Stability lecture 20

Stability lecture 20 - Hydrodynamic Stability Lecture 20...

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Hydrodynamic Stability Lecture 20 Numerical solution of the Orr-Sommerfeld equation. () 24 2 2R e Separate into IV iU C U AC B φ αφ α ′′  −+ = −−  = ± ± ± ± The most obvious approach would be to use 2 nd order accurate centered differences. But there is a complication: () () ( ) () ( ) 4 16 4 1 2 Is difficult to implement at 2, 1 IV iii i i x in x φφ ≈+− + + − − +− =− Consider a grid At , the 4 2 i = th derivative includes a point at 2 i that is undefined. It is possible to proceed this way. Using B.C.’s, to figure out relationships for a ghost point at . 0 i = It gives a nice pentadiagnol system to invert. I chose another approach. In effect, it does the exact same thing, but it breaks it into two steps. From: 1
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() () ( )() () () () 22 22 2 2 2 2 2 3 2 2 22 222 2 2 2 2 Re 12 1 1 11 2 1 2 ,, 1 1 Re i DD U D U D UC D ii i Ui Ui Ui i i zz iii i iC i i zzz z z z z αα φ α φφ φα  −− + − −= −  +− +  ∆∆ ++ + = + +  ∆∆∆ i ( ) 3 Ci αφ Separate into matrix coefficients ri Ai AC B += ±± ± ± B – Terms: 3 2 2 3 2 2 2 1 2 1 i i z Bi z z z  + +  =− +
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Stability lecture 20 - Hydrodynamic Stability Lecture 20...

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