Consider the Sturm-Liouville problem $y^{\prime\prime}+ \lambda y=0,y(0)=y( \pi)=0.$

Question

Consider the Sturm-Liouville problem $y^{\prime\prime}+ \lambda y=0,y(0)=y( \pi)=0.$ Which of the following statement(s) are true:

1. There exists only countably many characteristic values

2. There exists uncountably many characteristic values

3. Each characteristic function corresponding to the characteristic value $\lambda$ has exactly $\lfloor \sqrt\lambda \rfloor -1$ zeros in $(0, \pi)$

4. Each characteristic function corresponding to the characteristic value $\lambda$ has exactly $\lfloor \sqrt\lambda \rfloor $ zeros in $(0, \pi)$

My Approach: Solving for $\lambda >0$, $y=c_1 \cos \sqrt{\lambda}x+c_2 \sin \sqrt{\lambda}x$,

As $y(0)=0$, $c_1=0$; As $y(\pi)=0$, so either $c_2=0$ or $\sqrt{\lambda} \pi=n \pi$, $n \in \mathbb Z$, i.e. $\lambda_n=n^2$, $y_n=C_n \sin nx$

For $\lambda \le0$, $y=0$ is trivial solution. So, I guess option $1.$ is correct but I cannot understand about options $3.$ and $4.$ please help.

Answer 1

So after solving the given Sturm-Liouville problem $($for $\lambda=m^2>0$, as for $\lambda\le0$ we get trivial solution$)$, we have

$$y(x)=B\sin(nx)~,$$ which is a characteristic function corresponding to the characteristic value $~m=n\in\mathbb Z~.$

For option $\bf {(1)}$ and $\bf {(2)}$ : Since $~\mathbb Z~$ is countable, so there exist only countably many characteristic value.
Hence option $(1)$ is correct but option $(2)$ is incorrect.

For option $\bf {(3)}$ and $\bf {(4)}$ : Now we have $$y_n=B_n\sin(nx)$$ For $~n=1~,~~~~y_1=B_1\sin(x)~$ which has no solution (i.e., no zero) in $~(0,\pi)~.$
i.e., characteristic function corresponding to the characteristic value $~\lambda=n^2=1~$ has exactly $~\lfloor \sqrt\lambda \rfloor -1=1-1=0~$ zero in $(0, \pi)$.

For $~n=2~,~~~~y_2=B_2\sin(2x)~$ which has only one solution in $~(0,\pi),\big[$which is $~x=\pi/2~\big].$
i.e., characteristic function corresponding to the characteristic value $~\lambda=n^2=4~$ has exactly $~\lfloor \sqrt\lambda \rfloor -1=2-1=1~$ zero in $(0, \pi)$.

Similarly, we can verify for the other values of $~n\in\mathbb Z~.$

Hence each characteristic function corresponding to the characteristic value $~\lambda~$ has exactly $~\lfloor \sqrt\lambda \rfloor -1~$ zeros in $(0, \pi)$.

Therefore option $(3)$ is correct but option $(4)$ is incorrect.