I'm reading the Cheng's thesis ""Eigenvalue Comparison Theorems and Its Geometric Applications," and the author obtains an estimate of eigenvalues of the Laplacian based upon his theorem:
If $M$ is $n$-dimensional compact Riemannian manifold with Ricci curvature $\geq(n-1)k$, then $$\mu_m(M)\leq\lambda_1(V_n(k,\frac{d_M}{2m}))$$ where $d_M$ denotes the diameter of $M$.
Geometric meaning of comparing eigenvalues is what I am trying to learn these days, but I have a trouble in understanding how certain estimates come in (and also others here and there).
Let's assume the case Ricci curvature $\geq(n-1)(-k),\ k>0$, $V_n(k,r_0)$ denote the ball of radius $r_0$ in the space form of dimension $n$ with constant curvature $k$.
What I could understand: The first eigenfunction of $V_{n}(-1,r_0)$ is radial and satisfies the differential equation $$\frac{d^2\varphi}{dr^2}+(n-1)\coth (r)\frac{d\varphi}{dr}+\lambda\varphi=0.$$ $$\frac{d\varphi}{dr}(0)=\varphi(r_0)=0.$$ Performing a substitution $$s=\cosh(r)=\mathrm{ch}(r),$$ it transforms into $$(s^2-1)\frac{d^2\varphi}{ds^2}+ns\frac{d\varphi}{ds}+\lambda\varphi=0,\ s\geq 1.$$ In case of $n=2$, the eigenfunctions are the well-known Legendre functions $P_{-\frac{1}{2}+ip},\ p\in\mathbb{R}$ with $\lambda=\frac{1}{4}+p^2$. Requiring that $$\frac{d\varphi}{ds}(0)=\varphi(s_0)=0.$$ qualifies $p$ for only discrete values, and it is known that there is a zero within each interval $(\mathrm{ch}\frac{l\pi}{p}, \mathrm{ch}\frac{(l+1)\pi}{p})$, $l$ a positive integer. Using this information, we have $s_0=\mathrm{ch}(r_0)\leq\mathrm{ch}\frac{2\pi}{p}$, and thus, $$\lambda_1(V_2(-1,r_0))=\frac{1}{4}+p^2\leq\frac{1}{4}+\left (\frac{2\pi}{r_0}\right )^2.$$ Since its derivatives $\frac{d^m}{ds^m}(P_{-\frac{1}{2}+ip})$ satisfies $$(s^2-1)\frac{d^2\varphi}{ds^2}+2(m+1)s\frac{d\varphi}{ds}+\left ( \frac{(2m+1)^2}{4}+p^2\right )\varphi=0.$$ (It is another part that I don't understand, is it a well known fact?)
Similarly, within each interval $\left (\mathrm{ch}(\frac{l\pi}{p}),\mathrm{ch}(\frac{(l+2^m)\pi}{p})\right )$ there is a zero of $\frac{d^m}{ds^m}(P_{-\frac{1}{2}+ip})$, therefore, $$\lambda_1(V_{2(m+1)}(-1,r_0))\leq\frac{(2m+1)^2}{4}+\frac{(1+2^m)^2\pi^2}{r_0^2},\ m=0,1,2,\cdots.$$
What I couldn't understand:
For the case of odd dimension, let $\varphi$ be a function satisfying $$(s^2-1)\frac{d^2\varphi}{ds^2}+2(m+1)s\frac{d\varphi}{ds}+\left ( \frac{(2m+1)^2}{4}+p^2\right )\varphi=0.$$ Then $\psi=(s^2-1)^(-\frac{1}{4})\varphi$ satisfies $$(s^2-1)\frac{d^2\psi}{ds^2}+(2m+3)s\frac{d\psi}{ds}+\left ( \frac{(2m+2)^2}{4}+p^2+\frac{4m+1}{4}(s^2-1)^{-1}\right )\psi=0.$$ It is all good so far, but the next estimate suddenly came out: $$\lambda_1(V_{2m+3}(-1,r_0))\leq\frac{(2m+2)^2}{4}+\frac{(1+2^{2m})^2\pi^2}{r_0^2}+\frac{1}{\sinh(\frac{r_0}{1+2^{2m}})^2}.$$ This is the part that I would like to ask a question.
Some facts that may be needed are the following:
We are dealing with functions on the interval $[1,s_0]$, where $s_0=\mathrm{ch}(r_0)$, so we can assume $s\leq s_0$.
The identity $\sinh(x)^2=\cosh(x)^2-1.$
The Legendre differential equation is $$(1-x^2)\frac{d^2y}{dx^2}-2x\frac{dy}{dx}+l(l+1)y=0,$$ here $l$ equals to the value $-\frac{1}{2}+ip.$
The associated Legendre differential equation is $$(1-x^2)\frac{d^2y}{dx^2}-2x\frac{dy}{dx}+\left (l(l+1)-\frac{c^2}{1-x^2}\right )y=0.$$
The last fact looks helpful, but two things do not agree with the above situation. One is that the coefficient in the second term is $2x$, not $(2m+3)x$. Also, the coefficient in the third term is $\frac{1}{4}+p^2+\frac{1}{4}\frac{1}{s^2-1}$, not $\frac{(2m+2)^2}{4}+p^2+\frac{4m+1}{4}(s^2-1)^{-1}$. Well, even if I overcome this by differentiating the solution of the case $m=0$ mimicking as above, the other reamains, that is, I do not see what an eigenvalue is in the associated Legendre differential equation.
I overall felt that the author uses basic facts regarding Legendre functions without much explanations, so leaving references would really help me.
Thanks for reading it.