How do I solve $u = e ^ {-u}$? Is there a single solution? 
I need to solve this equation and I have no idea how to do it? $$u = e ^ {-u}$$

 A: $u=e^{-u}\Leftrightarrow u-e^{-u}=0$
Let $f(x)=x-e^{-x}$. Then $f'(x)=1+e^{-x}>0$, so the function is monotonically increasing. As a sum of two continuous functions it is also continuous. Now note $f(0)=-1$ and $f(1)=1-\frac{1}{e}>0$. So by the Intermediate Value Theorem there must be a $u$ for which $f(u)=0$ and that number would satisfy $u=e^{-u}$. Since the function is monotonically increasing, we can also see that there is only one such $u$ and furthermore that $0<u<1$.
This shows that your equation has a unique, real solution. However we don't know a "nice" expression for that number, but we can express it in terms of the Lambert W function (refer to orlp's answer).
A: Let $f(u)=u-e^{-u}$. The derivative is $1+e^{-u}$, which is always positive. Hence $f$ is monotonic.
As it is continuous and $f(0)<0,f(1)>0$, it has a single root, which is in range $(0,1)$.
The value of this root cannot be expressed by elementary functions (you need Lambert's $W$), so you can resort to numerical methods. Newton's iterations will work well.
$$u_{n+1}=u_n-\frac{u_n-e^{-u_n}}{1+e^{-u_n}}=\frac{u_n+1}{e^{u_n}+1}.$$
A good starting approximation can be obtained from the second order Taylor development of the exponential, leading to
$$x-\left(1-x+\frac{x^2}2\right)=0$$ and $$u_0\approx 2-\sqrt2=0.58$$
A: The given equation $u=e^{-u}$ is not algebraically solvable.
But you can solve it numerically with the Newton-method:
$f(u)=e^{-u}-u$
$f'(u)=-e^{-u}-1$
Now with the formula:
$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$ and some iterations, we get close to a solution.
First we have to find a $x_0$ to start the iteration.
We can find one by searching for a change in the sign by calculation a few.
f(0)=1, f(1)<0
Therefore we root has to be between 1 and 0 and we can choose $x_0=0.5$
Then
$x_1=0.5-\frac{f(0.5)}{f'(0.5)}\approx 0.56631$
$x_2=0.56714$
...
Which is already pretty close to exact value given by orlp.
A: There is only one solution.
Because , if you look at the function $f(u)=u-e^{-u}$ then $f(1)>0$ and $f(-1)<0$, so it must hit zero somewhere in $[-1,1]$ because it is continuous, so there is one.
Because function increases strictly that shows that it has at most one solution.
So it has only one.
