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Here you'll see a first example of how a map can arise from a system of differential equations and gain some first insight into the dynamics of 1D maps.

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Consider the system \\[ \\dot{x}=\\lambda_1 x\\\\ \\dot{y}=\\lambda_2y ,\\] where $\\lambda_1 < 0 < \\lambda_2$ and $D:=\\lambda_1+\\lambda_2 < 0$.

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The solution is given in the part \"solution\".

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The flow naturally defines a mapping from the half line $L_1 = \\{(1,y):y>0\\}$, to the half line $L_2=\\{(x,1):x>0\\}$. Indeed given a point $(0,\\eta) \\in L_1$ the orbit through $(1,\\eta)$ intersects $L_2$ in one and only one point $(\\xi(\\eta), 1) \\in L_2$. In this way define a mapping $\\xi : \\mathbb{R_+}\\to \\mathbb{R_+}$.

Find an expression for $\\xi(\\eta)$.
Show you can extend $\\xi$ continuously to $0$ by letting $\\xi(0)=0$.
Show that the derivative satisfies $\\xi'(0)=0$.

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There are several ways to proceed, here's one.

You can assume that for a solution in the first quadrant you can write $y$ as a function of $x$ (this follows since in this case $x(t)$ is injective). By the chain rule
\\[y'(x)=\\frac{\\dot{y}}{\\dot{x}}.\\]
This should give you solutions depending on an arbitrary constant we could call $c$.

The solution we are interested in, is the one where $y=\\eta>0$ when $x=1$. Use this to determine $c$.

Knowing $c$, you can find $\\xi=\\xi(\\eta)$ such that $y(\\xi)=1$.

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In the first part you should've shown that $\\xi(\\eta)=\\eta^a$, with $a = -\\frac{\\lambda_1}{\\lambda_2}$.

Show that $a > 0$, this should give you 2.

Show that $a > 1$ (since $D = \\lambda_1 + \\lambda_2 <0$), this should give you 3.

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Since for a solution in the first quadrant $\\dot{x}<0$, $x(t)$ is an injective mapping, so a solution curve can be written $(x,y(x))$.

Using the chain rule
\\[ y'(x)=\\frac{\\dot{y}}{\\dot{x}} = \\frac{\\lambda_2}{\\lambda_1} \\frac{y}{x}=-\\frac{1}{a}\\frac{y}{x} ,\\]
where we have set $a = -\\lambda_1 / \\lambda_2$.

Since $\\lambda_1 < 0 < \\lambda_2$, we have $a > 0$. Since $\\lambda+\\lambda_2 < 0$ we have $-\\lambda_1 > \\lambda_2$, so $a > 1$.

Separating the variables, we find that $y(x)= c x^{-1/a}$, where $c$ is a constant.

We look at the particular solution that satisfies $y(1)=\\eta$. This corresponds to $c=\\eta$, so the particular solution is $y(x)=\\eta x^{-1/a}$.

When does the solution curve cross $L_2$? It does so when $(y(x), x) = (1, \\xi)$ for some $\\xi>0$. So we look at the equation $y(x) = 1$: \\[\\eta x^{-1/a} = 1 \\Rightarrow x=\\eta^a .\\]

This means that the function we are looking for is $\\xi(\\eta)=\\eta^a$.

Since $a > 0$, we have $\\xi(\\eta) \\to 0$ when $\\eta \\to 0$. This shows 2.

We also have $\\xi'(\\eta) = a \\eta^{a-1} $, and since $a>1$ this is equal to zero when $\\eta = 0$. This shows 3.

End of exercise.

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Let $f:\\mathbb{R}\\to \\mathbb{R}$ be a mapping. This exercise explores how we can consider $f$ a dynamical system.

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The exercise should be somewhat self-explanatory. Ask me if you are in doubt.

