I will get you started, but there are lots of blanks to fill. We are
interested in the PDE
\begin{align*}
\min\left\{ -u_{t}+\Delta u,u-\varphi\right\}  & =0 & \text{in }(0,T]\times\mathbb{R}^{n}\\
u(0,\cdot)-\varphi(0,\cdot) & =0 & \text{in }\mathbb{R}^{n}
\end{align*}
where $\varphi$ is a smooth function of polynomial growth. Let $u^{0}$
be a classical solution of
\begin{align*}
-u_{t}^{0}+\Delta u^{0} & =0 & \text{in }(0,T]\times\mathbb{R}^{n}\\
u^{0}(0,\cdot)-\varphi(0,\cdot) & =0 & \text{in }\mathbb{R}^{n}
\end{align*}
that is unique in an appropriately picked space of functions of polynomial
growth. Define inductively $u^{k}$ as a classical solution of
\begin{align*}
-u_{t}^{k}+\Delta u^{k}+k\min\left\{ u^{k-1}-\varphi,0\right\}  & =0 & \text{in }(0,T]\times\mathbb{R}^{n}\\
u^{k}(0,\cdot)-\varphi(0,\cdot) & =0 & \text{in }\mathbb{R}^{n}
\end{align*}
that is unique in an appropriately picked space of functions of polynomial
growth. Now you would have to show that you can pass to limits ($k\rightarrow\infty$)
to obtain a solution of the original PDE.
**The limiting solution is not, in general, twice differentiable in space**. But, you should probably be able to establish sufficient conditions for it to be a **once differentiable viscosity solution**.

This technique is called a penalty method and is at least useful for establishing existence.
Uniqueness can be handled in the space of viscosity solutions using a comparison principle argument.
I wrote about this in an [expository post][1]. The setting there is slightly more general, but handles your case as well.


  [1]: http://parsiad.ca/post/comparison-principle-for-an-optimal-stopping-problem