# hw3.pdf - AMATH 586 SPRING 2020 HOMEWORK 3 u2014 DUE MAY 8...

• 7
• 100% (1) 1 out of 1 people found this document helpful

This preview shows page 1 - 3 out of 7 pages.

AMATH 586 SPRING 2020 HOMEWORK 3 — DUE MAY 8 ON GITHUB BY 11PM Be sure to do a git pull to update your local version of the amath-586-2020 repository. Problem 1: It is natural to ask if a neighborhood of z = 0 can be in the absolute stability region S for a LMM. You will show that this cannot be the case. Consider a consistent and zero-stable LMM r X j =0 α j U n + j = k r X j =0 β j f ( U n + j ) . Recall the characteristic polynomial π ( ξ ; z ) = ρ ( ξ ) - ( ξ ). Show: Consistency implies that π (1; 0) = 0. Stability implies that ρ 0 (1) 6 = 0. Suppose ξ = 1 + η ( z ) for z near zero so that π ( ξ ; z ) = π (1 + η ( z ); z ) = 0. Compute η 0 (0). Why does this imply that there must be an interval (0 , ] for some small > 0 that does not lie in the absolute stability region S . (a) To have consistency, we need to satisfy (5 . 48), hence π (1; 0) = ρ (1) = j α j = 0. (b) At z = 0: π ( ξ ; 0) = ρ ( ξ ) = ρ (1) + ρ 0 (1)( ξ - 1) + 1 2 ρ 00 (1)( ξ - 1) 2 + · · · . To have consistency, ρ (1) = 0. If ρ 0 (1) = 0 also, then the above Taylor expan- sion becomes π ( ξ ; 0) = ρ ( ξ ) = 1 2 ρ 00 (1)( ξ - 1) 2 + · · · . Hence, we can factor out a repeated root ( ξ - 1) 2 at ξ = 1 from π ( ξ ; 0), which violates the root condition and makes π ( ξ ; 0) unstable at z = 0. (c) Suppose π ( ξ ; z ) = π (1 + η ( z ); z ) = 0, then we have X j α j (1 + η ( z )) j - z X j β j (1 + η ( z )) j = 0 , X j ( α j - j )(1 + η ( z )) j = 0 . Take the derivative with respect to z using implicit differentiation: X j - β j (1 + η ( z )) j + ( α j - j ) j (1 + η ( z )) j - 1 η 0 ( z ) = 0 . 1
Since η (0) = 0 (because we know π ( ξ ; 0) has a root at ξ = 1), we can substitute η (0) = 0 and z = 0 into the above: X j - β j + j η 0 (0) = 0 . Hence, η 0 (0) = j β j j j ( a ) = 1 , where ( a ) follows from (5 . 48) for consistency. Since η (0) = 0 and η 0 (0) > 0, η is increasing at z = 0 and should be positive when z is a slightly greater than 0. In other words, this would mean that there is a small interval z (0 , ], such that η ( z ) > 0. This would mean that 1 + η ( z ) is slightly greater than 1 in this interval and π (1 + η ( z ); z ) = 0 would have a root slightly greater than 1, which violates the stability condition. Problem 2: Recall the test problem v 000 ( t ) + v 0 ( t ) v ( t ) - β 1 + β 2 + β 3 3 v 0 ( t ) = 0 , where β 1 < β 2 < β 3 . It follows that v ( t ) = β 2 + ( β 3 - β 2 )cn 2 r β 3 - β 1 12 t, s β 3 - β 2 β 3 - β 1 ! is a solution where cn( x, k ) is the Jacobi cosine function and k is the elliptic modulus. Some notations use cn( x, m ) where m = k 2 . The corresponding initial conditions are v (0) = β 3 , v 0 (0) = 0 , v 00 (0) = - ( β 3 - β 1 )( β 3 - β 2 ) 6 .
• • • 