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AMATH 586 SPRING 2020HOMEWORK 3 — DUE MAY 8 ON GITHUB BY 11PMBe sure to do agit pullto update your local version of theamath-586-2020repository.Problem 1:It is natural to ask if a neighborhood ofz= 0 can be in the absolute stability regionSfor a LMM. You will show that this cannot be the case. Consider a consistentand zero-stable LMMrXj=0αjUn+j=krXj=0βjf(Un+j).Recall the characteristic polynomialπ(ξ;z) =ρ(ξ)-zσ(ξ). Show:•Consistency implies thatπ(1; 0) = 0.•Stability implies thatρ0(1)6= 0.•Supposeξ= 1 +η(z) forznear zero so thatπ(ξ;z) =π(1 +η(z);z) = 0.Computeη0(0). Why does this imply that there must be an interval (0,] forsome small>0 that does not lie in the absolute stability regionS.(a) To have consistency, we need to satisfy (5.48), henceπ(1; 0) =ρ(1) =∑jαj=0.(b) Atz= 0:π(ξ; 0) =ρ(ξ) =ρ(1) +ρ0(1)(ξ-1) +12ρ00(1)(ξ-1)2+· · ·.To have consistency,ρ(1) = 0. Ifρ0(1) = 0 also, then the above Taylor expan-sion becomesπ(ξ; 0) =ρ(ξ) =12ρ00(1)(ξ-1)2+· · ·.Hence, we can factor out a repeated root (ξ-1)2atξ= 1 fromπ(ξ; 0), whichviolates the root condition and makesπ(ξ; 0) unstable atz= 0.(c) Supposeπ(ξ;z) =π(1 +η(z);z) = 0, then we haveXjαj(1 +η(z))j-zXjβj(1 +η(z))j= 0,Xj(αj-zβj)(1 +η(z))j= 0.Take the derivative with respect to z using implicit differentiation:Xj-βj(1 +η(z))j+ (αj-zβj)j(1 +η(z))j-1η0(z) = 0.1
Sinceη(0) = 0 (because we knowπ(ξ; 0) has a root atξ= 1), we can substituteη(0) = 0 andz= 0 into the above:Xj-βj+jαjη0(0) = 0.Hence,η0(0) =∑jβj∑jjαj(a)= 1,where (a) follows from (5.48) for consistency. Sinceη(0) = 0 andη0(0)>0,ηis increasing atz= 0 and should be positive when z is a slightly greater than0. In other words, this would mean that there is a small intervalz∈(0,], suchthatη(z)>0. This would mean that 1 +η(z) is slightly greater than 1 in thisinterval andπ(1 +η(z);z) = 0 would have a root slightly greater than 1, whichviolates the stability condition.Problem 2:Recall the test problemv000(t) +v0(t)v(t)-β1+β2+β33v0(t) = 0,whereβ1< β2< β3. It follows thatv(t) =β2+ (β3-β2)cn2rβ3-β112t,sβ3-β2β3-β1!is a solution where cn(x, k) is the Jacobi cosine function andkis the elliptic modulus.Some notations use cn(x, m) wherem=k2. The corresponding initial conditionsarev(0) =β3, v0(0) = 0, v00(0) =-(β3-β1)(β3-β2)6.