Assignment 8 solutions

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Unformatted text preview: MIT OpenCourseWare http://ocw.mit.edu 18.085 Computational Science and Engineering I Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. 18.085 - Mathematical Methods for Engineers I Prof. Gilbert Strang Solutions - Problem Set 8 Section 4.1 ⎡ ⎣1 ⎤0 � , |x| < 2 � < |x| < � , 2 f (x) � 3) f (x) = −� 1 a0 = 2� 1 = 2� = = � � � −� /2 f (x) dx � /2 � 1 bk = � 1 = � = � � �x � f (x) sin k x dx −� � /2 −� � /2 −� /2 1 · dx sin k x dx −� /2 1 [� /2 + � /2] 2� 1 2# � ��/2 1 cos k x − � k −� /2 � � �� 1 k� k� = − cos + cos − �k 2 2 = 0# 1 ak = � = = 1 � � � � f (x) cos k x dx −� � /2 cos k x dx −� /2 � ��/2 1 sin k x � k −� /2 � ⎦ � �� 1 k� = sin − sin k − �k 2 2 �� 2 k� = sin �k 2# 1 10) Boundary condition for Laplace’s equation � 1 , 0<λ<� u0 = 0 , −� < λ < 0 �� 1 1 1 a0 = 1 dx = (� − 0) = 2� 0 2� 2 � �� 1� 1� sin k � ak = cos k x dx = sin k x 0 = = 0 �k �0 �k �k � �� 1� 1� 1 bk = sin k x dx = − cos k x 0 = [1 − cos k � ] �0 �k �k ⎡ ⎣ 2 , k odd = �k ⎤0 , k even � u0 (λ) = � � 1 2 sin 3x sin 5x sin x + + +··· + 2� 3 5 � � r3 sin 3x r5 sin 5x 1 2 r sin λ + + +··· ∀ u(r, λ) = + 2� 3 5 # At origin, r = 0, λ = 0 u(0, 0) = 2 1 + [0 + 0 . . .] 2 � 1 = 2# F (x) � 1 13) width h = � , � a) |F (x)|2 dx � �/2 = 12 dx −�/2 = �# −� b) 1 Ck = 2� 1 = 2� = � � � −�/2 � /2 � �x F (x) e−ikx dx −� � /2 −�/2 − e ikx dx � ��/2 1 e−ikx 2� −ik −�/2 �� � −ik�/2 � 1 1 = −e + eik�/2 2� ik � � k� 1 = 2i sin 2�ki 2 2 C0 = = = = c) � 1 sin(k�/2) 2� (k/2) ⎡ ⎥1 ⎥ ⎥ � k , k = ±1, 5, 9, . . . ⎥ ⎥ ⎣ = − 1 , k = ±3, 7, 11, . . . ⎥ ⎥ �k ⎥ ⎥ ⎥ ⎤0 , k even = 1 2� 1 2� � � � F (x) dx 1dx −� � /2 −�/2 1 [�/2 + � /2] 2� 1 2# � −� |F (x)|2 dx � �� � � �2 � �2 � �2 � �2 2 1 1 1 1 1 + +− + +− +··· = 2� 2 � 3� � 3� � � � 4 1 1 =+ 1+ 2 + 2 +··· 2 � 3 5 � 2� � 4� =+ 2 �8 � �� � Observed from ramp series eqn (15) pg321 = �# �� 1 If h = 2�, C0 = 1 dx = 1 2� −� �� 1 1 sin(k� ) Ck = e−ikx dx = = 0 for all k 2� −� 2� (k/2) � F (x) = 1 a constant function # 18) Heat equation ut = uxx point