LMAM and School of Mathematical Sciences Peking University
Transcript of LMAM and School of Mathematical Sciences Peking University
Numerical Analysis
Zhiping Li
LMAM and School of Mathematical SciencesPeking University
Lecture 11: Numerical Solution for Nonlinear Equations
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��{Ä�g��´±��, � Newton {�«O3u^L�þü:����O�, Cq¦)�� x- ¶��:.
� xk−1, xk �1w¼ê f (x) ":��¥�ü�:, KLü:(xk−1, f (xk−1)), (xk , f (xk)) ���� x-¶�:��I´
xk+1 = xk −xk − xk−1
f (xk)− f (xk−1)f (xk).
ùÒ´��{�S��ª, §��±w�´3 Newton {¥ò���Ç f ′(xk) �����Ç
f (xk )−f (xk−1)xk−xk−1
����,
��{I�kü�Ð��UåÚ, Ïd¡�üÚ{. ù����ØÄ:S�{ØÓ.
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Lecture 11: Numerical Solution for Nonlinear Equations
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½nµ � f (x) 3Ù": x∗ �,���S��ëY��, �f ′(x∗) 6= 0. XJÐ� x0 6= x1 ¿©�C x∗, K��{½Â�S�S� {xk+1}∞k=1 Âñu x∗, �Ù(²þ)Âñ�Ý�±����
r = 1+√
52 ≈ 1.618 �.
y²µ Äk·�F"�Ñ x∗ �����, ¦�±T��¥?Ûü�ØÓ:�Ð�d��{S��ª�Ñ�:Eá3T��¥.
d®�^�§�3 x∗ ��� ∆1 = {x : |x − x∗| ≤ δ1}, ¦�f (x) ∈ C2(∆1), f ′(x) 6= 0, ∀x ∈ ∆1. P M1 =
maxx∈∆1|f ′′(x)|
2 minx∈∆1|f ′(x)| , �
δ2 < 1/M1, - δ = min{δ1, δ2}, ∆ = {x : |x − x∗| ≤ δ}. ·��y²�� ∆ ÷v�¦. 5¿§M = maxx∈∆ |f ′′(x)|
2 minx∈∆ |f ′(x)| ≤ M1, Mδ < 1.
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Lecture 11: Numerical Solution for Nonlinear Equations
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é?¿�Ð� x0, x1 ∈ ∆, P {xk}∞k=2 �d��S�{�)�S�. � xk−1, xk ∈ ∆, d x∗ ∈ ∆, f (x∗) = 0, Ú���§
P1(x) = f (xk)x − xk−1
xk − xk−1+ f (xk−1)
x − xkxk−1 − xk
(= f ��5��¼ê)�{��O (5¿§ùp x∗ Ø73 xk−1
Ú xk �m), �3 ξ1 ∈ ∆ ¦�
P1(x∗) = −1
2f ′′(ξ1)(x∗ − xk)(x∗ − xk−1).
,��¡, d���§k
P1(xk+1)−P1(x∗) =f (xk)− f (xk−1)
xk − xk−1(xk+1−x∗) = f ′(ξ2)(xk+1−x∗),
Ù¥ ξ2 3 xk−1, xk �m§Ïd ξ2 ∈ ∆.
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Lecture 11: Numerical Solution for Nonlinear Equations
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5¿� P1(xk+1) = 0, ¿P ej = |xj − x∗|, Kd±þüª�
ek+1 =∣∣∣ f ′′(ξ1)
2f ′(ξ2)
∣∣∣ekek−1 ≤maxx∈∆ |f ′′(x)|2 minx∈∆ |f ′(x)|
ekek−1 ≤ (Mδ)δ < δ.
ùÒy² xk+1 ∈ ∆. dþª�k
ek+1 ≤ (Mek−1)ek ≤ (Mδ)ek ≤ · · · ≤ (Mδ)ke1 ≤ (Mδ)kδ ≤ 1
M(Mδ)k+1.
dd9 Mδ < 1 � limk→∞ xk = x∗. Âñ5�y.
��©ÛÂñ�. P E0 = Me0, E1 = Me1, - Ek+1 = EkEk−1,k = 1, 2, · · · , K8B�y Ek+1 ≤ (Mδ)Ek ≤ (Mδ)kE1, ∀k ≥ 1,Ïd, limk→∞ Ek = 0. P yk = lnEk .
