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Finitedifferences

Numericalanalysis




|Tj| Ch2, j = 1, 2, . . . , n 1.




100/120

Part III

Numerical analysis

HOMEWORK


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Finitedifferences

Numericalanalysis




|Tj| Ch2, j = 1, 2, . . . , n 1.




101/120

Part III

Apply stability theorem

Recall the tridiagonal matrix entries ..

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20401



Finitedifferences

Numericalanalysis


1h2 w2h a

uj1 + 2h2 b

uj + 1h2 + w2h c

uj+1 = fj

To show that the approximation is stable we simply need tocheck that the tridiagonal coefficients satisfy (S).

a 0 1

h2+

w

2h 0 (?)

c 0

1

h2

w

2h 0

b a + c 2

h2

2

h2()

102/120

Part III



1 2 1

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Finitedifferences

Numericalanalysis


1h2 w2h a

uj1 + 2h2 b

uj + 1h2 + w2h c

uj+1 = fj


a 0 1

h2+

w

2h 0 (?)

c 0

1

h2

w

2h 0

b a + c 2

h2

2

h2()

103/120

Part III

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1 2 1

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Finitedifferences

Numericalanalysis


1h2 w2h a

uj1 + 2h2 b

uj + 1h2 + w2h c

uj+1 = fj


a 0 1

h2+

w

2h 0 (?)

c 0

1

h2

w

2h 0

b a + c 2

h2

2

h2()

104/120

Part III

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1 2 1

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Finitedifferences

Numericalanalysis

Finitedifferences forconvection

diffusionmodel

1h2 w2h a

uj1 + 2h2 b

uj + 1h2 + w2h c

uj+1 = fj


a 0 1

h2+

w

2h 0 (?)

c 0

1

h2

w

2h 0

b a + c 2

h2

2

h2()

105/120

Part III

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1 2 1

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Finitedifferences

Numericalanalysis


diffusionmodel

1h2 w2h a

uj1 + 2h2 b

uj + 1h2 + w2h c

uj+1 = fj


a 0 1

h2+

w

2h 0 (?)

c 0

1

h2

w

2h 0

b a + c 2

h2

2

h2()

106/120

Part III

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1 w 2 1 w

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Finitedifferences

Numericalanalysis


diffusionmodel

1h2 w2h a

uj1 + 2h2 b

uj + 1h2 + w2h c

uj+1 = fj


a 0 1

h2+

w

2h 0 (?)

c 0

1

h2

w

2h 0

b a + c 2

h2

2

h2()

107/120

Part III

20401

Condition for stability

Assume that w > 0 then

1 w

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Finitedifferences

Numericalanalysis


diffusionmodel

a 0 1

h2 +w

2h 0 ()

c 0 1

h2

w

2h 0 (?)

The centred approximation method is stable when

1

h2

w

2h 0

w h

2 1 h

2

w.

108/120

Part III

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Stability

Assume that w < 0 then

1 w

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Finitedifferences

Numericalanalysis


diffusionmodel

a 0

1

h2 +

w

2h 0 (?)

c 0 1

h2

w

2h 0 ()

Thus, the centred approximation method is stable whenever

1

h2+

w

2h 0

wh

2 1 h

2

w.

We deduce that a sufficient condition for stability is that

|w|h2 1.

The ratio |w|h2 is called the mesh Peclet number.

Computationally, if wh2 > 1 the centred difference solution109/120

Part III

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Stability


1 w

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Finitedifferences

Numericalanalysis


diffusionmodel

a 0

1

h2 +

w

2h 0 (?)

c 0 1

h2

w

2h 0 ()


1

h2+

w

2h 0

wh

2 1 h

2

w.


|w|h2 1.


Computationally, if wh2 > 1 the centred difference solutionh 110/120

Part III

20401

Stability


1 w

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Finitedifferences

Numericalanalysis


diffusionmodel

a 0

1

h2 +

w

2h 0 (?)

c 0 1

h2

w

2h 0 ()


1

h2+

w

2h 0

wh

2 1 h

2

w.


|w|h2 1.


Computationally, if wh2 > 1 the centred difference solutionh 111/120

Part III

20401

Solution of convection diffusionmodel with w = 100;

Central differencing h = 1/n with n = 20

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Finitedifferences

Numericalanalysis


diffusionmodel

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

2

4

6

8

10

12

14

16

x

u

Note the oscillation: the exact solution has no oscillation

112/120

Part III

20401

Alternative method: upwindfinite difference

Solution of convection with w = 100; h = 1/n with n = 20

12

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Finitedifferences

Numericalanalysis


diffusionmodel

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

2

4

6

8

10

x

u

There is no oscillation.The truncation error for this upwind finite differecing is O(h)

(first order) compared to the second order central differencingmethod.Even though truncation error bigger, the solution is better.

113/120

Part III

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Alternative method: Upwindfinite difference method

One sided finite difference approximation:

u(xj +h)

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Finitedifferences

Numericalanalysis


diffusionmodel

xj h xj xj +h

u(xj h)

u(xj)

u(xj) +u(xj)

1

hu(xj) u

(xj),1

h+u(xj) u

(xj),

where

u(xj) =u(xj) u(xj h)

+u(xj) =u(xj + h) u(xj).

114/120

Part III

20401

Upwind method

1 For u, use the centered difference: u(xj) 1h2

2u(xj)

2

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Finitedifferences

Numericalanalysis


diffusionmodel

2

u(xj) =[u(xj)]=u(xj + h) 2u(xj) + u(xj h)

2 For u, if w > 0, approximate u(xj) 1

h

u(xj)

whereu(xj) = u(xj) u(xj h).

or, if w < 0, approximate u(xj) 1h

+u(xj) where

+u(xj) = u(xj + h) u(xj).

Called the upwind difference approximation to u(x).

115/120

Part III

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Upwind method


2u(xj)

2

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Finitedifferences

Numericalanalysis


diffusionmodel

2



h

u(xj)



+u(xj) where



116/120

Part III

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Upwind method


2u(xj)

2

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Finitedifferences

Numericalanalysis


diffusionmodel

2



h

u(xj)



+u(xj) where



117/120

Part III

20401

Linear system of eqns

When w > 0 we obtain

1

(u 2u + u ) +w

(u u ) = f

8/3/2019 PDEs - Slides (3)

118/120



Finitedifferences

Numericalanalysis


diffusionmodel

h2(

j+1 j+

j1) +

h(

j j1)

j

1

h2

w

h

auj1 +

2

h2+

w

h

buj +

1

h2

cuj+1 = fj

Apply BCs u(0) = and u(1) = , this the following linearsystem

b c 0. . .

. . .

a b c. . .

. . .

0 a b

u1...

uj...

un1

=

f1 + a...fj...

fn1 + c

.

118/120

Part III

20401

Consistency and Stability forupwind

HOMEWORK


8/3/2019 PDEs - Slides (3)

119/120



Finitedifferences

Numericalanalysis


diffusionmodel

1 You show upwind method is 1st order consistentTaylorexpansions.

2 And the method is stable independent of h.

We knowconsistent + stable convergent

Hence the global erroror the error in approximating true solution u(x) is

|u(xj) uj| = O(h).

119/120

Part III

20401

Consistency and Stability forupwind

HOMEWORK


8/3/2019 PDEs - Slides (3)

120/120



Finitedifferences

Numericalanalysis


diffusionmodel

1 You show upwind method is 1st order consistentTaylorexpansions.

2 And the method is stable independent of h.

We knowconsistent + stable convergent

Hence the global erroror the error in approximating true solution u(x) is

|u(xj) uj| = O(h).

120/120

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