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.



Uniqueness of solutions to an ODE system
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 $\mathbf{u}(t) = 0$.