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Lecture 11: Numerical Solution for Nonlinear Equations
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K yk ÷v�©�§ yk+1 = yk + yk−1, k = 1, 2, · · · , ÙÏ)�L«� Fibonacci ê�yk = C1p
k1 + C2p
k2 , Ù¥ p1, p2 ´A��§
p2 = p + 1
��, p1 = 1+√
52 ≈ 1.618, p2 = 1−
√5
2 ≈ −0.618. u´k
Ek+1
Ep1k
= eC1pk+11 eC2p
k+12
eC1pk+11 eC2p
k2p1
= eC2pk2 (p2−p1). d |p2| < 1, k
limk→∞
Ek+1
Ep1
k
= 1.
dd� Ek Âñu"��Ý� p1 = 1+√
52 ≈ 1.618 ��.
dc©Û�, Mek ≤ Ek , Ïd, ek Âñu"��ÝØ$up1 ≈ 1.618 �.
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Lecture 11: Numerical Solution for Nonlinear Equations
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,��¡§�Ä Ek+1 = Ek Ek−1(1 + δ), P yk = ln Ek , K yk ÷v�©�§ yk+1 = yk + yk−1 + ln(1 + δ), k = 1, 2, · · · . T�§���A)� y = − ln(1 + δ), Ïd§�§�Ï)�
yk = C1pk1 + C2p
k2 − ln(1 + δ).
u´k Ek = Ek(1 + δ)−1. q p1 + p2 = 1, ¤±
limk→∞
Ek+1
Ep1
k
= (1 + δ)−p2 .
ù`²§é?¿� δ, Ek Âñu"��Ý�´ p1.
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Lecture 11: Numerical Solution for Nonlinear Equations
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é k � 1, M∗ek+1 ≈ (M∗ek)(M∗ek−1), Ù¥ M∗ =∣∣∣ f ′′(x∗)2f ′(x∗)
∣∣∣. Ïd§zk = ln(M∗ek) ìCÂñ��A� Fibonacci ê�. Ïd§ekÂñu"��Ý� p1 ≈ 1.618 �.
5 1µ � Newton {�'§��{Ø^O��ê�§ÏdO�þ�~§¦+Âñ�Ývk����§�E,´��5�"
5 2µ ��{�g��?�Úÿ2�^L�þn:��g�� x-¶��:�Ñ#S�:��Ô�{§Ù(²þ)Âñ�Ý´ p3 − p2 − p − 1 = 0 ���¢� ≈ 1.84. ¦+Âñ�ÝEvk����§��Ô�{�`("):��´µ=B�Ñn�¢�Щ� x0, x1, x2, �k�US�� f (x) �E�.
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Lecture 11: Numerical Solution for Nonlinear Equations
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�Ĺ n > 1 ���þ!n ��§��§|µf1(x1, x2, · · · , xn) = 0,
f2(x1, x2, · · · , xn) = 0,
· · · · · · · · · · · ·fn(x1, x2, · · · , xn) = 0,
Ù¥ fi : Rn → R, i = 1, 2, · · · , n, ¥��k��´��5�.
(J: ØO�þìO�§�æ��´1 "y°()�êÆnØ (�35!��5!· · · );
2 Nõ���¦)g�!�nÚ�{Ã{E��õ�;
] :1 ÛÜ�5z�{!¦)�5�§|�S�{;
2 Ø N��nÚØÄ:S����êÃ'�g���{.
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Lecture 11: Numerical Solution for Nonlinear Equations
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�¦)�5�§|� Jacobi S�{aq/, 31 i fÚ§3�½ xj , j 6= i �^�e, ò fi �� xi ���5¼ê¦�; é¤k1 ≤ i ≤ n, ѦÑ���§�å�# {xi}ni=1. �{Xeµ
for k = 0, 1, 2, · · ·for i = 1, 2, · · · , n) fi (x
(k)1 , · · · , x (k)
i−1, u, x(k)i+1, · · · , x
(k)n ) = 0 � u;
x(k+1)i = u;
endXJ÷vS�Ê�^�§Kª�Ì�¿ÑÑ x (k+1)
end
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Lecture 11: Numerical Solution for Nonlinear Equations
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�¦)�5�§|� Gauss-Seidel S�{aq/, 31 i fÚ§3�½ xj , j 6= i �^�e, ò fi �� xi ���5¼ê¦�, ¦���á=^Ù�# xi . �{Xeµ
for k = 0, 1, 2, · · ·for i = 1, 2, · · · , n) fi (x
(k+1)1 , · · · , x (k+1)
i−1 , u, x(k)i+1, · · · , x
(k)n ) = 0 � u;
x(k+1)i = u;
endXJ÷vS�Ê�^�§Kª�Ì�¿ÑÑ x (k+1)
end
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Lecture 11: Numerical Solution for Nonlinear Equations
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�¦)�5�§|� SOR S�{aq/, 31 i fÚ§3�½xj , j 6= i �^�e, ò fi �� xi ���5¼ê¦�, ¦���á=^Ù(ÜtµÏf ω �# xi . �{Xeµ
for k = 0, 1, 2, · · ·for i = 1, 2, · · · , n) fi (x
(k+1)1 , · · · , x (k+1)
i−1 , u, x(k)i+1, · · · , x
(k)n ) = 0 � u;
x(k+1)i = x
(k)i + ω(u − x
(k)i );
endXJ÷vS�Ê�^�§Kª�Ì�¿ÑÑ x (k+1)
end
5µ ù�{3�½^�eUÂñ§���Âñ�ú.
