Let u be a solution to $$u_t+cu_x=\epsilon u_{xx}$$ with $\epsilon>0$, $c \in \mathbb{R}$. Show that for any convex function $\mu \in C^2(\mathbb{R})$ the total entropy $\int_\mathbb{R}\mu(u(x,t))\,\text dx$ is a non increasing function in time. You may assume that $\mu(u)$ falls off at $|x|\to\infty $ as fast as you wish.
What I have so far: I multiplied the PDE with $ \mu_{u}$ to get $\mu_{u}u_t+\mu_{u}cu_x=\mu_{u}\epsilon u_{xx}$, which leads to $$ (\mu(u))_t + (\mu(u))_x-\epsilon ((\mu(u))_x)_x+((\mu(u))_x\, \mu_{uu}\,(\mu(u))_x)=0 . $$ And because $\mu$ is convex, $((\mu(u))_x)_x+((\mu(u))_x\,\mu_{uu}\,(\mu(u))_x)\geq 0$ and so i get the inequality $$ (\mu(u))_t + (\mu(u))_x-\epsilon ((\mu(u))_x)_x\leq 0 $$ I have a clue that I need to integrate this with respect to $x$ to get to $$ \frac{\text d}{\text d t}\int_{\mathbb{R}}\mu(u)\,\text dx\leq 0 $$ but why do the other terms disappear? And what exactly does "fall off" mean?
There are several algebra mistakes or typos in OP's derivation. From multiplying the PDE by the derivative $\mu'(u)$ of $u\mapsto \mu(u)$, we obtain $$ \mu(u)_t + c \mu(u)_x = \epsilon \left( \mu(u)_{xx} - \mu''(u)\, {u_x}^2 \right) . $$ Integrating over $x\in \left]-X, X\right[$, we have $$ \int_{-X}^X \mu(u)_t\,\text d x + c\, \left[\mu(u)\right]_{x=-X}^{x=X} = \epsilon\, \left[\mu(u)_x\right]_{x=-X}^{x=X} - \epsilon \int_{-X}^X \mu''(u)\, {u_x}^2\,\text d x . $$ If $\mu(u)$ and $\mu(u)_x$ vanish as $|x|\to +\infty$, i.e. $\mu(u)$ "falls off" fast enough, taking the limit $X\to \infty$ in the previous identity leads to $$ \frac{\text d}{\text d t}\int_{\Bbb R} \mu(u)\,\text d x = - \epsilon \int_{\Bbb R}\mu''(u)\, {u_x}^2\,\text d x < 0 , $$ due to the convexity of $\mu$.