Let $y_1(x)$ be a non trivial solution of the ODE $(p(x)y')'+q(x)y=\lambda y$, with $y(a)=y(b)=0$, and $p(x)>0$ in all the interval $[a,b]$. Prove/disprove: every solution $y_2$ is of the form $y_2=cy_1$.

I'v learnt only the definition of the Sturm-Liouville problem, so it should be something quite elementary, and yet I don't know how to relate $y_2$ to $y_1$, and somehow use the boundary conditions.


This question was answered here: Sturm-Liouville eigen value problem with one-dimensional eigenspace

The following is a rephrasing of the answer given there, without the operator-theoretic context which is given there.

Consider the ODE $$(py')' + qy = \lambda y.$$ Then, there exit unique solutions $\varphi_\lambda$ and $\psi_\lambda$ of this ODE with $$\varphi_\lambda(0)=0, \ p\varphi'_\lambda(0)=1; \ \psi_\lambda(0)=1, \ p\psi'_\lambda(0)=0.$$

Then, $\varphi_\lambda, \psi_ \lambda$ span the solution space of the ODE and every solution can be written in a unique linear combination of these two solutions: $y=A\varphi_\lambda + B\psi_\lambda$.

If there is some nontrivial $y$ which solves the ODE with $y(0)=y(1)=0$, then by $y(0)=A\cdot 0 + B \cdot 1$ we have $B=0$ and $y$ is a multiple of $\varphi_\lambda$ (furthermore, $\varphi_\lambda(1)=0$). Thus $y_1=A_1 \varphi_\lambda $ and $y_2=A_2\varphi_\lambda$ for some nonzero $A_1,A_2$ and we have found $c=\frac{A_2}{A_1}$.

  • $\begingroup$ Couldn't you take just $$ \varphi'_\lambda(0)=1; \psi'_\lambda(0)=0.$$ instead of $$ \ p\varphi'_\lambda(0)=1; \ \ p\psi'_\lambda(0)=0.$$? $\endgroup$ – Emolga Feb 19 '14 at 14:57
  • $\begingroup$ That depends on your definition of "solution". If you only consider functions which have a continuous first and second derivative, then you might drop the p. But in the general setting for Sturm-Liouville operators, you consider functions $y$ which are absolutely continuous (which only yields $y'\in L^1_{\text{loc}}$) s.th. $py'$ is absolutely continuous. And in this case, you can't evaluate $y'$, but you can evaluate $py'$. Cf. eg. G. Teschl, Mathematical Methods in Quantum Mechanics, Chapter 9 (mat.univie.ac.at/~gerald/ftp/book-schroe/schroe.pdf) $\endgroup$ – Roland Feb 19 '14 at 15:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.