# Why are the two ODE solutions linearly independent?

I notice that some second-order ODEs can be related to the triconfluent Heun's equation $$u''(z)-(3z^2+\gamma)u'(z)+(\alpha-(3-\beta) z)u(z)=0.$$ And people usually say the general solution of the original ODE contains two parts like [from this answer for the ODE $$y'' +(x^4 +x^2+x+c)y(x) =0$$] $$y( x ) ={C_1}\,{{\rm e}^{\frac{1}{6}\,ix \left( 2\,{x}^{2}+3\right)}}{\mathrm{HeunT}} \left(\alpha, \beta, \gamma, x \right) +{ C_2} {{\rm e}^{-\frac{1}{6}\,ix \left( 2\,{x}^{2}+3\right)}} {\mathrm{HeunT}} \left( \alpha,-\beta, \gamma, -x\right).$$ Such a solution form is also generated in many examples in Maple and Mathematica. E.g., this answer and another one. So I guess it's some known fact.

The two parts are indeed solutions as one can easily transform the original ODE to obtain. But they seem to just result from different transforms using $$y(x)={{\rm e}^{\pm\frac{1}{6}\,ix \left( 2\,{x}^{2}+3\right)}} u(x)$$. How to see they are linearly independent?

• The easiest way is to look at the behavior at $z\to\infty$. Your two solutions have different asymptotics at $\infty$ (one has a + under the exponent, another -). – Alexandre Eremenko Oct 11 at 14:36
• How is your HeunT function precisely defined? Cf. reference.wolfram.com/language/ref/HeunT.html – Iosif Pinelis Oct 11 at 16:01
• @AlexandreEremenko I feel that only looking at the exponential factor might not be enough because the $\mathrm{HeunT}$ here is in general not truncated to a polynomial. Am I wrong? – xiaohuamao Oct 11 at 18:49
• You are wrong because Heun has regular singularities, therefore its solutions grow at most like powers, and cannot thus interfere with exponential asymptotics – Alexandre Eremenko Oct 11 at 18:53
• @AlexandreEremenko But triconfluent Heun has an irregular singularity at $\infty$ (see (12) on this page). Will it matter? Thanks. – xiaohuamao Oct 11 at 20:11

Two functions will be linearly independent if their Wronskian is nonzero at some point.

The Wronskian of your functions $$u_1$$ and $$u_2$$ given by $$u_1(x):=\exp\Big\{\frac{ix(2x^2+3)}6\Big\}\,\text{HeunT}(\alpha,\beta,\gamma,x),$$ $$u_2(x):=\exp\Big\{-\frac{ix(2x^2+3)}6\Big\}\,\text{HeunT}(\alpha,-\beta,\gamma,-x)$$ is $$-i\ne0$$ at $$x=0$$. So, $$u_1$$ and $$u_2$$ are linearly independent.

Here are an image of a calculation of the Wronskian in Mathematica:

(Click on the image to enlarge it.)

• Thank you for the nice answer! Yes, I'm using the Maple convention. Looks that I can use Abel's identity to show the Wronskian is nonzero everywhere from your calculation. – xiaohuamao Oct 11 at 18:53