In order to solve a Physics problem, I have to model an axisymmetric well or pit on a plane. I am looking for a function $f(x)$ that is smooth and monotonic in the domain $(0,1)$, where the origin corresponds to the bottom of the pit $f(0)=-1$ and the pit meets the plane at 1 with $f(1)=0$. There are some necessary, must-have requirements:

  • The first derivative of the function at $0^+$ must be zero
  • The first and second derivatives of the function at $1^-$ must be zero

With these two requirements, I have come up with $$ f(x) = -\frac{1}{4} \left(1+\cos (\pi x) \right)^2 $$ See the plot in Wolfram Alpha

But ideally, I would like to have more control over the shape of the pit, especially I would like to

  1. Specify the second derivative of the function at $0^+$
  2. Have some kind of "transition lengthscale", meaning I can make it arbitrarily similar to a square-well, or on the other hand make it have a gentle rise.

Does anyone have any ideas? Thanks a lot.

  • $\begingroup$ You did not specifically require this, but I am assuming you want a monotonic function - since that makes most sense regarding a "single well" function in QM. Please confirm. $\endgroup$ – XYZT Dec 12 '14 at 3:12
  • $\begingroup$ Yes, I should have mentioned the word monotonic. Thank you. $\endgroup$ – m3tro Dec 12 '14 at 3:13

The typical mathematician's way to do this is by the process of convolution with an approximate identity. The convolution of a function $f$ with another function $g$ is defined by $$ (f*g)(x) = \int_{-\infty}^\infty f(y)g(x-y)~dy = \int_{-\infty}^\infty f(x-y)g(y)~dy, $$ provided the above integrals are well defined, which they are as long as $f$ and $g$ satisfy some appropriate decay conditions. Certainly if $f$ and $g$ are zero outside of a bounded domain.

The reason to consider such a thing is that if you choose the proper $g$, then convolution with $g$ tends to provide a smooth approximation to $f$. For example, suppose $g$ has the properties

  1. $g(x) \geq 0$ for all $x$,
  2. $\int_{-\infty}^\infty g(y)~dy = 1$,
  3. $g(x) = 0$ for all $x$ outside of a bounded domain.

Then define $g^\epsilon(x) = \frac{1}{\epsilon}g(\frac{x}{\epsilon})$. Then for all piecewise continuous $f$ such that $\int_{-\infty}^\infty f(y)~dy < \infty$,

$$ (f*g^\epsilon)(x) \to f(x) $$ as $\epsilon\to 0$, and moreover $f*g^\epsilon$ is a smooth function. We call such $g$ an approximation to the identity, or a mollifier. There are many commonly used mollifiers; the standard mollifier $\eta$ can be found at http://en.wikipedia.org/wiki/Mollifier#Concrete_example.

So to construct your desired smooth well function, start with a piecewise continuous function $f$ that has the properties you desire. I suggest something like $$ f_\delta(x) = \begin{cases} 0 & x < \delta, x > 1-\delta\\ -1 & \delta \leq x \leq 1-\delta \end{cases} $$ where $\delta>0$ is a small real number. (This is just a square well from $\delta$ to $1-\delta$, and it has zero derivatives at $0$ and $1$.) Convolve with the standard mollifier to get $f_\delta * \eta^\epsilon$. Then this function is smooth, and for $\epsilon$ very small it will be an excellent approximation to $f_\delta$; in particular you can check it will have $zero$ derivatives at $0$ and $1$. You can vary the $\delta$ to vary the length of the well, or modify the function $f_\delta$ appropriately to better suit your exact purpose. The point is to find a function $f$ that does what you want, and use convolution to smooth it out.


Example 4.1.b in the following link contains an example of a smooth (infinitely-differentiable) function $g: \mathbb{R} \to \mathbb{R}$, such that $g(t) = 0$ for all $t \leq 0$, and $g(t) = 1$ for all $t \geq 1$, and such that $g'(t) > 0$ for all $t \in (0,1)$.


You can easily alter this function to make it steeper, etc., by replacing $t$ with $kt$ for constant $k>0$ (horizontal stretch/shrink). Also, obviously subtract 1 (downward shift) to make $g(0) = -1$ and $g(1) = 0$.

For more information, search for "Bump function" on Google or on this site. Hope this helps you!


Say, there is a function $g(x)$ with the following properties:

$$ g(0) = 0 \\ g(1) = 0 \\ g'(1) = 0 \\ g(0 < x < 1) > 0 $$

Now you take the integral,

$$ f(x) = \int_{0}^{x}g(t)\,{\rm d}t $$

What you seek, dear friend, is $h\,f(x/R)$, a well of depth $h$ and a one-sided width of $R$. Since this is easily found using $g(x)$, I will begin by trying to find a fun and flexible example of it and then see where it takes us.


$$ g(x) = x^b (1 - x)^c $$ where $b > 0$ and $c > 1$.

It is quite trivial to show that this satisfies the conditions for $g(x)$ as prescribed above.

Of course, taking the integral of this is quite difficult in general, and there is no easily expressible function (in analytical terms), but if you look at incomplete Beta functions, you can derive the function for your well,

$$ f(x) = a\, {\rm Beta}(x; 1 + b, 1 + c) - d $$

where ${\rm Beta}()$ is an incomplete Beta function, and the four parameters allow you to tinker with the shape, width, depth, and position of the function.

This is but one simple example that helps you achieve your goal.

Firstly, it is simple to specify the second derivative of your well at $0^+$ by specifying $g'(0)$.

Secondly, it is easy to get arbitrarily close to a square well by keeping $c$ small (say, $c = 2$), and making $b$ arbitrarily large.

If this does not give you enough flexibility in shape, then you can always multiply some strictly positive function to $g(x)$ and tinker in that manner.

Have a nice day!


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