# 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 '10 at 12:03
Put $u(x)=0$ and $\Psi, w$ to whatever satisfies the conditions. – Per Alexandersson Nov 30 '10 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 '10 at 5:44
Is $\sin(x+y)\sin(x-y)$ a counterexample? – Will Sawin Nov 16 '11 at 21:15

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 $\Phi(\frac {u(f(x,y))}{u(x)})$