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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?

Thanks!

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1 Answer 1

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.

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