Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given the Gelfand triple $V \subset H \subset V^*$, we for for $y \in L^2(0,T;V)$ also $y \in L^2(0,T;V^*)$ and thus $y \in L^1(0,T;V^*)$.

I understand this because if $y(t) \in V$ then $y(t) \in V^*$ since $V$ is a subset. The compactness of $[0,T]$ gives us the $L^1$ result.

Therefore, it makes sense to require that $y$ has weak derivative $y_t \in L^1(0,T;V^*)$, and to impose the additional condition $y_t \in L^2(0,T;V^*)$.

Why does it therefore make sense? Please can someone explain. Thanks.

share|cite|improve this question

I find the remark about going through $L^1$ unnecessary. It would be more appropriate for the strong derivative (limit of $(y(t+h)-y(t))/h$ as $h\to 0$), which naturally takes values in the same vector space as $y$. But the weak derivative is different.

Weak derivatives are defined by an axiomatic version of integration by parts: $y_t$ is a function $g$ such that $\int_0^T g\varphi=-\int_0^T y\varphi'$ for all "test functions" $\varphi$. It's up to us to decide what the space of allowable "test functions" should be; the smaller it is, the more permissible is our definition of derivative. The natural choice in this setting is $\varphi\in C^{\infty}_0(0,T;V)$, because $V$ is the smallest Hilbert space we have here. Then $\int_0^T g\varphi$ makes sense whenever $g\in L^2(0,T, V^*)$ because the integral is parsed as $\int_0^T \langle g(t),\varphi(t)\rangle\,dt$, with angle brackets indicating the pairing between $V^*$ and $V$.

Here is another way to see why $y_t$ needs to take values in $V^*$. This setup is supposed to help us reformulate the heat equation $y_t=\Delta_x y$ (and other parabolic problems) in terms of functional analysis. According to the equation, first-order time derivative should belong to the same function space as the Laplacian. The appropriate Gelfand triple here is $H^{1}\subset L^2\subset H^{-1}$ in which the Laplacian is an operator from $V=H^1$ to $V^*=H^{-1}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.