Htykuut

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
$$mu_m(M)leqlambda_1(V_n(k,frac{d_M}{2m}))$$
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
$$frac{d^2varphi}{dr^2}+(n-1)coth (r)frac{dvarphi}{dr}+lambdavarphi=0.$$
$$frac{dvarphi}{dr}(0)=varphi(r_0)=0.$$
Performing a substitution $$s=cosh(r)=mathrm{ch}(r),$$ it transforms into
$$(s^2-1)frac{d^2varphi}{ds^2}+nsfrac{dvarphi}{ds}+lambdavarphi=0, sgeq 1.$$
In case of $n=2$, the eigenfunctions are the well-known Legendre functions $P_{-frac{1}{2}+ip}, pinmathbb{R}$ with $lambda=frac{1}{4}+p^2$. Requiring that $$frac{dvarphi}{ds}(0)=varphi(s_0)=0.$$ qualifies $p$ for only discrete values, and it is known that there is a zero within each interval $(mathrm{ch}frac{lpi}{p}, mathrm{ch}frac{(l+1)pi}{p})$, $l$ a positive integer. Using this information, we have $s_0=mathrm{ch}(r_0)leqmathrm{ch}frac{2pi}{p}$, and thus, $$lambda_1(V_2(-1,r_0))=frac{1}{4}+p^2leqfrac{1}{4}+left (frac{2pi}{r_0}right )^2.$$
Since its derivatives $frac{d^m}{ds^m}(P_{-frac{1}{2}+ip})$ satisfies
$$(s^2-1)frac{d^2varphi}{ds^2}+2(m+1)sfrac{dvarphi}{ds}+left ( frac{(2m+1)^2}{4}+p^2right )varphi=0.$$
(It is another part that I don't understand, is it a well known fact?)

Similarly, within each interval $left (mathrm{ch}(frac{lpi}{p}),mathrm{ch}(frac{(l+2^m)pi}{p})right )$ there is a zero of $frac{d^m}{ds^m}(P_{-frac{1}{2}+ip})$, therefore,
$$lambda_1(V_{2(m+1)}(-1,r_0))leqfrac{(2m+1)^2}{4}+frac{(1+2^m)^2pi^2}{r_0^2}, m=0,1,2,cdots.$$

What I couldn't understand:

For the case of odd dimension, let $varphi$ be a function satisfying
$$(s^2-1)frac{d^2varphi}{ds^2}+2(m+1)sfrac{dvarphi}{ds}+left ( frac{(2m+1)^2}{4}+p^2right )varphi=0.$$
Then $psi=(s^2-1)^(-frac{1}{4})varphi$ satisfies
$$(s^2-1)frac{d^2psi}{ds^2}+(2m+3)sfrac{dpsi}{ds}+left ( frac{(2m+2)^2}{4}+p^2+frac{4m+1}{4}(s^2-1)^{-1}right )psi=0.$$
It is all good so far, but the next estimate suddenly came out:
$$lambda_1(V_{2m+3}(-1,r_0))leqfrac{(2m+2)^2}{4}+frac{(1+2^{2m})^2pi^2}{r_0^2}+frac{1}{sinh(frac{r_0}{1+2^{2m}})^2}.$$
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 $sleq s_0$.




• The identity $sinh(x)^2=cosh(x)^2-1.$




• The Legendre differential equation is $$(1-x^2)frac{d^2y}{dx^2}-2xfrac{dy}{dx}+l(l+1)y=0,$$ here $l$ equals to the value $-frac{1}{2}+ip.$




• The associated Legendre differential equation is $$(1-x^2)frac{d^2y}{dx^2}-2xfrac{dy}{dx}+left (l(l+1)-frac{c^2}{1-x^2}right )y=0.$$

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.