ODE system with lower triangular coefficient matrix

Question

Let $v_k$ satisfies the following equation

$\frac{d}{dt} v_k(t)=\sum_{i=0}^{k-1}v_k(t)$

with initial data $v_0(t=0)=1$ and $v_j(t=0)=0$.

Notice that the coefficient matrix $A$ of this ODE system (probably infinite dimensional) is a lower triangular matrix with entries one.

I was trying to find an upper bound of $v_k(t)\leq P(t)e^{at}$ with $P$ some polynomial of $t$ (maybe with coefficient depends on $k$ or maybe not) and $a$ a constant that does not depend on $n$.

(This is equivalent to the original question I asked, but to avoid confusion I write the ODE in recursive form.)

Answer 1

We are given the system of ODE's $$ \frac{dv_k}{dt} = \sum_{i=0}^{k-1} v_i(t),\qquad k \geq 1, $$ with $v_0(t)$ some integrable function in $[0,\infty)$, with initial condition $v_k(0) = 0 $ for all $k\geq 1$, and $v_0(0)=1$.

Let $||v_0||_t = \int_0^t |v_0(\tau)|d\tau$.

Claim. The estimate \begin{equation} |v_k(t)| \leq ||v_0||_t + (2^k-k-1)e^t ||v_0||_t,\qquad k \geq 1 \end{equation} holds for all $t\geq 0$.

Proof. By induction in $k$. First note that $$ \frac{dv_{k+1}}{dt} = \frac{dv_{k}}{dt} + v_{k},\qquad k\geq 1. $$ Integrating on both sides, and using the given initial conditions, $$ v_{k+1}(t) = v_{k}(t) + \int_0^t v_k(\tau) d \tau,\qquad k\geq 1. $$ For $k=1$, the inequality obviously holds. For $k\geq 1$, \begin{align} |v_{k+1}(t)| &\leq |v_k(t)| + \int_0^t |v_k(\tau)| d\tau \\ \\ &\leq \left(1 + (2^k-k-1)e^t \right) ||v_0||_t + \int_0^t \left[ 1 + (2^k-k-1)e^\tau \right] ||v_0||_\tau d\tau \\ &\leq \left( 1 + 2^k e^t -k e^t - e^t + t + (2^k-k-1)(e^t-1) \right) ||v_0||_t \\ &\leq \left( 1 + (2^{k+1}-(k+1)-1) e^t \right) ||v_0||_t. \end{align} Where I used $t\leq e^t-1$ from the third to last line (and some simplifications).

For the exact solution of the system I got \begin{equation} v_k(t) = \int_0^t v_0(\tau) d \tau + \sum_{j=1}^{k-1} \sum_{m=1}^j \binom{j}{m} \frac{1}{m!} \int_0^t (t-\tau)^m v_0(\tau) d \tau, \end{equation} $k\geq 1$, $t\geq 0$. This can be obtained by the Laplace method.

Answer 2

We can actually solve the system explicitly to order $t^2$, obtaining $$ v_1(t) = 1, \; v_j(t) \ge t + \dfrac{j-2}{2} t^2 \ \text{for}\ j \ge 2, \; t \ge 0$$ (i.e. $v_j(t)$ is a polynomial in $t$ with nonnegative coefficients and lowest-order terms $t + ((j-2)/2) t^2$). Note also that $v_j(t)$ is the same for all $n \ge j$. Thus it is impossible to get a bound that is uniform in $n$: for any fixed $a$ and $t > 0$, $e^{-at} v_j(t) \to +\infty$ as $j \to \infty$.