Define the n'th iterate of f by
\\[ f^{\\circ n}(x)= \\begin{cases} x & \\text{when } n = 0 ; \\\\ f(f^{\\circ (n-1)}(x)) & \\text{when } n =1,2,3,\\ldots \\end{cases} \\]

When $f$ is invertible, when can also define $f^{\\circ n}$ for $n = -1, -2, \\ldots$, by letting $f^{\\circ -k}$ denote the $k$'th iterate of the inverse map, when $k \\in \\mathbb{N}$.

We can think of $f$ as defining a flow, by letting $\\phi_n(x)=f^{\\circ n}(x)$. Notice, that in this setting \"time\" n is discrete, $n \\in \\mathbb{N}$ (or $n \\in \\mathbb{Z}$ in the invertible case).

Given $x_0$, we can let $x_1 = f(x_0)$, $x_2 = f(x_1)$, ... Then $x_n = f^{\\circ n}(x_0)$. The points $z_0, z_1, z_2, \\ldots,$ form what is called the forward orbit of $z_0$.

A point $x^*$ where $f(x^*) = x^*$ is called a fixed point.

How many fixed points do the mapping $x \\mapsto x^3$ have: [[0]].

Let the diagonal in $\\mathbb{R}^2$ be the points $\\{(x,x):x\\in\\mathbb{R}\\}$. Consider the following statement:

The fixed points of f are the values of x where the graph of f intersects the diagonal.

Is this statement true? [[1]]

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Now consider the simple linear case $f(x) = \\lambda x$, where $\\lambda$ is a constant. Then $x^*=0$ is a fixed point.

Determine a condition that is equivalent (meaning sufficient and necessary) to the following:
For any $x_0 \\in \\mathbb{R}$ it holds that $f^{\\circ n}(x_0) \\to 0$ when $n\\to \\infty$.

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Suppose $f : \\mathbb{R} \\to \\mathbb{R}$ is differentiable at the point $x^*$, and that $f(x^*)=x^*$.

Let $\\lambda = f'(x^*)$, and suppose $\\lvert \\lambda \\rvert < 1$.

Prove that there exists an interval open $I = (x^*-\\delta, x^*+\\delta)$, so that when $x_0 \\in I$, it holds that $f^{\\circ n}(x_0) \\to x^*$, when $n \\to \\infty$.

Hint: Given any $\\kappa$ with $\\kappa > \\lvert \\lambda \\rvert$, there exists an open interval around $x^*$ where $\\lvert f(x)-f(x^*)\\rvert \\leq \\kappa \\lvert x-x^* \\rvert$.

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We can assume $x^*=0$ with no loss of generality.

Pick $\\kappa$ with $\\lvert \\lambda \\rvert < \\kappa < 1$.

Since $f'(0)=\\lambda$, we know $\\frac{f(x-0)}{x-0}=\\frac{f(x)}{x}\\to \\lambda$ when $x\\to 0$.

It follows that $\\left\\lvert \\frac{f}{x}\\right\\rvert \\to \\lvert \\lambda\\rvert$ when $x\\to 0$. It particular, there exists $\\delta > 0$ such that $\\lvert \\frac{f(x)}{x} \\rvert \\leq \\kappa$ when $0 < \\lvert x \\rvert <\\delta$.

We deduce that $\\lvert f(x) \\rvert \\leq \\kappa \\lvert x\\rvert$, when $x \\in I =(-\\delta, \\delta)$.

By induction, $\\lvert f^{\\circ n}(x) \\rvert \\leq \\kappa^n \\lvert x \\rvert \\to 0$ when $n \\to \\infty$, for any $x \\in I$.

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We will later talk about asymptotic stability, and basically, what we have just seen, is that $\\lvert f'(x^*) \\rvert < 1$ is a sufficient condition for a fixed point $x^*$ to being asymptotically stable.

End of exercise.

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These questions are two prepare you for understanding the homoclinic bifurcation. Again, this is not an exam, no scores are recorded anywhere.

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