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 requires a maximum principle argument.