How Can Gronwall's Inequality Help Prove a Mathematical Lemma?

[kalish1]

I need to prove the following with the help of Gronwall's inequality:

If, for $t \in [a,b]$, $$\phi(t) \leq \delta_2(t-a) + \delta_1 \int_{a}^{t}\phi(s)ds + \delta_3,$$ where $\phi$ is a nonnegative continuous function on $[a,b]$, and $\delta_1>0, \delta_2 \geq 0$, and $\delta_3 \geq 0$ are constants, then $$\phi(t) \leq (\frac{\delta_2}{\delta_1}+\delta_3)e^{\delta_1(t-a)} - \frac{\delta_2}{\delta_1}.$$

Here is the version of Gronwall's inequality that I am using:

**Gronwall:** Suppose that $a<b$ and let $\alpha, \phi,$ and $\psi$ be nonnegative continuous functions defined on the interval $[a,b]$. Moreover, suppose that $\alpha$ is differentiable on $(a,b)$ with nonnegative continuous derivative $\dot\alpha$. If, for all $t \in [a,b]$, $$\phi(t) \leq \alpha(t) + \int_{a}^{t}\psi(s)\phi(s)ds,$$ then $$\phi(t) \leq \alpha(t)e^{\int_{a}^{t}\phi(s)ds}$$ for all $t \in [a,b]$.**My attempts:**

I've tried everything with around 6 pages of work...

$1.$ I set $\alpha = (\frac{\delta_2}{\delta_1}+\delta_3)$, and $\phi(t) = \delta_1$ as per the formula, and tried to match the formulae, but without success.

$2.$ I used the following link: Averaging Methods in Nonlinear Dynamical Systems - Jan A. Sanders, Ferdinand Verhulst, James Murdock - Google Books

and followed their suggestion (Lemma $1.3.3$), but that did not work. It looks like something is messed up in their proof of Lemma $1.3.3$.

$3.$ I set the top two inequalities (in my post) equal to each other and then solved for $\phi(t)$, but that yielded me something $\alpha(t) = \frac{\delta_2}{\delta_1}+\delta_3$, which does not work.

$4.$ I set $\phi(t) =$ R.H.S. of the top inequality in my post and then differentiated, and tried to derive the result by an approach similar to the proof of Lemma 1.3.1, but that gave me unsatisfactory results.

What more can I do?

I have crossposted this question here: homework - Need help with proving a lemma - Mathematics Stack Exchange

 

[Opalg]

I need to prove the following with the help of Gronwall's inequality:

If, for $t \in [a,b]$, $$\phi(t) \leq \delta_2(t-a) + \delta_1 \int_{a}^{t}\phi(s)ds + \delta_3,$$ where $\phi$ is a nonnegative continuous function on $[a,b]$, and $\delta_1>0, \delta_2 \geq 0$, and $\delta_3 \geq 0$ are constants, then $$\phi(t) \leq (\frac{\delta_2}{\delta_1}+\delta_3)e^{\delta_1(t-a)} - \frac{\delta_2}{\delta_1}.$$

Here is the version of Gronwall's inequality that I am using:

**Gronwall:** Suppose that $a<b$ and let $\alpha, \phi,$ and $\psi$ be nonnegative continuous functions defined on the interval $[a,b]$. Moreover, suppose that $\alpha$ is differentiable on $(a,b)$ with nonnegative continuous derivative $\dot\alpha$. If, for all $t \in [a,b]$, $$\phi(t) \leq \alpha(t) + \int_{a}^{t}\psi(s)\phi(s)ds,$$ then $$\phi(t) \leq \alpha(t)e^{\int_{a}^{t}{\color{red}{\psi}}(s)ds}$$ for all $t \in [a,b]$.

You have quoted Grönwall's inequality wrongly. The $\phi$ in the exponential should be $\psi$, as indicated in red above.

You can write the term $\delta_2(t-a)$ as \(\displaystyle \delta_1\int_a^t \frac{\delta_2}{\delta_1}\,ds.\) Now replace the function $\phi(t)$ by the new function $\phi(t) + \dfrac{\delta_2}{\delta_1}$. Take $\alpha(t)$ to be the constant $\dfrac{\delta_2}{\delta_1} + \delta_3$ and $\psi(t)$ to be the constant $1$, and apply Grönwall's inequality.

 

[kalish1]

You have quoted Grönwall's inequality wrongly. The $\phi$ in the exponential should be $\psi$, as indicated in red above.

You can write the term $\delta_2(t-a)$ as \(\displaystyle \delta_1\int_a^t \frac{\delta_2}{\delta_1}\,ds.\) Now replace the function $\phi(t)$ by the new function $\phi(t) + \dfrac{\delta_2}{\delta_1}$. Take $\alpha(t)$ to be the constant $\dfrac{\delta_2}{\delta_1} + \delta_3$ and $\psi(t)$ to be the constant $1$, and apply Grönwall's inequality.

Sneaky! How did you know to make those substitutions and transformations??

 

[kalish1]

Sneaky! How did you know to make those substitutions and transformations??

By the way, I believe $\psi(t) = \delta_1$.

 

[Opalg]

Sneaky! How did you know to make those substitutions and transformations??

By looking at the answer and working backwards! The answer includes the exponential of $\delta_1(t-a)$. Comparing that with the exponential of $\int_a^t \psi(s)\,ds$ that occurs in the conclusion of Grönwall's inequality, you can guess that $\int_a^t \psi(s)\,ds = \delta_1(t-a)$, so that $\psi(t) = \delta_1$. You can guess which functions to take for $\alpha(t)$ and $\phi(t)$ in a similar way.

Just one of those tricks of the trade. (Evilgrin)