Introduction
Suppose we have a convex, real function $f(x)$. We can define a tangent line to this function $t(x,s)$. Then, we can find the intersection of $t(x,s)$ with the $x$ axis. Let's call this point $o(x)$. Then we define $l(x)$ as $o(x)-x$ and $r(x)^2=l(x)^2+f(x)^2$. See my poorly drawn graphic for a visual explanation:
The simple questions we can ask ourselves are:
What are $l(x)$ and $r(x)$ for a given $f(x)$?
The answer is
$$l(x)=\frac{f(x)}{f'(x)}\qquad\qquad r(x)=\sqrt{\frac{f(x)^2}{f'(x)^2}+f(x)^2}$$
A slightly more difficult question is to ask:
What are $f(x)$ and $r(x)$, if $l(x)$ is known?
The answer is:
$$f(x)=c\exp\left(\int_1^x \frac{1}{l(s)}\text{d}s\right)\label{f(l)}\qquad\qquad r(x)=\sqrt{c^2\exp\left(2\int_1^x \frac{1}{l(s)}\text{d}s\right)+l(x)^2}$$
A real challenge is to find
What are $f(x)$ and $l(x)$, if $r(x)$ is known?
I have no answer to that question. A simple way of deriving the necessary ODE's can be found from:
$$ \arctan(f')=\arctan\left(\frac{f}{l}\right)\\ \arctan(f')=\arcsin\left(\frac{f}{r}\right)\\ \arctan(f')=\arccos\left(\frac{l}{r}\right)$$ which you can deduce from the figure above. If we find the solution for $f(x)$ in terms of $r(x)$ we can automatically find $l(x)$. In short, my question reduces to
Solve the following ODE for $f(x)$ knowing $r(x)$: $$f'(x)=\frac{f(x)}{\sqrt{r(x)^2-f(x)^2}}$$
I hope you find the project interesting and I'm looking forward to your collaboration.
Suggested approaches
Try it with Maple. There is evidence that Mathematica doesn't handle this problem correctly, whereas Maple might.
Compare the ODE to one of the forms in HANDBOOK OF EXACT SOLUTIONS for ORIDINARY DIFFERENTIAL EQUATIONS by Polyanin and Zaitsev, or any other similar source.
Use Mathematica or similar software to generate a series expansion of the function in question and try to guess the pattern.
The code in Mathematica is:
sol1 = AsymptoticDSolveValue[{y[x]^2*y'[x]^2 +
y[x]^2 - (r[x])^2*y'[x]^2 == 0}, y[x], {x, 0, n}]
where n is the order of the expansion (n=4 recommended for start).
- Using split-quaternions $(\mathbb{P})$ factorisation.
We can rewrite the problem as: $$-r(x)^2 y'(x)^2+y(x)^2 y'(x)^2+y(x)^2=0$$ Notice that a polynomial $p\in \mathbb{R}[a,b,c], p=a^2+b^2-c^2$ can be factor in $\mathbb{P}[a,b,c]$ as
$$p=(a+bi +cj)(a-bi-cj)$$
This suggests that we can factor $$y(x)^2 y'(x)^2+y(x)^2-r(x)^2 y'(x)^2=0\\ (y(x)y'(x) +y(x) i +r(x)y'(x) j)(y(x)y'(x) -y(x) i -r(x)y'(x) j)=0$$ and solve independently $$y(x)y'(x) +y(x) i +r(x)y'(x) j=0\\ y(x)y'(x) -y(x) i -r(x)y'(x) j=0$$ Here we discussed how this technique was applied to a simpler problem with success. Whether this technique is legitimate is still unclear to me.
- Solve for $r(x)$ as a function of $l(x)$ and then convert to $f(x)$.
One can use any of the techniques described above. Refer to this for one of the forms I obtained in terms of $l(x)$. (Note that following the link $l(x)$ is replaced by $f(x)$)
Physical applications
Here we discuss the possible applications in the field of physics.
Physical interpretation of $l(x)$
Imagine there is an object that you cannot see. However, this object casts a shadow on the surface of Earth, since the sun is shining on it. You can only measure the length of this shadow. Can you deduce the shape of the object by measuring the shadow as the sun progresses over the Earth?. The answer is yes, and if we denote this shadow by $l(x)$ we can use the formulas discussed in this question to find $f(x)$. This easily generalizes to 3D but this is not interesting for us. In the real world this technique is applied in spectroscopy.
Physical interpretation of $r(x)$
When $r(x)=const.$ we obtain an equation of a tractrix. For a generin $r(x)$ can this equation can be interpreted as an equation of a tractrix with a variable chain length?
Still in progress.
Other ways to contribute
Properties of the solution for $f(x)$ or $l(x)$
If solving the ODE is out of reach, we could still try to find some of it's properties. Try to discuss the existence and uniqueness of the solution given that $r(x)>0$, $l(x)>0$ and $f(x)$ is concave.
Examples:
The solution should be a function of $x+const.$
Constructing a table of special cases of the solution
A good way of understanding the problem is to evaluate $r(x)$ for various $f(x)$. Below is a table of some examples.
The code for generating this in Mathematica is
fs = {f[x], x, x^2, x^n, a + b x, a + b x + c x^2, Sin[x], Cos[x],
Tan[x], Sinh[x], Cosh[x], Tanh[x], ArcSin[x], ArcCos[x], ArcTan[x],
ArcSinh[x], ArcCosh[x], ArcTanh[x], Exp[x], Log[x]}
ys = {};
xs = {};
For[ii = 1, ii <= Length[fs], ii++,
g[x] = fs[[ii]];
AppendTo[ys,
Simplify[Sqrt[g[x]^2 + g[x]^2/D[g[x], x]^2]] // Refine];
AppendTo[xs, fs[[ii]] // Refine]]
Text@Grid[Prepend[Transpose[{xs, ys}], {"f[x]", "r[x]"}],
Background -> {None, {Lighter[Yellow, .9], {White,
Lighter[Blend[{Blue, Green}], .8]}}},
Dividers -> {{Darker[Gray, .6], {Lighter[Gray, .5]},
Darker[Gray, .6]}, {Darker[Gray, .6], Darker[Gray, .6], {False},
Darker[Gray, .6]}}, Alignment -> {{Left, Right, {Left}}},
Frame -> Darker[Gray, .6], ItemStyle -> 14,
Spacings -> {Automatic, .8}]
PS I've been passively working on this problem for 5 years... I started it in my first year of University. A year ago my friend with whom I've shared this idea challenged me to solve this question before I turn 25 years of age. I must accept my defeat as today I turned 25 and hence I'm making this project public.