# Is this question solvable? $2$ non-linear equations and the proof that the solution is unique

As mentioned in the title I want to show the uniqueness of the solution to $2$ non-linear equations. However, it seems that I can not solve this question with my current mathematical knowledge. More than this I am not aware about the difficulty of the problem. Is this question solvable by a very good mathematician at all? This is also what I wonder very much.

Let me introduce my problem:

I have orgininally $4$ equations as given below.

$$\int_{-\infty}^{y_l}\frac{b}{eb^{\mu_1}}f_1(y)\mathrm{d}y+\int_{y_l}^{y_u}K_1 l(y)^{\frac{-\ln(b)}{\ln(a b)}}f_1(y)\mathrm{d}y+\int_{y_u}^{\infty}\frac{1}{eb^{\mu_1}}f_1(y)\mathrm{d}y=1$$

$$\int_{-\infty}^{y_l}\frac{1}{ea^{\mu_0}}f_0(y)\mathrm{d}y+\int_{y_l}^{y_u}K_0 l(y)^{\frac{\ln(a)}{\ln(a b)}}f_0(y)\mathrm{d}y+\int_{y_u}^{\infty}\frac{a}{ea^{\mu_0}}f_0(y)\mathrm{d}y=1$$

$$\small \int_{-\infty}^{y_l}\frac{b}{eb^{\mu_1}}\ln \left(\frac{b}{e b^{\mu_1}}\right)f_1(y)\mathrm{d}y+\int_{y_l}^{y_u}K_1 l(y)^{\frac{-\ln(b)}{\ln(a b)}}f_1(y) \ln \left( K_1 l(y)^{\frac{-\ln(b)}{\ln(a b)}}\right)\mathrm{d}y+\int_{y_u}^{\infty}\frac{1}{eb^{\mu_1}}\ln \left(\frac{1}{eb^{\mu_1}}\right)f_1(y)\mathrm{d}y=\epsilon_1$$

$$\small \int_{-\infty}^{y_l}\frac{1}{ea^{\mu_0}}\ln \left(\frac{1}{e a^{\mu_0}}\right)f_0(y)\mathrm{d}y+\int_{y_l}^{y_u}K_0 l(y)^{\frac{\ln(a)}{\ln(a b)}}f_0(y) \ln \left( K_0 l(y)^{\frac{\ln(a)}{\ln(a b)}}\right)\mathrm{d}y+\int_{y_u}^{\infty}\frac{a}{e a^{\mu_0}}\ln \left(\frac{a}{ea^{\mu_0}}\right)f_0(y)\mathrm{d}y=\epsilon_0$$

Given:

$\rightarrow f_0$ and $f_1$ are some density functions

$\rightarrow l(y)=\frac{f_1(y)}{f_0(y)}$ is an increasing function

$\rightarrow b=l(y_u),\quad a=\frac{1}{l(y_l)}$

$\rightarrow K_0=\frac{e^{\frac{\ln^2(a)}{\ln (ab)}}}{e a^{\mu_0}},\quad K_1=\frac{e^{\frac{\ln^2(b)}{\ln (ab)}}}{e b^{\mu_1}}$

As you can see the first two equations can be inserted into the third and fourth equation via $e a^{\mu_0}$ and $e b^{\mu_1}$. As a result we can obtain

$$z_0=\int_{-\infty }^{y_l} f_0(y) \, \mathrm{d}y+\int _{y_l}^{y_u}e^{\frac{\text{ln}^2(a)}{\text{ln}(ab)}}l(y)^{\frac{\text{ln}[a]}{\text{ln}(ab)}}f_0(y)\mathrm{d}y+\int _{y_u}^{\infty }a f_0 (y)\mathrm{d}y$$

