Let $\mathcal{E}/\mathbb{Q}(t)$ be given by $$y^2=x^3+A(T)x+B(T)$$ for some $A(T),B(T)\in\mathbb{Q}[T]$ and assume $\mathcal{E}$ is non-isotrivial (the $j$-invariant $\frac{6912 A(T)^3}{4A(T)^3 + 27B(T)^2}$ is not constant). Does there necessarily exist $t\in\mathbb{Q}$ such that the Mordell-Weil group of $\mathcal{E}_t/\mathbb{Q}:y^2=x^3+A(t)x+B(t)$ has positive rank?

An unconditional proof or explicit counterexample would be wonderful, but if that's not possible, I would be okay with a conditional proof assuming standard conjectures (e.g. BSD), or a proof for as wide a class of curves as possible, or a discussion of some properties a hypothetical counterexample would have.

Here are my thoughts so far:

- We expect that for "most" families, at least $50\%$ of all specializations (ordering $t$ by height) are positive rank. One result in this direction is by Helfgott, who shows (assuming some standard conjectures) that if $\mathcal{E}$ has a finite place of multiplicative reduction, then half of the specializations $\mathcal{E}_t$ ordered by height have root number $-1$, and therefore have positive rank assuming the parity conjecture.
- In contrast to Helfgott's work, there exist non-isotrivial families of curves with constant root number: for example, $W(\mathcal{E}_t)\equiv -1$ due to Rizzo and $W(\mathcal{E}_t)\equiv 1$ due to Bettin, David, and Delaunay (EDIT: These results only hold for all $t\in\mathbb{Z}$, not $t\in\mathbb{Q}$, so are not directly relevant; see comments for discussion). It turns out that in both of these families, $100\%$ of the specializations have positive rank (again assuming the parity conjecture), but it certainly may be possible to have a family with generic rank $0$ and constant root number $1$. Even in this case, though, I would still expect some rank $\geq 2$ specializations.
- Joe Silverman conjectures that in fact any non-isotrivial family should have
*infinitely many*positive rank specializations, but notes that it's not clear how one would prove this. My question is weaker (I'm only asking for a single specialization), and perhaps naively I would hope this makes it more tractable. - For any particular family, it is often possible to explicitly construct a rank $1$ subfamily (as Joe mentions in the answer I cited above, and Siksek demonstrates). There likely isn't any way to turn this into a general construction guaranteed to work for all families (if there were, it would prove Joe's conjecture).