sln2 - 7224 Homework Solution 1 Problem 1 tt u ( x,y,t ) =...

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Unformatted text preview: 7224 Homework Solution 1 Problem 1 tt u ( x,y,t ) = ru- (1 + 2 ) 2 u- u 3 . Rewrite second-order PDE as t u = u t u = ru- (1 + 2 ) 2 u- u 3 . The Jacobian matrix: J = f u | u = u b = 1 r- (1 + 2 ) 2- 3 u 2 b ! For u b = 0, the- 3 u 2 b term is dropped off.We obtain: t u u ! = 1 r- (1 + 2 ) 2 ! u u ! . The eigenfunctions should be: u q = a q ( t ) e i q x The eigenvalues of the Jacobian q : 2 q = r- (1- q 2 ) 2 where q 2 = ( q 2 x + q 2 y ). The maximum growth rate q = Re ( r ) at q = 1, so for r < 0 the growth rate vanished because all eigenvalues are imaginary and the state is marginally stable and for r > 0, there are positive eigenvalues and the uniform state is unstable. Although it is stationary and the instablity occurs at a finite wave number,the instability is not type-I,because the growth rate is not locally parabolic. Here is a sketch of growth rate at r = 1 and r = 0 . 1. For r < the growth rate is zero. 0.5 1 1.5 2 2.5 3 0.2 0.4 0.6 0.8 1 1 Problem 2 t u = au + 2 x u- b 2 x v- ( u 2 + v 2 )( u + cv ) t v = av + b 2 x u + 2 x v- ( u 2 + v 2 )( u + cv ) The Jacobian matrix:...
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sln2 - 7224 Homework Solution 1 Problem 1 tt u ( x,y,t ) =...

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