I wish I had more time to explore this. I found a passing comment in S. V. Meleshko's Methods for Constructing Exact Solutions of Partial Differential Equations about envelopes of curves. I will get to his statement at the end. Basically suppose we have a family of curves like the blue ones below

This is an example of a one parameter family of curves. Specifically, I made this by finding lines where the intercepts always add to 10,

$$y(x,a)= \frac{a - 10}{a} x + 10 - a.$$

This begs the question of finding the envelope of the family of curves shown in black in the figure. What is the equation that describes the envelope? How do we find it?

I have always been fascinated by these types of curves ever since I did a string art project in middle school. I wondered, as I do now, what types of curves you could produce.

This has been addressed before in various places, and I gather that the topic used to be quite common in Calculus courses. However, in case you don’t see it right away, here is a way to obtain the envelope.

Let’s fix $x$. Now for this $x$, say $x_0$, we have a relation between $y$ and $a$ through the equation for a curve in the family. We can think of the envelope as an extremum of the relationship between $y$ and $a$. Illustrating this graphically, if we consider the intersection of the vertical line $x_0$ and our family of curves $y(x_0,a)$, then as we scan through $a$, we will see that there are special points where the intersection point reaches a maximum. This can be seen in the animation below (right click on the image below and select 'open in a new tab').

As we scan through $a$, we can see the intersection point (in red). When it slows down and changes direction, this is a point of the envelope.

This is exactly $g(x_0)$. We can then describe the envelope equation as

$$g(x) = y ( x, a_{*} (x) ) $$

where $a_{*}(x)$ is found by finding the extremum of $y(x,a)$ with respect to $a$. Thus $a_{*}(x)$ is the solution to

$$\frac{\partial y(x,a)}{\partial a}=0.$$

For our case explicitly, the solution is

$$g(x)=10 \pm 2 \sqrt{10} \sqrt{x} + x.$$

A graph of these two curves can be seen in black in the first figure.

Note also from the way we have written the envelope that every point of $g(x)$ necessarily lies on a specific curve in the family of curves. This means that at each point a particular curve from our family is tangent to the envelope. Likewise, we also know that the envelope is tangent to one of our curves in the family. This is a good thing to recognize because from this we may construct a differential equation. We know that

$$g(x) = \frac{a_{*}-10}{a_{*}} x + 10 - a_{*},$$

$$g'(x)=\frac{a_{*} - 10}{a_{*}}.$$

From this, we may eliminate $a_*$ to obtain a differential equation for $g(x)$. In this case we get

$$g'(x)(10-x+x g'(x))=g(x)(g'(x)-1).$$

This is a nonlinear, first-order differential equation. The solution for this differential equation is then the envelope function. If we plug the envelope in, we find that it is, in fact, a solution.

The interesting point is that the two branches of the envelope are not the only solutions! All of the curves in the family are also solutions to this nonlinear differential equation. Since it is not a linear differential equation, there is not a uniqueness or existence theorem. We can see that the one parameter family of curves satisfies this equation since they satisfy the equations that the differential equation was derived from. I was just shocked that a family of lines could satisfy such an unruly differential equation.

Oftentimes we are confronted with a nonlinear differential equation where we are excited when we are able to get one solution to the problem. This is an example of an infinite number of solutions to a complicated looking differential equation. But most (an infinite number) of the solutions are just straight lines!

The converse of the above is actually easier to prove and we take it out of Meleshko (not the proof -- although it is not hard). He says it in passing, and for two dimensional surfaces. Basically, he says (on page 9) that if you have a family of curves that is a solution of a differential equation, then the envelope is also a solution! Amazing!

Anyway, I know this is too long for anyone to read, but I will end with some interesting families of curves and their envelopes.

Here is a family of lines with sin waves on top

$$y_{1}(x,a)= \frac{a - 10}{a} x + 10 - a + \sin(10 x).$$

Here is a family of exponentials

$$y_{2}(x,a) = a e^{\frac{x}{a}}$$

Here is a family of gaussians

$$y_{4}(x,a) = a e^{-(x-a)^2}$$

Here is another family of gaussians

$$y_{5}(x,a) = \frac{1}{\sqrt{\pi} a} e^{-\left(\frac{x}{a}\right)^2}$$

Here is one where I was playing with lines again

$$y_{5}(x,a) = \frac{a^2 - 1}{a} \left( \frac{x}{a}+ \frac{1}{2} \right) - a$$

Two and three dimensions should be fairly straightforward to do (although I have not). What I would like to do is to be able to go backwards. I have played enough to make a

__: any reasonable curve can be an envelope to a family of lines.__

*Conjecture*I want to make this explicit, but that is a topic for another day.

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