The Wronskian of parabolic cylinder functions

Question

Suppose I have the second order differential equation $$ y''(t) + (k^{2} + m^{2}t^2)y(t) = 0, \quad 0< t_{0} < t < \infty $$ The solution of this equation is parabolic cylinder functions, namely $$ y(t) = c_{1}D_{-\frac{ik^{2}}{2m} + \frac{1}{2}}\left( e^{\frac{i \pi}{4}}\sqrt{m}(t - t_{0})\right) + c_{2}D_{\frac{ik^{2}}{2m} - \frac{1}{2}}\left( e^{\frac{3i \pi}{4}}\sqrt{m}(t - t_{0})\right) $$ $$ \equiv c_{1}D_{1}(t - t_{0}) + c_{2}D_{2}(t - t_{0}) $$ Is it possible to compute the wronskian $$ W[D_{1}(t - t_{0}), D_{2}(t - t_{0})] \equiv \dot{D}_{1}D_{2} - \dot{D}_{2}D_{1} $$ in terms of elementary functions? If yes, how?

Answer 1

For any two solutions $y_a,y_b$, the wronskian $W_{ab}:=W\left[y_a(t),y_b(t)\right]$ satisfies the equation $$W'_{ab}=\ddot{y}_a y_b -y_a \ddot{y}_b=0,$$ hence it is necessarily a constant. The actual value of this constant can be found using e.g. the asymptotics of the basis of solutions at some point (for instance, $t_0$). In your case, correcting the typo (sign in front of $\frac12$) in the first solution, this gives $W_{ab}=-\sqrt m\,e^{-\frac{\pi k^2}{4m}}$.