source u(x, 0) = � (x) with free boundary conditions u � (�, t) = u � (−�, t) = 0 These boundary conditions require cosine series u(x, t) = b0 (t) + � ⎪ � � = 2� |C0 |2 + |C1 |2 + |C−1 |2 + · · · bn (t) cos nx n=1 3 Substitute into heat equation � b0 (t) + � � ⎪� ⎪ bn (t) cos nx = − bn (t)n2 cos nx n=1 n=1 since b � (t) 0 = 0, bn (t) = e −n2 t bn (0), b0 (t) = constant Initial condition � ⎪ 1 1 b0 (0) + bn (0) cos nx = � (x) = + [cos x + cos 2x + cos 3x + · · · ] 2� � n=1 Comparing coefficient b0 (0) = b0 (t) = 1 , 2� 1 2� bn (0) = 1 � since b0 is a constant # � u(x, t) = � ⎪ 1 −n2 t 1 + e cos nx 2� n=1 � Section 4.2 a) F = quarter square = � 1 , 0 � x � �, 0 � y � � 0 , −� < x < 0 or − � < y < 0 F (x, y ) � � � � 1� �y � � �� � � � � � � �x −� −� � � � 1 2� 1 2� 1 2� �2 � �2 � �2 � � −� � 0 � 0 Cmn = = � � � −� � F (x, y )e−imx e−iny dx dy e−imx e−iny dx dy � 0 �� e−imx dy −im 0 0 � �2 � � �� � 1 1� = e−iny 1 − e−im� dy 2� im 0 � �2 � � � � � e−iny � 1 1� −im� = 1−e 2� im −in 0 = e−iny = � 1 2� �2 � 1 im �� 1 in � � 1 − e−im� �� 1 − e−in� 4 � For m, n ≥= 0 ⎡ ⎣ − 1 , for m and n both odd � 2 mn Cmn = ⎤ 0 , for m even or n even C� = � 1 2� 1 2� 1 4 �2 � �2 � 0 � � 1 dx dy 0 = = � � � �2 # � 0 � C0n = = 1 2� 1 2� �2 � �2 � � � � e−iny dx dy 0 Cm0 �2 � � � � 1 = e−imy dx dy 2� 0 0 ⎡ 1 ⎥ ⎣ , for m odd 2�im = ⎥ ⎤0 , for m even # � 1 0 �� e−iny dx −in 0 0 � �2 � � � � � � 1 1 1 − e−in� dx = 2� in 0 � �2 ⎦ � � 1 �� = 1 − e−in� 2� in ⎡ ⎥ 1 , for n odd ⎣ 2�in = ⎥ ⎤0 , for n even # � b) F = checkerboard = if xy > 0 if xy < 0 −� <x�� −� <y �� y � � ����� � �� � �� ��� �� � �� ��� � � �� �� � ��� � −� ���� � � � �� � � � � ��� � � � F =1 ��� �� � �� �� �� ����� −� � x 5 Cmn = � � � � � 1 2� 1 2� 1 2� 1 2� 1 2� �2 � �2 � � −� � 0 � � � � −� � F (x, y ) e−imx e−iny dx dy � 0 = e−imx e−iny dx dy + � e−imx −im �� 0 0 1 2� �2 � 0 −� � 0 −� e−imx e−iny dx dy �0 � = � 2 �� e−iny 1 im � dy + 0 � e−iny −� � e−imx −im dy −� = �2 �� �2 � � 1 − e−im� �� � � � �� � �� � e−iny � � e−iny 0 1� −1 + eim� + −in 0 im −in −� �� �� �� � 1 − e−in� + −1 + eim� −1 + ein� � = 1 im �� 1 in 1 − e−im� For m, n ≥= 0, ⎡ ⎥ − 2 , for m, n odd ⎣ � 2 mn Cm = ⎥ ⎤ 0 , if m even or n even C� = � 1 2� � 2 �� � 0 � � dx dy + 0 � 0 −� � 0 −� � dx dy = 1 2# � � � � � 1 2� 1 2� 1 2� 1 2� 1 2� � 2 �� � C0n = 0 � 0 � � � e−iny dx dy + �� 0 0 � 0 0 −� � 0 −� � e−iny dx dy �0 dx 1 in � ⎩ (−1 + ein� )dx � = � 2 �� � 2 �� �2 � �2 ⎦ � e−iny −in 1 in � dx + � −� � e−iny −in � 0 −� −� = 0 � (1 − e−in� )dx + � = 1 in �� � (1 − e−in� )� + (−1 + ein� )� = � � �� −in� in� (1 − e ) + (−1 + e ) in Cm0 = 0# � � 2 �� � � � � �0�0 1 −imx −imx = e dx dy + e dx dy 2� 0 0 −� −� = 0# 6 2) −S (x, y ) = S (−x, −y ) odd function will have double sine series ��bmn sin mx sin ny For e.g. S (x, y ) = � 1 −1 if xy � 0 , −� < x � � if xy < 0 , −� < y � � S (−x, y ) = −S (x, y ) S (x, −y ) = −S (x, y ) ⎩ are also odd functions Orthogonality ���� (sin k x sin ly )(sin mx sin ny )dx dy = 0 −� −� for k = m and l = n on the square 20) −� � x � � − � � y � � # Sturm-Liouville eigenvalue problem (pu � ) � + q u + �wu = 0 � � u1 (pu2 � ) � + q u2 + �1 wu2 = 0 � � u2 (pu1 � ) � + q u1 + �2 wu1 = 0 � 1 � 2 � 3 � −� 1 2 � �0 u1 (pu2 � ) � − u2 (pu1 � ) � + q (u���� 1 ) + wu1 u2 (�1 − �2 ) = 0 1 u2 − u 2 u � u1 (pu2 � ) � − u2 (pu1 � ) � + wu1 u2 (�1 − �2 ) = 0 � � b a � u2 (pu1 ) � dx where a, b are boundary � b �0 � b ��b � � � � �) − = u2 (� 1 a pu u2 (pu1 )dx = − u2 (pu1 )dx � a a Boundary � b a � u1 (pu2 ) � dx �b �0 � b � ��b � � � � �� = u1 (pu2 ) a − u1 (pu2 )dx = − u1 (pu2 )dx � a a Boundary From � 3 � �b � b� �� �� u1 (pu2 ) − u2 (pu1 ) dx = (�2 − �1 )wu1 u2 dx a a �b �b � b � � � � − u1 (pu2 )dx + u2 (pu1 )dx = (�2 − �1 )wu1 u2 dx a a a 7 � � b a (�2 − �1 )wu1 u2 dx = 0 If (�2 − �1 ) = 0 ≥ � b a wu1 u2 dx = 0 # weighted orthogonality Section 4.3 2) fj = N −1 ⎪ k=0 ck w j k ∀ f = F N c w 2j . . . w j (N −1) ] w2(N −j ) . . . w(N −j )(N −1) ] row j th of F = [1 w j row (N − j )th of F = [1 w(N −j ) w j = e−2�ij/N wN −j = e(2�i/N )(N −j ) = e2�i e−2�ij/N = 1 · e−2�ij/N = wj Similarly w 2j = w2(N −j ) . . . w j (N −1) = w(N −j )(N −1) � row j th