Upper bound on partial derivatives gives exponential upper bound on function

Question

Problem Statement

Let $q : \mathbb{R} \rightarrow \mathbb{R}^n$ be such that its componentwise derivatives are bounded:

$$\frac{d}{dt}q_i(t) \leq \beta A q_i(t)$$

And let $\lambda$ be the largest eigenvalue of $A$.

I want to show that $\sum_i q_i(t) \leq C e^{\beta \lambda t} \sum_i q_i(0)$ for some constant $C > 0$.

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What I've tried

I think that since $\lambda$ is the largest eigenvalue of $A$, we can say that $\|Ax\|_1 \leq \|\lambda x\|_1$, where $\|\cdot\|_1$ is the L1 norm.

This gives us $\frac{d}{dt}q(t) \leq \beta \lambda \|q(t)\|$.

Now, I want to use an exponential function to bound $q(t)$ in terms of $q(0)$. My rationale is this: $e^{x}$ is the greatest function that has a derivative no greater than $e^{x}$.

Does it seem like I'm on the right track, or am I missing something?

Answer 1

This is a vector-valued generalization of something called Grönwall's inequality:

If $u:I\to\mathbb{R}$ is differentiable in the interior of the closed interval $I = [a,b]$ and satisfies the differential inequality $u'(t) \leq \beta(t)u(t)$ for a continuous function $\beta$, then $u$ is bounded by the solution of the differential equation $v'(t) = \beta(t)v(t)$, i.e. $u(t) \leq u(a)\exp\left(\int_a^t \beta(s)ds \right)$.

The idea of the proof is basically to do the same thing for the differential inequality as you would for the differential equation. $v'(t) = \beta(t)v(t)$ is, among other things, a first-order linear equation, and that means you can solve it by introducing the integrating factor $\mu(t) = \exp(-\int_a^b \beta(s)ds)$. Now, if we try the same thing for the inequality $u'(t) \leq \beta(t)u(t)$, and rearrange a little, we obtain: $$ (\mu(t)u(t))' = \mu(t)u'(t) - \beta(t)\mu(t)u(t) = \mu(t)(u'(t)-\beta(t)u(t)) \leq 0. $$ Now, you can integrate this inequality from $a$ to $t$, and use the fundamental theorem of calculus with the leftmost term to conclude: $$ \mu(t)u(t) - \mu(a)u(a) \leq 0. $$ But $\mu(a) = 1$ from its definition, so we can rearrange to obtain $$ u(t) \leq \frac{1}{\mu(t)}u(a) = u(a)\exp\left( \int_a^t \beta(s)ds\right). $$

The idea for your problem should basically be to generalize this proof to the vector-valued case. The question is basically to find the correct integrating factor; I would expect it to be a type of matrix exponential. You should be able to get a more precise estimate using this rather than working with norms as you've done so far. You'll also need to do some further analysis to see how the largest eigenvalue enters the picture. I would suggest trying this first in the case where $A$ is diagonalizable and $q$ can be decomposed along an eigenbasis to see what's going on - the generalization from there should be relatively straightforward using a decomposition into generalized eigenvectors.