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Consider the following initial value problem for a parabolic PDE : $$\begin{cases} \textrm{div}\big(A\,\nabla u(t,x) + b(x)\, u(t,x)\big) \,=\,\partial_t u(t,x) \quad x\in\mathbb R^d\,,\ t>0 \\[4pt] u(0,x) \,=\, v(x) \end{cases}$$ where $A$ is a constant positive-definite matrix, $b$ is a smooth vector field, $v$ is a suitable initial datum with $$v>0\,,\ \int_{\mathbb R^d} v(x)\,d x=1\,,\ \int_{\mathbb R^d} v(x)\,|\log v(x)|\,d x\,<\infty\,. $$ I am interested in the probability solution $u$ of the problem, i.e., $u>0\,,\ \int_{\mathbb R^d} u(t,x)\,d x=1$ for all $t>0\,$. I know that it exists, is unique, belongs to $C^\infty((0,\infty)\times\mathbb R^d)$, and moreover $$ \int_{\mathbb R^d} |b(x)|^2\,u(t,x)\,d x\,<\infty$$ for all $t\geq0\,$. I would like to find conditions on the initial datum $v$ that guarantee continuity of the entropy at $t=0$, i.e.,

$$\int_{\mathbb R^d} u(t,x)\,\log u(t,x)\,d x \,\xrightarrow[t\to0]{}\, \int_{\mathbb R^d} v(x)\,\log v(x)\,d x \,.$$

I've found only standard results that guarantee $$\int_{\mathbb R^d} |u(t,x)-v(x)|\,d x \,\xrightarrow[t\to0]{}\, 0 $$ which seems to be a weaker condition.

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    $\begingroup$ Something's a bit odd: your equation will not preserve the spatial integral of the solution, in general, since $b(x)$ depends on $x$. If you want the integral to be preserved you either need $b$ to be constant or, more generally, you need to replace $b(x) \cdot \nabla u(t,x)$ with $\operatorname{div}( b(x) u(t,x))$. $\endgroup$
    – Jochen Glueck
    Commented Oct 14, 2021 at 6:35
  • $\begingroup$ @JochenGlueck thank you, there was a mistake in my equation: I've fixed it now $\endgroup$
    – tituf
    Commented Oct 14, 2021 at 8:47

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