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). 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?


1 Answer 1


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.

  • $\begingroup$ Thanks for the demonstration. Do you have any idea, if I had relevant assumptions, what general techniques I could possibly use to show the solution is unique? (provided such techniques don't rely on the specific assumptions too much...) $\endgroup$
    – D. Dring
    Oct 9, 2014 at 7:59
  • $\begingroup$ The general idea to proofs of that type you want is the construction of a monotonicity law. In the question you linked to in your post, the answer Igor gave was to show that your system is dissipative, in that the presence of the $a$ term forces the energy to be monotonically decreasing. While energy may not work for your system, you should still try to find a quantity that is monotonic, which will contradict the periodicity of the solutions. $\endgroup$ Oct 14, 2014 at 7:24

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