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Unformatted text preview: Problem 1 (a) Linearization is trivial (just like that of the standard SwiftHohenberg equation) and we get so we immediately get the growth rate of an infinitesimal solution u = e iqx e t . For u b = 0 the real part of the growth rate is the same as for the standard SHE: so expanding it about the bifurcation point r c = 0, q c = 1 (i.e., its a typeI instability) we get Since the bifurcation occurs at r c = 0 we can just take = r , which immediately determines the characteristic time scale = 1 and correlation length c = 2 by comparing with the canonical expression for a typeI instability. For nonzero s , the instability is oscillatory, i.e., typeI o with critical frequency c = Im q  q = 1 = s . (b) Substituting the growth rate into the solution of the linearized equation we get which by a change of variables reduces to the corresponding expression for a standard SHE. This change is equivalent to a translation of the growing pattern with speed s ! It is trivial to check that this change of variables reduces the modified SHE to the standard form. , 3 ) 1 ( 2 2 2 u u u u s u r u b x x t + + = isq q u r b q + = 2 2 2 ) 1 ( ) 3 ( , ) 1 ( Re 2 2 q r q q = = . ) 1 ( 2 ) 1 ( ) 1 ( 2 2 2 2 L + = + = q r q q r q ] ) ) 1 ( exp[( )] ( exp[ 2 2 t q r st x iq e e u t iqx + = = st x y x + = (c) For the two nontrivial solutions the growth rate has a real part which is negative for all q when r > 3/2 where these solutions are stable. In the range 1 < r < 3/2 these solutions will be unstable as Re q is not signdefinite there. These solutions will not be selected at the onset of primary instability since they dont even exist there....
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This note was uploaded on 08/25/2011 for the course PHYS 7268 taught by Professor Staff during the Spring '11 term at Georgia Institute of Technology.
 Spring '11
 Staff

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