Heun Method - University of South Floridanatasha.eng.usf.edu/.../experiments/cstr/2_heun_2_correction_example.pdf · inc Heun Method . The simplest example of a predictor corrector

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Heun Method

The simplest example of a predictor corrector method

A “marching” method for obtaining ordered pairs starting with an initial value set

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Heun Example

The application of the Heun method using a know form for the differential equation,

(2)

Notation Examples:

2 f(z) = z +4z + 12z -75 3

This function could be labeled f(z) since there was no F(z) term in the expression. (1) f(z, F(z) ) = z +4z + 12z -75 2 3

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Heun Example

This function can not be labeled f(z) since there is a F(z) term in the expression.

(2)

Notation Examples:

2 f(z) = z +4z + 12z -75 3

This function could be labeled f(z) since there was no F(z) term in the expression. (1) f(z, F(z) ) = z +4z + 12z -75 2 3

0.8x f(x, y) = 4 e - 0.5 F(x)

Additional Required Information initial ordered pair, (0,2)

The application of the Heun method using a know form for the differential equation,

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Example: f(x, F(x) ) = 4 e - 0.5 F(x) 0.8x (1 , ?) desired ordered pair (0 , 2) Initial ordered pair Two correction Heun calculation

For predictor/corrector marching methods it is a two step process for each corrected “answer”

The objective is to obtain the other value in the ordered pair (1,?). After that value is obtained at an approximation level that satisfies the uncertainty of measurement system you “march” on to obtain the two values in the next ordered pair in the sequence.

= 2 + (∆ x) f(x,F(x))

F(0) = 2

dx dy

X= 0 + 2∆x F(2) = F(1) + (∆ x)

= 2 + (∆ x) f(x,F(x)) dx dy

X= 0 F(1) = 2 + (∆ x)

= 2 + (∆ x) f(x,F(x)) dx dy

X= 0 + 1∆x F(2) = F(1) + (∆ x)

X= 0

X= 0 + 1∆x

X= 0 + 2∆x

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Example: f(x, F(x) ) = 4 e - 0.5 F(x) 0.8x (1 , ?) desired ordered pair (0 , 2) Initial ordered pair Two correction Heun calculation

Two step process for each corrected “answer”

Step #1: Use the Euler to get a prediction of the “answer”. solution value using Euler

current value of F(x) 0 F(x) = F(x ) + f(x ,F(x ) 0 0 ](h) [

(1 , ?)

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Example: f(x, F(x) ) = 4 e - 0.5 F(x) 0.8x

Two correction Heun calculation


Step #1: Use the Euler to get a prediction of the “answer”. solution value using Euler


F(1) = 2 + f(0,2) (1) = ? ] [ (1 , ?)

(1 , ?) desired ordered pair (0 , 2) Initial ordered pair

f ( 0, 2 ) = 4 e - 0.5 (2) = 3 [ ] 0

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When x = 1 the F(x) value is suppose to equal 5.

solution value using Euler


F(1) = 2 + f(0,2) (1) = ? ] [ (1 , ?)

F (1) = 2 + 3(1) = 5

value for F(1) obtained using Euler ( the predicted value)

equation based on initial Euler method results.)

(A predicted solution to differential 0

(1 , ?) desired ordered pair

(1 , 5 ) predicted ordered pair

(0 , 2) Initial ordered pair

Step #1: Use the Euler to get a prediction of the “answer”.

f ( 0, 2 ) = 4 e - 0.5 (2) = 3 [ ] 0

(1, 5 )

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F(1) = 2 + f(0,2) (1) = ? ] [ (1 , ?)

(1, 5 )

F (1) = 2 + 3(1) = 5

value for F(1) obtained using Euler ( the predicted value)

equation based on initial Euler method results.)

(A predicted solution to differential 0

(1 , ?) desired ordered pair




f ( 0, 2 ) = 4 e - 0.5 (2) = 3 [ ] 0

If this predicted value isn’t good enough, it is time to move on to the correction step!

When x = 1 the F(x) value is suppose to equal 5.

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Example:

f(x, F(x) ) = 4 e - 0.5 F(x) 0.8x



Step #2: the correction to the Euler prediction of the “answer”.



F(1) = 2 + f(0,2) (1) = ?

f ( 0, 2 ) = 4 e - 0.5 (2) = 3

] [

[ ] 0

(1 , ?)

f (1, 5) = 4 e - (0.5) 5 = 6.402 0.8(1) P ] [

(A predicted solution to differential equation when x = 1

F (1) = 2 + 3(1) = 5 0

desired ordered pair

predicted ordered pair

Initial ordered pair


(1 , ?) (1 , 5 )

(0 , 2)

f (1, 5) = 4 e - (0.5) F(1) = ? 0.8(1) P ] [

(1, 5 )

(1, 5 )

f(x, F(x) ) = 4 e - 0.5 F(x) 0.8x

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Example: f(x, F(x) ) = 4 e - 0.5 F(x) 0.8x (1 , ?)

(1 , ? )



(0 , 2) Initial ordered pair Two correction Heun calculation



Heun first correction to solution value, F(1).

