Let $\Omega = (0,1)$. Suppose $\rho \in L^{\infty}(0,T; L^{\infty}(\Omega))$ satisfies the following weak formulation of the continuity equation

$$ \int_{0}^{T} \int_{\Omega} \rho(t,x) \partial_{t}\varphi + \rho(t,x)u(t,x) \partial_{x}\varphi(t,x)~dxdt = 0, $$ for all $\varphi \in C^{\infty}_{c}((0,T) \times \Omega)$. Also assume $\rho u \in L^{2}(0,T; L^{2}(\Omega))$.

Apparently in the book I am reading it follows that the function $$ t \mapsto \int_{\Omega} \rho(t,x)\varphi(x)~dx$$ is absolutely continuous in $[0,T]$. But I don't see why.

My attempt is to show that $\partial_{t} \int_{\Omega} \rho \varphi~dx \in L^{1}(0,T)$. But this would require us to integrate by parts in the first term of the weak formulation. However $\rho$ is not differentiable so we cant do that in the classical sense. But if we understand the derivative in the weak sense then we could say

$$ \int_{\Omega} \partial_{t} \rho \cdot\varphi~dx = \partial_{t} \int_{\Omega} \rho \varphi~dx \in L^{1}(0,T)$$ using the weak formulation and integrability of $\rho u$. But then I am unsure if this swapping of the derivative and integral can be justified rigorously or not.


1 Answer 1


I believe the following works (but be sure to check it!):

Define, for fixed $\varphi_2\in C_c^\infty(\Omega)$, $$ F(t):=\int_{\Omega} \rho(t,x)\varphi_2(x)\, dx $$ We want to show that $F$ is absolutely continuous on $[0,T]$; we will in fact show that $F\in W^{1,2}(0,T)$.

First we note that $F$ is bounded: For a.e. $t\in (0,T)$, $$ |F(t)|\leq \Vert \rho\Vert_{L^\infty((0,T);L^\infty( \Omega))} \Vert \varphi_2\Vert_{L^1(\Omega)}<\infty. $$

Next, to show that $F$ has a weak derivative in $L^2(0,T)$, it's enough to show (this is in Brezis' functional analysis book) that there exists a constant $C>0$ such that for every $\varphi_1\in C_c^\infty(0,T)$ we have $$ \left|\int_0^T F(t)\varphi_1'(t)\, dt \right| \leq C\| \varphi_1\|_{L^2(0,T)}. $$ However, note that if we define $\varphi(t,x):=\varphi_1(t)\varphi_2(x)$, then $\varphi$ is an admissible test function in the weak formulation and $$ \int_0^T F(t)\varphi_1'(t)\, dt = \int_0^T\int_\Omega \rho(t,x)\partial_t \varphi(t,x)\, dx dt = -\int_0^T\int_\Omega \rho(t,x)u(t,x)\partial_x\varphi_2(x)\varphi_1(t)\, dxdt. $$ Using the Cauchy-Schwarz inequality, we see (remember that $\rho, u, \varphi_2$ are fixed!) $$ \left|\int_0^T F(t)\varphi_1'(t)\, dt \right|\leq \| \rho u\|_{L^2((0,T);L^2(\Omega))} \| \partial_x\varphi_2\|_{L^2(\Omega)}\| \varphi_1\|_{L^2(0,T)}=:C\| \varphi_1\|_{L^2(0,T)}. $$

  • $\begingroup$ Yes I understand and agree, thank you! I did not know this result from Brezis. $\endgroup$
    – duelspace
    Sep 13, 2023 at 6:03

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