Let $a,b$ be constants with $a>0.$ Consider the IVP $$u_t=a \Delta u+b |\nabla u|^2=0~~in~~R^n \times (0,\infty), u=g~~on~~R^n \times \{ t=0 \}$$ using the transformation $w=\phi(u)$ with proper choice of $\phi,$ and then $(ii)$ use the solution formula for $w.$
Can someone explain me how to use the solution formula to solve this problem. Any help is much appreciated.
If we set $$u = \psi(w) \quad (1)$$ we can differentiate $u$, substitute the result into our PDE and in turn construct a $\psi = \phi^{-1}$ that returns the linear heat equation (assuming a few conditions and that the inverse function even exists; see the link I put in the comments). Differentiating gives
\begin{align} u_{t} &= \psi_{w} w_{t} \\ \nabla u &= \psi_{w} \nabla w \\ \Delta u &= \nabla (\nabla u) \\ &= \nabla (\psi_{w} \nabla w) \\ &= \nabla(\psi_{w}) \nabla w + \psi_{w} \nabla (\nabla w) \\ &= \psi_{ww} \lvert \nabla w \rvert^{2} + \psi_{w} \Delta w \end{align}
and so our PDE becomes
\begin{align} \psi_{w} \nabla w &= a(\psi_{ww} \lvert \nabla w \rvert^{2} + \psi_{w} \Delta w) + b \lvert \psi_{w} \nabla w \rvert^{2} \\ &= \lvert \nabla w \rvert^{2}(a \psi_{ww} + b \psi_{w}^{2}) + a \psi_{w} \Delta w \end{align}
In order to obtain the linear heat equation, we require
$$a \psi_{ww} + b \psi_{w}^{2} = 0$$
which is a nonlinear second order ODE. This can be solved by setting $z = \psi_{w}$, separating and integrating (setting the integration constant to be zero for convenience) which gives
$$\psi = \frac{b}{a} \ln \left ( \frac{a}{b} w \right )$$
Hence,
$$u = \frac{b}{a} \ln \left ( \frac{a}{b} w \right ) \implies w = \frac{b}{a} \exp \left ( \frac{a}{b} u \right ) = \phi(u)$$
which satisfies
$$w_{t} = a \Delta w$$