Does this algebraic relation between the functions imply linear dependence?

Question

While working on a geometric problem, I reached to the following uniqueness question:

Let $s(t),\tilde s(t), b(t),\tilde b(t)$ be continuous* real-valued functions, and suppose that $$ \bigg(\frac{\tilde s}{s}\bigg)^2+\bigg(\frac{s}{\tilde s}\bigg)^2+\bigg( \frac{\tilde b}{s}-\frac{b}{\tilde s}+\frac{c}{s}\bigg)^2 $$ is independent of $t$, where $c \in \mathbb{R}$ is some given non-zero constant.

Is it true that $\tilde b = \alpha b, \tilde s = \frac{1}{\alpha} s$ for some constant scalar $\alpha$?

If this is the case, then one easily sees that $s(t)$ must be constant.

*I am fine with assuming higher regularity of the functions, i.e. that all the functions are $C^1$.

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I guess we could differentiate the equation, but this doesn't look too simple.

Answer 1

Let $s(t)=2+\sin t$, $\tilde s(t)=\frac{e^t}{e^t+e^{-t}}+1$, $c=17$, $b(t)=2+\cos t^3$ or similar mostly arbitrary functions bounded from above and bounded sufficiently away from $0$ and as smooth as you desire. Then $-\frac b{\tilde s}+\frac cs$ is also bounded from above and away from $0$, and the same holdes for $\frac s{\tilde s}$ and $\frac {\tilde s}s$. Let $$M> \left(\frac{\tilde s}{s}\right)^2+\left(\frac{s}{\tilde s}\right)^2+\bigg( \frac{\color{red}0}{s}-\frac{b}{\tilde s}+\frac{c}{s}\bigg)^2$$ be a strict upper bound and define the smooth function $\tilde b$ as $$ \tilde b=s\cdot \sqrt{M-\left(\frac{\tilde s}{s}\right)^2-\left(\frac{s}{\tilde s}\right)^2}+\frac {bs}{\tilde s}-c$$

Answer 2

Let's call $\hat b=\tilde b+c$, $\hat s=\tilde s$.

Notice that necessarily $\left(\frac{\hat s}{s}\right)^2+\left(\frac{s}{\hat s}\right)^2\ge2$, so let's say that that whole expression of yours is constant equal to $2h+2$, for some $h\ge0$.

For $h>0$ you have tricks to devise real-analytic solutions unlike the ones you describe: for instance, by imposing simultaneously $$\begin{cases}\left(\frac{\hat s}{s}\right)^2+\left(\frac{s}{\hat s}\right)^2=2+h+h\cos^2t\\\frac{\hat b\hat s-bs}{\hat ss}=\sqrt h\sin t\end{cases}$$

which is certainly satisfied by the condition $$\begin{cases}\widehat s=\sqrt{\frac{2+h+h\cos^2t+\sqrt{(2+h+h\cos^2t)^2-4}}2}\cdot s\\ \widehat b=s\sqrt h\sin x+\frac{bs}{\hat s}=s\sqrt h\sin t+\sqrt{\frac2{2+h+h\cos^2t+\sqrt{(2+h+h\cos^2t)^2-4}}}\cdot b\end{cases}$$

And you can really choose $s$ and $b$ to be anything you want, as long as $s$ has no zeros (take, say, $s=1+t^2$ and $b=\arctan t$).

For $h=0$, the only solutions are with $s=\pm s$ and $\hat s\hat b=bs$.