# Why $\frac{d}{dt}(u(t),v(t))_{H} = \langle u'(t), v(t)\rangle_{V^*,V} + \langle v'(t), u(t)\rangle_{V^*,V}$

For a Hilbert triple $V \subset H \subset V^*$, and for $u,v \in L^2(0,T;V)$ with $u',v' \in L^2(0,T;V^*)$

Why is it true that

$$\frac{d}{dt}(u(t),v(t))_{H} = \langle u'(t), v(t)\rangle_{V^*,V} + \langle v'(t), u(t)\rangle_{V^*,V}?$$

I just saw this result on MO and tried to look in Evans but he does it for the $L^2$ case. Can someone tell me the proof? Thanks.

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Have you tried following the proof of $(uv)'=u'v+v'u$ and if so, where have you encountered a problem? – user53153 Jan 20 '13 at 18:00
@5PM But $uv$ is not defined in general, right? We don't know what kind of space $H$ is. – george.s Jan 20 '13 at 18:02
I meant the ordinary calculus proof for real valued functions $u,v$. It proceeds through things like $u(t+h)v(t+h)-u(t)v(t) = (u(t+h)-u(t))v(t)+(v(t+h)-v(t))u(t+h)$, which apply equally well to all bilinear forms, not just scalar multiplication. – user53153 Jan 20 '13 at 18:03
@5PM Thankyou, I will try it out. – george.s Jan 20 '13 at 18:31
@5PM I tried it and I get the RHS with the pairing turned into inner product in $H$. I think $\frac{d}{dt}$ is weak derivative, so we cannot do this way maybe. – george.s Jan 20 '13 at 18:37

## 1 Answer

I would suggest the following:

1. Prove the equation for $u,v \in H^1(0,T, V)$.

2. Recall that $$(u,v)_H = (i \, u, v)_{V^*,V}$$ for all $u,v \in V$. Here, $i : V \to V^*$ is the embedding from $V$ into $V^*$ defined via the Hilbert triplet.

3. Now, a density argument shows that the formula holds for $u,v$ in the desired space.

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