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Let us consider a system of ordinary differential equations: \begin{equation} \frac{d f}{dt}=F(f(t)) \text{ with } f(0)=v \end{equation} where $f:[0,T] \to \mathbb{R}^n$ is a vector-valued function on a compact interval and $F :\mathbb{R}^n \to \mathbb{R}^n$ is assumed to be smooth (not necessarily linear). $v \in \mathbb{R}^n$ is some specified vector.

Let us assume that the above ODE has a unique strong solution on whole of $[0,T]$, denoted as $f_{str}$.

Now, I am aware that a weak solution is defined to be any $f:[0,T] \to \mathbb{R}^n$ such that \begin{equation} -\int_{0}^T f(t) \cdot g'(t) dt = \int_0^T F(f(t))g(t) dt+v \cdot g(0) \end{equation} for any differentiable $g : [0,T] \to \mathbb{R}^n$ with $g(T)=0$.

Now, I wonder if $f$ is a weak solution satisfying the above formula, then if it is in fact equal to the unique strong solution $f_{str}$ and thus differentiable with respect to $t$.

I suspect this is the case, but I am not sure..Could anyone please help me?

Keith
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    I suspect we need uniqueness of a weak solution for this statement to hold true, and for this we might need some more regularity properties of $F$, c.f. https://math.stackexchange.com/questions/922710/existence-theorem-on-weak-solutions-of-ordinary-differential-equations?rq=1. – stange May 28 '23 at 22:18

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