$$z_1=\int _{-\infty }^{y_l}b f_1(y)\mathrm{d}y+\int _{y_l}^{y_u}e^{\frac{\text{ln}^2(b)}{\text{ln}(ab)}}l(y)^{\frac{-\text{ln}[b]}{\ln (ab)}}f_1(y)\mathrm{d}y+\int_{y_u}^{\infty } f_1(y) \, \mathrm{d}y$$

$$h_0=-\text{ln}(z_0)+\frac{1}{z_0}\left(\int _{y_l}^{y_u}e^{\frac{\text{ln}^2(a)}{\text{ln}(ab)}}l(y)^{\frac{\text{ln}(a)}{\text{ln}(ab)}}\text{ln}\left(e^{\frac{\text{ln}^2(a)}{\text{ln}(ab)}}l(y)^{\frac{\text{ln}(a)}{\text{ln}(ab)}}\right)f_0(y)\mathrm{d}y+\int _{y_u}^{\infty }a\text{ln}(a)f_0 (y)\mathrm{d}y\right)=\epsilon_0$$

$$h_1=-\text{ln}(z_1)+\frac{1}{z_1}\left(\int _{y_l}^{y_u}e^{\frac{\text{ln}^2(b)}{\text{ln}(ab)}}l(y)^{-\frac{\text{ln}(b)}{\text{ln}(ab)}}\text{ln}\left(e^{\frac{\text{ln}^2(b)}{\text{ln}(ab)}}l(y)^{-\frac{\text{ln}(b)}{\text{ln}(ab)}}\right)f_1(y)\mathrm{d}y+\int _{-\infty}^{y_l }b\text{ln}(b)f_1 (y)\mathrm{d}y\right)=\epsilon_1$$

Now I have two functions $h_0(y_l,y_u)=\epsilon_0$ and $h_1(y_l,y_u)=\epsilon_1$.

My conjecture:

There exists a unique solution to $\{h_0(y_l,y_u)=\epsilon_0,\,h_1(y_l,y_u)=\epsilon_1\}$ with some $\{y_l,y_u\}\in\mathbb{R}$ for $\{\epsilon_0\in[0,h_0(y_l\rightarrow -\infty,y_u \rightarrow \infty)],\epsilon_1\in[0,h_1(y_l\rightarrow -\infty,y_u \rightarrow \infty)]\}$

My Question:

Prove or disprove if my conjecture is correct.


Extra Information:

When $y_l\rightarrow -\infty$ and $y_u\rightarrow \infty$, the first $4$ equations above simplifies considerably. Then, for the equation $h_0$ we have a single parameter $0<x=\frac{\ln(a)}{\ln(a b)}<1$. If we take the derivative with respect to $x$, we can show that it is positive. The latter part of this proof (to show that it is positive) can be seen here Prove or disprove that the given expression is "always" positive , thanks to @julien. I havent done by my own but I would guess that the similar story also applies to $h_1$, and this determines the set $\{\epsilon_0\in[0,h_0(y_l\rightarrow -\infty,y_u \rightarrow \infty)],\epsilon_1\in[0,h_1(y_l\rightarrow -\infty,y_u \rightarrow \infty)]\}$ for which a unique solution is sought for.

• Please don't put information about a bounty in the title; it already automatically gets displayed alongside the title. (Also the question currently has a bounty of $50$ points, not $500$ points.) – joriki Mar 25 '13 at 12:19
• @joriki I want to make something out of standard. So I want to offer asymmetric bounty. I hope it is not a problem. I want to give $500$ for the proof and $50$ for disproof and I know how I can give it. – Seyhmus Güngören Mar 25 '13 at 12:22
• This sounds like giving 500 points for a proof of the Riemann hypothesis and 50 points for its refutation $\ldots$ – Christian Blatter Mar 29 '13 at 19:56
• @ChristianBlatter funny). The proof should be much more difficult than finding an extreme and probably simple example which can disprove it. It also reflects that a proof makes me happier than this proof i mean disproof )). – Seyhmus Güngören Mar 29 '13 at 20:05