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Differentiate the backward equation for $V$ with respect to the stock state, and put $w=V_S$. The product derivative of $a^2/2$ is $a{a}^{\prime }$, givingThe assumed smoothness and bounded derivatives put $w$ in the uniqueness class for the equation defining $U$. Therefore $U=V_S$, and ${\pi }_t=V_S(t,S_t)$ is the option delta. Substituting in the Itô integral from part (a) and using $d{S}_t=a(S_t)d{W}_t$ gives the self-financing representationThe boundedness hypotheses justify both stochastic integrals. This derivation does not divide by $a$ and remains valid even at states with zero diffusion amplitude.

There need not be a weak solution of a stochastic differential equation $d{X}_t=\sqrt{2\theta (X_t)}d{B}_t$, $X_0=0$, even when $\theta$ is a bounded Borel measurable function strictly positive everywhere. For the displayed example, any solution would be a continuous local martingale with $[X{]}_t=2{\int }_0^t\theta (X_s)ds\le 2t$, hence a square-integrable martingale. The occupation-times formula applied to $1_{\{0\}}$ gives $0={\int }_0^t{1}_{\{X_s=0\}}d[X{]}_s=2{\int }_0^t{1}_{\{X_s=0\}}ds$. Since $\theta (x)\le x^2$ away from zero, $\mathbb{E}{X}_t^2\le 2{\int }_0^t\mathbb{E}{X}_s^{2}ds$. The Gronwall inequality makes $X\equiv 0$, a contradiction. Thus a diffusion amplitude positive at a single problematic point cannot repair arbitrarily slow neighbouring motion. A local positive lower bound, as in the Brownian time change for a variable diffusion speed, rules out this example.