Proving uniqueness for a system of nonlinear equations

Question

How do I prove that this system has a unique solution? $(1-x)e^y = -e^{-x} ; y(1-x)e^x = e^{-y}$

I just don't even know where to start, should I try using limits?

Answer 1

Given $(1-x)e^y = -e^{-x} ; y(1-x)e^x = e^{-y} \to $

\begin{align}(1-x)e^{x+y} = -1 ; y(1-x)e^{x+y} = 1, y=-1,(x-1)e^{x-1}=1\end{align}

Set $f(x)=(x-1)e^{x-1},f'(x)=((x-1)e^{x-1})'=x\cdot e^{x-1} $.

$f(x)$ is a decreasing function when $x\in(-\infty,0)$ , increasing function when $x\in (0,+\infty)$.

$f(-\infty) \to 0 , f(0)=-\frac{1}{e} \to $if x $\in (-\infty,0), x\neq 1$. Similarly, $f(0)=-\frac{1}{e}, f(+\infty)\to +\infty\to $ if x$\in (0,+\infty)$, there is only one x value satisfying $f(x)=1$.

Therefore, this system has only a unique solution of $(x,y).$

Answer 2

You do not need any limits.

The system is equivalent to

\begin{eqnarray*} (1-x)e^{x+y} & + & 1 & = & 0 \\ y(1-x)e^{x+y} &- & 1 & = & 0 \end{eqnarray*}

Adding both equations yields

$$e^{x+y}(1-x)(1+y)=0$$ So, you have either $x=1$, which doesn't give a solution.

For $y=-1$ you get the equation

$$(x-1)e^{x-1}-1=0 \stackrel{t:= x-1}{\Leftrightarrow} te^t =1$$

which has a unique solution since $te^t$ is strictly increasing for $t \geq 0$ and $te^t <0$ for $t<0$.