Let $\Phi (x) := \int_{-\infty}^x \frac{1}{\sqrt{2\pi}} e^{-\frac{y^2}{2}} dy$ the distribution Function of the standard normal distribution and $T,a>0$ and $c \in \left( 2 \Phi (\frac{-a}{\sqrt{T}}), 1\right)$. I would like to know whether one can find an explicit solution (by choosing the variables $T,a,c$) of the function $f:[0,T) \to (0,\infty)$ defined by
$$\Phi \left( \frac{f(t) - a}{\sqrt{T-t}} \right) + \Phi \left( \frac{-f(t)-a}{\sqrt{T-t}} \right) =c$$
I have done the following until now:
This function also fulfils by transformation
$$f(t) = a + \sqrt{T-t} \left(\Phi^{-1} \left( c - \Phi\left( \frac{-f(t)-a}{\sqrt{T-t}} \right) \right) \right)$$
By derivation one can infer that
$$f' (t) = \left( {f'(t)} + \frac{f(t) +a}{2(T-t)} \right) \exp\left( -\frac 1 2\frac{(f(t)+a)^2}{T-t} + \frac 1 2 \left(\Phi^{-1} (c-\Phi (\frac{-a-f(t)}{\sqrt{T-t}})) \right)^2\right)\\ - \frac{1}{2(T-t)} (f(t) -a) $$
Furthermore one can choose $c= \frac 1 2+\Phi (\frac{-2a}{\sqrt{T}})$, which implies $f(0) = a$ and thus $$f' (0) = \frac a T \frac {1}{\exp (\frac{2a^2} T ) +1}$$