# A question about PDE argument involving monotone convergence theorem and Sobolev space

I'm reading this paper. In it there is the following argument (see page 240). Firstly, what precisely does the author mean by the displayed equation after 66? The PDE in (65) only holds weakly.. $$b(\overline{v}_n)$$ is only in $$H^1(0,T;H^{-1})\cap L^\infty(0,T;L^s)$$ and $$\overline{v}_n \in L^2(0,T;H^1)$$. He just integrates it in time.. I don't follow.

Secondly, I am lost with the spaces when he says that $$\int_0^T \overline{v}_n(s)\;ds$$ is a Cauchy sequence in the space $$L^\infty(0,T;W^{1,r}_0)$$. And then he uses the monotone convergence theorem to say that $$\overline{v}_n$$ convreges strongly in $$L^1(\Omega \times (0,T))$$ to some $$\overline{v}$$. Any explanation appreciated.

I posed this here on MSE but did not receive an answer.

• Yes, also taking into account initial data.
– riem
Jul 7, 2014 at 19:31
• There's no issue with applying the comparison principle. I think you misread my OP. My problem is with the displayed equation after (66). The PDE does not hold pointwise a.e. because it isn't smooth enough (the exact spaces it lies in are written in the paper just above where I cut the image but I reproduced it in my post) so I don't understand what the author did.
– riem
Jul 7, 2014 at 20:38
• What's wrong with writing it in a distributional sense? $b(u)$ is in the right space $L^\infty(0,T,L^1(\Omega)$ it is written before. Jul 7, 2014 at 20:54
• How does something like $\int_0^t f_n(s)$ make sense though? I have never seen any distributional equation with terms like that.
– riem
Jul 8, 2014 at 20:56

Hy

First: take $\varphi_n(t) w(x)$ where $\varphi(t)\in C_0([0,T))$ and $w\in H^1_0(\Omega)$. If $\varphi_n$ is well chosen and converges to the characteristic function of the set $[0,t]$ you will understand the formulation.

Second: to be short, you have to mix the fact that $\int_0^t f_n(s,x)$ is a Cauchy sequence in $L^1(\Omega)$, that ($n$ fixed) $\int_0^t f_n(s,x)$ is continuous in $L^1(0,T; L^1(\Omega))$ and the De Giorgi regularity result. Since $\bar{v}_n$ is monotonically inscreasing in $n$ you have almost convergence to some $\bar{v}$ and then proving that $\bar{v}_n$ converges strongly to $v$ in $L^1(0,T;L^1(\Omega))$ is equivalent to prove that $\int_0^T\int_\Omega \bar{v}_n dxdt$ goes to $\int_0^T\int_\Omega \bar{v} dxdt$. From the previous result (the Cauchy sequence in $L^\infty(0,T;W^{1,q}_0(\Omega))$) you obtain the result.

• Thank you for answering, let me think about it for a bit.
– riem
Jul 15, 2014 at 16:31
• Do you have any references for arguments like this (eg in some paper)?
– riem
Jul 18, 2014 at 7:35
• No, I have no reference in mind. I think it is classical and with the explanation of the paper you can complete all the arguments.
– O.G.
Jul 18, 2014 at 12:06