Background
When applying exponential time difference (ETD) schemes to parabolic problem
\begin{equation}\left\{
\begin{aligned}
&u_t = (\partial_{xx} – \kappa) u + f(u), \quad x \in [a, b], t \in (0, T], \kappa \geq 0\\
&u(x, t=0) = u_0,\\
&u(x = a, t) = u(x = b, t)
\end{aligned}
\right.
\end{equation}
we usually need to calculate the $\varphi$ functions
\begin{align}
\varphi_j(\tau (D – \kappa I)) = \frac{1}{\tau^j}\int_0^\tau\mathrm{e}^{(\tau – s)(D – \kappa I)} \frac{s^{j-1}}{(j-1)!} \mathrm{d} s,
\end{align}
where
\begin{equation}
D = \frac{1}{h^2} \left[
\begin{array}{ccccc}
-2 & 1 & 0 & \cdots & 1 \\
1 & -2 & 1 & \cdots & 0 \\
\vdots & 1 & -2 & 1 & \vdots \\
1 & \cdots & 0 & -2 & 1
\end{array}
\right]
\end{equation}is the second order finite difference discretization for $\partial_{xx}$ and $I \in \mathbb{R}^{N\times N}$ is the identity matrix.
Let $\varphi_0(z) = \mathrm{e}^z$, it can be easily calculated that
\begin{align}
&\varphi_{k+1}(z) = \frac{\varphi_k(z) – \frac{1}{k!}}{z}, \quad \varphi_k(0) = \frac{1}{k!}, \\
&\varphi_1(z) = \frac{\mathrm{e}^z – 1}{z}, \\
& \varphi_2(z) = \frac{\mathrm{e}^z – z – 1}{z^2}.
\end{align}
Since $(D – \kappa I)$ is a diagonally dominant matrix with negative diagonal entries, from Lemmas 3.1 and 3.2 on p. 8 in Maximum principle preserving exponential time differencing schemes for the nonlocal Allen-Cahn equation.
we obtain
\begin{align}
\|\mathrm{e}^{(D – \kappa I) \tau } \|_{l^\infty} \leq \mathrm{e}^{-\kappa \tau},
\end{align}
Problem 1. For the matrix $D$ in Eq. (1.3), is it true that $\|\mathrm{e}^{(D – \kappa I) \tau } \|_{l^\infty} = \mathrm{e}^{-\kappa \tau}$?
Problem 2. For those $\varphi$ functions, is it true that $\|\varphi_k((D – \kappa I) \tau)\|_{l^\infty} = \varphi_k(-\kappa \tau)$?
I wrote a short matlab code to verify
N = 20;
column = [-2 1 zeros(1, N-3) 1];
h = 0.1;
D = toeplitz(column, column) / h^2;
I = eye(N);
kappa = 2; tau = 0.1;
phi_0 = @(z) expm(z);
phi_1 = @(z) (expm(z) - eye(size(z))) / z;
phi_2 = @(z) (expm(z) - z - eye(size(z))) / z^2;
phi_0(-kappa * tau)
norm(phi_0((D - kappa*I)*tau), inf)
phi_1(-kappa * tau)
norm(phi_1((D - kappa*I)*tau), inf)
phi_2(-kappa * tau)
norm(phi_2((D - kappa*I)*tau), inf)
It seems the statements in Problems 1 and 2 are ture.
Best Answer
The statement in Problem 1 is indeed true:
Suppose the size of $D$ is $n×n$. The sum of all entries of $e^{(D-\kappa I)\tau}$ is given by
$$\begin{align} v\, e^{(D-\kappa I)\tau} \,v^T = \sum_{k=0}^∞ \frac{\tau^k}{k!} \; v(D-\kappa I)^Tv^k &= \sum_{k=0}^∞ \frac{\tau^k}{k!} \; v\;\underbrace{(D-\kappa I)^k v^T}_{=\left(-\kappa\right)^kv^T } \\&= \sum_{k=0}^∞ \frac{(-\kappa\tau)^k}{k!} \; \underbrace{v v^T}_{=n} \\&= n e^{-\kappa\tau}. % % \end{align}$$
Now, from the Lemma quoted by the OP, the $l^∞$ operator norm of the matrix is at most $e^{-\kappa\tau}$ : that is to say, each row sums to at most $e^{-\kappa\tau}$ — thus by the pigeonhole principle, each row sums to this value.
In particular, not only is the norm equal, but more can be said: it follows for any $w$ in $\mathbb R^{n,1}$,
$$\|e^{(D-\kappa I)\tau} w\|_{l^∞} = e^{-\kappa\tau}\|w\|_{l^∞}.$$
We can also do a similar thing with columns (corresponding to the $l^1$ norm).
Problem 2:
For this reason, the answer to Problem 2 is also affirmative: let $f_i\in \mathbb R^{n,1}$ denote the column vector with a one in the $i$th place and zeroes everywhere else. The sum of the $i$th row of $\phi_j\left(\tau(D-\kappa I))\right)$ is given by
$$\begin{align} f_i\cdot\phi_j\left(\tau(D-\kappa I))\right)\cdot v^T &= \frac 1 {\tau^j}\int_0^\tau \underbrace{f_i \cdot e^{(\tau-s)(D-\kappa I)}\cdot v^T}_{= e^{-\kappa\tau}}\frac{s^{j-1}}{(j-1)!} ds \\&= \frac 1 {\tau^j}\int_0^\tau e^{-\kappa\tau}\frac{s^{j-1}}{(j-1)!} ds \\&= \phi_j(-\kappa\tau) \end{align}$$
Maximising over $i$ gives the desired result:
$$\|\phi_j\left(\tau(D-\kappa I))\right)\|_{l^∞} = \phi_j(-\kappa\tau).$$