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My quesion involves the weak time derivative. In the book: 'Partial Differential Equations' by Evans the time derivative $u'$ of a function $u: [0,T] \rightarrow H^1_0(U)$ is defined by an element $u_t \in L^2(0,T;H^{-1}(U))$ such that $$ \forall \phi \in C_0^{\infty}([0,T]): \int_{[0,T]} u \phi' dt = -\int_{[0,T]} u' \phi dt $$ Where $u'\phi$ and $u \phi'$ has to be seen as scalar multiplications.

1) First I want to ask why we have to consider $u'(t)$ to belong to $H^{-1}(U)$ for every $t$ and not as an element of $H_0^1(U)$?

2) Secondly I want to ask how to interpret the above equation. The left-hand side belongs to $H_0^1(U)$ and the right-hand side to $H^{-1}(U)$. Does the equality in the definition means that $$ \forall \phi \in C_0^{\infty}([0,T]), \forall w \in H_0^1(U): \int_U \left( \int_{[0,T]} u \phi' dt \right) w \: dx = \left( \int_{[0,T]} u' \phi dt \right) (w) $$ ? Where I've let each side act on $w$ in the natural way.

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For a slightly clearer approach, I recommend consulting the first pages of the book on evolution equations by Cazenave and Haraux. Summarizing, the point is that any function $u\in L^1(0, T;X)$ (here $X$ is a Banach space) can be seen as an $X$-valued distribution (definition (1)). In the space of distributions one can differentiate at will.

Now if $X$ is embedded in a weaker space $Y$, then $X$-valued distributions are embedded into the space of $Y$-valued distributions. In particular, it makes perfect sense to require that $u'\in L^1(0, T; Y)$. This is what happens here with $X=H^1_0$ and $Y=H^{-1}$.

See also John Hunter's notes: https://www.math.ucdavis.edu/~hunter/pdes/ch6A.pdf

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  • $\begingroup$ Thanks! Is the embedding $J: H_0^1 \rightarrow H^{-1}$ in my case: $u \rightarrow Ju$ with $Ju(v) = \int_{U} uv dx$? $\endgroup$ – abcdef Mar 29 '16 at 15:52
  • $\begingroup$ Yes, that's it. $\endgroup$ – Giuseppe Negro Mar 29 '16 at 20:16
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  1. Typically you may study the heat equation $$u_t- \Delta u = 0.$$ We can't always expect the solution $u$ to be twice weakly differentiable, so we ask that $\Delta u$ in fact means $$(\Delta u)(v) = -\int \nabla u \nabla v$$ that is, it is a linear functional $\Delta: H^1_0 \to H^{-1}$ which is bounded. Hence from $u_t = \Delta u$ (the equation) we see that it is sensible to ask $u_t$ also belonging to $H^{-1}$ for a.a. $t$.

  2. The equality makes sense since $H^{1} \subset H^{-1}$, and the equality is n $H^{-1}$.

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When I read the book the first time, I have the same questions.

As for your questions,

2). The basic question is how to understand $H^{1}_0\subset H^{-1}$. The way to understand it is to use $L^2$ space and use $L^2$ inner product $(\cdot,\cdot)$ instead of Riesz representation of $H^{-1}$ using $H^1_0$ inner product: For any $v\in H^{1}_0$, since it is in $L^2$ also, we get $v\in H^{-1}$ defined as $<v,u>_{H^{-1}, H^{1}_0} = (v, u)$ for any $u\in H^1_0$. (In general, it is called gelfand triplpe $V\subset H\subset V'$ with pivot space $H$.) This convection is used in the definition of weak derivative you gave.

1) Take heat equation as an exmaple. $u'$ is in $H^{-1}(U)$ not everywhere, but only a.e. because of the existence theory of ODE (page 376). As for the distribution derivative, one has to use $H^1_0$ and it is defined as $u'\in \mathcal{L}(\mathcal{D}(U), H^1_{0})$, see (1.17) of Non-Homogeneous Boundary Value Problems and Applications: Vol. 1. But to show the derivative of the solution $u'$ is in $L^2$ over time with value in some space, one has to use $H^{-1}$. This is because for Galerkin approximation of heat equation, the energy estimate is on $H^{-1}$ norm (page 378)

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