If $\alpha \cosh (x/\alpha)$ is a solution to $f'f''^2=f'''(1+f'^2)$, can I conclude that this is the only answer?

Question

While solving a physics problem, this equation was the result.

$$f'f''^2=f'''(1+f'^2)$$

The problem wanted the function describing the shape of a hanging chain. By seeing $1+f'^2$ I immediately thought about the identity $\cosh^2x=1+\sinh^2x$ and that $(d/dx)\cosh x=\sinh x$. So I tested the function $A\cosh(Bx)$ to see if this can be the answer:

After plugging in, you will see this:

$$A^2B^2\cosh^2x=1+A^2B^2\sinh^2x$$

So if I choose $AB:=1$, then $A\cosh(Bx)$ is an answer. So for all $\alpha$, $\alpha\cosh(x/\alpha)$ is an answer to the above differential equation.

I know that I must use a theorem to conclude uniqueness. But, this equation is describing a natural phenomenon which is unique, If you shake the chain many times, after a long time, the final shape will not change!

In such cases, provided that the answer is reasonable (here it is like a parabola which is reasonable for a hanging chain.), can we conclude the uniqueness of the answer without using any theorem?

Answer 1

$$f'f''^2=(1+f'^2)f'''$$ Let $f'(x)=y(x) \quad\to\quad yy'^2=(1+y^2)y''$ $$\frac{y''}{y'}=\frac{yy'}{1+y^2}$$ $$2\ln|y'|=\int \frac{2y}{1+y^2}y'=\ln(1+y^2)+C$$ $$y'^2=c_1^2(1+y^2) \quad \text{with}\quad c_1=\pm e^{C/2}$$ $$y'=c_1\sqrt{1+y^2}$$ $$\frac{dy}{\sqrt{1+y^2}}=c_1dx$$ $$\sinh^{-1}(y)=c_1x+c_2$$ $$y=\sinh(c_1x+c_2)$$ $f(x)=\int ydx$ $$f(x)=\frac{1}{c_1}\cosh(c_1x+c_2)+c_3$$ Let $c_1=\alpha \quad\to\quad f(x)=\frac{1}{\alpha}\cosh(\alpha x+c_2)+c_3$

Thus $\quad f(x)=\frac{1}{\alpha}\cosh(\alpha x)\quad$ isn't the only solution since they are an infinity of other solutions with $c_2\neq 0$ and/or $c_3\neq 0$

But if some well posed boundary conditions are specified which allow to determine a unique value of $c_2$ and a unique value of $c_3$, in that case the solution $\quad f(x)=\frac{1}{\alpha }\cosh(\alpha x+c_2)+c_3\quad $ is unique.

Moreover, if those boundary conditions are particular so that they lead to $c_2=0$ and $c_3=0$ , then the solution $\quad f(x)=\frac{1}{\alpha }\cosh(\alpha x)\quad $ is unique.

One cannot answer about the uniqueness of the solution if the boundary conditions are not explicitly given.

Of course, all particular solutions belonging to the general solution of the ODE are related one to another by translation. This means that the pattern is unique for one $\alpha$ given.

Answer 2

No, you cannot conclude that way because you have raised the order of ODE.

As a simple example, if given

$$ y= x^2, \text{then } ~y'= 2x, y''= 2 $$

If after this differentiation $$ y''=2, $$

is given then on integration there appear two sets with different constants of integration / focal lengths along with original parabola appearing as one first case only:

$$ y= x^2+c_1x+ c_2$$

Similarly the case of a catenary ( given in the question) by such differentiation /integration process it would introduce new curves which are not catenaries:

Let $ ( u=f').$ By Quotient Rule of differentiation for a catenary

$$ \frac{u'}{\sqrt{1+u^2}}=\frac{1}{c}=\frac{u''\sqrt{1+u^2}}{uu'} \tag1$$

Cross multiplying the first and third fractions you got I believe

$$u u^{'2}=u^{''}(1+u^2)$$

which is same as the given ODE

$$ f'f''^2=f'''(1+f'^2) $$

Plugging the first part of catenary ODE 1) into the second part, we have a mixed ODE

$$ ~\frac{u''}{u}-\frac{1}{c^2}=0; \tag2$$

The ODEs are different. The superset includes after numerical integration constants of integration which are not catenary curves. This is evident from the latter's new solution graph in green. (Arbitrary initial conditions are assumed).

EDIT1:

I find it interesting to take solutions of mixed ODE (2) further.

$$ u= C_3 e^{x/c}+C_4 e^{-x/c} \tag 3 $$

Recast it in terms of hyperbolic functions:

$$ u= C_1 \cosh (x/c)+C_2 \sinh (x/c) $$ Integrating

$$ f = c (C_1 \sinh (x/c)+C_2 \cosh (x/c))+ C_5 \tag 4 $$

So to answer your question.. a modification of ODE as you did results in solutions that can be either displaced & scaled cosh functions or such sinh functions.

In particular the mixed ODE 2) satisfies even and odd pure hyperbolic functions viz.,

$$ f= c \cosh (x/c), ~~f= c \sinh (x/c); \tag 5 $$

The latter new option ( non-hanging cable) is graphed in green, it is the sinh type with an inflection point. It is found by integration of mixed ODE 2) with some assumed numerical values for constants.

If 4 points are given constants $(c,C_1,C_2,C_5)$ can be determined.

These solutions are unique for a set of given boundary conditions.