This question is about mathematics, but it came up in a very physical setting. When studying Sakurai's Modern Quantum Mechanics, problem 2.38 asks us the following:
Show how one may obtain the correct wave function for a plane wave by starting with the solution of the classical Hamilton–Jacobi equation with $V(x)$ set equal to zero. Why do we get the exact wave function in this particular case?
We assume the wave function is proportional to $\exp\left[i S(x,t)/ \hbar\right]$, where $S(x,t)$ is a phase whose spatial derivative can be identified with momentum. The Hamilton-Jacobi equation in this case is $$ H\left(x, \frac{\partial S}{\partial x}, t\right) + \frac{\partial S}{\partial t} = \frac{1}{2m}\left( \frac{\partial S}{\partial x} \right)^{2} + \frac{\partial S}{\partial t} = 0. $$
The standard solution of this differential equation is obtained by letting $S(x,t) = X(x) + T(t)$, which then gives $$ S(x,t) = \pm \sqrt{2 m \alpha}x - \alpha t, \tag{1}$$ where $\alpha > 0$ is some parameter, recovering the plane wave solution as $$ \psi(x, t) \propto \exp\left[ \pm \frac{i}{\hbar} \left(\sqrt{2 m \alpha} x - \alpha t \right)\right]. $$
However, my first attempt was the ansatz $S(x,t) = X(x)T(t)$, which also yielded a solution
$$ S(x,t) = \frac{\left( \sqrt{m\alpha / 2} \, x + c_1\right)^2}{\alpha t + c_2}, \tag{2}$$
corresponding to $$ \psi(x,t) \propto \exp\left[ \frac{i \left( \sqrt{m\alpha / 2} \, x + c_1\right)^2}{\hbar\left(\alpha t + c_2\right) } \right]. $$
The two solutions above are completely different (I observed that my solution looks very similar to the free particle propagator, but that may just be a happy coincidence). How is it possible that the same Hamilton-Jacobi equation gave two different solutions? Is there any uniqueness theorem for nonlinear differential equations?
Also, what is the interpretation of the alternate solution I got, since it does not correspond to a plane wave?
