Let $V \subset H$ be a continuous, compact and dense embedding with $V$ and $H$ Hilbert spaces.


Let $\beta\_j:[0,T] \to \mathbb{R}$ be functions for each $j$, and let $v\_j$ be a basis of $V\_0$.


Suppose that for each $j$, we have
$$\frac{d}{dt}h(t;\sum\_{i=1}^\infty \beta\_i(t)v\_i, v\_j) + a(t;\sum\_{i=1}^\infty \beta\_i(t)v\_i, v\_j) = 0$$
$$\beta\_j(0) = \beta\_j(T)$$
where $h(t;\cdot,\cdot):H \times H \to \mathbb{R}$ is an inner product on $H$ for each $t$, and $a(t;\cdot,\cdot):V \times V \to \mathbb{R}$ is an inner product on $V$ for each $t$.


I wish to show that this equation and this information implies that $\beta\_j = 0$ (a.e) for all $j$.


I have tried a lot of things. The problem is I don't know how to tackle the first term in the equation, since the basis cannot be orthonormal wrt. $h(t;\cdot,\cdot)$ since $t$ is variable. So I cannot use the results of this thread ([Uniqueness of solutions to an ODE system](https://mathoverflow.net/questions/180506/uniqueness-of-solutions-to-an-ode-system)). And I can't multiply the equation by $\beta\_i$ and sum up because I wouldn't know what to do with the first term.


Does anyone have any ideas? 



ODE system has zero as the only solution?
You can't prove it because it ain't true. 


Let $V = H = \mathbb{R}$. Let $h(t; x,y) = \frac{xy}{t+1} $ and $a(t;x,y) = \frac{xy}{(t+1)^2}$. Let $v\_1 = 1$ your equation reads 


$$ \frac{d}{dt} \frac{\beta(t)}{t+1} + \frac{\beta(t)}{(1+t)^2} = 0 $$


which happens to be solved by any $\beta \equiv C$ and verifies $\beta(0) = \beta(T)$. 


You will need more conditions if you want what you want to be true. 


