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Solution of Non-linear Equation Systems

In this lecture, we shall look at the mixed symbolic and numerical solution of algebraically coupled non-linear equation systems. The tearing method lends itself also to the efficient treatment of non-linear equation systems .

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Solution of Non-linear Equation Systems

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  1. In this lecture, we shall look at the mixed symbolic and numerical solution of algebraically coupled non-linear equation systems. The tearing method lends itself also to the efficient treatment of non-linear equation systems. The numerical iteration of the non-linear equation system can be limited to the tearing variables. Solution of Non-linear Equation Systems

  2. Non-linear equation systems Newton iteration Newton iteration with tearing Newton iteration of linear equation systems Table of Contents

  3. Non-linear Equation System: An Example I

  4. Sluice Reservoir p p q q q p Consumer II 3 1 2 0 2 1 Consumer I Environment pressure Non-linear Equation System: An Example II

  5. q p q = k · sign(p ) · p Non-linear Equation System: An Example III q p q: Flow rate p: Pressure reduction  p = sign(q) · q2 / k2

  6. p2 = 100 p0 = 1 fS(q1 ,p1 ,p2) = 0 fI(q2 ,p0 ,p1) = 0 fII(q3 ,p0 ,p1) = 0 q1 = q2 + q3 Sluice Reservoir Consumer II p q q p p q Consumer I 0 2 3 2 1 1 Environment pressure Non-linear Equation System: An Example IV

  7. p2 = 100 p0 = 1 fS(q1 ,p1 ,p2) = 0 fI(q2 ,p0 ,p1) = 0 fII(q3 ,p0 ,p1) = 0 q1- q2- q3= 0 p2 = 100 p0 = 1 fS(q1 ,p1 ,p2) = 0 fI(q2 ,p0,p1) = 0 fII(q3 ,p0,p1) = 0 q1- q2- q3= 0 Non-linear equation system in 4 unknowns Non-linear Equation System: An Example V 

  8.  f(x) H(x) =  x Newton Iteration I f(x) = 0 x   n f  n Non-linear equation system: x 0 Initial guess: x i+1 = x i - Dx i Dx   n Iteration formula: Dx i = H(x i )-1 · f(x i ) Increment: H   n  n Hessian matrix:

  9. p1 q1 q2 q3 x = p2- p1 - sign(q1) · q12 /k12 p1–p0 - sign(q2) · q22 /k22 p1–p0 - sign(q3) · q32 /k32 q1- q2- q3 = 0 f(x) = -2|q1 |/k12 0 -1 1 1 0 0 0 -2|q2 |/k22 0 0 1 H(x) = -2|q3 |/k32 0 -1 -1 Newton Iteration : Example I

  10. Dx i = H(x i )-1 · f(x i )  H(x i ) · Dx i = f(x i )  Linear equation systemin the unknownsDx  Dx   n Newton Iteration II Computation of increment:

  11. p2 = 100 p0 = 1 fS(q1,p1 ,p2) = 0 fI(q2 ,p0,p1) = 0 fII(q3 ,p0,p1) = 0 q1-q2-q3= 0 p2 = 100 p0 = 1 fS(q1,p1,p2) = 0 fI(q2,p0,p1) = 0 fII(q3,p0,p1) = 0 q1-q2-q3= 0 Choice Newton Iteration with Tearing I p2 = 100 p0 = 1 fS(q1 ,p1 ,p2) = 0 fI(q2 ,p0,p1) = 0 fII(q3 ,p0,p1) = 0 q1- q2- q3= 0  

  12. p2 = 100 p0 = 1 fS(q1,p1,p2) = 0 fI(q2,p0,p1) = 0 fII(q3,p0,p1) = 0 q1-q2-q3= 0 p2 = 100 p0 = 1 q1 = q2+q3 p1 = f1(q1,p2) q2 = f2(p0,p1) q3 = f3(p0,p1) q1 = f2(p0,p1) + f3(p0,p1) = f2(p0, f1(q1,p2)) + f3(p0, f1(q1,p2)) Newton Iteration with Tearing II 

  13. f(x) = q1-f2(p0, f1(q1,p2)) - f3(p0, f1(q1,p2)) = 0 x = q1  Linear equation systemin the unknownDx  H(x i ) · Dx i = f(x i )  Dx   1 q1 = f2(p0,p1) + f3(p0,p1) = f2(p0, f1(q1,p2)) + f3(p0, f1(q1,p2)) Newton Iteration with Tearing III

  14. p2 = 100 p0 = 1 q1 = q2+q3 p1 = p2 - sign(q1 ) · q12 / k12 q2 = k2 · sign(p1 -p0) · p1 -p0 q3 = k3 · sign(p1 -p0) · p1 -p0 pq1q1 = 1 pp1q1 = - 2|q1| / k12 pq2q1 = k2 / ( 2 ·  p1 -p0)· pp1q1 pq3q1 = k3 / ( 2 ·  p1 -p0)· pp1q1 f =q1-q2-q3 h =pq1q1-pq2q1-pq3q1  The symbolic substitution of expressions is hardly ever worthwhile. It is much better to iterate over all equations and to differentiate every equation separately in the determination of the partial derivatives. Newton Iteration : Example II

  15. q1 = Initial guess dx = 1 whiledx> dxmin p1 = p2 - sign(q1 ) · q12 / k12 q2 = k2 · sign(p1 -p0) · p1 -p0 q3 = k3 · sign(p1 -p0) · p1 -p0 pp1 = - 2|q1| / k12 pq2 = k2 / ( 2 ·  p1 -p0)· pp1 pq3= k3 / ( 2 ·  p1 -p0)· pp1 f =q1-q2-q3 h =1-pq2-pq3 dx = h \ f q1 = q1–dx end  The iteration is carried out over all equations. However, the internal linear equation system is only solved for the tearing variables. Newton Iteration : Example III

  16. The Newton iteration converges here in a single iteration step Newton Iteration for Linear Systems A·x = b Linear system:  f(x) = A·x – b = 0  H(x) = f(x)/  x = A  A·Dx = A·x – b Dx = x – A-1·b x 1 = x 0 – (x 0 – A-1·b) = A-1·b

  17. The tearing method is equally suitable for use in non-linear as in linear systems. The Νewton iteration of a non-linear equation system leads internally to the solution of a linear equation system. The Hessian matrix of this equation system needs only to be determined for the tearing variables. The Νewton iteration can also be used very efficiently for the solution of linear systems in many variables, since it converges (with correct computation of the H(x) matrix) in a single step. In practice, the H(x) matrix is often numerically approximated rather than analytically computed. Yet, symbolic formula manipulation techniques can be used to come up with symbolic expressions for the elements of the Hessian matrix. Summary

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