Nonlinear Algebraic Equations Example

Nonlinear Algebraic Equations Example Continuous Stirred Tank Reactor (CSTR). Look for steady state concentrations & temperature. In: N s spieces with concentrations ci

(in )

,

) (in) heat capacities c(in and temperature T p,i

Inside: N r reactions with stoichiometric coefficients σ α ,i and reaction constants rα . Out: N s spieces with concentrations ci (c-s may be equal to zero), heat capacities cp,i and temperature T. November 2001

10.001 Introduction to Computer Methods

(in )

Nonlinear Algebraic Equations Example Nr

(F/ V) ci(in) − ci  + ∑σα ,i rα = 0

i = 1, 2,..., Ns

α =1

Mass balancefor spieces1, 2,..., Ns Ns

Nr

i =1

α =1

(in) (F/ V)∑ ci(in)c(in) T − ci cp,i T − ∑ ∆Hα rα = 0 p,i

Energy balance Column of unknown variables :x = [c1 ,c2 ,...,cNs ,T]

T

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Nonlinear Algebraic Equations Example Each of the above equations may be written in the general form: f i (x1 , x 2 ,..., x N ) = 0 : In vector form: f(x)=0

f1 (x1 , x 2 ,..., x N ) = 0

f1 (x)  f (x)   f(x)=  2 ...    f (x)  N 

f 2 (x1 , x 2 ,..., x N ) = 0 ... f N (x1 , x 2 ,..., x N ) = 0 ∧

Let x be the solution staisfying f(x)=0. ∧

We do not know x and take x[0] as initial guess. November 2001


Nonlinear Algebraic Equations We need to form a sequence of estimates to the solution: [1]

[2]

∧

[3]

x , x , x ,... that will hopehully converge to x . Thus we want: lim x m →∞

[m]

∧

=x

∧

lim x − x[m] = 0

m →∞

Unlike with linear equations, we can’t say much about existence or uniqueness of solutions, even for a single equation. November 2001


Single Nonlinear Equation ∧

We assume f(x) is infinitely differentiable at the solution x . ∧

Then f(x) may be Taylor expanded around x : ∧

∧ ∧ 1 1 2 ’’ f (x) = f (x) + (x − x)f (x) + (x − x) f (x) + (x − x)3 f ’’’(x) + ... 2 3!

Assume x

∧

[0]

f (x ) ≈ (x [0]

’

∧

being close to x so that the series may be truncated:

[0]

∧

− x)f ’(x[0] ),

or

f (x[0] ) ≈ (x[0] − x[1] )f ’(x[0] )

[0] f (x ) [1] [0] x = x − ’ [0] − f (x )

− Newton ’s method Example : f (x) = (x − 3)(x − 2)(x − 1) = x 3 − 6x 2 + 11x − 6 November 2001


Single Nonlinear Equation

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Single Nonlinear Equation

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Single Nonlinear Equation For f(x)=(x-3)(x-2)(x-1)=x3-6x2+11x+6=0, we se that Newton’s method converges to the root at x=2 only if 1.60 such that J(y)-J(z) ≤ L y − z or there is some upper bound on the "stretching" effect of J. 1

f (x + sp) =

∫ [ J(x + sp) − J(x) ]p ds 0

1

≤

∫ [ J(x + sp) − J(x) ] ds 0

1

and

∫ [ J(x + sp) − J(x) ] ds

≤ L s p , so in this case

0

1

f (x + sp) =

∫ [ J(x + sp) − J(x) ] pds ≤ ( L s p ) p 0

( )

≤ (L s ) p = O p 2

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2

p

Systems of Nonlinear Equations Thus if we are at distance p from the solution, our error scales as ||p||2. What about convergence? How the error scales with the number of iterations? The answer is: ∧ 2  x[i +1] − x = O  x[i] − x  - local quadratic convergence,   a very fast one! Works when you are close enough to the solution. ∧

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Systems of Nonlinear Equations ∧

x − x : 0.1 = 10−1 [i]

x x x

[i +1]

[i + 2]

[i + 3]

∧

− x : 0.01 = 10−2 ∧

− x : 0.00001 = 10−4 ∧

− x : 0.000000001 = 10−8

Works if we are close enough to an isolated  ∧  (non-singular) solution: det  J( x )  ≠ 0.   November 2001


Systems of Nonlinear Equations, Example Let’s examine how Newton’s method works for a simple system : f1 = 3x + 4 x 2 − 145 = 0 ∧ 3  x=  2 3 f 2 = 4 x1 − x 2 + 28 = 0 4 3 1

2

The Jacobian is :  ∂f1  ∂x J= 1  ∂f 2  ∂x1 November 2001

∂f1  2   9 x1 ∂x 2 = ∂f 2   8x 1 ∂x 2 

8x 2  2 − 3x 2 


Systems of Nonlinear Equations, Example At each step we have the following system to solve :

( )

[i ] 2 [i ]   9 x1 8x 2 [i] J =  [i ] [i ] 2 − 3 x 2   8x1 3 x [i ] 3 + 4 x [i ] 2 − 145 1 2 [i ]   f = 4 x [i ] 2 − x [i ] 3 + 28  2  1  [i] [i ] [i ] [ i +1] [i ] [i ] = x + ∆x J ∆x = −f x

( ) ( ) ( ) ( ) ( )

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Systems of Nonlinear Equations, Example Let us examine performance of Newton’s method with the convergence

criterion f = max{f1 , f 2 } < δ abs

δ bs = 10

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−10


Newton’s Method Newton’s method works well close to the solution, but otherwise takes large erratic steps, shows poor performance and reliability. Let’s try to avoid such large steps: employ reduced step Newton’s algorithm. “Full” Newton’s step gives x[i+1]=x[i]+p[i]. We’ll use only fraction of of the step: x[i+1]=x[i]+λi p[i], 0