Solution to the parabolic cylinder equation

Question

In the Gradshteyn & Ryzhik (7th ed.) the differential equation (9.255) leading to parabolic cylinder functions is $$\frac{d^2u}{dz^2}+(p+\frac{1}{2}-\frac{z^2}{4})u=0.$$ The solutions are $u=D_p(z),D_p(-z),D_{-p-1}(iz),D_{-p-1}(-iz)$, where $D_p(z)$ is the parabolic cylinder function. These four solutions are linearly dependent. My question is why is there four solutions to the second order ODE? In my case $p$ is complex, and Mathematica gives solution in the form $C_1 D_p(z)+C_2 D_{-p-1}(iz)$.

Answer 1

This second order ODE has not four solutions as you wrote, but has an infinity of solutions.

Don't write

<< The solutions are $u_1=D_p(z),u_2=D_p(-z),u_3=D_{-p-1}(iz),u_4=D_{-p-1}(-iz)$ >> ,

better write

<< Some solutions are $u_1=D_p(z),u_2=D_p(-z),u_3=D_{-p-1 }(iz),u_4=D_{-p-1}(-iz)$ >>.

For example $u_5=3u_1-7u_3$ is also a solution of the ODE. You will not say "Why is there five solutions to the second order ODE ? ".

In fact, among the infinity of solutions we can only find COUPLES of linearly INDEPENDENT solutions. Not TRIPLET.

All solutions can be defined on the form of a linear combination of any couple of independent solution.

Thus the general solution expressed on the form $$u(z)=C_1 D_p(z)+C_2 D_{-p-1}(iz)$$ is equivalent to $$u(z)=C_3 D_p(-z)+C_4 D_{-p-1}(iz)$$ or equivalent to $$u(z)=C_5 D_p(-z)+C_6 D_{-p-1}(-iz)$$ Etc.

Of course the coefficients are generally not the same: For example $C_2\neq C_6$.