# Finding a Topological Conjugacy Using Hartman-Grobman

Here is the problem statement:

Find a homeomorphism $h$ in a neighborhood of $0$ that establishes a topological conjugacy between the flow of the differential system and the flow of the linearized system, i.e.,

$\qquad\qquad\qquad\qquad\qquad\qquad\qquad h(\psi(t, x)) = e^{tA}h(x)$

where $A = Df(0)$ and $\boldsymbol{\psi}(t, x_0)$ is the solution of $\mathbf{\dot{x}} = f(\mathbf{x})$ and $\mathbf{x}(t_0) = x_0$, the nonlinear system given by

$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad \dot{x} = −x \\ \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\dot{y} = −y + xz \\ \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\dot{z}=z .$

Without loss of generality, I let $t_0=0$ and found that the solutions for both the linear and non-linear systems, respectively, should be

$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\mathbf{x}=\begin{pmatrix} x_0e^{-t} \\ y_0e^{-t} \\ z_0e^t \end{pmatrix}$

and

$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\boldsymbol{\psi} = \begin{pmatrix} x_0e^{-t} \\ x_0z_0+(y_0-x_0z_0)e^{-t} \\ z_0e^t \end{pmatrix} .$

We can calculate the matrix exponential since $A=\begin{pmatrix}-1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1\end{pmatrix}$:

$\qquad\qquad\qquad\qquad\qquad\qquad\qquad e^{tA} = \begin{pmatrix} e^{-t} & 0 & 0 \\ 0 & e^{-t} & 0 \\ 0 & 0 & e^t \end{pmatrix} .$

I think that my misunderstanding arises from my confusion in what is required of the homeomorphism $h$. Must $h$, in fact, be a matrix transformation which is time-independent? Also, in the right-hand side of the above equality I wish to show with $h$, should $x=\mathbf{x}$, the solution of the linear system? What I understand is that I am showing that flowing in the non-linear space and mapping by $h$ is the same as mapping by $h$ and then flowing in the linear space.

• The “non-linear” system you have written is in fact linear, so it coincides with its linearization. So trivially the identity map $h(x,y,z)=(x,y,z)$ works. Are you sure you copied the question right? – Hans Lundmark Nov 27 '17 at 6:23
• I just edited the question to include the nonlinear piece. Thank you! – Rebecca Hardenbrook Nov 27 '17 at 23:27
• You did not solve the DE correctly. You are missing the particular solution in $y$. – MrYouMath Nov 28 '17 at 6:48
• Hi! I believe that I did find the particular solution. $xz=x_0z_0$, so for $\mathbf{\phi}$, $y_p(t) = x_0z_0$. Am I missing something else? – Rebecca Hardenbrook Dec 3 '17 at 22:56

It might be easier to rename the variables so that the nonlinear system still lives in $xyz$-space, but the linear system lives in $uvw$-space instead.

So you have the solution of the nonlinear system, $$(x(t),y(t),z(t)) = (x_0 \, e^{-t}, x_0 z_0 + (y_0 - x_0 z_0) \, e^{-t}, z_0 \, e^t) ,$$ and the solution of the linear system, $$(u(t),v(t),w(t)) = (u_0 \, e^{-t}, v_0 \, e^{-t}, w_0 \, e^t) .$$ What you want to find is a function $$(u,v,w)=h(x,y,z)$$ (independent of $t$) such that $$h(x(t),y(t),z(t))$$ is the same thing as $$(u(t),v(t),w(t)) \qquad \text{with} \quad (u_0,v_0,w_0) = h(x_0,y_0,z_0) .$$

Now, the first thing one notices when comparing the nonlinear and linear solutions is that $x(t)$ and $z(t)$ are doing the same thing as $u(t)$ and $w(t)$, which suggests that we can just map $x$ and $z$ to $u$ and $w$: $$(u,v,w) = h(x,y,z) = (x, [?], z) .$$ To figure out what's going to be in the middle slot we need to be a bit more clever. But we may notice that in the nonlinear solution, the quantity $y(t) - x(t) z(t)$ is just decreasing exponentially from $y_0-x_0 y_0$ to $0$, as $t$ goes from $0$ to $\infty$. And this is similar to what $v(t)$ is doing in the linear solution: it decreases exponentially from $v(0)$ to $0$. So we could try with $$(u,v,w) = h(x,y,z) = (x, y-xz, z) ,$$ perhaps?

I'll leave it to you to check that this really works (and that it's actually a global diffeomorphism, not just a local homeomorphism).

(One may also view it as follows: since $x(t) z(t) = x_0 z_0$ is constant, we have $(xz)'=0$, so the nonlinear system can be written as $$x'=-x ,\quad (y-xz)'=-(y-xz) ,\quad z'=z ,$$ and thus the change of variables $u=x$, $v=y-xz$, $w=z$ actually performs the linearization.)

• Hi Hans! Thank you for this comment; it has been really helpful. I am still curious...does $h$ necessarily have to be a global diffeomorphism? I thought that Hartman-Grobman only required a local homeomorphism. – Rebecca Hardenbrook Dec 3 '17 at 22:59
• It was just a remark to point out that in this particular case you actually happen to get more than you asked for. As you say, the theorem only guarantees that that a local homeomorphism exists. – Hans Lundmark Dec 4 '17 at 6:58
• Oh, I see! Thank you for your response to my question. I think I understand the Hartman-Grobman theorem much more than I did going into this example. – Rebecca Hardenbrook Dec 4 '17 at 21:08