operational amplifier - Problem understanding when the OPAMP is switching

I have a problem with following oscillator circuit:

Assuming an ideal op-amp. I want to draw the output of Uout, Uc, U+.

First of all I want: $Ua=f(U_+)$

That is a simple voltage divider: $U_+ = U_{out}\cdot\dfrac{R_1}{R_1+R_2}$

After that I thought about what values can $U_{out}$ have and when?

$U_{out} = 12V$ when $U_+ > U_-$, $U_{out} = -12V$ when $U_- > U_+$

The capacitor will charge to $U_{+}$ because of an ideal opamp will hold $U_D=U_+-U_-$ to zero.

so with $u_C(t=0)=0$ I get:

$u_C(t)=U_+(1-e^{-t/\tau})$ with $\tau = 9.1k\cdot100nF$

So my question is first of all:

1. 
   

   In case that NO energy is in my system, does it oscillate? -> Falstad shows an oscillation without any source connected (maybe noise?)

   
   


2. 
   
   

   Since $U_c$ is decr/increasing to $U_+$, $u_C$ never reaches $U_+$, so WHY does the Comparator-Output Change? I thought that an Comparator works like this:

   
   

if $U_+ > U_- -> U_{out}=Vcc_+$

if $U_- > U_+ -> U_{out}=Vcc_-$

Maybe someone can help me to understand this? I have to learn a little bit more about oscillator circuits with opamps, so I have to learn the basic idea behind it. Hopefully someone can explain me that intuitively.

edit: I love it, maybe I got the answer myself after questioning.

$Uc(t->\inf) = U_{out}$ not $Uc(t->\inf) = U_{+}$

so there is an active case when $U_- > U_+$ so the Comp is switching.