I seem to have some very basic misunderstanding about the following probably simple problem when I approach it naively, but I'm unsure where (please excuse my ineptitude...).
We are given an ODE of the form $\dot y_t = (\dot \alpha_t + f(\alpha_t))y_t$ with $y\in [0,\infty)$ and $t\in [0,T]$. The goal is to find $\alpha^*$, such that $-y_T^{\alpha^*}$ (the solution under this control) is maximized for any starting value $y_0$. Then, to get rid of the control derivative, it seems reasonable to set $Y = (y, \alpha)$ and $\beta := \dot \alpha$ and instead look at the two-dimensional ODE $$\dot Y_t = \begin{pmatrix} \beta + f(Y_{2,t}) \\ \beta \end{pmatrix}.$$
The value function for this new problem is then $V(t,x_1) = \sup_{\beta\in \mathcal A(t)} -Y_{1,T}^{t,\beta}(x_1)$ (it does not depend on $x_2$).
This leads to the HJB equation
$$0 = - \partial_t V(t,x_1) - \sup_{b\in \mathbb R} \partial_{x_1} V(t,x_1)(b + f(x_2)) = - \partial_t V(t,x_1) - \partial_{x_1} V(t,x_1) \inf_{b\in \mathbb R} b + f(x_2).$$
This seems non-sensical, though. It's clear that $\beta$ must be bounded if its assumed to be continuous, but even then, when we restrict to the infimum over $[\inf \beta, \sup \beta]$ instead of $\mathbb R$ this would suggest that $\dot \alpha^*$ is constant, which is much simpler then I would expect and completely ignores $f$.
Where am I going wrong with this?
The problem is $$ \min_{\alpha} Y_T $$ subject to $\dot{Y}_t = Y_t (\dot{\alpha}_t + f(\alpha_t))$ with $t \in [0,T]$, $y_0$ given, $Y_t \ge 0$.
Well, $$ \dfrac{\dot{Y}_t}{Y_t} = \dot{\alpha}_t + f(\alpha_t) $$ so that $$ D_t \log(Y_t) = \dot{\alpha}_t+ f(\alpha_t) $$ and $$ \log(Y_t) = \int_{0}^{t} \dot{\alpha}_z + f(\alpha_z)dz + K $$ or $$ Y_t = exp \left\lbrace \int_0^{t} \dot{\alpha}_z + f(\alpha_z)dz \right\rbrace e^K. $$ From the boundary conditions, $Y_0 = e^0 e^k$, so $e^k = Y_0$.
So pick $\alpha_z$ to minimize $$ Y_T = Y_0 * \exp \left\lbrace \int_0^{T}\dot{\alpha}_z + f(\alpha_z)dz \right\rbrace = Y_0 * \exp \left\lbrace \alpha_T - \alpha_0 + \int_0^{T} f(\alpha_z)dz \right\rbrace $$
Can we achieve the minimum of $Y_T = 0$? Only if $\int_0^T \dot{\alpha}_z + f(\alpha_z)dz = -\infty$. If you can do that, the problem is somewhat trivial, since it looks like $\alpha_T \rightarrow -\infty$. In fact, you are picking $\{\alpha_t\}_{t \in [0,T]}$ to minimize $$ \alpha_T - \alpha_0 + \int_0^{T} f(\alpha_z)dz. $$
Without more information on $f()$ and the constraint set for $\alpha_t$ or $\dot{\alpha}_t$, I can't really say much else. As usual, necessary conditions are not sufficient, and verifying them in optimal control settings can be painful (e.g., Arrow-Mangasarian Sufficient Conditions). I was going to suggest looking up Bang Bang solutions for linear controls and the Optimal Rocket Problem, which has a discontinuous solution in a norm maximization problem, which might give you some ideas.