Bump Functions in Dynamical Systems.

Question

Define

$B(x) = e^{-1/x^2}$ for $x > 0, B(x) = 0 $ otherwise.

1. sketch the graph of B(x);

2. prove that B'(0) = 0.

When $x = 0$, $B(x) = 0$. It follows that the rate of change of a constant function is 0 and hence B'(x) = 0.

3. Inductively prove that $B^{(n)}(0) = 0$ for all n. Conclude that $B(x)$ is a $C^{\infty}$ - function.

The base case is trivially true from the previous part of the question. Assume that $B^{(n)}(0) = 0$ for some n > 0. We must show that $B^{(n +1 )}(0) = 0$ is true to complete the inductive phase.

$B^{(n+1)}(0) = (B^{(n)})' (x)$

since $B^{(n)}(0) = 0$ we can conclude by saying that the derivative of a constant function is 0. (?)

by the induction hypothesis,$B^{(n+1)}(0) = 0 $ is true. Hence by mathematical induction the result follows.

That last part i think is wrong, but my REAL issue is what comes next.

3.Modify $B(x)$ to construct a Bump function $D(x)$ on the interval $[a,b]$ that is:

$D(x) = 1 , x \in [a,b]$

$D(x) = 0 $for$ x < \alpha $and$ x > \beta,$ where $ \alpha < a $and$ \beta > b.$

$D'(x) \ne 0$ on the intervals $(\alpha,a)$ and $(b,\beta)$.

I have sketched something that resembles a bump and satisfies all of these conditions. and it looks like a bump!.

$D(x)= e^0 $for $x \in [a,b]$

$D(x) = 0 $for$ x < \alpha $and$ x > \beta,$ where $ \alpha < a $and$ \beta > b.$

$D(x) = e^{-1/x^2} $ for $x\in[\alpha,a),$

but i am unsure how to cover the interval $(b,\beta)$

Then there is the next part.

4.Use a bump function to construct a diffeomorphism $f:[a,b] \rightarrow [c,d]$ which satisfies $f'(a) = f'(b) = 1$ and $f(a) = c, f(b) = d$.

this part I have not attempted because I'm really stuck. I know it doesn't specify that I use the previous "bump" function, but I don't really know what a bump function is anyway

Thanks in advance.

Answer 1

Alright... let's start with part 2.

When x=0, B(x)=0. It follows that the rate of change of a constant function is 0 and hence B'(x) = 0.

I'll try to explain why this reasoning is flawed with a different example. Consider the following function:

$$f(x)=\begin{cases} 0 & x >0\\ x & x = 0\\ 0 & x <0\\ \end{cases}$$

Now, if you think about this for a second, you should realize that this is nothing more than a convoluted way of writing $f(x)=0$. If this is not immediately obvious, try drawing the graph. So, we know $f'(x)=0$, and, in particular, $f'(0)=0$.

But what happens if we try to apply your previous logic?

When $x=0$, $f(x)=x$. It follows that the rate of a change of a linear function is equal to the slope, and hence $f'(0)=1$.

The reason this doesn't work is that you are only using the information at a single point, but derivatives are defined by the limiting behavior of the function in a neighborhood around a point. To compute the derivative, we need to use this definition.

$$B'(0)=\lim_{h\rightarrow0}\frac{B(h)-B(0)}{h}=\lim_{h\rightarrow0}\frac{e^{-1/h^2}}{h}=\lim_{h\rightarrow0}\frac{1/h}{e^{1/h^2}}$$

We apply Hôpital's rule to get

$$B'(0)=\lim_{h\rightarrow0}\frac{-1/h^2}{-2(1/h^3)e^{1/h^2}}=\frac{1}{2}\lim_{h\rightarrow0}\frac{h}{e^{1/h^2}}=0$$

For the sake of completeness, we should also note that the limit approaching from the left is also zero, so the derivative is well defined.

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Now, on to part 3. We again apply the definition of a limit.

$$B^{(n+1)}(0)=\lim_{h\rightarrow0}\frac{B^{(n)}(h)-B^{(n)}(0)}{h}=\lim_{h\rightarrow0}\frac{B^{(n)}(h)}{h}$$

So we need some sort of expression for $B^{(n)}$, preferably in terms of $B$.