of F is the same as row (N − j )th of F # � 1 1 1 −1 1 1 w4 w8 1 w4 w8 1 w w2 −1 −w −w 2 1 w ⎢� 11 1 w2 1 w4 1 w4 w8 ⎢� ⎢ 100000 0 0 1 0 0 0⎧ ⎧ 0 0 0 0 1 0⎧ ⎧ ⎧ ⎧ 0 1 0 0 0 0⎧ ⎧ 0 0 0 1 0 0⎨ 000001 �� � P � � � � 6) F6 = � � �1 � � � 1 −w 1 w3 w6 ⎧� ⎧� w2 ⎧ � ⎧� ⎧� ⎧� ⎧� ⎧� ⎨� −w 2 1 w5 w10 −1 −w5 −w10 0 0 ⎢� 11 1 w2 1 w4 ⎧ ⎧ ⎧ ⎧ ⎧ ⎧ 1⎧ ⎧ w4 ⎨ w8 ⎢ ⎧ ⎧ ⎧ ⎧ ⎧ ⎧ ⎧ ⎧ ⎨ � � � � =� � �1 1 � � 1 w2 1 w4 11 1 w2 1 w4 1 1 1 1 1 1 −1 −w3 −w6 ⎧� ⎧� ⎧� ⎧� ⎧� ⎧� ⎧� ⎧� ⎨� � � � � � � � � � � 8 � � � =� � � � � � 1 1 1 w 1 w2 1 −1 1 −w 1 −w 2 1 1 1 1 1 1 1 w w2 w3 w4 w5 1 w2 w4 1 w2 w4 1 w2 w4 w6 w8 w10 1 w3 w6 −1 −w 3 −w 6 1 w3 w6 w9 w12 w15 1 w4 w8 1 w4 w8 1 w4 w8 w12 w16 w20 1 w5 w10 −1 −w 5 −w10 1 w5 w10 w15 w20 w25 ⎢ ⎧ ⎧ ⎧ ⎧ ⎧ ⎧ ⎨ ⎢ ⎧ ⎧ ⎧ ⎧ ⎧ ⎧ ⎨ Where D = � � � 1 w w2 ⎢ ⎨ � � � =� � � � 8) 11 F3 = � 1 w 2 1 w4 ⎢ 1 w4 ⎨ w8 # verified c = (1, 0, 1, 0), c � = (1, 1) f � = F2 c � �� � � 1 1 1 = 1 −1 1 �� 2 = 0 First half, fj = fj� + (wN )j fj�� f1 = � 2 0 � + � 1 i �� 0 0 N =4 c �� = (0, 0) f �� = F2 c �� �� � � 1 1 0 = 1 −1 0 �� 0 = 0 � = � 2 0 � Second half, fj +M = fj� − (wN )j fj�� f2 = � 2 0 � − � 1 i �� 0 0 � � = � 2 0 � ⎢ 2 �0⎧ � Combine f = [f1 ; f2 ] = � ⎧ �2⎨ 0 For c = (0, 1, 0, 1) c � = (0, 0) f � = F2 c � � �� � 1 1 0 = 1 −1 0 �� 0 = 0 # c �� = (1, 1) f �� = F2 c �� � �� � 1 1 1 = 1 −1 1 �� 2 = 0 9 First half, ��� 0 1 f1 = + 0 Second half, ��� 1 0 f2 = − 0 � Combine � ⎢ 2 � 0⎧ ⎧ f =� � −2 ⎨ 0 i �� �� 2 0 � � = � � 2 0 � � i 2 0 = −2 0 # 10) w = e2�i/64 ≤ w2 = e2�i/32 w = e2�i/128 ≤ � w2 and w are among the 32 and 128 roots of 1 # 15) � 1i � 1 i2 =� � 1 i3 11 11 �1 i =� � 1 i2 1 i3 � 0 �0 � �0 1 � 1 0 0 0 0 1 0 0 i2 i4 i6 1 1 i2 i4 i6 ⎢� 0 11 0 ⎧� 1 i ⎧� 1 ⎨ � 1 i2 0 1 i3 ⎢ i3 i6 ⎧ ⎧ i9 ⎨ 1 1 i2 i4 i6 ⎢ 1 i3 ⎧ ⎧ i6 ⎨ i9 � �1 = 1, �2 = i, ⎢� 1 1 i3 ⎧ � i ⎧� i6 ⎨ � i9 i 2 ⎢ i3 ⎧ ⎧ ⎨ � 3 = i2 , �4 = i3# 10 ...
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This note was uploaded on 04/12/2011 for the course MATH 18.085 taught by Professor Staff during the Fall '10 term at MIT.

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