$\nabla u+u \nabla V=u \nabla\left(\log u+V\right)$

Question

In the book Entropy Methods for Diffusive Partial Differential Equations, §2.1, p. 20 , the steady state solution for

$$u_{t}=\operatorname{div}(\nabla u+u \nabla V) \quad in \quad \mathbb{R}^{d}, t>0, \quad u(0)=u_{0}$$

is derived by using $$0= \nabla u_{\infty}+u_{\infty} \nabla V=u_{\infty}\nabla\left( \log u_{\infty}+V\right).$$

Can somebody explain the transformation after the second equal sign?

Answer 1

Note that $$ \frac{\partial}{\partial x_i} \log u_\infty(x)=\frac{1}{u_\infty(x)}\frac{\partial}{\partial x_i} u_\infty(x) $$ thus $$ \nabla \log u_\infty(x)= \frac{1}{u_\infty(x)} \nabla u_\infty(x) $$ and this implies $$ u_\infty(x)\nabla(\log u_\infty(x) +V(x))= \nabla u_\infty(x) + u_\infty(x)\nabla V(x). $$