ordinary differential equations (ode’s)faculty.kfupm.edu.sa/che/binoushousam/files/lecture... ·...

1

Numerical Methods

Process Systems Engineering

ORDINARY DIFFERENTIAL

EQUATIONS (ODE’S)

Numerical methods in chemical

engineering

Edwin Zondervan

2

Numerical Methods


OVERVIEW

• A common problem encountered in engineering is the

initial value problem: how does the state of a system

vary with time, given some starting conditions, when is it

governed by an ODE (Ordinary differential equation)?

• We will look at some simple solvers and numerical

schemes for solving ODE’s

• We will investigate stability so that we are able to choose

a suitable solving method

• In the end we will take a look at systems of coupled

ODE’s

3

Numerical Methods


EULER’S METHOD

• For the differential equation:

• With the initial condition

• We can generate an estimate of x at t =

t+dt, by a Taylor expansion:

),( txfdt

dx=

0)0( xtx ==

tttxftx

tdt

dxtxttx

d

dd

)),(()(

)()(

+=

+=+

(6-1)

(6-2)

(6-3)

4

Numerical Methods


MARCHING FORWARD IN

TIME

t t+dt

x(t)

x(t+dt)

The true

solution!

With Euler’s method we can step forward in time …

5

Numerical Methods


CODE FOR EULER’S METHODfunction[x,t] = Euler(MyFunc,InitialValue,Start,Finish,Nsteps)

x(1) = InitialValue;

t(1) = Start;

dt = (Finish-Start)/Nsteps;

for i =1:Nsteps

F = feval(MyFunc,x(i),t(i));

t(i+1) = dt +t(i);

x(i+1) = F*dt + x(i);

end

t = t’;

x = x’;

return

6

Numerical Methods


TEST CASE

Aprod.

( ) 10 ==

-=

tx

kxdt

dx

A batch reactor, where a reactant is consumed by a first

order reaction:

An analytical solution to this equation will give you an exponential decay

with a time constant t = 1/k.

(6-4)

7

Numerical Methods


PERFORMANCE OF EULER’S METHOD

function [f] = TestFunction(x,t)

f = -1*x;

return

For 100 steps, the numerical solution gives a good match!

0 1 2 3 4 5 6 7 8 9 100

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

t

x

Exact

Numerical

8

Numerical Methods



25 STEPS

0 2 4 6 8 10 120

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

t

x

Exact

Numerical

25 Steps: solution OK, small deviation

9

Numerical Methods



6 STEPS:

Oscillates, but tends to

correct asymtpote

0 1 2 3 4 5 6 7 8 9 10-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

t

xExact

Numerical

6 Steps: solution follows trend, starts

oscillating

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Numerical Methods



5 STEPS:

Solution becomes

unstable

0 1 2 3 4 5 6 7 8 9 10-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

t

x

Exact

Numerical

5 Steps: solution becomes unstable

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Numerical Methods



3 STEPS: Solution is

diverging from the true

solution

0 1 2 3 4 5 6 7 8 9 10-14

-12

-10

-8

-6

-4

-2

0

2

4

6

t

x

Exact

Numerical

3 Steps: solution runs away from actual

solution

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Numerical Methods


ACCURACY OF THE EULER SCHEME

• The error in Euler’s method can be found

in the Taylor expansion, where we

neglected all terms starting from O(dt2)

and higher…

• An estimate of the accumulated error is:

– N O(dt2) = Dt/dt O(dt2)

• Euler method is accurate to O(dt) (a low

order method)… NOT VERY

IMPRESSIVE

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Numerical Methods


STABILITY OF THE EULER

SCHEME• The exact solution to:

lxdt

dx=

And x(t=0) = 1

is: x = exp(lt)For l<0, the solution should decay to zero. Our numerical solution

should do the same…

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Numerical Methods


STABILITY OF EULER’S

METHOD

)1(

)()()(1

tx

txx

txftxttx

i

ii

iii

ld

dl

dd

+=

+=

+=++

|(1+ldt)|<1 i.e. -ldt< 2, for the

solution to decay to zero …

So for dt <2 (for l=-1)… Euler is conditionally stable

(6-5)

15

Numerical Methods


THE IMPLICIT EULER

METHOD• The implicit Euler method uses a gradient

in the future, rather than at a current time

step:

tttxfxx iiidd ),( 11

++=++

(6-5)

