Flow Around a Cylinder with Circulation

    We can combine the previous two phenomena and obtain a flow that flows around a cylinder with circulation. The resulting vector field is
        Overscript[v, ⇀] = U(1 - R^2 (x^2 - y^2)/(x^2 + y^2)^2, R (-2x y)/(x^2 + y^2)^2) + γ/(2π) (-y/(x^2 + y^2), x/(x^2 + y^2))  defined for (x^2 + y^2)^(1/2) ≥R
with the parameters
        R - radius of the obstacle,
        U - velocity of the fluid far from the obstacle, and
        γ - which represents the strength of the vortex.
    In this case the general structure of the flow varies depending on the values of the parameters R, U, and γ. Indeed, we use the following Mathematica commands to view the field.

Needs["Graphics`PlotField`"] U = 1/2 ; R = 1 ; γ = 4 ;  ... sp;           (* Show the result *)

[Graphics:../HTMLFiles/index_36.gif]

    To see how changing parameters affect the vector field, we examine the following graphs, where we fix U = 1/2 and R = 1, and let γ vary in the range -10≤γ≤10.

Needs["Graphics`Arrow`"] Needs["Graphics`PlotField`"] Off[ ... bsp;  (* Here is the loop for Table *)       

Varying Gamma

Stagnation points

    One important thing to notice are the points where Overscript[v, ⇀] = (0, 0). These points are called stagnation points, and are points where the fluid is stationary. We can determine these points by simply solving the equation Overscript[v, ⇀] = (0, 0).

Clear[U, R, γ] ; Solve[U {1 - R^2 (x^2 - y^2)/(x^2 + y^2)^2, R^2 (-2 x y)/(x^2 + y^2)^2} + γ/(2π) {-y/(x^2 + y^2), x/(x^2 + y^2)}  {0, 0}, {x, y}]

{{x0, y (γ - (-16 π^2 R^2 U^2 + γ^2)^(1/2))/(4 π U)}, {x&# ... 62754; (16 π^2 R^2 U^2 - γ^2)/U^(1/2)/(4 π U^(1/2)), yγ/(4 π U)}}

The solutions are
        x = 0, y = (-γ ± (γ^2 - (4π R U)^2)^(1/2))/(4 π U)
and
        x = ± ((4 π R U)^2 - γ^2)^(1/2)/(4 π U)  , y = γ/(4π U).  
Clearly, the behavior depends on the sign of γ^2 - (4π R U)^2. If γ is small enough that γ^2 - (4π R U)^2<0, then only the last two solutions are real, and the resulting pair of stagnation points lie on the obstacle. Indeed, direct calculation shows that in this case x^2 + y^2 = R^2:

Clear[U, R, γ] ; Simplify[(((4 π R U)^2 - γ^2)^(1/2)/(4 π U))^2 + (γ/(4 π U))^2]

R^2

On the other hand, if γ is large enough that γ^2 - (4π R U)^2>0, then only the first two solutions are real. Exactly one of these solutions lies outside the obstacle, yielding precisely one stagnation point; if U>0 it will be above the obstacle if γ>0 and below the obstacle if γ<0. Indeed, if γ>0 and U>0 we have
        y_ ± =[-γ/(4 π R U) ± ((γ/(4 π R U))^2 - 1)^(1/2)] R.
Thus, if we make the change of variables ζ = γ/(4 π R U), we see that y_ ± =[-ζ ± (ζ^2 - 1)^(1/2)] R while the requirement γ^2 - (4π R U)^2>0 implies that ζ>1. One can verify algebraically that f_ ±(ζ) = -ζ ± (ζ^2 - 1)^(1/2)satisfies -1<f_ +(ζ) <0 and f_ -(ζ) < -1 for ζ>1; indeed

f_ +[ζ_] := ζ + (ζ^2 + 1)^(1/2) ; f_ -[ζ_] := ζ - (ζ^2 - 1)^(1/2 ...                           +                                                                     -

[Graphics:../HTMLFiles/index_73.gif]

The motion of the stagnation points as γ varies can be clearly seen in the animations.