Is there a nice expression that encapsulates the solution to all three cases of the Sturm-Liouville problem?

Question

When encountering a second-order linear ODE with an unknown constant, such as $X'' = \lambda X$, it is typical to traverse the three cases where $\lambda$ is positive, negative, and zero separately. However, the positive and negative cases are easy to express together, since linear combinations of $e^{-\sqrt{\lambda} x}$ and $e^{\sqrt{\lambda} x}$ could be exponential or sinusoidal depending on $\lambda$.

However, I cannot think of a way to incorporate the extra powers of $x$ introduced by the resonance in the $\lambda = 0$ case into the same solution. Is there a way to incorporate this case into a single, caseless expression for the solution to such a problem?

Answer 1

You can form a basis of solutions: $$ X''=\lambda X,\;\;\; X(0)=0, X'(0)=1 \\ Y''=\lambda Y,\;\;\; Y(0)=1, Y'(0)=0 $$ The solutions are $$ X_{\lambda}(x)=\frac{\sinh(\sqrt{\lambda}x)}{\sqrt{\lambda}},\;\; Y_{\lambda}(x)=\cosh(\sqrt{\lambda}x). $$ $X_{\lambda}$ remains valid for $\lambda=0$ in the limit as $\lambda\rightarrow 0$, where the result is $X_0(x)=x$. This is generally true because the solutions for fixed conditions will vary analytically in $\lambda$, which means that the special cases hold in the limit. The solutions for fixed conditions are holomorphic in $\lambda$.

Answer 2

If you choose suitable linear combinations, you can have a set of basis solutions where the case $\lambda=0$ is included as limit, e.g.: $$\lim_{\lambda\to0}\frac{1}{2}\left(e^{\sqrt{\lambda}x}+e^{-\sqrt{\lambda}x}\right)=1$$ $$\lim_{\lambda\to0}\frac{1}{2\sqrt{\lambda}}\left(e^{\sqrt{\lambda}x}-e^{-\sqrt{\lambda}x}\right)=x$$.