I been struggling with a nonlinear equation for a couple of weeks (months). Maybe you can give me a hint.
I need to prove the following.
Given the equation
\begin{equation} \tag{1}\label{eq} (\sigma\beta)^2(-1+x/\beta+e^{-x/\beta })=y \end{equation} and fixed values of $\beta>0$ and $\sigma>0$ two pair of points $(x_a, y_a)$ and $(x_b, y_b)$ which verify Eq.(\ref{eq}) define the parameters $\beta$ and $\sigma$ univocally. I want to show that there are no other combinations of $\beta$ and $\sigma$ that can generate the the same two pairs of points $(x_a, y_a)$ and $(x_b, y_b)$. In other words, given two pair of points, there is only one curve of the form of Eq.(\ref{eq}) that passes thought the points $(x_a, y_a)$ and $(x_b, y_b)$.
The domains are $x\in(0,\infty)$ for $\beta>0$ and $\sigma>0$ we have $y\in(0,\infty)$.
I solved the system numerically for really a lot of combinations of parameters and I always found what I am trying to prove, but still, I need to show it. Thanks ;)
Well, I am not a mathematician and all the approaches that I tried were trying to solve the real system, which is not needed. I just need to find that the solution is unique. Yesterday after posting the question I realised this and I tried a couple of approaches.
I insist, I am not a mathematician, but maybe you can give me a hint if the procedure I used to "proof" this is fine. Here we go:
Given $\sigma>0$, $\beta>0$ lets consider two sets of points, $(x_1,y_1)$ and $(x_2,y_2)$, that verify \begin{eqnarray} \beta ^2 {\sigma}^2 ({x_1} /\beta - 1+ e^{- x_1/\beta}) = y_1 \label{eq:proof_1} \\ % \beta ^2 {\sigma}^2 ({x_2} /\beta - 1+ e^{- x_2/\beta}) = y_2 \label{eq:proof_2} \end{eqnarray} , where $y_1 \neq y_2$ and $y_1,y_2>0$. First note that $\beta ^2 {\sigma}^2 (x /\beta - 1+ e^{- x/\beta})$ is a monotonically increasing function as as a function of $x>0$, $\beta>0$, $\sigma>0$, then also $x_1 \neq x_2$. Now lets suppose that there are other two values $\tilde{\sigma}>0$, $\tilde{\beta}>0$ that verify \begin{eqnarray} \tilde{\beta}^2 {\tilde{\sigma}}^2 ({x_1} /\tilde{\beta} -1+ e^{- x_1/\tilde{\beta}}) = y_1 \label{eq:proof_1a} \\ % \tilde{\beta}^2 {\tilde{\sigma}}^2 ({x_2} /\tilde{\beta} -1+ e^{- x_2/\tilde{\beta}}) = y_2 \label{eq:proof_2a} \, . \end{eqnarray} Then these equations verify \begin{eqnarray} \beta ^2 {\sigma}^2 ({x_1} /\beta - 1+ e^{- x_1/\beta}) = \tilde{\beta}^2 {\tilde{\sigma}}^2 ({x_1} /\tilde{\beta} -1+ e^{- x_1/\tilde{\beta}}) \label{eq:proof_aux_d1} \tag{1}\\ \beta ^2 {\sigma}^2 ({x_2} /\beta - 1+ e^{- x_2/\beta}) = \tilde{\beta}^2 {\tilde{\sigma}}^2 ({x_2} /\tilde{\beta} -1+ e^{- x_2/\tilde{\beta}}) \label{eq:proof_aux_d2} \tag{2} \end{eqnarray} Dividing \ref{eq:proof_aux_d1} by \ref{eq:proof_aux_d2} we have \begin{eqnarray} \frac{ ({x_1} /\beta - 1+ e^{- x_1/\beta})} { ({x_2} /\beta - 1+ e^{- x_2/\beta})} = \frac{ ({x_1} /\tilde{\beta} -1+ e^{- x_1/\tilde{\beta}})} { ({x_2} /\tilde{\beta} -1+ e^{- x_2/\tilde{\beta}})} \end{eqnarray} and by defining \begin{equation} g(\beta;x_1,x_2) \equiv \frac{ ({x_1} /\beta - 1+ e^{- x_1/\beta})} { ({x_2} /\beta - 1+ e^{- x_2/\beta})} \end{equation} we have that \begin{equation} g(\beta;x_1,x_2) = g(\tilde{\beta};x_1,x_2) \quad \, \forall x_1, x_2 \, . \label{eq:proof_g_eq} \tag{3} \end{equation} Then in order to satisfy \ref{eq:proof_g_eq}, $\beta = \tilde{\beta}$.
Replacing in $\beta = \tilde{\beta}$ \ref{eq:proof_aux_d1} \begin{equation} \beta ^2 {\sigma}^2 ({x_1} /\beta - 1+ e^{- x_1/\beta}) = {\beta}^2 {\tilde{\sigma}}^2 ({x_1} /{\beta} -1+ e^{- x_1/{\beta}}) \end{equation} that yields ${\sigma}^2 = {\tilde{\sigma}}^2$, and since ${{\sigma}}^2, {\tilde{\sigma}}^2>0$ by definition, then we have that ${\sigma} = {\tilde{\sigma}}$.