The exponential functions are essential ingredients of Exponential Time Difference (ETD) integrators. Let us introduce following functions \begin{equation*} \varphi_k(z) = \int_0^1 \mathrm{e}^{(1 - s) z} \frac{s^{k-1}}{(k-1)!} \mathrm{d} s > 0, \quad k \geq 1, \end{equation*} which satisfy the recursion \begin{equation} \varphi_k(z) = \frac{\varphi_{k-1}(z) - \frac{1}{(k-1)!}}{z}, k \geq 1; \quad \text{~with~} \varphi_0(z) = \mathrm{e}^z. \end{equation} and we have $\varphi_1(z) = \frac{\mathrm{e}^z - 1}{z}$, $\varphi_2(z) = \frac{\mathrm{e}^z - z - 1}{z^2}$.
Consider a matrix $L \in \mathbb{R}^{N \times N}$ which is weak diagonally dominant in the form: \begin{equation*} L = \left[ \begin{array}{cccccc} -2 & 1 & 0 & \cdots & 0 & 1 \\ 1 & -2 & 1 & 0 & \cdots & 0\\ 0 & 1 & -2 & 1 & \ddots & \vdots \\ \vdots & \vdots & \ddots & \ddots & \ddots & \vdots \\ 0 & 0 & \cdots & 1 & -2 & 1 \\ 1 & 0 & \cdots & 0 & 1 & -2 \end{array} \right] \end{equation*} Let $L_\kappa = L - \kappa I $ where $I \in \mathbb{R}^{N\times N}$ is the identity matrix and $\kappa > 0$.
When $\tau > 0$, it can be derived that every entry of $\varphi_k(\tau L_\kappa)$ is positive and each row sums to $\varphi_k(-\tau \kappa)$ and $\|\varphi_k(\tau L_\kappa)\|_\infty = \varphi_k(-\tau \kappa)$.
Consider the function $f(z) = \varphi_1(z) - z \varphi_1(z) \varphi_2(z)$, then its matrix form is: $$f(\tau L_\kappa) = \varphi_1(\tau L_\kappa) - \tau L_\kappa \varphi_1(\tau L_\kappa) \varphi_2(\tau L_\kappa) = 2 \varphi_1(\tau L_\kappa) - \varphi_1(\tau L_\kappa)^2,$$ it can be derived that each row of $f(\tau L_\kappa)$ sums to $f(-\tau \kappa)$.
Question: Is it true that all entries of the matirx $f(\tau L_\kappa)$ are non-negative?
If so, then the infinity norm $f(\tau L_\kappa)$ also equals $f(-\tau \kappa)$, that is $\|f(\tau L_\kappa)\|_{\infty} = f(-\tau \kappa)$.
Here is the matlab code to do the verification.
phi_1 = @(z) (expm(z) - eye(size(z))) / z;
f = @(z) 2*phi_1(z) - phi_1(z)^2;
kappa = 1;
L = [-2 1 0 0 1; 1 -2 1 0 0; 0 1 -2 1 0; 0 0 1 -2 1; 1 0 0 1 -2];
L_k = L - kappa * eye(size(L));
norm(f(L_k), inf)
sum(f(L_k), 2)
f(-kappa)
Thank you very much for any clues!
Here is a start for the solution. It is getting late here. I will update with the rest tomorrow when I will have a clearer idea on how to do it.
Firstly, note that the matrix $L$ is Metzler, therefore $\exp(Lt)$ is nonnegative for all $t\ge0$.
Moreover, the matrix $L_\kappa$ is Hurwitz stable with $-\kappa$ as dominant eigenvalue. Moreover, since
$$\varphi_1(\tau L_\kappa)=\int_0^1\exp(\tau L_\kappa s)ds$$ and the integrand is nonnegative, then $\varphi_1(\tau L_\kappa)$ is nonnegative as well.