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Unformatted text preview: 4.6 The method of images 4.6.1 A vortex next to a wall xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx 1 2 b Γ 4 πb ψ = 0 Consider the motion of a vortex located at some position z = a + ib in the upper half of the complex plane. The x axis is the location of a solid wall. Thus we have to satisfy the boundary condition of no flow through the wall: v = 0 for y = 0. Simply taking the flow field of the vortex, u iv = i Γ 2 π x a i ( y b ) ( x a ) 2 + ( y b ) 2 , (39) does not satisfy this condition: the flow field must be modified by the presence of the wall. The method of images is a technique that permits to find solutions to Laplace’s equa tion which satisfy the right boundary condition on the solid wall. Near the vortex at z , the solution should look like (39). The method of solution is indicated in the figure: imagine replacing the wall by another image vortex that is placed at an equal distance on the other side of the wall. Its position is where an image in a mirror would appear. And equally like a mirror image, the sense of rotation of the image vortex is reversed. By symmetry it is clear that the v component of the image vortex has opposite sign, so on the line of symmetry (the locus of the wall), the two contributions cancel and v = 0. The boundary condition on the wall is satisfied and we have solved our problem! In other words, the flow problem we want to solve ( one vortex in the presence of a wall) is exactly the same as the flow problem of two vortices (the original vortex and its image) without the wall. Let’s check if our reasoning was correct. It is easiest to do the calculation using complex notation. If the vortex is at z , it’s image is at z . Since the image also has the opposite sense of rotation, there is a minus sign in front of it. The total potential (vortex + image) becomes w ( z ) = i Γ 2 π log( z z ) + i Γ 2 π log( z z ) . We have to verify that Im { w } = ψ = const for z = x real! Now since ln( z ) = ln( z ) (why?) we have w ( z ) = i Γ 2 π log( z z ) i Γ 2 π log( z z ) . But if z is real, z = z , and thus w ( x ) w ( x ) = 0, and so ψ = 0 along the wall. In fact, we have shown that the problem of a vortex near a wall is mathematically equivalent to the problem of two counterrotating vortices of equal strength, which in the previous section we showed to move at constant speed in the direction perpendicular to the line connecting them, in other words, parallel to the wall. Let us check this fact as well, using complex notation. The vortex in question moves in the flow field of its image, i.e. u iv = i Γ 2 π x a i ( y + b ) ( x a ) 2 + ( y + b ) 2 , evaluated at z = a + ib . Thus u iv = i Γ 2 π parenleftbigg 2 ib 4 b 2 parenrightbigg = Γ 4 πb , and the vortex moves with a horizontal speed of Γ 4 πb , which corresponds to the earlier result with h = 2 b ....
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This note was uploaded on 01/20/2012 for the course MATH 33200 taught by Professor Eggers during the Fall '11 term at University of Bristol.
 Fall '11
 Eggers

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