A simple geometric problem, solving $f'(x)=\frac{f(x)}{\sqrt{r(x)^2-f(x)^2}}$, given $r(x)$.

Question

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

1. Try it with Maple. There is evidence that Mathematica doesn't handle this problem correctly, whereas Maple might.

2. 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.

3. 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).

4. 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.

5. 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.

Answer 1

Here we assume in general that the plane curve is $C\in\textbf{C}^{(1)}[a,b]$ and of the form $$ C:\overline{x}(s)=\left\{A(s),B(s)\right\},\tag 1 $$ where $s$ is its canonical parameter i.e. $$ (A'(s))^2+(B'(s))^2=1.\tag 2 $$ If $P_0\in C$ (a point of the curve) and $t(P_0)$ is the tangent line in $P_0$ and $O_1=(x_1,0)$ is the intersection of $t(P_0)$ with the $x-$axis, then $r(s_0)=|P_0O_1|$. Hence $$ t(P_0):y-B(s_0)=\frac{B'(s_0)}{A'(s_0)}(x-A(s_0)) $$ and $O_1:y_1=0$. Hence
$$ -B(s_0)=\frac{B'(s_0)}{A'(s_0)}(x_1-A(s_0))\Leftrightarrow x_1-x_0=-B(s_0)\frac{A'(s_0)}{B'(s_0)}. $$ $$ r(s_0)=\sqrt{(x_1-x_0)^2+(y_1-y_0)^2}=\sqrt{B(s_0)^2\frac{(A'(s_0))^2}{(B'(s_0))^2}+(B(s_0)^2)}\Leftrightarrow $$ $$ r(s_0)=\frac{|B(s_0)|}{|B'(s_0)|}\sqrt{(A'(s_0))^2+(B'(s_0))^2}=\left|\frac{B(s_0)}{B'(s_0)}\right|. $$ Hence when $s_0$ varies, we get $$ \frac{B'(s)}{B(s)}=\frac{\pm 1}{r(s)}\Rightarrow B(s)=\exp\left(\pm\int\frac{ds}{r(s)}\right). $$ Also form (2) we get $$ A(s)=\pm\int\sqrt{1-(B'(s))^2}ds=\pm\int\sqrt{1-\frac{B(s)^2}{r(s)^2}}ds= $$ $$ =\pm\int\sqrt{1-\frac{\exp\left(\pm2\int r^{-1}(s)ds\right)}{r(s)^2}}ds $$ Hence for a given $r(s)$, the curve can be parametrized as $$ \overline{x}(s)=\left\{\pm\int\sqrt{1-\frac{1}{r(s)^2}\exp\left(\pm2\int\frac{ds}{r(s)}\right)}ds,\exp\left(\pm\int\frac{ds}{r(s)}\right)\right\},\tag 3 $$ for $s\in[a,b]$ and $s$ being its canonical parameter.

Answer 2

Solve the following ODE for $f(x)$ knowing $r(x)$: $$f'(x)=\frac{f(x)}{\sqrt{r(x)^2-f(x)^2}}$$

I tried to simplify this equation by different substitutions.

1) Multiplying both its sides by $f(x)$ and putting $g(x)=f(x)^2$, we obtain $$g’(x)=\frac{2g(x)}{\sqrt{r(x)^2-g(x)}}.$$

2) Putting $h(x)= r(x)^2-g(x)$, we obtain

$$2r(x)r’(x)-h’(x)=\frac{2 r(x)^2-2h(x)}{\sqrt{h(x)}},$$ $$h’(x)=2\left({\sqrt{h(x)}}-\frac{r(x)^2}{\sqrt{h(x)}}+r(x)r’(x)\right).$$

3) Putting $v(x)=\sqrt{r(x)^2-f(x)^2}$, we obtain

$$v’(x)=\frac{r(x)r’(x)-f(x)f’(x)}{v(x)}= \frac{r(x)r’(x)-\tfrac{f(x)^2}{v(x)}}{v(x)}=$$ $$\frac{r(x)r’(x)-\tfrac{r(x)^2-v(x)^2}{v(x)}}{v(x)}= 1+\frac{r(x)r’(x)}{v(x)}-\frac{r(x)^2}{v(x)^2}.$$

4) Putting $u(x)=\frac 1{v(x)}$, we obtain

$$-\frac{u’(x)}{u(x)^2}=1+r(x)r’(x)u(x)-r(x)^2u(x)^2,$$

$$u’(x)= r(x)^2u(x)^4-u(x)^2-r(x)r’(x)u(x)^3.$$

We can try to solve the last two equations in terms of power series for $r$, $u$, and $v$.

Answer 3

Just another idea. Have you tried to look for parametrized solutions? That is, starting with $$r(x)^2=f(x)^2\left(1+\frac{1}{f'(x)^2}\right)$$ and instead of looking for solutions $(x,f(x))$ you reparametrize $x(t)=h(t)$ for some function $h$. In particular, for example, assuming $f(x)$ is invertible on some domain and choosing $h(t)=f^{-1}(t)$, you then obtain the ode for $x(t)$, so $$r(x)^2 = t^2 \left(1+x'(t)^2\right) \, .$$

The solution would then be given by $(x(t),t)$ for some range of the parameter $t$. Unfortunately, except for the simple case $r(x)={\rm const}$, this still does not allow separation of variables, but maybe you can work from there with a different ansatz for $h(t)$.

Answer 4

This is just offering a different approach, which produce a different ode.
Let's put the problem in the Tractrix perspective.

Consider the two reference systems, one fixed and one attached the "tractor" at $\xi$ and which is moving at constant speed $v$ along the $x$ axis.

Assume that we know $r$ as function of $\xi$.

The speed of $P$ in the moving frame has the polar components $r', \, r \alpha '$, and in the fixed frame $v$ is added to those.
The resultant speed normal to the curve and thus to $r$ shall be null, so $$ \left\{ \matrix{ r\alpha ' = v\cos \left( {\alpha - \pi /2} \right) = v\sin \alpha \hfill \cr r = r(\xi ) = r(vt) \hfill \cr} \right. \tag {1}$$ we can rewrite the first expression separating the variables $$ {{d\alpha } \over {\sin \alpha }} = {v \over {r(vt)}}dt $$ which means $$ {{d\alpha } \over {\sin \alpha }} = {1 \over 2}d\ln \left( {{{1 - \cos \alpha } \over {1 + \cos \alpha }}} \right) = d\ln \left( {\tan \left( {\alpha /2} \right)} \right) = {v \over {r(vt)}}dt = {{d\xi } \over {r(\xi )}} $$

So under the assumption of $r=r( \xi )$, we arrive at a closed parametric equation for the curve as seen from the moving frame $$ \left\{ \matrix{ d\ln \left( {\tan \left( {\alpha /2} \right)} \right) = {{d\xi } \over {r(\xi )}} \hfill \cr r = r(\xi ) \hfill \cr} \right. \tag {2}$$ (of course provided that the integral of $ {{d\xi } \over {r(\xi )}}$ be known in closed form).