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Lecture 11: Numerical Solution for Nonlinear Equations
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Newton S�{9ÙU?.�{
ÛÜ�5z���5�§|� Newton S�{
� f : Rn → Rn �1w¼ê�, d Taylor Ðmªk
f(x∗) = f(x(k)) + ∇f(x(k))(x∗ − x(k)) + O(‖x∗ − x(k)‖2),
Ù¥ ∇f(x(k)) =(∂fi (x
(k))
∂x(k)
)´ f � Jacobi Ý.
dd£¡�ÛÜ�5z¤=�Cq�5�§|
∇f(x(k))(x∗ − x(k)) ≈ −f(x(k)).
½Â��5�§|� Newton S�S�µé k = 0, 1, · · ·1 )�5�ê�§| ∇f(x(k))y(k) = −f(x(k));
2 - x(k+1) = x(k) + y(k).
¢SO��, ��±Ú\�5|¢ minλ ‖f(x(k) + λy(k))‖.
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Lecture 11: Numerical Solution for Nonlinear Equations
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Newton S�{9ÙU?.�{
Newton S�{�C/ Broyden �{
����/aq§Newton S�{���kÛÜÂñ5§é1w¼ê�ü�§Âñ�Ý����;
zÚÑIO� Jacobi Ý, $�þã�, cÙ´p��/.
XÛ;�O� Jacobi Ý, ��±$�þ��CqO�´U? Newton S�{�Ä�Ñu:.
Broyden �{3 ∇f(x(0))−1 �Ä:þ§zÚ�I�é�AgÝ��þ¦{=¼�e�ÚS�¤I�Cq_Ý, l ^é���d¼�S�?��þ y(k).
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Lecture 11: Numerical Solution for Nonlinear Equations
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Newton S�{9ÙU?.�{
Broyden �{�nØ| — Sherman-Morrison Ún
Únµ � A � n ��_Ý, x, y ∈ Rn. XJyTA−1x 6= −1,KA + xyT ��_, �
(A + xyT )−1 = A−1 − A−1xyTA−1
1 + yTA−1x.
� y(k−1) = x(k) − x(k−1) é��, Cq/k
∇f(x(k))y(k−1) = ∇f(x(k))(x(k)−x(k−1)) ≈ f(x(k))−f(x(k−1)) =: g(k−1).
·�F"é�UCq�O ∇f(x(k)) �Ý A(k), ¦�A(k)y(k−1) = g(k−1), � A(k) = A(k−1) + u(k−1)(y(k−1))T , Ù¥ u(k−1) �½.
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Lecture 11: Numerical Solution for Nonlinear Equations
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Newton S�{9ÙU?.�{
Broyden �{�g�9�EL§
d^��
g(k−1) = A(k)y(k−1) = A(k−1)y(k−1) + u(k−1)(y(k−1))Ty(k−1).
dd� u(k−1) = g(k−1)−A(k−1)y(k−1)
(y(k−1))T y(k−1) , u´
A(k) = A(k−1) +g(k−1) − A(k−1)y(k−1)
(y(k−1))Ty(k−1)(y(k−1))T .
ØJ�yÚn�^� (y(k−1))T (A(k−1))−1u(k−1) 6= −1 �du (y(k−1))T (A(k−1))−1g(k−1) 6= 0 (Ø�b�o¤á).
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Lecture 11: Numerical Solution for Nonlinear Equations
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Newton S�{9ÙU?.�{
Broyden �{�g�9�EL§
- A(0) = ∇f(x(0)), � (A(0))−1 ®�. dÚn 4.3.1 �
(A(k))−1
= (A(k−1))−1−[(A(k−1))−1g(k−1)−y(k−1)
](y(k−1))T (A(k−1))−1
(y(k−1))T (A(k−1))−1g(k−1).
k (A(k))−1, Ò�±O�1 k Ú�S�?��þ
y(k) = (A(k))−1f (x(k)).