Suppose $f(x)$ is a rational function, and suppose $B'(x)=g(x)B(x)$, for some rational function $g(x)$. (We derived $g$ in step 2, so we know this is true). We can obtain the following.

$$\frac{d}{dx}[f(x)B(x)]=f'(x)B(x)+f(x)B'(x)=[f'(x)+f(x)g(x)]B(x)$$

Since rational functions are closed under derivation, addition, and multiplication, we know $f'(x)+f(x)g(x)$ is also a rational function. Thus, in general, we know $B^{(n)}(x)=q(x)B(x)$ for some rational function $q$.

And so, for some rational function $R_1$., we know $B^{(n+1)}(0)=\lim_{h\rightarrow0}\frac{R_1(h)}{e^{1/h^2}}$

Define a new rational function $R_2(h)=R_1(1/h)$, to get $B^{(n+1)}(0)=\lim_{h\rightarrow \infty}\frac{R_2(h)}{e^{h^2}}$.

As the argument tends to infinity, any rational function is necessarily bounded by some sufficiently high power of $x$. I.e. for some $H>0$, $h>H \implies -h^{2m}< R_2(h)<h^{2m}$, for some sufficiently large $m$.

Making the substitution $\delta=h^2$, we get $\lim_{\delta\rightarrow \infty}-\frac{\delta^m}{e^\delta}\leq B^{(n+1)}(0)\leq\lim_{\delta\rightarrow \infty}\frac{\delta^m}{e^\delta}$

Repeated application of Hôpital's rule shows both bounds go to zero, as desired.

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On to part 3! (your question has two part 3s. How cool is that?)

This might not be the most elegant way to solve this part, but it works.

Before worrying about arbitrary intervals, let's try to construct a bump function on $[-1,1]$.

$B(x)$ is $0$ for $x\leq0$, so $B(x-1)$ is zero for $x\leq1$, and positive everywhere else.

Thus, $B(x^2-1)$ is $0$ for $-1\leq x\leq1$, and positive everywhere else.

We know $B((2)^2-1)\approx 0.89$, so $0.5-B(x^2-1)$ is positive on some interval $[-\delta,\delta]$, $\delta>1$, negative outside $[-\delta,\delta]$, and $0.5$ on $[-1,1]$.

Therefore, we can conclude $\frac{B(0.5-B(x^2-1))}{B(0.5)}$ is $1$ on $[-1,1]$, $0$ outside some larger interval $[-\delta,\delta]$. Moreover, we know $\frac{d}{dx}B(f(x))=0$ if and only if $f(x)<0$ or $f'(x)=0$. By construction, $0.5-B(x^2-1)$ only satisfies these conditions on the interval $[-1,1]$ and outside $[-\delta,\delta]$, so we can conclude that $\frac{B(0.5-B(x^2-1))}{B(0.5)}$ has a non-zero derivative everywhere else.

Let's define a new function. $F(x)=\frac{B(0.5-B(x^2-1))}{B(0.5)}$.

Since $F(x)=1$ iff $x\in[-1,1]$, we conclude that $F(\frac{x-b}{a})=1$ iff $x\in [-a+b,a+b]$.

To find the correct pair $(a,b)$, we just need to solve the system of equations $\alpha = b-a$, $\beta = b+a$, which is trivial.

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Part 4!

I don't have a full proof for this bit. I'm not sure that my construction in part 3 is necessarily the one you want.

However, $x-kF(x)$ takes $[0,2]$ to $[-k,2]$. It has slope $1$ at both endpoints. You can probably apply some suitable transformation to map arbitrary intervals to each other in the desired way.

Is this a diffeomorphism? I don't know. I suspect it might be. Maybe someone else can tell you.

Answer 2

I am new to M.SE and degree levels maths but i will have an attempt at this.

The induction proof seems ok. $B^{(n+1)}(0) = (B^{(n)})'(0) $ makes sense. You should add a bit more detail in your inductive phase.

However, a more simple attempt at modifying $B(x)$, would give:

\begin{equation} f(x)=\begin{cases} 1, & x \in [a,b] \text{ }.\\ 0, & x < \alpha. \text{ and } x > \beta\\ x, & x \in (\alpha,a)\\ -x,& x \in (b, \beta)\\ \end{cases} \end{equation}

where $ \alpha < a$ and $\beta > b$

This would satisfy the conditions.

for the last part you have to know what a diffeomorphism is.

It is a function that is bijective, continuous such that its inverse is also continuous. Also the function and its inverse has to be a $C^r$. This means that the r-th derivative exists and is continuous.

Now there are some added requirements in the question. that require $f'(a) = f'(b) = 1$ and that $f(a) = c, f(b) = d$.

I would suggest thinking in terms of the exponential function. Anyone please feel free to destroy my answer.