MIT OpenCourseWare http://ocw.mit.edu 18.085 Computational Science and Engineering IFall 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 , |x| < 2 3) f(x) = � ⎤0 , < x < � 2 | | f(x) x −� −�/2 �/2 � 1 � � 1 � � a0 = f(x) dx bk = f(x) sin kx dx 2� −� � −� 1 � �/2 1 � �/2 = 1 dx = sin kx dx 2� −�/2 · � −�/2 1 1 � cos kx ��/2 = [�/2 + �/2] = 2� � − k −�/2 1 1 � k� � k� �� = 2 # = �k − cos 2 + cos − 2 = 0 # 1 � � ak = f(x) cos kx dx � −� 1 � �/2 = cos kx dx � −�/2 1 � sin kx ��/2 = � k −�/2 1 � k� ⎦ � �� = �k sin 2 − sin k − 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� 1 � � 1 � �� sin k� ak = � 0 cos kx dx = �k sin kx = �k = 0 �k0 1 � � 1 � �� 1 bk = � 0 sin kx dx = �k − cos kx = �k [1 − cos k�]0 ⎡ 2 ⎣ , k odd = �k ⎤0 , k even 2 � sin 3x sin 5x � � u0(λ) = + sin x + + + � 3 5 ··· 1 2 � r3 sin 3x r5 sin 5x � ∀ u(r, λ) = + r sin λ + + + 2 � 3 5 ··· # At origin, r = 0, λ = 0 2 u(0, 0) = + [0 + 0 . . .]2 � 1 = 2 # 13) width h = �, F (x) a) �|F (x)| 2dx � �/2 = 12dx −�/2 = � # 1 x −� −�/2 �/2 � 1 � � b) Ck = F (x) e −ikx dx 2� −� 1 � �/2 −ikx dx= e 2� −�/2 1 � e−ikx ��/2 = 2� −ik −�/2 1 � 1 �� −ik�/2 ik�/2� = −e + e2� ik 1 � k� � = 2i sin 2�ki 2 2 1 2 1 2� � � � � � � � � � � � � �� � = � # � 1 sin(k�/2) = 2� (k/2) = ⎡ ⎥⎥⎥⎥⎥⎣ ⎥⎥⎥⎥⎥⎤ 1 , k = ±1, 5, 9, . . . �k 1 − �k , k = ±3, 7, 11, . . . 0 , k even 1 C0 = F (x) dx 2� −� 1 � �/2 = 1dx 2� −�/2 1 = [�/2 + �/2]2� 1 = 2 # c) F (x)2 dx −� | |2 2 2= 2� |C0| C1C−1+ + +| | | | ··· �� 1�2 � 1 �2 �2 � 1 �2 �21 1 = 2� 2+ � + − � +3� + − 3� + ··· � 4 1 1 = + 1 + + + 2 � 32 52 ··· � 4 = + �2� 2 � 8 Observed from ramp series eqn (15) pg321 1 � � If h = 2�, C0 = 1 dx = 1 2� −� 1 � � −ikx dx = 1 sin(k�)Ck = e = 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) + bn(t) cos nx n=1 3� � � � � ���� ���� ���� � � Substitute into heat equation b0�(t) + ⎪ b � (t) cos nx = − ⎪ (t)n2 cos nxnbnn=1 n=1 2since b0�(t) = 0, bn(t) = e−n tbn(0), b0(t) = constant Initial condition ⎪ 1 1 b0(0) + bn(0) cos nx = �(x) = + [cos x + cos 2x + cos 3x + ] n=1 2� � ··· Comparing coefficient 1 1 b0(0) = , bn(0) = 2� � 1 b0(t) = since b0 is a constant 2� 21 ⎪ 1 −n t � u(x, t) = + 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) �y 1 � x 0 �−� −� � 1 �2 � � � � Cmn = F (x, y)e −imx e −iny dx dy 2� −� −� � 1 �2 � � � � = e −imx e −iny dx dy 2� 0 0 � 1 �2 � � � e−imx �� = e −iny dy 2� −im00 = � 1 �2 � � e −iny � 1 ��1 − e −im�� dy2� 0 im � 1 �2 � 1 � � −iny �� �−im�� e= 2� im 1 − e−in 0 � 1 �2 � 1 �� 1 � �−im���−in�� = 1 − e 1 − e2� im in 4� � For m, n = 0≥⎡ 1 Cmn = ⎣ − �2mn , for m and n both odd ⎤ 0 , for m even or n even � 1 �2 � � � � C� = 1 dx dy2� 0 0 � 1 �2 = �2 2� 1 = 4 # � 1 �2 � � � � C0n = e −iny dx dy 2� 0 0 � 1 �2 � � � e−iny �� = dx 2� 0 −in 0 � 1 �2 � 1 �� � �−in� � =2� in 0 1 − e dx � 1 �2 ⎦ � ��−in� � = 1 − e2� in ⎡ 1 ⎥, for n odd⎣ 2�in= ⎥⎤ 0 , for n even # � 1 �2 � � � � Cm0 = e −imy dx dy 2� 0 0 ⎡ 1 ⎥, for m odd⎣ 2�im= ⎥⎤ 0 , for m even # b) F = checkerboard = 1 if xy > 0 − � < x � � 0 if xy < 0 − � < y � � y � � � � � � � � � � � � � � � � � � � � � � ����� � � � � � � � � � ��� �� � � � � � � � � ��� ��� � � � � � � � � ����� ���−� −� � F = 1 x 5� 1 �2 � � � � Cmn = F (x, y) e −imx e −iny dx dy 2� −� −� � 1 …