Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Consider the regular Sturm-Liouville Problem:
$$-\frac{d}{dx} \Bigg( p(x)\frac{dv}{dx} \Bigg)=\lambda \rho (x)v$$ $$\alpha _1v(0)-\beta _1v'(0)=0$$ $$\alpha _2v(L)-\beta _2v'(L)=0$$ with $p(x),\rho (x)$ positive on $[0,L]$, $p,p',\rho $ continuous on $[0,L]$ and for $i=1,2$, we have $\alpha _i,\beta _i $ non-negative with $\alpha _i+\beta _i >0$.

Show $\lambda=0$ is an eigenvalue iff $\alpha _1=\alpha _2=0$.

Using the Rayleigh quotient, we have $$\lambda =\frac{-p(x)v(x)v'(x)\Bigg|^L_0 +\int^L_0 p(x)(v'(x))^2dx}{\int^L_0\rho(x)(v(x))^2dx}$$

It's clear that $\alpha _1=\alpha _2=0$ implies $\lambda=0$ is an eigenvalue, but I'm not seeing the reverse implication.

Any help?


share|cite|improve this question

I do not think the reverse implication is true.

As a counterexample, take $p(x)=1$, $L=3/2$, $\alpha_1=\beta_2=2$, $\alpha_2=\beta_1=1$. Then the problem at $\lambda=0$ becomes $$ v''=0,\;\;2v(0)-v'(0)=0,\;\;v(3/2)-2v'(3/2)=0. $$ The equation $v''=0$ gives $v=ax+b$. Then the boundary conditions imply $a=2b$. Thus, we have the eigenfunction $$ v=2x+1, $$ for the eigenvalue $\lambda=0$.

Also, another way to prove that $\alpha_1=\alpha_2=0$ implies that $\lambda=0$ is an eigenvalue is just to realize that in that case, the eigenfunction is the constant one.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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