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ENGR 516 Computational Methods for Graduate Students Catholic University of America Assignment #7 Elliptic PDE ENGR516 Assignment #5: Two Dimensional PDEs Problem #1: Alternating Direction Implicit (ADI) Method Research and write a brief explanation of the ADI Method and solve: Two Dimensional LaPlace Equation: Electric Potential Over a Flat Plate with Point Charge ! ! ! ∇ 2u(x, y) = f (x, y) for -1 ≤ x ≤ 1, -1 ≤ y ≤ 1 boundary conditions: u(x,y) = 0 for all boundaries f (0.5, 0.5) = −1 f (−0.5, −0.5) = 1 elsewhere : f (x, y) = 0 Two Dimensional Temperature Diffusion: ⎛ ∂ 2u(x, y,t) ∂ 2u(x, y,t) ⎞ ∂u(x, y,t) 10 −4 ⎜ + ⎟⎠ = ∂x 2 ∂y 2 ∂t ⎝ ! ! ! for 0 ≤ x ≤ 4, 0 ≤ y ≤ 4 0 ≤ t ≤ 5000 u(x, y, 0) = 0 u(x, y,t) = e y cos x − e x cos y for x = 0, x = 4, y = 0, y = 4 ! ! Present results for t= 5000 Problem #2: Crank-Nicolson Problem Solve the Two-Dimensional Temperature Problem above using Crank-Nicolson Method.
ENGR 516 Computational Methods for Graduate Students Catholic University of America Assignment #6 EigenValue Problems 1.) The moment of inertia, Ix, Iy, and the product of inertia Ixy of the cross sectional area shown in the figure are: 1 y inertia I xy of the 20 mm 4 mm O 4 mm s of the matrix 3 mm 2 mm 24 mm on of the eigen- solving the charprincipal axes of ). x 4.23 The moment of inertia I x , I y , and the product of inertia I xy of the cross-sectional area shown in the figure are: I x = 5286 mm4, I y = 4331 mm4, and I xy = 2914 mm4 The principal moments of inertia are the eigenvalues of the matrix y 20 mm O 24 mm 5286 2914 , and the principal axes are in the direction of the eigen2914 4331 vectors. Determine the principal moments of inertia by solving the characteristic equation. Determine the orientation of the principal axes of inertia (unit vectors in the directions of the eigenvectors). 2 mm Solution 2 mm The eigenvalues of the matrix 5286 2914 are the principal moments of inertia. They are dete 2914 4331 the roots of the characteristic equation: 1 ( 5286 – λ ) ( 4331 – λ ) – ( 2914 ) ( 2914 ) = 0 ipal moments of inertia. are determined 2.) They The structure of anfrom acetylene molecule may be idealized as four masses1 connected by which simplifies to 14402270 – 9617λ + λ 2 = 0 . 4.50 The structure of the C H (acetylene) molecule may be ide- ( 2914 ) ( 2914 ) = 0 2 two springs shown in2 the figure below. Applying thequadratic equation of motion; we can The equation is solved by using formula: alized as fouras masses connected two springs (see discussion in the 4.50 The structure of by the C 2H2 (acetylene) molecule may be idegenerate system model forequation the amplitudes of vibration of each atom: Problem a4.25). By applying the motion, thediscussion following 2 alized as four masses connected byoftwo springs (see in or λ 1 = 7761.36 and genvalue must satisfy: (i) = λi 9617 ( 9617 ) – 4 ( 1 ) ( 14402270 ) 1 Problem By applying equation ofλmotion, the following --------------------------------------------------------------------------------------- or λ 1 = 7761.36 and λ 2 = 1855.64 system of equations can4.25). be written forthethe amplitudes vibration 1 2 = of 2 system of equations can be written for the amplitudes of vibration of each atom: of each atom: 4.50 Thekstructure ofkthe C2H2 (acetylene) molecule may be idek-------k CH CH CH CH By - – ω2 – -------0 eigenvectors corresponding to each eigenvalue must satisfy: 2 --------- – ωalized 0 definition, 0 the as– -------four by0two springs (see discussion in m - masses connected mH mH m HH A 1 the 0following k CC ( k CH + k CC ) k CH By 4.25). applying the equation of motion, (i) (i) – -------- ---------------------------- – ω2 – -------0 A λ 2k = Problem 1855.64 5286 2914 u 1 = λ u 1 kwritten ( k CH +ofk CC m)C mfor mcan 1A 2 00vibration C C CH system CC equations be the amplitudes of 2 = – --------- ----------------------------- – ω – -------0 i 0 k CC k CH + k CC ) k CH A A 3 (i) m C of eachmatom: m C(---------------------------2914 4331 u ( i ) C 0 – -------- – ω 2 – -------- 2 = 0 u2 2 0 mC mC mC A4 