# sln2 - 7224 Homework Solution 1 Problem 1 tt u(x y t = ru(1...

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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 0 t u 0 = ru - (1 + 2 ) 2 u - u 3 . The Jacobian matrix: J = ∂f ∂u | u = u b = ˆ 0 1 r - (1 + 2 ) 2 - 3 u 2 b 0 ! For u b = 0, the - 3 u 2 b term is dropped off. We obtain: t ˆ δu δu 0 ! = ˆ 0 1 r - (1 + 2 ) 2 0 ! ˆ δu δu 0 ! . The eigenfunctions should be: δu q = a q ( t ) e i q · x The eigenvalues of the Jacobian σ q : σ 2 q = r - (1 - q 2 ) 2 where q 2 = ( q 2 x + q 2 y ). The maximum growth rate σ 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 < 0 the growth rate is zero. 0.5 1 1.5 2 2.5 3 0.2 0.4 0.6 0.8 1 1

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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: J = ˆ a + 2 x - ( u
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