lmam and school of mathematical sciences peking university
TRANSCRIPT
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�
�¦)�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�
�¦)�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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����{{0— ÓÔ�{
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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ÓÔ�{¢yL§¥�UÑy��¹
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!