{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

ps5sol

# ps5sol - ACM 95a/100a Problem Set 5 Solutions Prepared by...

This preview shows pages 1–4. Sign up to view the full content.

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 Φ( z ) = φ ( x, y ) + ( x, y ), where φ and ψ are harmonic functions representing the velocity potential and stream function, respectively. Sketch and describe the flow defined by Φ( z ) = α log z , where 0 arg z < 2 π and α is a complex constant. Solution 1 Let α α 1 + 2 . Then, Φ( z ) = α log( z ) = ( α 1 + 2 )(Log( r ) + ) = α 1 Log( r ) - α 2 θ + i ( α 2 Log( r ) + α 1 θ ) (1) The streamlines are ψ = α 2 Log( r ) + α 1 θ = const -→ r = (const) e - α 1 α 2 θ . So, α 1 α 2 finite -→ “spiral” flow (2) α 1 = 0 -→ “vortical” flow: r = const (3) α 2 = 0 -→ “radial” flow: θ = const (4) See Figure 1 for plots and the Matlab code which generates the plots of “spiral” flow. Now, φ = ∂φ ∂r b i r + 1 r ∂φ ∂θ b j θ = α 1 r b i r - α 2 r b j θ (5) Therefore, at z = 0 we have a source if α 1 > 0 and a sink if α 1 < 0 . 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
-1 -0.5 0 0.5 1 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 -1 -0.5 0 0.5 1 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 Figure 1: Top left: The streamlines for α 1 α 2 = 1 Top right: The streamlines for α 1 α 2 = - 1 Bottom left: The streamlines for α 1 = 0 Bottom right: The streamlines for α 2 = 0 2
Matlab Code for the Figure (by courtesy of Nawaf Bou-Rabee): theta = linspace(0, 2*pi-eps, 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 iπ/ 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 π 4 , r 0, but non-zero on the interior of the region we would like a function like sin(4 θ ). Consider z 4 = r 4 (cos(4 θ ) + i sin(4 θ )) Since z 4 is entire, f ( r, θ ) = r 4 sin(4 θ ) must be harmonic since it is the imaginary part of an analytic function; however, it also satisfies the boundary conditions and is non-zero for 0 < θ < π 4 . In rectangular coordinates we would have f ( x, y ) = xy ( x 2 - y 2 ) = Im( z 4 4 ).

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 10

ps5sol - ACM 95a/100a Problem Set 5 Solutions Prepared by...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online