## Solving Functional Equation

Continue with my previous question “Regarding Kolmogorov's Superposition Theorem”, here are some further questions:

Question-1 Is it true: for any given $C^1$ continuous real function $f(x, y): \Re^2 \to \Re^1$, there exist $C^1$ continuous real functions $\Psi(x), w(x,y), u(x)$ such that: $u(f(x,y))=u(x)\Psi(\frac{w(x, y)}{u(x)})$ where $\frac{du(x)}{dx}\ge 0$; $\frac{\partial w(x, y)}{\partial y}\ge 0$

Question-2 Can these functions $\Psi(x), w(x,y), u(x)$ be solved, analytically or approximately?

It is found that constraint $\frac{\partial w(x,y)}{\partial y}\ge 0$ is hard to deal with. Without this condition, there is one obvious solution $u(x)=x, w(x, y)=f(x,y), \Psi(x)=x$, where $\frac{du(x)}{dx}=1>0$, but $\frac{\partial w(x, y)}{\partial y}\ge 0$ may not satisfy.

One may find by letting $w(x,y)=w(y)$, this becomes the question we asked in "Regarding Kolmogorov's Superposition Theorem" but since the answer is NO found by AgCl, we want to remove the constraint $w(x,y)=w(y)$ hoping solution exist...

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@Wang Tao: I fixed your LaTeX for you. Also, there's no need to sign your posts as the software does so automatically. Now, if AgCl has given a satisfactory answer to your previous question, perhaps you should consider "accepting" that answer by ticking the little check mark next to the answer in the previous question. Ah! It seems that you have two separate accounts: one used to post the previous question and one for this one. I'll file a merge request for your accounts. Please make sure in the future to log-in using the same credentials everytime. – Willie Wong Nov 30 2010 at 12:03
Put $u(x)=0$ and $\Psi, w$ to whatever satisfies the conditions. – Per Alexandersson Nov 30 2010 at 13:25
Yes. This yields one solution. However any more functions other this? I gave one example in belowing post, hope to have your advices? – Wang Tao Dec 1 2010 at 5:44
Is $\sin(x+y)\sin(x-y)$ a counterexample? – Will Sawin Nov 16 2011 at 21:15

So far what I can think of on this question is:

Rewrite $u(f(x,y))=u(x)\Psi(\frac{w(x,y)}{u(x)})$ as: $\frac {u(f(x,y))}{u(x)}=\Psi(\frac {w(x,y)}{u(x)})$ Since $\Psi$ can be arbitrary, we can further rewrite it as: $\Phi(\frac{u(f(x,y))}{u(x)})=\Gamma(\frac {w(x,y)}{u(x)})$ where $\Psi=\Phi^{-1}(\Gamma)$ Above equation shows that any $w(x,y),u(x)$ which makes $(\frac{w(x,y)}{u(x)})$ span into $(\Phi(\frac{u(f(x,y))}{u(x)})$ is a solution.

For example, given $f(x,y)=ax+by$, select $u=x$, then: $\Phi(\frac{u(f(x,y))}{u(x)})= \Phi(a+b\frac{y}{x})$ which can be spanned by $\frac {y}{x}$, so $w=y$ is a solution.

Question: how to find the appropriate $w(x,y),u(x)$ to span into $\Phi(\frac {u(f(x,y))}{u(x)})$

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JFYI: New information on the problem is best added by editing the question, rather than posting the information as an answer. For one thing, some people pay less attention to questions that already received an answer, so posting your own comments as an answer may decrease the exposure of your question. – Willie Wong Dec 1 2010 at 0:27
Hi Willie, Thanks for the advise. I intended to do so but found the words exceeded the limit, so I changed to use the "Add Another Answer". But I will keep your advice in mind. – Wang Tao Dec 1 2010 at 5:39