I was looking at previous year exam papers and was stuck on the following problem:

For the boundary value problem, $\,\,y''+\lambda y=0; y(0)=0,y(1)=0, \,\,\exists$ an eigenvalue $\lambda$ for which there corresponds an eigenfunction in $(0,1)$ that

  1. does not change sign
  2. changes sign
  3. is positive
  4. is negative

Now,I have to decide which of the aforementioned options are correct ?

My Attempt:

The general solution is :
$y(x)=a\cos(\sqrt \lambda x)+b\sin(\sqrt \lambda x).$

Then we apply the boundary values. From $y(0)=0$,we get $a=0.$

From $y(1)=0,a=0\,\,$ we get $\,\,0=b\sin(\sqrt \lambda )$.

We assume that $b \neq 0$ so that $\,\,0=b\sin(\sqrt \lambda )$.Then we have,$\sqrt \lambda=n \pi \implies \lambda =(n \pi)^2$.Let's write $\lambda_n=(n \pi)^2.$ Since we have $a=0,$ only sine term remains,so eigenfunctions are $y_n=\sin(\sqrt \lambda_n x)\,\,$ with eigenvalues $\lambda_n=(n \pi)^2,\,\,\,\, n=1,2,3, \dots$.

Now,when I look at the options ,I can not progress about which way to go. Can someone point me in the right direction keeping in mind for deciding which options are correct?


You've got it! What can you say about the function $y_1(x)=\sin(\pi x)$ on the interval $(0,1)$? What about scalar multiples for various scalars? And what about $y_2(x)$? P.S. I find the problem confusing. I interpret it as $4$ separate T/F questions.


You missed the case $\lambda<0$, which nevertheless does not have an eigenvalue!


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.