s4_95-100aSOLUTION

# s4_95-100aSOLUTION - ACM95a/100a Last updated Problem Set...

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Unformatted text preview: ACM95a/100a Last updated October 28, 2009 Problem Set IV Solutions Prepared by Faisal Amlani 1. (50 pts.) With reference to pages 129 through 131 of the Saff and Snider text, find solutions to the five boundary value problems for the Laplace equation depicted in Figures 3.11, 3.12, 3.13, 3.14 and 3.16, and evaluate them at the points (0 , 0), (0 , 0), (1 , 1), (2 , 3) and (0 , 0), respectively. SOLUTION : (a) We want a harmonic function for the annulus centered at 1 + i , i.e. φ = A Log | z − (1 + i ) | + B. The inner and outer boundaries ( | z − (1+ i ) | = 1 , 2 respectively) give the system 0 = A Log 1 + B 10 = A Log 2 + B. The first equation gives B = 0 which from the second implies A = 10 / Log2 . Thus we have the solution φ = 10 Log2 Log | z − (1 + i ) | . At (0 , 0), we have | z − (1 + i ) | = | − 1 − i | = √ 2 . This gives φ (0 , 0) = 10 Log 2 Log √ 2 = 5 . (b) We cannot use the harmonic φ = A Arg( z − (1 + i )) + B since the principal branch of the Log contains a cut through the domain. Instead, we can use a different branch such that the cut is outside, e.g. the branch defined as θ = arg − π/ 4 ( z − (1 + i )) from − π/ 4 to 7 π/ 4 . This choice places the cut in the middle of the white plane, and the function is still harmonic in the shaded region. We then have from the boundary conditions the system 10 = A (0) + B 0 = A (3 π/ 2) + B. The first equation clearly implies that B = 10 which from the second equation gives A = − 20 / (3 π ) . Thus we have the solution φ = − 20 3 π arg − π/ 4 ( z − (1 + i )) + 10 , arg from − π/ 4 to 7 π/ 4 . At (0 , 0) we have an angle of 5 π/ 4 from the arg definition. This gives φ (0 , 0) = − 20 3 π 5 π 4 + 10 = 5 3 . (c) We are looking for a function that is harmonic between the lines y = − x +3 , y = − x − 3 ....
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s4_95-100aSOLUTION - ACM95a/100a Last updated Problem Set...

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