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Unformatted text preview: ACM 95a/100a Problem Set 5 Solutions Prepared by: Zhiyi Li 11/06/2007 Total: 110 points Include grading section: 2 points Problem 1 (10 points) Consider potential flow described by a complex potential ( z ) = ( x, y ) + i ( x, y ), where and are harmonic functions representing the velocity potential and stream function, respectively. Sketch and describe the flow defined by ( z ) = log z , where 0 arg z < 2 and is a complex constant. Solution 1 Let 1 + i 2 . Then, ( z ) = log( z ) = ( 1 + i 2 )(Log( r ) + i ) = 1 Log( r ) 2 + i ( 2 Log( r ) + 1 ) (1) The streamlines are = 2 Log( r ) + 1 = const r = (const) e 1 2 . So, 1 2 finite spiral flow (2) 1 = 0 vortical flow: r = const (3) 2 = 0 radial flow: = const (4) See Figure 1 for plots and the Matlab code which generates the plots of spiral flow. Now, = r b i r + 1 r b j = 1 r b i r 2 r b j (5) Therefore, at z = 0 we have a source if 1 > 0 and a sink if 1 < . 110.5 0.5 110.80.60.40.2 0.2 0.4 0.6 0.8 110.5 0.5 110.80.60.40.2 0.2 0.4 0.6 0.8 1 Figure 1: Top left: The streamlines for 1 2 = 1 Top right: The streamlines for 1 2 = 1 Bottom left: The streamlines for 1 = 0 Bottom right: The streamlines for 2 = 0 2 Matlab Code for the Figure (by courtesy of Nawaf BouRabee): theta = linspace(0, 2*pieps, 100); c = linspace(.1,.1,40); for i=1:length(c) x=c(i)*exp(theta).*cos(theta); y=c(i)*exp(theta).*sin(theta); figure(1); hold on; plot(x,y); end axis equal axis([1 1 1 1]) % notice the different range of the constant c c = linspace(40,40,40); for i=1:length(c) x=c(i)*exp(theta).*cos(theta); y=c(i)*exp(theta).*sin(theta); figure(2); hold on; plot(x,y); end axis equal axis([1 1 1 1]) Problem 2 (10 points) Consider a wedge region bounded on two sides by the rays z = r and z = re i/ 4 for r 0. Find a function that is harmonic in this region that vanishes on the rays but is nonzero in the interior of the region. Solution 2 Since f must be zero on the rays z = r and z = re i 4 , r 0, but nonzero on the interior of the region we would like a function like sin(4 ). Consider z 4 = r 4 (cos(4 ) + i sin(4 )) Since z 4 is entire, f ( r, ) = r 4 sin(4 ) must be harmonic since it is the imaginary part of an analytic function; however, it also satisfies the boundary conditions and is nonzero for 0 < < 4 . In rectangular coordinates we would have f ( x, y ) = xy ( x 2 y 2 ) = Im( z 4 4 )....
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 Fall '06
 NilesA.Pierce

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