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Hi could anyone help with a solution for problem 7 Evans PDE chapter 5? enter image description here

I think it is basically about checking which $p$ allows $$\int_{\Omega} |u|^p dx+\int_{\Omega}|Du|^p dx<\infty$$ ? But I tend to generate messy output as I always do...

Please help...

Boar
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math101
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1 Answers1

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You first have to check that $u$ has a weak derivative. You have an obvious candidate for the weak gradient (just take the gradient separately in the four pieces), and you just have to check that it works. You might also have a theorem that saves you from calculations at this point, having a weak derivative should still be checked somehow.

Then you have explicit formulas for $u$ and $\nabla u$. Observe that $|u|$ and $|\nabla u|$ are bounded (actually they are both bounded by $1$).

You either find an upper bound for the integrals you have or just state that $u\in L^p(\Omega)$ and $|\nabla u|\in L^p(\Omega)$. For $p=\infty$ the norm does not have an integral representation, so you have to resort to the second one. If you do the integrals, you don't have to do them explicitly and messily; a very rough estimate will suffice.

Can you finish with these ideas?

Checking weak differentiability: If you know that the pointwise minimum of two $W^{1,p}$ functions is again in $W^{1,p}$, you can use that. You can also do it by hand. In the four pieces (where $u$ is defined), it is easy to calculate the strong gradient $\nabla u$. It takes the four values $(\pm1,0)$ and $(0,\pm1)$. Now you have $\nabla u$ defined almost everywhere. You will just have to check that for any $\eta\in C_0^\infty(U)$ you have $$ \int_Uu\nabla\eta=\int_U\eta\nabla u. $$

Joonas Ilmavirta
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  • Step 1, I have worked out $Du=v=1$ for the first term, and $v=-1$ for the rest of the 3 terms. By inspection that $u$ is clearly bounded by 1. But, I am not sure what you meant in your second last paragraph...could you illustrate the idea more explicitly? I am inexperienced of writing this sort of thing in full mathematics sentences... – math101 Jul 10 '15 at 14:48
  • @math101, you have a function $u\in L^1$ that has a weak gradient $Du\in L^1$. (Note that the weak gradient is vector valued, so $Du=1$ does not make sense. In the first piece $DU=(-1,0)$.) Since $u\in L^p$ and $Du\in L^p$, then by definition of the Sobolev space, $u\in W^{1,p}$. This holds for any $p$, so $u\in W^{1,p}$ for all $p\in[1,\infty]$. – Joonas Ilmavirta Jul 10 '15 at 15:05
  • Yes I just realised $\mathbb{R}^2$, so the next 3 pieces should all be $Du=(1,0). The rest of your comment also make sense to me now. Appreciate that... – math101 Jul 10 '15 at 15:12
  • By the way, I have another vector calculus type question urgently need some help but no one answer, http://math.stackexchange.com/questions/1355970/a-vector-calculus-problem-when-coping-with-problem-9-chapter-5-evans but never mind if you are not interested – math101 Jul 10 '15 at 15:15
  • @math101, no, the weak gradient is different in the four pieces. It's values are (-1,0), (1,0), (0,-1) and (0,1), in the order of the definition. I don't have the time to look into your other question now (it did look like a fine question, though). I might have a look later, but I won't promise anything now. – Joonas Ilmavirta Jul 10 '15 at 15:33
  • okay...I think I am lost, so how did you get those three different...feel free to answer any of my question at any time of your convenience, if you wish. It is almost 2am here, will be back in the next 10 hours... – math101 Jul 10 '15 at 15:39
  • @math101, I was just calculating the gradient in the usual (classical, strong, whatever you call it) way in each of the components. In the third component we have $u(x)=1-x_2$ so $\nabla u(x)=(0,-1)$. The others are similar. – Joonas Ilmavirta Jul 10 '15 at 15:42
  • llmavirta, I was silly enough...trying to plug things into the weak derivative definition (formula) $\int_U u\phi'dx=-\int_0^2 v\phi dx$. So, to compute the weak derivative here, all I need is $Du=v?$ – math101 Jul 10 '15 at 23:37
  • sorry, i wasn't able to get name display property here... – math101 Jul 11 '15 at 06:26
  • @math101, it seems that you should perhaps ask a separate question about the weak derivative (for this function or another one and link here if needed). A complete answer would not be very short, so I would prefer not to include it here. I added some details to the end of my answer. – Joonas Ilmavirta Jul 11 '15 at 08:09
  • I am not sure what separate question I need, this is currently the only question relates to weak derivative. What bothering me now are my other unanswered questions… – math101 Jul 11 '15 at 08:14
  • @math101, if the thing is clear, then there is no need for a new question. I hope my answer was of help to you. :) – Joonas Ilmavirta Jul 11 '15 at 09:04
  • Your answer seems neat…But I am struggling with some other bugs so I will check again later :-) – math101 Jul 11 '15 at 09:13
  • Actually, I am looking at https://sunlimingbit.wordpress.com/2012/11/25/one-example-related-to-weak-derivative-2/ a bit confused when it reached the last line... – math101 Jul 11 '15 at 14:49
  • @math101, the blog post seems well explained and clear. The blog shows that the function defined in the last equation indeed satisfies the definition of weak partial derivative in (1). If this needs further clarification, I would suggest asking a new question and explaining in detail what is it that confuses you. This comment discussion is getting quite long. – Joonas Ilmavirta Jul 11 '15 at 15:02
  • I still have some confusion in the posted answer found in the blog, could you check my newly posted question http://math.stackexchange.com/questions/1358450/computing-weak-derivatives-on-an-open-square? – math101 Jul 12 '15 at 14:59
  • @math101, that looks like a good question. I might look into that if I have the time; I only had the time to read and vote now. – Joonas Ilmavirta Jul 12 '15 at 15:06
  • i have ticked your answer. By the way, have you got a moment to look at my other post last night? – math101 Jul 13 '15 at 14:49
  • @math101, I did give it a look, but I might not have the time to write a solution, as it may take a while to do so properly. – Joonas Ilmavirta Jul 13 '15 at 20:38