I am very new to partial differential equations, so would be very grateful if somebody could explain the following.
Given the equation $$ a_1 F + a_2 x F_x + a_3x^2F_{xx} + a_4tF_t=0 $$
and the conditions $$\begin{aligned} F(x_*,t_*)&=V(x_*,t_*)\\ F(0,t_*)&=0\\ F_x(x_*,t_*)&=V_x(x_*,t_*)\\ F_t(x_*,t_*)&=V_t(x_*,t_*)\\ \end{aligned} $$
where $a_1...a_4$ are constants and $V$ is a known function, how would one go about solving for $F$ and obtaining $x_*$ and $t_*$? After some looking around, I saw that this equation is very similar to the Black Scholes Equation, which is of the form: $$ a_1 F + a_2 x F_x + a_3x^2F_{xx} + F_t=0 $$ The only difference is that the $F_t$ term in my equation has the non-constant coefficient $a_4t$. How does having this non-constant coefficient change the solution of the PDE? As I said, I am very new to PDEs, so would be grateful if anyone could offer a simple step-by-step walkthrough of how to solve it.
For your information, I am trying to solve an optimal stopping problem involving two variables. The conditions given above are the value matching and smooth pasting conditions.
$$ a_1 F + a_2 x F_x + a_3x^2F_{xx} + a_4tF_t=0 \tag 1 $$
and the conditions $$\begin{aligned} F(x_*,t_*)&=V(x_*,t_*)\\ F(0,t_*)&=0\\ F_x(x_*,t_*)&=V_x(x_*,t_*)\\ F_t(x_*,t_*)&=V_t(x_*,t_*)\\ \end{aligned} $$ Change of variable : $$t=e^\tau$$ The problem is transformed into :
$$ a_1 F + a_2 x F_x + a_3x^2F_{xx} + a_4F_\tau =0 \tag 2 $$
and the conditions with $t_*=e^{\tau_*}$ $$\begin{aligned} F(x_*,\tau_*)&=V(x_*,\tau_*)\\ F(0,\tau_*)&=0\\ F_x(x_*,\tau_*)&=V_x(x_*,\tau_*)\\ F_t(x_*,\tau_*)&=V_\tau(x_*,\tau_*)\\ \end{aligned} $$ Equation $(2)$ is the Black-Scholes equation that you can treat thanks to the background related to this kind of equation.
Finally come back to the solution of the modified Black-Scholes equation $(1)$ in remplacing $\tau$ by $\ln(t)$.
Having a non-constant coefficient $a_4t$ changes the solution of the PDE : One can expect that the behaviour as a function of time roughly becomes more of the power kind than of the exponential kind. That's all we can say because the initial condition is changed as well : For example if $t_*=0$ then $\tau_*=-\infty$ which is something else on analytical viewpoint.