For a free particle, the Hamiltonian is
$$H(p)=\frac{p^2}{2m}.$$
The corresponding H-J equation thus can be written as
$$\frac{1}{2m} \left(\frac{\partial S(q,t)}{\partial q}\right)^2=- \frac{\partial S(q,t)}{\partial t}$$
To solve the partial differential equation of $S(q,t)$, i assume there are two possible forms of solution:
(a) $S(q,t)=F(q)+G(t)$
then the H-J equation is
$$\frac{1}{2m}\left(\frac{dF}{dq}\right)^2=-\frac{dG}{dt}=\alpha$$
we can obtain
$$F(q)=\pm \sqrt{2m\alpha}q+c_1 \\ G(t)=-\alpha t+c_2$$
Then, the $S(q,t)$ is the form
$$S(q,t)=\pm \sqrt{2m\alpha}q-\alpha t+c \tag{1}$$
(b) $S(q,t)=F(q)G(t)$
then the H-J equation is
$$\frac{1}{2m}\left(\frac{dF}{dq}\right)^2\frac{1}{F}=-\frac{dG}{dt}\frac{1}{G^2}=\alpha$$
we can obtain
$$F(q)=\left(\pm\frac{q}{2}\sqrt{2m\alpha}+c_1\right)^2 \\ G(t)=\frac{1}{\alpha t+c_2}$$
Then, the $S(q,t)$ is the form
$$S(q,t)=\left(\pm\frac{q}{2}\sqrt{2m\alpha}+c_1\right)^2\frac{1}{\alpha t+c_2} \tag{2}$$
So, there is the question: The $S(q,t)$ given in (a) and (b) are obviously different! Which is the right $S(q,t)$ for the free particle?