F (x) = c 1


f (1, 5) = 4 e - (0.5) 5 = 6.402 0.8(1) P ] [

f (1, 5) = 4 e - (0.5) F(1) = ? 0.8(1) P ] [

(1, 5 )

First correction (1 , 5 )

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(1 , ? )


First correction


(0 , 2) Initial ordered pair Two correction Heun calculation


{ } (1) ? F (x) = 2 + c 1


f (1, 5) = 4 e - (0.5) 5 = 6.402 0.8(1) P ] [

f (1, 5) = 4 e - (0.5) F(1) = ? 0.8(1) P ] [

(1, 5 )


(1 , 5 )

current value of F(x)


inc


(1 , ? )




F (x) = 2 + c 1




? { } (1) = 2 [4 e - 0.5(2)] 0.8(0) [4 e - 0.5(5)] 0.8(1) +

First correction


f (1, 5) = 4 e - (0.5) 5 = 6.402 0.8(1) P ] [

f (1, 5) = 4 e - (0.5) F(1) = ? 0.8(1) P ] [

(1, 5 )


(1 , 5 )


inc


(1 , ? )




2 { } F (x) = 2 + [ ] (1) = 2 + { } (1)= ? c

1 3 + 6.402



current value of F(x) 0 F(x) = F(x ) + f(x ,F(x ) 0 0 ](h) [ First correction

Step #2: the correction to the Euler prediction.

f (1, 5) = 4 e - (0.5) 5 = 6.402 0.8(1) P ] [

f (1, 5) = 4 e - (0.5) F(1) = ? 0.8(1) P ] [

(1, 5 )


(1 , 5 )


inc


(1 , 6.702)


(1 , ? ) Second correction



f( 1, 6.702) =[ 4 e - 0.5 (6.702)] 0.8(1) 2

{ } F (x) = 2 + [ ] (1) = 2 + {4.702} (1)= 6.702 3 + 6.402





f (1, 5) = 4 e - (0.5) 5 = 6.402 0.8(1) P ] [

f (1, 5) = 4 e - (0.5) F(1) = ? 0.8(1) P ] [

(1, 6.702 )


c 1

(1, 5 )

When x = 1 the F(x) value is now suppose to equal 6.702.

First correction

inc


(1 , 6.702)





f( 1, 6.702) =[ 4 e - 0.5 (6.702)] 0.8(1) 2

{ } F (x) = 2 + [ ] (1) = 2 + {4.702} (1)= 6.702 3 + 6.402





f (1, 5) = 4 e - (0.5) 5 = 6.402 0.8(1) P ] [

f (1, 5) = 4 e - (0.5) F(1) = ? 0.8(1) P ] [

(1, 6.702 )


c 1

(1, 5 )

If this corrected value still isn’t good enough, it is time to move on to a second correction step!!

When x = 1 the F(x) value is now suppose to equal 6.702.

First correction

inc


(1 , 6.702)





f( 1, 6.702) =[ 4 e - 0.5 (6.702)] 0.8(1) 2

{ } F (x) = 2 + [ ] (1) = 2 + {4.702} (1)= 6.702 3 + 6.402




(1, 6.702 )


c 1

(1, 5 ) Original predicted value of F(x) =5 when x =1

Heun second correction to the predicted value of F(1)

? { } (1) 2

F (x) = 2 + c 2

F (x) = 2 + c 2

First correction

continue to use the current value of F(x)

inc


(1 , 6.702)





f( 1, 6.702) =[ 4 e - 0.5 (6.702)] 0.8(1) 2

{ } F (x) = 2 + [ ] (1) = 2 + {4.702} (1)= 6.702 3 + 6.402




(1, 6.702 )


c 1



? { } (1) 2

F (x) = 2 + c 2

F (x) = 2 + c 2 { }

2 [4 e - 0.5(2)] 0.8(0) [4 e - 0.5(6.702] 0.8(1) + (1) = ?

First correction

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(1 , 6.702)


(1 , 6.382 ) Second correction



f( 1, 6.702) =[ 4 e - 0.5 (6.702)] 0.8(1) 2

{ } F (x) = 2 + [ ] (1) = 2 + {4.702} (1)= 6.702 3 + 6.402




(1, 6.382 )


c 1



? { } (1) 2

F (x) = 2 + c 2

F (x) = 2 + c 2 { }

2 [4 e - 0.5(2)] 0.8(0) [4 e - 0.5(6.702] 0.8(1) + (1) = 6.382

(1, 6.702 )

First correction

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Example: f(x, F(x) ) = 4 e - 0.5 F(x) 0.8x (1 , 6382) desired ordered pair (0 , 2) Initial ordered pair

f( 1, 6.702) =[ 4 e - 0.5 (6.702)] 0.8(1) 2

{ } F (x) = 2 + [ ] (1) = 2 + {4.702} (1)= 6.702 3 + 6.402




(1, 6.382 )


c 1



? { } (1) 2

F (x) = 2 + c 2

F (x) = 2 + c 2 { }

2 [4 e - 0.5(2)] 0.8(0) [4 e - 0.5(6.702] 0.8(1) + (1) = 6.382

(1, 6.702 )

When x = 1 the F(x) value is now really suppose to equal 6.382

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f(x, F(x) ) = 4 e - 0.5 F(x) 0.8x

(1 , 6.195)

(1 , 6.702)

correct ordered pair

1st correction

(1 , 6.382) 2nd correction



The analytical solution of this example differential equation gives the correct ordered pair as (1, 6.195 ).

Plot the relative error.

2st corrected value

Predicted value

1st corrected value

100%

0%

Correct value

rela

tive

erro

r


Write a MATLAB routine that will do this two correction numerical method.

Have the routine also calculate and plot the relative errors as well.

Have the routine check the results provided to the left.

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End of Presentation


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Heun Method - University of South Floridanatasha.eng.usf.edu/.../experiments/cstr/2_heun_2_correction_example.pdf · inc Heun Method . The simplest example of a predictor corrector