
#1
Sep607, 03:20 AM


#2
Sep607, 06:24 AM

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It's been a long time since I have done problems like that but here's how I would do it (perhaps awkwardly):
In the complex plane, writing z= x+ iy, the velocity function, [itex]q(z)= dx/dt+ i dy/dt= \overline{z}+ 6i= x iy+ 6i= x+ (6y)i[/itex] which gives the two differential equations dx/dt= 3x and dy/dt= 6 3y. You can integrate those directly or divide one by the other to eliminate t and get dy/dx= (63y)/3x. I would be inclined to go ahead and cancel "3"s. dy/dx= (2y)/x. That is a separable equation: dy/(2y)= dx/x and integrating ln(2y)= ln(x)+ C or 1/(2y)= C'x so x(2y)= C', a family of hyperbolas. That is, except for a factor of 3 which is now incorporated in the constant, what you have for the stream lines. The "equipotential" lines are always orthogonal to the stream lines so they are given by dy/dx= x/(2 y). (2y)dy= xdx so 2y y^{2}/2= x^{2}/2+ C That is the same as x^{2}/2 y^{2}/2+ 2y= C or x^{2} y^{2}+ 4y= C' (C'= 2C), the family of hyperbolas orthogonal to the stream lines. That is also, except for the factor of 3, what you have. To answer (c), determine C' in x(2y)= C'. For example, the stream line through (0,0) has both x and y= 0 so 0(20)= C'. C'= 0 so x(2y)= 0 which is only satisfied by (0,0) (that's your "degenerate" stream line). The stream line through 1+i= (1,1) has x=y= 1 so 1(21)= C'. C'= 1 and the stream line is the hyperbola x(2y)= 1. From dx/dt= x, we see that, at x= 1> 0, x is increasing. The direction of flow is to the right. 


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