By Mats Gyllenberg, Lars-Erik Persson

ISBN-10: 0585328285

ISBN-13: 9780585328287

ISBN-10: 0824792173

ISBN-13: 9780824792176

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**Additional info for Analysis, Algebra and Computers in Mathematical Research**

**Sample text**

If either of the asymptotic matrices A± (λ) is not hyperbolic, then the range of the operator L∞ − λ : H n (R) ⊂ L2 (R) → L2 (R) is not closed, and the operator is not Fredholm. In particular, λ ∈ σess (L∞ ). Proof. We argue by contradiction. If the range Rλ := R(L∞ −λ) is closed, then the restricted operator Lr := (L∞ − λ) ker(L ∞ −λ) ⊥ : ker(L∞ − λ)⊥ → Rλ will have no kernel. Since the restricted operator is closed, it is Fredholm with index zero. By the closed graph theorem we conclude that Lr has a bounded inverse on Rλ .

We call these operators exponentially asymptotic and they typically arise as the linearizations of a nonlinear partial diﬀerential equation (PDE) about a heteroclinic (pulse) or homoclinic (front) solution of the equilibrium equations. 6, in terms of the noninvertibility of a particular matrix. In particular, the matrix is square precisely when the Fredholm index of the operator is zero. We show that in exponentially weighted spaces the essential spectrum is shifted, and characterize the absolute spectrum as the leftmost possible shift of the boundary of the essential spectrum.

The Fredholm index of a Fredholm operator is defined by ind(L) = dim[ker(L)] − codim[R(L)]. The operator L is Fredholm if and only if La is, and the indices are related via ind(L) = − ind(La ). If λ ∈ σ(L) is an isolated eigenvalue with ma (λ) < ∞, then λI − L is a Fredholm operator with index 0. It is easy to see that the range of L must be orthogonal to the kernel of La ; indeed, if v ∈ ker(La ) and Lu = f then f , v = Lu, v = u, La v = 0. 1 (Fredholm alternative). Suppose that X is a Hilbert space with inner product ·, · , and L : D(L) ⊂ X → X is a closed Fredholm operator with domain D(L) ⊂ X dense in X-norm.

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Categories: Differential Equations