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January 20, 2006, 06:53 |
Taylor problem
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#1 |
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Hi everybody
I found in a paper a test case called the Taylor problem. It's used to check the accuracy of a numerical scheme for the two dimensional incompressible viscous flows. In a square [2Pi][2Pi], the initial velocity field is given by : u(x,y,0) = -cos(x)sin(x) v(x,y,0) = sin(x)cos(x) And the analytical solution is : u(x,y,t) = -cos(x)sin(y)exp(-2t/Re) v(x,y,t) = sin(x)cos(y)exp(-2t/Re) P(x,y,t) = -0.25[cos(2x)+cos(2y)]exp(-4t/Re) Unfortunaly, the authors didn't give any reference to this problem, so I tried to find by myself on the net but no way. So please, if anyone have any informations about it, i'll be very grateful. |
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January 20, 2006, 06:55 |
Re: Taylor problem
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#2 |
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Sorry, i've made a mistake int the initial values:
u(x,y,0) = -cos(x)sin(y) v(x,y,0) = sin(x)cos(y) |
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January 23, 2006, 15:46 |
Re: Taylor problem
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#3 |
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The original work by Taylor: \bibitem{tay23} G.~I. Taylor. Stability of a viscous liquid contained between two rotating cylinders. {\em Phil. Trans. Roy. Soc. Lond A}, 223:289--343, 1923.
You have to have the boundary conditions time dependent. I had a (0,1)(0,1) square horizontal velocity: v_x=-\cos{(2\pi x)}\sin{(2\pi y)}e^{-\frac{8\pi^2}{Re}t} vertical velocity v_y=+\cos{(2\pi y)}\sin{(2\pi x)}e^{-\frac{8\pi^2}{Re}t} vorticity: \omega=4\pi\cos{(2\pi x)}\cos{(2\pi y)}e^{-\frac{8\pi^2}{Re}t} stream function \psi=\frac{\omega}{8\pi^2} with best regards, Jure. iepoi.uni-mb.si/ravnik/research.html |
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