16

Numerical Methods


A SMALL CHANGE IN THE CODEfunction[x,t] = EulerImplicit(MyFunc,InitialValue,Start,Finish,Nsteps)

x(1) = InitialValue;

t(1) = Start;


for i =1:Nsteps

t(i+1) = dt +t(i);

x(i+1) = fsolve(@FunToSolve,x(i),[],x(i),t(i+1),MyFunc,dt);

end

t = t’;

x = x’;

return

function residual = FunToSolve(x,xo,t,MyFunc,dt)

residual = xo + feval(MyFunc,x,t)*dt-x;

return

17

Numerical Methods


THE IMPLICIT METHOD

5 STEPS: with the implicit

Euler scheme

0 1 2 3 4 5 6 7 8 9 100

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

t

x

Exact

Numerical

18

Numerical Methods


STABILITY OF THE IMPLICIT METHOD

)1(

)1(

)()()(

1

1

t

xx

xtx

txx

txftxttx

i1i

i1i

ii

ii1i

ld

ld

dl

dd

+=

=+

+=

+=+

+

+

+

++

lxdt

dx-=The implicit Euler scheme for:

is:

For l<0, the numerical

solution will decay to zero

for any stepsize

(6-6)

19

Numerical Methods


SYSTEMS OF ODE’S

• We often want to solve coupled ODE’s:

...

),...,,,(

),...,,,(

),...,,,(

3213

3212

3211

txxxfdt

dx

txxxfdt

dx

txxxfdt

dx

=

=

=

(6-7)

20

Numerical Methods


JUST SMALL CHANGES IN THE EXPLICIT CODE

function[x,t] = EulerCoupled(MyFunc,InitialValue,Start,Finish,Nsteps)

x(:,1) = InitialValue’;

t(1) = Start;


for i =1:Nsteps

F = feval(MyFunc,x(:,i),t(i));

t(i+1) = dt +t(i);

x(:,i+1) = F*dt + x(:,i);

end

t = t’;

x = x’;

return

Similarly you can adjust the Implicit Euler code

to deal with ODE systems!!!

21

Numerical Methods


USING IT ON A TEST PROBLEM

1)0()0(

2

21

212

211

==

-=

--=

xx

xxdt

dx

xxdt

dx function [dx_dt] = TestFunction2(x,t)

dx_dt(1) = -1*x(1) – 1*x(2)

dx_dt(2) = 1*x(1) - 2*x(2);

dx_dt = dx_dt’

return

A linear ODE system:

(6-8)

0 1 2 3 4 5 6 7 8 9 10-0.2

0

0.2

0.4

0.6

0.8

1

1.2

t

x

x1(t)

x2(t)

22

Numerical Methods


STABILITY OF ODE SYSTEMS

• Let’s evaluate our example:

• For linear systems we can get an analytical solution by factorizing M:

• M = U-1LU

-

--=

2

1

2

1

21

11

x

x

x

x

dt

d

M

(6-9)

23

Numerical Methods



L=

L=

L=

-

-

2

1

2

1

2

11

2

1

2

11

2

1

x

xU

x

xU

dt

d

x

xUUU

x

x

dt

dU

x

xUU

x

x

dt

d

=

L=

=

2

1

2

1

2

1

2

1

2

1

2

1

2

1

0

0

y

y

y

y

dt

d

y

y

y

y

dt

d

x

xU

y

y

l

l

A diagonal matrix: equations are no

longer coupled, and a solution can be

found!!(6-10)

(6-11)

24

Numerical Methods



=

=

-

)exp(

)exp(

)exp(

)exp(

2

11

2

1

2

1

2

1

t

tU

x

x

t

t

y

y

lb

la

lb

la

Matrix of eigenvectors l1 and l2 are

eigenvalues

Eigenvalues can be complex numbers: l = Re + Im j

If eigenvalues have an imaginary part, the solution will

oscillate. An the real part determines wether a solution will go

to a steady value or explode to infinity

(6-12)

25

Numerical Methods



• Back to our example:

l1 = -1.5 + j 3 /2

l2 = -1.5 - j 3 /2

It has imaginary parts oscillation, but it

will decay to a steady value

(6-13)

-

-=

21

11M

-

26

Numerical Methods


STABILITY CRITERION

• For the explicit Euler scheme, we found

earlier: |1+ldt|< 1, for our ODE system we

can represent this criterion in the complex

plane:

Re(ldt)

Im(ldt)

-1

stable

UnstableFor our example: dt<1

(See if you can check

this yourself!)

