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I take course in PDE and I'm a little bit puzzled with space $W^{1,p}(0,T,X)$.

Evans defines in §5.9.2 $W^{1,p}(0,T,X)$ as follows $$ W^{1,p}(0,T,X) = \{ u \in L^p(0,T,X): u'\in L^p(0,T,X)\} $$ but solution $u$ to the parabolic system is defined in §7.1.1 as $$ u \in L^2(0,T,H^1_0) \qquad u' \in L^2(0,T,H^{-1}) $$ so $u\notin W^{1,2}(0,T,H^1_0)$. Why does he define $ W^{1,p}(0,T,X)$ as he does when the solution does not belong to this space?

And with Evans definition we have $$ W^{1,p}(0,T,X) \subseteq C(0,T,X) $$ I will address to this later.

In our course we define $W^{1,p}(0,T,X)$ differently $$ W^{1,p}(0,T,X) = \{ u \in L^p(0,T,X): u'\in L^{p'}(0,T,X^*)\} $$ definition of solution is the same, so for us the solution $u$ does belong to the space $W^{1,p}(0,T,X)$. Is it natural that the time derivative belongs to $X^*$ or is it just technical condition?

We also define rigged Hilbert space(Gelfand triple) $X\subseteq H \subseteq X^*$. It seams to me that you have to define rigged Hilbert space in order to make sense of our definition of $W^{1,p}(0,T,X)$. Because weak time derivative is defined by following equality $$ \int_0^T u'(t) \psi(t) = - \int_0^T u(t) \psi'(t) dt \qquad \psi \in D(0,T) $$ where by definition left hand side is in $X^*$ and right hand side is in $X$, so to check equality you have to use the embedding of $X$ in $X^*$ on right hand side. The question is, do you really need to rigged Hilbert space to make sense of our definition of $W^{1,p}(0,T,X)$?

And with our definition apparently the embedding in continuous functions does not hold in original form $$ W^{1,p}(0,T,X) \not\subseteq C(0,T,X) $$ but this holds $$ W^{1,p}(0,T,X) \subseteq C(0,T,H). $$ Is there a function $u\in W^{1,p}(0,T,X)$ which is not $C(0,T,X)$?

I was thinking about function in the form $$ u(t,x) = t^\alpha sin\left(\frac{x^\beta}{t\gamma}\right) $$ where I would pick $\alpha,\beta,\gamma \geq 0$ so that $u(t)\rightarrow 0$ as $t\rightarrow 0$ in $L^2$ but not in $W^{1,2}$, but I think it is not possible to pick such $\alpha,\beta,\gamma \geq 0$. Thus something more sophisticated is needed.

You don't have to answer to all of the questions, any answer which would shed a little bit of light on space $W^{1,p}(0,T,X)$ and its two different definitions.

tom
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1 Answers1

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(I pick $p=2$ for simplicity).

  1. It is natural to require the time derivative $u' \in L^2(0,T;X^*)$ because typically you have a parabolic equation of the form $u' + Au =0$ where $A$ is an elliptic operator like $A=-\Delta$. Then $\langle -\Delta u, v \rangle := \int \nabla u \nabla v$ implies that $-\Delta u$ lies in the dual space of $L^2(0,T;X)$, i.e., $-\Delta u \in L^2(0,T;X^*)$. Then rearranging the equation shows that $u' = -Au$ so that $u' \in L^2(0,T;X^*)$ too.

  2. Yes you do need the rigged Hilbert triple to make sense of "your $W^{1,2}(0,T;X)$." For example you need it to make sense of formulae like $$\frac{d}{dt}(u(t),v)_H = \langle u'(t), v \rangle = (u'(t), v)_H$$ whenever $u'(t) \in H$.

  3. I don't know why Evans defines the space like he does. Maybe because it avoids the need for discussing the Hilbert triple because it is simpler.

C_Al
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