I am confused about the usage of the terms eigenfunctions and eigenvalues for ODE’s. I am used to thinking of eigenvalues for some linear operator $D$ as the solutions $$D y=\lambda y$$ for some eigenfunction $y$. But I’ve seen in places that say an eigenvalue $\lambda$ satisfies something such as $$y''+ \lambda y=0,$$ but wouldn’t the eigenvalue be $- \lambda$ in this case for linear operator $$\frac{d^2}{dx^2}$$ not $$\frac{d^2}{dx^2} + \lambda$$ with eigenvalue $\lambda?$ This seems like a minor nuance, but is the second usage simply an informality that simplifies the analysis of ODE’s or is there an alternative I am missing?

  • 1
    $\begingroup$ You could just as well think of $\lambda$ as the eigenvalue of $-\frac{d^2}{dx^2}.$ Incidentally, I notice you have the boundary-value-problem tag associated with this question. The boundary values are always considered to be part of the operator, as they can affect eigenvalues drastically. $\endgroup$ – Adrian Keister Aug 14 at 15:15
  • $\begingroup$ @AdrianKeister I do and I know they can but in this context they don’t affect the definition as the boundary values (as far as I understand them) only restrict the possible values from the existing set with no boundaries. Either way, the possible eigenvalues before the boundary value restrictions are unaffected by the boundary values themselves. I included the tag because this is, as you pointed out, an important topic in boundary value problems. But in this case, the Boundary values don’t affect what I’m trying to understand. $\endgroup$ – Colin Hicks Aug 14 at 15:46
  • $\begingroup$ You can convert this linear second-order ODE into a system of first-order equations by introducing a new function $u=y'$. The characteristic equation of the resulting coefficient matrix is $x^2-\lambda=0$. $\endgroup$ – amd Aug 14 at 20:11
  • $\begingroup$ @amd how did you get that? $$u_1^{‘}=u_2, u_2^{‘}=-\lambda u_1$$ then we have matrix $$0 1$$ $$-\lambda 0$$ which has has characteristic equation $$x^{2}+\lambda$$ where $$u_1=y$$ and $$u_2=y^{‘}$$ $\endgroup$ – Colin Hicks Aug 14 at 20:59
  • $\begingroup$ Right. Good catch. I made a sign error. The point either way is that you can connect the operator $D^2+\lambda$ to the characteristic polynomial $x^2+\lambda$ of the equation via this process of reducing the order of the equation by making it a system of equations. $\endgroup$ – amd Aug 15 at 0:25

Your Answer

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

Browse other questions tagged or ask your own question.