0 A3 k CH + k CC ) k2 k CH CC k CH(---------------------------k CH –2k-------kthe CHwith CH 2 Starting eigenvalue: 0 -–ω - - – ωfirst 0 –0-------- – ----------------------- – ω - 0 mH 0 m C – -------mmCH 0A 4 mC mH mH 2 2 2 are the restoring ( 1 )force spring ( 1 ) (1) frequency, k CHkCH =k5.92 k× 10 kg/s22 and kCC A= 115.8 × 100 kg/s ( k CHω+iskthe k CH where u1 u1 CC ) CC CH 5286 2914 u 0-----------------------------------C–H - – and ---------bonds, – ω respectively, – -------- constants - – ω 2–the -------0 representing C–C and m H = 1amu and m C = 112amu=areλ the = 7761.36 1 mH mC m C m H–27 A2 mC (1) (1) = 0 2914 4331 u ( 1 ) masses of the atoms ( 1amu = 1.6605 × 10 kg). u2 u2 2 2 2 0 ( + k ) A k k k 2πc- (where 2 andCH 2 are the CC CH × 10 CC(frequencies), 3corresponding (a) Determine the eigenvalues ω and the wavelengths λ = -------2 where ω is the frequency, kg/s k = 5.92 k = 15.8 × 10 kg/s restoring force spring 0 – -------- CH ----------------------------- – ω – ---------CC (ω 1) (1) 0 mC 8 m C areA 4redundant m C two equations These and yield u = 1.1772u . u1 0 (i) u2 m/s isand the speed light). respectively, and m H = 1amu and 1m C = 12amu 2are the c = the 3 × 10C–H constants representing C–Cofbonds, (1) u1 (1) (1) = 7761.36 u2 (1) u2 (1) 772u 2 1. u1 . k – 27 k to the eigenvalues found in part (a). From the eigenvectors, (b) Determine the eigenvectors corresponding CH CH 2 (1) 2 0the relative -------- – ω 2 moving toward, or away from = 1.6605 × –10 masses of the0atomsdeduce ( 1amu kg). motion of-------the- atoms (i.e., Since the eigenvector a unit vector: ( u (11 ) )each + other?). ( u2 ) = 1 . m m Hare theyis H Solution (a) Determine the eigenvalues ω (frequencies), and2 the corresponding2 wavelengths λ = 2πc --------- (where 2 2 ω force spring where ω is the frequency, k CH = 5.92 × 10 kg/s and k CC = 15.8 × 10 kg/s are the restoring 8 The following script file solves this problem: representing theof C–H and C–C bonds, respectively, and m H = 1amu and m C = 12amu are the is the speed light). c = constants 3 × 10 m/s – 27 (b) Determine eigenvectors corresponding = 1.6605 × 10to the masses the of the atoms ( 1amu kg).eigenvalues found in part (a). From the eigenvectors, clear, motion clc deduce the relative of the atoms (i.e., are theyand moving toward, or away from eachλother?). (a) Determine the eigenvalues ω (frequencies), the corresponding wavelengths = 2πc --------- (where Excerpts from this work may be reproduced by instructors for distribution on a not-forω for testing of light).or instructional purposes only to students enrolled in courses for which t c = 3 × 10 m/s is the speed M(1,1)=kCH/mH; M(1,2)=-M(1,1); (b) Determine the eigenvectors corresponding to the eigenvalues found in part (a).or From the eigenvectors, has been adopted. Any other reproduction translation of this work beyond that p The following script file solves this problem: M(2,1)=-kCH/mC; M(2,2)=(kCH+kCC)/mC; M(2,3)=-kCC/mC; deduce the relative motion of the atoms (i.e., are they toward, or awayStates from each other?). Act without the permis Sections 107 or 108 ofmoving the 1976 United Copyright M(3,2)=M(2,3); M(3,3)=M(2,2); M(3,4)=M(2,1); copyright owner is unlawful. Solution M(4,3)=M(1,2); M(4,4)=-M(4,3); Solution kCH=5.92e2; kCC=15.8e2; mH=1.6605e-27; mC=12*1.6605e-27; c=3e8;8 M=zeros(4); clcfollowing script file solves this problem: The ructors for distributionclear, on a not-for-profit basis kCC=15.8e2; mH=1.6605e-27; mC=12*1.6605e-27; nts enrolled in courseskCH=5.92e2; for which the textbook Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis c=3e8; M=zeros(4); for testing or instructional purposes only to students enrolled in courses for which the textbook slation of this work beyondclear, that permitted by Any other reproduction or translation of this work beyond that permitted by has adopted. clcbeen M(1,1)=kCH/mH; M(1,2)=-M(1,1); Sections of 107 the or 108 of the 1976 United States Copyright Act without the permission of the Copyright Act without the permission

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