How to construct a diffeomorphic function using another function with certain properties

Question

A $C^{\infty}$ function $f(x)$ on the interval $[a, b]$ satisfies the following 3 properties:

1) $f(x) = 1$ for $a \leq x \leq b$

2) $f(x) = 0$ for $x < \alpha$ and $x > \beta$ where $\alpha < a$ and $\beta > b$

3) $f'(x) \neq 0$ on the intervals $(\alpha, a)$ and $(b, \beta)$

I have to construct a diffeomorphism $g: [a, b] \to [c, d]$ which satisfies $g'(a) = g'(b) = 1$ and $g(a) = c, g(b) = d$.

One $f(x)$ I think that works is $$f(x) = \left\{ \begin{array}{lr} 1 & : x \geq 1 \\ e^{-\frac{1}{(\frac{x}{1 - x})^2}} & : 0 < x < 1\\ 0 & : x \leq 0 \end{array} \right. $$

I can't figure out an analytic way to find such a function, and guessing-and-checking hasdn't been fruitful for me so far.

Answer 1

I'm not giving an exact closed formula, but the idea to get one. The problem is to find a diffeo $g$ among the intervals $[a,b]$ and $[c,d]$ with prescribes derivatives at the extremes, in this case $g'(a)=g'(b)=1$. So it's better to look at the graphs. The picture I include summarizes the situation.

First we see the obvious homeomorphism, and we must smoothen the angles. The one marked ? is amplified, to see what we want: the red curve, smooth of course at the joints. By an affinity that fixes the corner and the right joint and moves up the left joint to the right we are reduced to find a function like $g$ in the lower left picture. Note that in our problem we cannot use an horizontal segment, because we want a bijection. This affine change makes the construction very simple using a bump function like the $f$ in the question. In fact one only needs half of it, as I have depicted with an $\varepsilon, \delta$: my $f$ is $\equiv 1$ for $x\ge\delta$ and the solution is $g(x)=\int_{-\infty}^xf(t)dt$.

A remark in any case. This is very common in differential topology, and really useful in higher dimensions and arbitrary manifolds. However to produce diffeos among intervals with prescribed derivatives at the extremes there is a very explicit method using splines. However the problem asked explicitely for the bump functions smooting technique!

Answer 2

Want to give an idea and some functions, you can try and work with: $$\varphi (x){\text{ = }}\left\{ {\begin{array}{*{20}{c}} {{e^{ - \frac{1}{{1 - {x^2}}}}}}&{ - 1 < x < 1} \\ 0&{otherwise} \end{array}} \right.$$ $$\psi (i,x){\text{ = }}\varphi (x - i)$$ $$\phi (i,x){\text{ = }}\frac{{\varphi (x - i)}}{{\psi ( - 2,x) + \psi ( - 1,x) + \psi (0,x) + \psi (1,x)}}$$ $$F(x){\text{ = }}\sum\limits_{k = - 1}^1 \phi (k,x)$$ Here are plots of $\phi ( - 1,x),\phi (0,x),\phi (1,x)$

This is a plot together with $F(x)$

And here $F(x)$ alone: