The proof follows by direct computation using evolution formula (4) in Lemma
3.1. Let
\(f(t_{0})\) and
\(\lambda (t_{0}) = \lambda (f(t _{0}),t)\) be an eigenpair. Then, for a smooth function
\(f(t)\), we can set
$$ \lambda \bigl(f(t), t\bigr) = - \int _{M} f(t) \Delta _{p,\phi } f(t) \,d\mu . $$
Then
$$\begin{aligned} \frac{d}{dt} \lambda _{1}(t)\bigg|_{t=t_{0}} = \frac{\partial }{\partial t}\lambda _{1}\bigl(f(t), t\bigr) = -\frac{\partial }{\partial t} \int _{M} f(t) \Delta _{p,\phi } f(t) \,d\mu . \end{aligned}$$
(4.10)
Using evolution formula (4) of Lemma
3.1, we obtain
$$\begin{aligned} &\frac{\partial }{\partial t} \int _{M} f \Delta _{p,\phi } f \,d\mu \\ &\quad = \int _{M} \frac{\partial }{\partial t} ( \Delta _{p,\phi } f ) f \,d \mu + \int _{M} \Delta _{p,\phi } f \frac{\partial }{\partial t} ( f\,d \mu ) \\ &\quad = 2 \int _{M} \mathcal{S}^{ij} \nabla _{i} ( Z \nabla _{j} f )f \,d\mu + \int _{M} g^{ij} \nabla _{i} ( Z_{t} \nabla _{j} f ) f \,d\mu + \int _{M} g ^{ij} \nabla _{i} (Z \nabla _{j} f_{t} ) f \,d\mu \\ &\qquad {}- 2 \alpha \int _{M} Z (\Delta \phi ) g^{ij} \nabla _{i} \phi \nabla _{j} f f \,d\mu - \int _{M} Z_{t} \langle \nabla \phi , \nabla f \rangle f \,d\mu - \int _{M} Z \langle \nabla \phi _{t}, \nabla f \rangle f \,d\mu \\ &\qquad {}- \int _{M} Z \langle \nabla \phi , \nabla f_{t} \rangle f \,d\mu - 2 \int _{M} S^{ij} Z \nabla _{i} \phi \nabla _{j} f + \int _{M} \Delta _{p, \phi } f \frac{\partial }{\partial t} ( f\,d\mu ). \end{aligned}$$
(4.11)
We now apply integration by parts formula on the first three terms on the right-hand side of (
4.11). For the first term
$$ \begin{aligned}[b] 2 \int _{M} S^{ij} \nabla _{i} ( Z \nabla _{j} f) f \,d\mu &= - 2 \int _{M} Z \nabla _{i} f \nabla _{j} \bigl(S^{ij}f e^{-\phi } \bigr)\,dv = -2 \int _{M} Z S^{ij} \nabla _{i} f \nabla _{j} f \,d\mu \\ &\quad {}- 2 \int _{M} Z\nabla _{i} S^{ij} \nabla _{j} f f \,d\mu + 2 \int _{M} Z S ^{ij}\nabla _{i} f \nabla _{j} \phi f \,d\mu . \end{aligned} $$
(4.12)
The second term on the right-hand side of (
4.12) can be written as (see the computation in Lemma
A.1 below):
$$\begin{aligned} - 2 \int _{M} Z\nabla _{i} S^{ij} \nabla _{j} f f \,d\mu & = \int _{M} S( \Delta _{p,\phi } f) f \,d\mu + \int _{M} S \vert \nabla f \vert ^{p} \,d\mu \\ &\quad {}+ 2 \alpha \int _{M} Z \Delta \phi \langle \nabla \phi , \nabla f \rangle \,d\mu . \end{aligned}$$
(4.13)
Substituting (
4.13) into (
4.12) we have
$$ \begin{aligned}[b] 2 \int _{M} S^{ij} \nabla _{i} ( Z \nabla _{j} f) f \,d\mu &= -2 \int _{M} Z S^{ij}\nabla _{i} f \nabla _{j} f \,d\mu + \int _{M} S(\Delta _{p,\phi } f) f \,d\mu \\ &\quad {}+ \int _{M} S \vert \nabla f \vert ^{p} \,d\mu+ 2 \alpha \int _{M} Z \Delta \phi \langle \nabla \phi , \nabla f \rangle \,d\mu\\&\quad{} + 2 \int _{M} Z S^{ij} \nabla _{i} f \nabla _{j} \phi f \,d\mu . \end{aligned} $$
(4.14)
For the second term on the right-hand side of (
4.11), integration by parts implies
$$ \begin{aligned}[b] \int _{M} g^{ij}\nabla _{i}(Z_{t} \nabla _{j} f) f\,d\mu &= - \int _{M} Z _{t} \nabla _{j} f \nabla ^{i}\bigl(fe^{-\phi }\bigr) \,dv \\ &= - \int _{M} Z_{t} \vert \nabla f \vert ^{2} \,d\mu + \int Z_{t} \langle \nabla f, \nabla \phi \rangle f \,d\mu . \end{aligned} $$
(4.15)
Similarly, the third term on the right-hand side of (
4.11) implies
$$\begin{aligned} \int _{M} g^{ij} \nabla _{i}(Z\nabla _{j} f_{t}) f \,d\mu = - \int _{M} Z \langle \nabla f_{t}, \nabla f \rangle \,d\mu + \int _{m} Z \langle \nabla f_{t}, \nabla \phi \rangle f \,d\mu . \end{aligned}$$
(4.16)
Putting (
4.14), (
4.15), and (
4.16) into (
4.11), we obtain
$$ \begin{aligned}[b] \frac{\partial }{\partial t} \int _{M} f \Delta _{p,\phi } f \,d\mu &= -2 \int _{M} Z S^{ij}\nabla _{i} f \nabla _{j} f \,d\mu + \int _{M} S \Delta _{p,\phi } f f \,d\mu \\ &\quad {}+ \int _{M} S \vert \nabla f \vert ^{p} \,d\mu - \int _{M} Z_{t} \vert \nabla f \vert ^{2} \,d \mu - \int _{M} Z \langle \nabla f_{t}, \nabla f \rangle \,d\mu \\ &\quad {}- \int _{M} Z \langle \nabla \phi _{t}, \nabla f \rangle f \,d\mu + \int _{M} \Delta _{p,\phi } f \frac{\partial }{\partial t} ( f\,d\mu ). \end{aligned} $$
(4.17)
Using the evolution formula (2) of Lemma
3.1 into (
4.17) yields
$$ \begin{aligned}[b] \frac{\partial }{\partial t} \int _{M} f \Delta _{p,\phi } f \,d\mu &= -p \int _{M} Z S^{ij}\nabla _{i} f \nabla _{j} f \,d\mu + \int _{M} S \Delta _{p,\phi } f f \,d\mu \\ &\quad {}+ \int _{M} S \vert \nabla f \vert ^{p} \,d\mu - (p-1) \int _{M} Z \langle \nabla f _{t}, \nabla f \rangle \,d\mu \\ &\quad {}- \int _{M} Z \langle \nabla \phi _{t}, \nabla f \rangle f \,d\mu + \int _{M} \Delta _{p,\phi } f \frac{\partial }{\partial t} ( f\,d\mu ). \end{aligned} $$
(4.18)
A straightforward computation also yields
$$ \begin{aligned}[b] &- (p-1) \int _{M} Z \langle \nabla f_{t}, \nabla f \rangle \,d\mu \\ &\quad = (p-1) \int _{M} \nabla _{i}\bigl(Z\nabla _{j} f e^{-\phi }\bigr) f_{t} \,dv \\ &\quad = (p-1) \int _{M} \nabla _{i}(Z\nabla _{j} f)\,d\mu - (p-1) \int _{M} Z\langle \nabla f, \nabla \phi \rangle f_{t} \,d\mu \\ &\quad = (p-1) \int _{M}\Delta _{p,\phi } f f_{t} \,d\mu \end{aligned} $$
(4.19)
and
$$ \begin{aligned}[b] \int _{M} Z \langle \nabla \phi _{t}, \nabla f \rangle f \,d\mu &= - \int _{M} \phi _{t} \nabla _{i}\bigl(Z \nabla _{j} f f e^{-\phi }\bigr) \,dv \\ &= - \int _{M} \phi _{t} \Delta _{p,\phi } f \,d\mu - \int _{M} \phi _{t} \vert \nabla f \vert ^{p} \,d\mu . \end{aligned} $$
(4.20)
Putting (
4.19) and (
4.20) into (
4.18) yields
$$\begin{aligned} &\frac{\partial }{\partial t} \int _{M} f \Delta _{p,\phi } f \,d\mu \\ &\quad = -p \int _{M} Z S^{ij}\nabla _{i} f \nabla _{j} f \,d\mu + \int _{M} S \Delta _{p,\phi } f f \,d\mu \\ &\qquad{} + \int _{M} S \vert \nabla f \vert ^{p} \,d\mu + (p-1) \int _{M} \Delta _{p,\phi } f f_{t} \,d\mu + \int _{M} \phi _{t} \Delta _{p,\phi } f f \,d\mu \\ &\qquad{} + \int _{M} \phi _{t} \vert \nabla f \vert ^{p} \,d\mu + \int _{M} \Delta _{p,\phi } f \frac{\partial }{\partial t} ( f\,d\mu ) \\ & \quad = -p \int _{M} Z S^{ij}\nabla _{i} f \nabla _{j} f \,d\mu + \int _{M} S \Delta _{p,\phi } f f \,d\mu + \int _{M} S \vert \nabla f \vert ^{p} \,d\mu \\ &\qquad{} + \int _{M} \phi _{t} \Delta _{p,\phi } f f \,d\mu + \int _{M} \phi _{t} \vert \nabla f \vert ^{p} \,d\mu + \int _{M} \Delta _{p,\phi } f \biggl((p-1) f_{t} \,d \mu - \frac{\partial }{\partial t} ( f\,d\mu ) \biggr). \end{aligned}$$
Using the facts that
$$\begin{aligned}& \frac{\partial }{\partial t} \int _{M} f \Delta _{p,\phi } f \,d\mu = - \frac{ \partial }{\partial t} \lambda _{1} \bigl(f(t),t\bigr)\bigg|_{t=t_{0}}, \\& \Delta _{p,\phi } f = - \lambda _{1}(t_{0}) \vert f \vert ^{p-2} f \end{aligned}$$
and the normalization condition
\(\int _{M} |f|^{p} \,d\mu =1\), which implies
$$\begin{aligned} 0 &= \frac{\partial }{\partial t} \biggl( \int _{M} \vert f \vert ^{p} \,d \mu \biggr) = \frac{\partial }{\partial t} \biggl( \int _{M} \vert f \vert ^{p-1} f \,d \mu \biggr) \\ & = \int _{M} \vert f \vert ^{p-2} f \biggl[(p-1) f_{t} \,d\mu + \vert f \vert ^{p-1} \frac{ \partial }{\partial t} (f \,d \mu ) \biggr], \end{aligned}$$
we arrive at
$$\begin{aligned} \frac{d}{dt} \lambda _{1}\bigl(f(t),t\bigr)\bigg|_{t=t_{0}} & = \lambda _{1}(t_{0}) \int _{M} S \vert f \vert ^{p} \,d\mu - \int _{M} S \vert \nabla f \vert ^{p} \,d\mu + \lambda _{1}(t _{0}) \int _{M} \phi _{t} \vert f \vert ^{p} \,d\mu \\ & \quad{}- \int _{M} \phi _{t} \vert \nabla f \vert ^{p} \,d\mu + p \int _{M} \vert \nabla f \vert ^{p-2} S^{ij} \nabla _{i} f \nabla _{j} f \,d\mu , \end{aligned}$$
which is what we wanted to prove. □