5. The Hartman-Grobman Theorem

This theorem was proved by Philip Hartman in 1960 [4]. It had been previously announced by Grobman in 1959 [1], who published his proof in 1962 [2]. The book [3] gives an up to date treatment of the subject. The cited references are listed below.

  1. David M. Grobman: Homeomorphisms of systems of differential equations, Doklady Acad. Nauk SSR 128 (1959) 880 - 881.
  2. David M. Grobman: Topological classification of meighbourhoods of a singularity in n-space, Mat. Sbornik 56, no 98 (1962), 77 - 94.
  3. Martin M. Guterman, Zbigniew H. Nitecki: Differential Equations -- A First Course, 3rd ed. Saunders 1992.
  4. P. Hartman: On local homeomorphism of Euclidean spaces, Bol. Sc. Mat. Mexicana (2) 5 (1960), 220 - 241.

(A) Statement of the Theorem

Definition. A matrix is hyperbolic if it has no zero or purely imaginary eigenvalues.

Hartman-Grobman Theorem. Let f:Rn → Rn be a C1 diffeomorphism with f (0) = 0 a hyperbolic fixed point (that is A = Df (0) is hyperbolic). Then there exists a neighbourhood U of 0 and a homeomorphism h from U to some other neighbouhood of 0 such that A○h = h○f. (A and f satisfying such a condition are said to be conjugate.)

Remark.In terms of systems of ODE, this means that the homeomorphism h transforms the system dx/dt = f(x) into dy/dt = A y. (The proof of this assertion uses the chain rule.) On account of continuity, h can be approximated by the identity and f by A in a small neighbourhood of 0. Thus, in the case of a hyperbolic equilibrium point at the origin, the system dx/dt = f(x) can be approximated for small values of x by the linear system dx/dt = A x, where a = Df(0).

The nonhyperbolic case is more complicated; some information may be found in the Scholarpedia article on Equilibrium.

(B) An Auxiliary Result

Definition. Let E be a Banach space. L:E → E is said to be Lipschitz if there exists k < ∞ such that ||Lz1 - Lz2|| ≤ k ||z1 - z2|| for all z1, z2 ∈ E.

Theorem. If φ1,φ2:Rn → Rn are continuous bounded functions with Lipschitz constant K < ε = (1 - a)/||A-1|| for some hyperbolic matrix A, then A + φ1 and A + φ2 are conjugate in Rn.

(This is Theorem 4.4 of the page The Theorem of Hartman-Grobman at warwick.ac.uk.)

(C) Proof of Hartman-Grobman Theorem

Write f(z) = Az + ψ(z) where A = Df(0), ψ(0) = 0 and Dψ(0) = 0. Choose a smooth function α:R → R which is equal to unity for |t| ≤ 1/2, vanishes for |t| ≥ 1 and is monotonic elsewhere. Then there exists k > 1 such that |α(t)| ≤ k for all t. For e > 0, write g(z) = Az + α(||z||/e) φ(z). Then g(z) = f(z) for ||z|| ≤ e/2 and g(z) = Az for ||z|| ≥ e. It is enough to show A conjugate to g.

Choose ε > 0 so that ||Dψ(x)|| < ε/2k for ||x|| ≤ e. If φ(z) = α(||z||/e)ψ(z), then

||φ(z1) - φ(z2)||

≤ ||(α(||z1||/e) - α(||z2||/e))ψ(z1)|| + ||α(||z2||/e)(ψ(z1) - ψ(z2))||

≤ k ||ψ(z1)|| |(||z1|| - ||z2||)/e| + ||Dψ(x)|| ||z1 - z2||

By Mean Value Theorem,
x on line from z1 to z2

≤ k(ε/2k)||z1|| ||z1 - z2||/e + (ε/2k)||z1 - z2||

≤ ε||z1 - z2||

The result now follows fom Theorem (B).

Exercise: Fill in the gaps in the proof.

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