[@RaSh]. (S3) is a sufficient condition for $\mu_i^n(\lambda_i)$ to be the principal eigenvalue follows from [@Cov Theorem 1.1] and Proposition \[spectrum-Hi\].

\(2) It follows from Proposition \[spectrum-Hi\].

The next proposition is essential to prove our main theorems. Each property in the proposition is also of independent interest.

\[property-PSP-prop\] Let $1\le i\le 3$.

-   Assume $m_i, m_{i, k} \in \mathcal X_i$ with $\lambda_i^km_{i,k}\to \lambda_im_i$ as $k\to\infty$ in $\mathcal{X}_i$. Then $\mu_i^n( \mathcal L_i+\lambda_i^k m_{i,k})\to\mu_i^n(\mathcal L_i+\lambda_i m_i)$ as $k\to\infty$.

-   We have $\mu_i^n(0)<0$ for $i=1$, and $\mu_i^n(0)=0$ for $i=2, 3$.

-   If $\lambda_i^1m_{i}^1(t, x)\leq \lambda_i^2m_{i}^2(t, x)$, then $\mu_i^n(\mathcal L_i+\lambda_i^1m_{i}^1)\leq \mu_i^n(\mathcal L_i+\lambda_i^2m_{i}^2)$. If, in addition, $\lambda_i^1 m_i^1(t,x)\not \equiv \lambda_i^2 m_i^2(t,x)$, then $\mu_i^n(\mathcal L_i+\lambda_i^1m_{i}^1)< \mu_i^n(\mathcal L_i+\lambda_i^2m_{i}^2)$ .

-   If $\mathcal P(m_i)=\int_0^T\max_{x\in \bar D_i}m_i(t, x)dt>0$, then $\mu_i^n(\mathcal L_i+\lambda_i m_i)>0$ for $\lambda\gg1$.

-   The mapping $\lambda_i\in \R^+\mapsto \mu_i^n(\mathcal L_i+\lambda_i m_i)\in \R$ is convex.

\(1) follows from [@RaSh Proposition 3.11].

\(2) In the case of $\lambda_i=0$, $\mu_i^n(0)$ is the principal eigenvalue of $\mathcal L_i$, since the condition (S2) holds. And by Proposition \[spectrum-autonomous\] (1), $\mu_i^n(0)$ is the principal eigenvalue of $K_i-B_i$, associated with an eigenfunction $\phi_i\in  X_i^{++}$, such that \[lambda1=0\] \_D (y-x)\_1( y)dy-\_1( x)=\_1\^n(0)\_1( x) in the Dirichlet boundary condition case, \[lambda2=0\] \_D (y-x)\_2( y)dy-\_Dk(y-x)dy\_2(x)=\_2\^n(0)\_2( x) in the 