From the following model problem:
$$\frac{\partial u}{\partial t}+V \cdot \operatorname{grad}(u)=\operatorname{div}(K \cdot \operatorname{grad}(u))+f(x, t)$$
where $K$ characterize the diffusion, and $V$ is transport speed of the scalar field $u$
I can't find by using the change of variable $\tilde{u}=u(\tilde{x}, \tilde{t})$ such as $\tilde{x}=x-Wt$ and $\tilde{t}=t$ why $\tilde{u}$ verify
$$\frac{\partial \tilde{u}}{\partial t}+(V-W) \cdot \operatorname{grad}(\bar{u})=\operatorname{div}(K \cdot \operatorname{grad}(\tilde{u}))+f(\tilde{x}, \tilde{t})$$
This is a travelling wave Ansatz. Using the chain rule for $u = \tilde u (\tilde x,\tilde t)$, we find \begin{aligned} \partial_t u &= \frac{\partial \tilde u}{\partial \tilde x} \frac{\partial \tilde x}{\partial t} + \frac{\partial \tilde u}{\partial \tilde t}\frac{\partial \tilde t}{\partial t} = \partial_{\tilde t} \tilde u - W \partial_{\tilde x} \tilde u \end{aligned} and similarly $\partial_x u = \partial_{\tilde x} \tilde u$. Thus, the gradient is unchanged, only the time derivative is modified. The proposed PDE then follows directly.