I am trying to divide $\mathbb{R}^+\times \mathbb{R}^+$ into some curves so that the integration of the function $ h(x)h(y) $(where $h(x)$ is a $C^1$ function from $\mathbb{R}^+\to \mathbb{R}^+$ that we can choose) on each of those curves are the same. It seemed to be me the integration might not be well-defined at least without further clarification of what I want exactly. So, more specifically, I am trying to find a function $ \phi: \mathbb{R}^+\times \mathbb{R}^+ \to \mathbb{R}^+ $ so that $ (c,t)\to (t,\phi(c,t)) $ is bijective map from $ \mathbb{R}^+\times \mathbb{R}^+ \to \mathbb{R}^+\times \mathbb{R}^+ $ and $C^1$ so that $$\int\limits_{0}^{\infty} h(t)h(\phi(c,t))\sqrt{1+(\frac{\partial \phi(c,t)}{\partial t})^2} dt $$ is finite and independent of $ c $. Also, preferable I would like the curves $ t\to (t,\phi(c,t)) $ to be symmetric wrt the $ x=y $ line, i.e. $\phi(c,\phi(c,t))=t $.

To do this I assumed that $h(t)h(\phi(c,t))\sqrt{1+(\frac{\partial \phi(c,t)}{\partial t})^2} $ is independent of $ c $ and this leads to solving differential equation of the form $ h(y)\sqrt{1+y'^2}=p(x) $. But I haven't solved this type of differential equation before.

For further clarification, neither $ \phi $ nor $ h $ is given. I am trying to find a pair $ \phi $ and $ h $ which will satisfy the above properties.

Edited: Earlier when I posted the question, I chose $ h(x)=x $, but after thinking about it, it seems it won't be possible to do this in that case. I think, choosing $ h(x)=e^{-x} $ or some function so that $ h(x)=O(\frac{1}{x^{1+\epsilon}}) $ would lead to a solution.

If anyone could give some reference where I can find differential equations of this type so that I might get some idea on how $ \phi $ might be or an example of such function $ \phi $, it would be really helpful.

Edit:- In the above question, if we fix $ \phi(c,t)=\frac{c}{t} $ then is it possible to get a function $ h $ which satisfies the property?

In this case, if we take $ g(t)=h(e^t) $ where $ g: \mathbb{R}\to \mathbb{R}^+ $, then the integration turns out to be $$ \int\limits_{-\infty}^{\infty} g(t)g(x-t)\sqrt{e^{2t}+e^{2(x-t)}} dt $$ So we need a function $ g $ for which the integral is independent of $ x $.

I am not sure if this is even possible. So any comment would be helpful.

  • $\begingroup$ There's something wrong in your formulation. If $\phi:\mathbb{R}^+\times \mathbb{R}^+\to \mathbb{R}^+\times \mathbb{R}^+$ is bijective and $C^1$, then $\phi(c,t)$ is an element of $\mathbb{R}^+\times\mathbb{R}^+$, but, in your integral you treat it as an element of $\mathbb{R}$ when you write $h(\phi(c,t))$ and when you write the partial derivative expression. Check your formula or your definition and fix this, or nobody will treat your question seriously. $\endgroup$ – Robert Bryant Apr 10 '16 at 10:44
  • $\begingroup$ Sorry, edited it. $\endgroup$ – DLN Apr 10 '16 at 10:50
  • $\begingroup$ That's better, but you still need some clarification: Are you looking for a single $\phi$ that will work for all $C^1$ functions $h$ (highly unlikely to exist) or are you looking for some pair of functions $h$ and $\phi$ so that the integral you have written down is independent of $c$? You need to make it clear which things are given and which are to be solved for. $\endgroup$ – Robert Bryant Apr 10 '16 at 10:59
  • $\begingroup$ Edited it. I just want a pair $ \phi $ and $ h $, neither of them are given already. $\endgroup$ – DLN Apr 10 '16 at 11:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.