Let $1\le i\le 3$.

-   If $m_i(t,x)\equiv m_i(x)$, then the principal spectrum point of equals to the principal spectrum point of .

-   If $m_i(t,x)\equiv m_i(t)$, then $\mu_i^n(\mathcal{L}_i+\lambda_i m_i)=\mu_i^n(0)+\lambda_i  \widehat m_i$.

\(1) It follows from [@ShXi1 Proposition 3.3] and Proposition \[spectrum-limit\].

\(2) It follows from Proposition \[spectrum-limit\] and the fact that $$\Phi_i(t,0;m_i)=e^{\lambda_i \int_0^t m_i(s)ds}\Phi_i(t,0;0).$$

Recall that $h_i(t, x)$ is defined in . Set \[hi(x)\] H\_i: (\_i)X\_i, (H\_i u)(t, x)=-\_t u(t, x)+h\_i(t, x)u(t, x) with $\mathcal D(\mathcal H_i)=\mathcal D(\mathcal L_i)$. Note that $\mathcal H_i+K_i=\mathcal L_i+\lambda_i m_i$. Hence, we may use $\mu_i^n(\mathcal H_i+K_i)$ or equivalently $\mu_i^n(\mathcal L_i+\lambda_i m_i)$ in different situations to denote the principal spectrum point of the eigenvalue problem . We use $\mathcal Re\{\sigma(\mathcal H_i)\}$ to denote the real part of the spectrum set of $\mathcal{H}_i$.

\[spectrum-Hi\] Let $1\le i\le 3$.

-   $\mathcal Re\{\sigma(\mathcal H_i)\}=[\widehat h_{i,\min},\widehat h_{i,\max}]$.

-   $\mu_i^n(\mathcal{L}_i+\lambda_i m_i)\ge \mu_i^n(\mathcal{L}_i+\lambda_i \widehat m_i)\ge \widehat h_{i,\max}$.

-   $\mu_i^n(\mathcal{L}_i+\lambda_i m_i)$ is the principal eigenvalue if and only if $\mu_i^n(\mathcal{L}_i+\lambda_i m_i)>
    \widehat h_{i,\max}$.

\(1) It follows from Lemma 3.7 in [@HuShVi].

\(2) $\mu_i^n(\mathcal{L}_i+\lambda_i m_i)\ge \mu_i^n(\mathcal{L}_i+\lambda_i \widehat m_i)$ follows from [@RaSh Theorem C], and $\mu_i^n(\mathcal{L}_i+\lambda_i \widehat m_i)\ge \widehat h_{i,\max}$ follows from Proposition \[spectrum-autonomous\] (1) and [@ShXi0 Proposition 3.9].

\(3) It follows from [@RaSh Theorem 