I would like to find out what boundary/final conditions i should be using to find the formula for a European asset-or-nothing call option, as i feel that is where I'm making my mistake.
I've read that the formula should be: $$ Se^{-r (T-t)}\Phi(d_1) $$ however, I've ended up with, $$ S\Phi(d_1) $$
I have final payoff function:
$$ C_{AoN}(S,T) = \left\{ \begin{array}{lll} 0 & \mbox{S$<$E};\\ S/2 & \mbox{S=E};\\ S & \mbox{S$>$E }\end{array} \right. $$
and I've set final and boundary conditions;
$$ u(x,0) =\hat{C}(\hat{S},1)=max(\hat{S},0)=max(e^x,0), \:\: x \in \mathbb{R}, $$
$$ \left. \ \begin{array}{cc} & u(x,\tau)\rightarrow 0 \:as \: x\rightarrow -\infty \\ & u(x,\tau) \sim e^x \: as \: x \rightarrow \infty \end{array} \right \} 0<\tau<\alpha$$
I've used the transformation: $$u(x,\tau)=e^{\lambda x +\mu \tau}v(x,\tau)$$ with, $$ \left \{ \begin{array}{cc} & \lambda = \frac{1-\nu}{2} \\ & \mu = -\frac{(\nu +1)^2}{4} \end{array} \right. $$
This part is where i think I've gone wrong;
I now try to solve the heat equation: $$ v_{\tau}-v_{xx}=0, \:\: x \in \mathbb{R},\: 0<\tau<\alpha $$ with the initial condition: $$v(x,0)= e^{-\lambda x}u(x,0)=e^{-\lambda x}\max(e^x,0)$$ $$ = \max(e^{-\lambda x}e^{ x},0)= \max(e^{\frac{\nu +1}{2}x},0)=:v_0(x) $$
The solution to out equation is given by
$$v(x,\tau) = \frac{1}{\sqrt{4 \pi\tau}}\int_{-\infty}^{\infty}e^{\frac{-(x-y)^2}{4 \tau}}v_0(y)dy$$
Using $ z=\frac{y-x}{\sqrt{2 \tau}}$, we have $$ v(x,\tau) =\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty}e^{-\frac{z^2}{2}}v_0(x+\sqrt{2 \tau}z)dz.$$
From here, carrying on with black scholes formula proof, I have ended up with $$ S\Phi(d_1) $$.
Any advice would be appreciated. This is a new subject for me.