27

Numerical Methods


SOME RESULTS

0 2 4 6 8 10-0.2

0

0.2

0.4

0.6

0.8

1

1.2

t

x

Explicit Euler 100 STEPS

x1

x2

0 2 4 6 8 10-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

t

x


x1

x2

0 2 4 6 8 10-4

-3

-2

-1

0

1

2

3

4

5

t

x


x1

x2

0 2 4 6 8 10-0.2

0

0.2

0.4

0.6

0.8

1

1.2

t

x

Implicit Euler 100 STEPS

x1

x2

0 2 4 6 8 10-0.2

0

0.2

0.4

0.6

0.8

1

1.2

t

x


x1

x2

0 2 4 6 8 10-0.2

0

0.2

0.4

0.6

0.8

1

1.2

t

x


x1

x2

28

Numerical Methods


STABILITY OF NON LINEAR SYSTEMS

• You remember the Jacobian, when we were

solving non linear equations ..

• The eigenvalues of the Jacobian determine the

stability of a system on non linear ODE’s

• Stiffness, is defined as the ratio of the largest to

the smallest eigenvalue of the Jacobian. A

system is called stiff if this ratio is much greater

than unity. For stiff problems you need an

implicit solver!!

29

Numerical Methods


HIGHER ORDER METHODS

• All routines for solving ODE’s work in a

similar way as the Euler method,

• With the Euler method we will have a poor

approximation to our solution, if we

evaluate the function at more points over

the interval we can have beter estimates.

This idea forms the basis for a class of

methods called the Runge-Kutta methods.

30

Numerical Methods


RUNGE-KUTTA SCHEMES

• Runge-Kutta schemes use weighted

trajectories, given as:

– ki = f(xiestimated,t+cdt)dt; 0<c<1

• We now update our solution:

– xdt = x0 +w1k1 +w2k2 + …

• The estimated values for x at the i-th

trajectory are calculated from:

– xiestimated = x0 +ai,1k1 …

Quite

Fuzzy!!

(6-15)

(6-16)

(6-17)

31

Numerical Methods


RK 2ND ORDER TIME STEPPING SCHEME

• For the 2nd order RK scheme we need to

evaluate 2 trajectories:

• xdt = x0 +w1k1 +w2k2

• Where k1 = f(x0,t)dt

• And k2 =f (x0 + a2,1k1,t+c2dt)dt

• We must choose the weights (w1,w2, a2,1)

and the position c2 such that we get an

error of O(dt3) over asingle time step!

(6-18)

(6-19)

(6-20)

32

Numerical Methods


RK 2ND ORDER TIME STEPPING SCHEME

• After expanding a function x into a Taylor

series we have to compare the following

outcomes:

)(!2

)()),((),(

)()(

32

000

32

22

2

1,2210

tOt

ftxffttxfxx

tOtfcwtffawwwfxx

ix

ix

w

dd

d

ddd

++++=

+++++=

21

22

21

1,22

21 1)(

=

=

=+

cw

aw

ww

Three eqn. With

4 unknowns

Set c2 =1 w1 = w2 = ½ a2,1 =1

Crank-Nicholson scheme

Set c2 = ½ w1 = 0 w2 = 1 a2,1 = ½

Euler mid point scheme

(6-21)

(6-22)

(6-23)

33

Numerical Methods


MATLAB CODE FOR 4TH

ORDER RKfunction [x,t] = RungeKutta_4(MyFunc,InitialValues,Start,Finish,Nsteps)

x(:,1) = InitialValues'

t(1) = Start;

dt = (Finish - Start)/Nsteps;

for i = 1:Nsteps

k1 = feval(MyFunc,x(:,i),t(i))*dt;

k2 = feval(MyFunc,x(:,i) + k1/2,t(i) + dt/2)*dt;

k3 = feval(MyFunc,x(:,i) + k2/2,t(i) + dt/2)*dt;

k4 = feval(MyFunc,x(:,i)+k3,t(i) + dt)*dt;

t(i+1) = dt + t(i);

x(:,i+1) = (k1+2*k2+2*k3+k4)/6 + x(:,i);

end

t = t';

x = x';

return

34

Numerical Methods


COMPARING EULER AND RK

0 1 2 3 4 5 6 7 8 9 100

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

t

x4th Order Runge-Kutta, 10 Steps

0 1 2 3 4 5 6 7 8 9 100

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

t

x

Euler, 10 Steps

Exact

Numerical

35

Numerical Methods


SUMMARY

• We looked at the Euler method, explicit and implicit

• We saw that an explicit method is conditionally stable

• We found that eigenvalues say something about stability

• Stiffness tells you if you can use an explicit or implicit

method (small or large time steps)

• Higher order methods provide large savings in

computational time.

• MATLAB has good solvers for ODE systems, e.g.

ODE15 and ODE45

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