Intersection of two non-linear equations?

Question

Graphically it is clear that $$1 + 2e^{{(x-y)}^2}(x-y) = 0$$ $$e^{{(x-y)}^2} - y = 0$$ has a unique solution, but how do I solve this analytically? If this is not possible, then what would be an appropriate technique to try and estimate the answer.

Answer 1

If you let $t = (x-y)^2$, the equations say

$$ \eqalign{1 \pm 2 e^{t} \sqrt{t} &= 0\cr y &= e^{t}\cr}$$ From the first equation, $$ 1/2 = 2 t e^{2t} $$ so $2t$ must be one of the branches of $W(1/2)$ where $W$ is the Lambert W function. If you're interested in real solutions, there is only one real branch. Moreover we must have $x-y < 0$ in the first equation so $x - y = -\sqrt{t}$. Thus $$ \eqalign{y &= e^t = e^{W(1/2)/2} = \dfrac{1}{\sqrt{2 W(1/2)}}\cr x &= y - \sqrt{t} = \dfrac{1}{\sqrt{2 W(1/2)}} - \sqrt{W(1/2)/2}\cr}$$