# When does this non linear 2 equation system have solutions? What is the solution?

I need to solve the following system:

$$\begin{cases} a x_0^2 = \exp{ \left( -\dfrac{x_0^2}{4 \sigma^2} \right) } +a r^2 \\ \exp{ \left( -\dfrac{x_0^2}{4 \sigma^2} \right) } + 4 a \sigma^2 = 0 \end{cases}$$

where the varaibles are $x_0, a$. You can also take $x_0 >0$

The solution for $|r|\ge 2\sigma$ is
$$\begin{cases} x_0 = \sqrt{r^{2}-4\sigma ^{2}}>0 \\[2ex] a=-\dfrac{1}{4\sigma ^{2}}\exp \left( -\dfrac{r^2-4\sigma ^{2}}{4\sigma ^{2}}\right)=-\dfrac{1}{4\sigma ^{2}}\exp \left( -\dfrac{x_0^2}{4\sigma ^{2}}\right). \end{cases}$$
Since $\exp ( -\frac{x_{0}^{2}}{4\sigma ^{2}})>0$, the second equation implies that $a< 0$. For convenience write
$$E=\exp \left( -\dfrac{x_{0}^{2}}{4\sigma ^{2}}\right).$$
$$\begin{cases} ax_0^2 = E+ar^2 \\[2ex] E+4a\sigma^2=0 \end{cases} \Leftrightarrow \begin{cases} ax_0^2 = -4a\sigma^2+ar^2 \\[2ex] E=-4a\sigma^2 \end{cases} \Leftrightarrow \begin{cases} x_0^2 = -4\sigma^2+r^2 \\[2ex] a=-\dfrac{1}{4\sigma ^{2}}E. \end{cases}$$