Hello! I'd like to give some hints in details.
1. The electric current $I$ at the collector is (micro mechanism):
I = n s e v
Here, $n$ is a constant relatively (depends on the material and temperature of the negative emitter), $e$ the constant of charge of an electron, $S$ a constant relatively (the cross section of the tube). Hence
I = I(v)
where $v$ is the independent variable, which is determined by the accelerating voltage in the tube, and the small decelerating voltage(U_2) near the destination collector.
2. The magnitudes of velocity of the electrons initially emitted are not the same, the electrons' emitting speeds form a Maxwell distribution, with one peak (most probably values). Some has the very speed (energy-eV) to be absorbed, while some not.
Now, let's turn to the $U_1$ vs $I_a$ graph.
1. Between a valley (on the left with smaller U_1) and a nearest peak on its right, the slope goes upwards and $I_a$ increases.
Because: Most electrons' speeds (energy) are improper to be absorbed. They keep accelerating. The larger $v$, the larger $I_a$.
2. As to the valleys, where most electrons' speeds are proper and therefore their energy are absorbed, and they could no longer go through the decelerating voltage(U_2) to arrive the collector. However, the minor proportion of the electrons don't suffer from the absorbing of kinematic energy, and could cross the decelerating voltage(U_2) to generate $I_a$ in the destination collector. At the valleys, it is this minor electrons, whose motion are not perturbed, that make a difference. Suppose that the proportion is relatively constant (for a not so long interval of U_1 where there are still several valleys and peaks), then $n(minor)$ is constant, then $I_a(valley)$ depends on $v(valley)$. Just as above, the larger $U_1$, the larger $I_a$ at the valleys. So, minimums increase, too.