5 1µ ùÒ´ Broyden �{�Ä�Ú½. dd��§ Broyden�{�O�þ�(é�. �±y² Broyden �{, � f ÷v�½^��, ´ÛÜ��5Âñ�.
5 2µ U?.� Newton S�{�kNõ. ��Ñäk�½^�e�ÛÜ��5Âñ5. éJ'�§��`�.
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Lecture 11: Numerical Solution for Nonlinear Equations
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ÓÔ�{— ���Âñ�{��«kÃ}Á
±þ0���«��5�§Ú��5�§|�S��{Ñ�kÛÜÂñ5§=���Âñ�S�S�§7L�Ñl) x∗ ¿©�C�Ð� x0, ù3¢SA^¥ Ø´��. Ï~�³/²�!�E}Á!±9N$í.
é,AÏa.���5�§½�§|§�±�Eäk�ÛÂñ5��{. ÓÔ{£homotopy¤, �¡�òÿ{£continuation¤,Ò´ù�¡�«kÃ�}Á.
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Lecture 11: Numerical Solution for Nonlinear Equations
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ÓÔ�{�Ä�g�
�¦) F(x) = 0, ��ЩXÚ G(x) = 0, Ù�®�� x0.
½ÂÓÔ¼êH(x, λ) = λF(x) + (1− λ)G(x).
w,k H(x, 0) = G(x), H(x, 1) = F(x).
·�F"H(x, λ) = 0 k) xλ, �§ëY/�6uëêλ ∈ [0, 1].
ù�, �� 0 = λ0 < λ1 < · · · < λn = 1, ¦� λi+1 − λi ¿©�,xi , xλi Ò�±�� xi+1 , xλi+1
�v�S��.
^ÛÜÂñ�S�{�gCq¦) xi , i = 1, 2, · · · , n, �ª��F(x) = 0 �) xn.
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Lecture 11: Numerical Solution for Nonlinear Equations
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ÓÔ�{�¢y
Ú 1: � G(x) = 0 ��®�� x0;Ú 2: é i = 1, 2, · · · , n, ± xi−1 �Ð�,
^ Newton {¦)��5�§| H(x, λi ) = 0,��Cq) xi ;
Ú 3: ��� xn =�¤¦.
5 1µ Ú� λi − λi−1 �>�>½§=g·A�N�.
5 2µ Newton {�±��Ù§?Û�«ÛÜÂñ�S�{.
5 3µ é i = 1, 2, · · · , n − 1, Ø7�Ñp°Ý�ê�). 3Ú�À�·���¹e, xi �°ÝU��
15 |xi+1 − xi | Òv.
5 4µ ý��]Ô5gu G(x) �À�.
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Lecture 11: Numerical Solution for Nonlinear Equations
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éuØÓ�ÓÔ¼ê H(x, λ) ÓÔ�{�U¬�ѱeA«ØÓ�(Jµ
1 �{�l�^n��ÓÔ�§�ª�ÑF(x) = 0 ����.
2 �{3,� λ0 ?uѧ= limλ→λ0−0 ‖xλ‖ =∞.
3 �{3,� λ0 ?Ñy=ò:, d�, ¤¦�� xλ0 Ø3?Û xλ, λ > λ0, �vC���¥.
4 �{3,� λ0 ?Ñy©�, d�, ¤¦�� xλ0 Ø3?Û½� xλ, λ > λ0, �vC���¥.
3���/ (2), (3), (4) �§�l��¬�}§d�I���ÓÔ¼ê.
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Lecture 11: Numerical Solution for Nonlinear Equations
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�{ü�ÓÔ¼ê�{
1 H(x, λ) = F(x) + (λ− 1)F(x0).
2 H(x, λ) = λF(x) + (1− λ)A(x− x0), Ù¥A ��_Ý.
5µ ��5`§ÓÔ¼ê�À�vk�½�5. ØL§éõ�ª¼ê®²/¤�@k���{£ë�ë�©z[26]¤.
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Lecture 11: Numerical Solution for Nonlinear Equations
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��5�§|¦���`z�{
�¦) F(x) = 0, - G (x) = F(x)TF(x). K¦�¯K�du¦¼ê G (x) ����:. Ïd§�A^�`z�nØÚ�{¦).~^��{k
1 ��eü{�FÝ.�{§ÛÜ�5z�{.
2 Úî{!&6�{§ÛÜ�g%C��{.
3 �[ò»{.
4 · · · · · · · · ·
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SKoµ3, 6, 8; þÅSKoµ3, 4, 5 (2), (5).
Thank You!