Uniqueness of solutions to an ODE system For each $i$ (up to infinity), let $u_i \in C^1(0,T)$ satisfy
$$\frac{d}{dt}u_i(t) + \sum_{j=1}^\infty b(t;w_j,w_i)u_j(t) = 0$$
$$u_i(0) = u_i(T)$$
where $b(t;\cdot,\cdot)$ is an inner product on some (infinite-dimensional) Hilbert space $H$ (which has an inner product $(\cdot,\cdot)_H$) for each $t$ and $w_i$ are basis functions on $H$.
Does anyone know how show that there is only one solution $u_i$ to this system of equations? Clearly $u_i(t) \equiv 0$ for all $i$ is a solution. I want to know that it is unique.
 A: I'll presume that $H$ is finite dimensional. Write $b(t;\mathbf{u},\mathbf{v}) = (\mathbf{u},B(t)\mathbf{v})_H$ for some positive symmetric linear operator $B(t)\colon H\to H$. $B(t)$ is positive symmetric because $b(t;\cdot,\cdot)$ is itself an inner product. Then your system of equations is equivalent to $$\frac{\partial}{\partial t} \mathbf{u}(t) = - B(t)\mathbf{u}(t),$$ where $\mathbf{u}(t) = \sum_i u_i(t) w_i$, with elementary consequence $$\frac{\partial}{\partial t} \mathbf{u}^2 = - (\mathbf{u},B(t)\mathbf{u}),$$ where $\mathbf{u}^2 = (\mathbf{u},\mathbf{u})_H$. By diagonalizing $B(t)$ we can find positive functions $c(t)$ and $C(t)$ such that $c(t) \mathbf{v}^2 \le (\mathbf{v},B(t)\mathbf{v}) \le C(t) \mathbf{v}^2$, for any $\mathbf{v} \in H$. In other words, there is the inequality $$-c(t) \ge (\mathbf{u}^2)^{-1} \frac{\partial}{\partial t} \mathbf{u}^2 = \frac{\partial}{\partial t} \log \mathbf{u}^2 \ge -C(t).$$ From here, it is an easy conclusion that $\mathbf{u}^2(t)$ must decrease with time by a factor between $\exp(-\int_0^t c(t) dt)$ and $\exp(-\int_0^t C(t) dt)$. On the other hand, your boundary conditions imply that the $\mathbf{u}^2(t)$ is the same at times $t=0$ and $T$. Hence the only solution that satisfies it is $\mathbf{u}(t) = 0$.
A: If you assume that $b(t;\mathbf{v},\mathbf{v}) \approx (\mathbf{v},\mathbf{v})_H$ for every $\mathbf{v}\in H$, and if you assume that the vector $\mathbf{u} = \sum u_i(t) w_i \in H$, then Igor's answer goes through essentially unchanged in the infinite dimensional case. 
But let me give you a counterexample if the assumptions are not verified. 
Let the index $i$ run from $0, 1, \ldots$. 
Let $b(t; w_i, w_j)$ be given by the matrix
$$ \begin{pmatrix}
1 & -1 \\ 
-1 & 2 & -1 \\
& -1 & 2 & -1 \\
&& -1 & 2 & -1 \\
&&& \ddots & \ddots & \ddots\end{pmatrix} $$
Assume that $w_i$ forms an orthonormal basis of $H$. We have that
$$ b(\mathbf{v}, \mathbf{v}) = v_0^2 + 2 \sum_{i = 1}^{\infty} v_i^2 - 2 \sum_{i = 0}^{\infty} v_{i} v_{i+1} = \sum_{i = 0}^\infty (v_i - v_{i+1})^2  $$
Since the right hand side is non-negative, we see that the bilinear form is positive definite, since 
$$ b(\mathbf{v},\mathbf{v}) = 0 \implies v_i = v_{i+1} $$
which for $\mathbf{v} \in H$ requires $v_i \equiv 0$. 
So $b(t) = b$ give inner products. 
But the same argument above also shows that for any constant $c$, 
$$ u_i(t) \equiv c \quad \forall i $$
is a solution to the infinite system of ODEs
$$ \partial_t u_i = \sum b(t;w_i,w_j) u_j $$
and so you don't have uniqueness of your solution. Note, of course, the vector $\mathbf{u} = \sum u_i w_i \not\in H$ whenever $c \neq 0$. 
