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Suppose we have $u \in L^2(0,T;H^1(\Omega))$, and $v \in L^2(0,T;H^{-1}(\Omega))$ is the weak time derivative of $u$, so by definition it satisfies $$\int_0^T u(t)\phi'(t) = -\int_0^T v(t)\phi(t)$$ for all $\phi \in C_c^\infty(0,T)$.

My question is how to interpret the RHS. Should I think of $v(t)\phi(t)$ as $v(t)(\phi(t)$) (as $v(t) \in H^{-1}$)?

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    by definition, we have $v(t)\in H^{-1}(\Omega)$ for every $t$ and it satisfies some integrability condition. So, $v(t)$ acts on the space $H^1$, of which $\phi$ is clearly a member. Therefore you interpretation was absolutely right, that $v(t)\phi(t)$ is actually $v(t)(\phi(t))$2012-12-04

2 Answers 2