Prove the eigenfunctions nth have $n-1$ zero.

Question

Consider the problem of eigenvalues: $$x^2y''+xy'+\lambda y=0\tag1$$

with boundary condition:

$$$y(1)=y(b)=0$$

Prove the eigenfunctions nth have $n-1$ zero.

My attempt

I know the Sturm-Liouville form of the equation is:

$$xy''+y'+\frac{\lambda}{x}y=0\tag2$$

With eigenvalues: $$\lambda_k=\frac{k^2\pi^2}{\ln(b)^2}\tag3$$

and eigenfunction of the form:

$$y_k(x)=\sin\frac{k\pi\ln(x)}{\ln(b)}$$

I could prove with help of other user, the orthogonality of two functions $y_n,y_m$ with weight function $$w(x)=\frac{1}{x}\tag4$$

But here i'm stuck. Can somenone help me?

Answer 1

Solve $\sin\dfrac{k\pi\ln(x)}{\ln(b)}=0$ so is $\dfrac{k\pi\ln(x)}{\ln(b)}=m\pi$ with $m\in\mathbb Z$ and $x\in[0,b]$

$\ln x=\dfrac{m}{k}\ln b$ And because $x\geq1$, $m\geq0$

$x=b^{m/k}$

Now, as $b>1$, $b^{m/k}\leq b$ if $\dfrac{m}{k}\leq 1$ or being $m\leq k$

$m=0,1,2,\dots,k$

Including the values $x=1$ and $x=b$, $y_k$ has $k+1$ zeros.

Answer 2

You can directly verify the orthogonality with respect to the weight function $1/x$. Start with $$ \int_1^b \sin\left(\frac{n\pi \ln x}{\ln b}\right)\sin\left(\frac{m\pi\ln x}{\ln b}\right)\frac{dx}{x}, $$ and set $$ y = \frac{\ln x}{\ln b},\;\; \ln x = y\ln b, \;\ x=e^{y\ln b}=(e^{\ln b})^y=b^y. $$

Then $\ln x = y\ln b$ or $\frac{dx}{x}=\ln b dy$. The first integral becomes $$ \int_0^1\sin(n\pi y)\sin(m\pi y)dy\cdot \ln b $$ Orthogonality is easy to verify for $n\ne m$.