Proving Gronwalls Extended Lemma

Let $r(t)$ and $v(t)$ be continuous for $|t-t_0|\leq\delta$, where $v(t)\geq0$, and suppose that $$0\leq r(t) \leq \varepsilon + \delta \left\| \int_{t_0}^t v(s)r(s)\,ds \right\|.$$ Show that $$r(t)\leq\varepsilon e^{\delta \left| \int_{t_0}^t v(s) \, ds \right|}.$$

This is a problem out of Grimshaw's Nonlinear ODE book, he refers to it as Gronwall's Extended Lemma. I'm having a hard time showing this result. Here is the work I have done so far:

It is sufficient to consider the case $t\geq t_0$. The proof for $t\leq t_0$ is analogous. Let $$R(t):= \int_{t_0}^t v(s)r(s)\,ds.$$ Then, $R(t)$ is continuous and differentiable since $r$ and $v$ are continuous on a $\delta$ interval about $t_0$. Then, by the chain rule, product rule, and fundamental theorem of calculus we have that: $$R'(t) = r(t) - \int_{t_0}^t v(s)r(s)ds. \hspace{1cm} \text{for some } s \in I: = (t_0-\delta,t_0+\delta)$$ Hence, our inequality takes the form: $$0\leq R'(t) \leq \varepsilon + \delta \|\ R(t) \|\ .$$ Now we'll multiply both sides by the non-negative function, $\exp\left(-\delta \left| \displaystyle\int_{t_0}^t v(s) \, ds \right| \right)$, so that $$0 \leq \left[ R(t)e^{-\delta \left| \int_{t_0}^t v(s)\,ds \right|} \right]' \leq \varepsilon e^{-\delta \left| \int_{t_0}^t v(s) \, ds \right|}.$$ Integrating from $t_0$ to $t$ yields: $$0\leq R(t)e^{-\delta \left| \int_{t_0}^t v(s) \, ds \right| } \leq \delta \int_{t_0}^t v(s) \left( e^{-\delta \left| \int_{t_0}^t v(q) \, dq \right|} \right) \, ds.$$ Using the definition of $R(t)$ from the first step, and then this inequality we obtain: $$0 \leq \int_{t_0}^t v(s)r(s) \left( e^{-\delta\left| \displaystyle\int_{t_0}^t v(q) \, dq \right|} \right) \, ds \leq \delta \int_{t_0}^t v(s) \left( e^{-\delta \left| \int_{t_0}^t v(q) \, dq \right|} \right) \, ds.$$ Everything from this point that I've tried does seem to work (i.e. get me the desired the inequality).

EDIT: New proof:

It is sufficient to consider the case $t\geq t_0$. The proof for $t\leq t_0$ is analogous. Let $$\phi(t):= \ln\left(\epsilon + \delta \int_{t_0}^t v(s)r(s)\right)$$ Then, $\phi(t)$ is continuous and differentiable since $r$ and $v$ are continuous on a $\delta$ interval about $t_0$. Then, by fundamental theorem of calculus we have that: $$\phi'(t) = \dfrac{\delta v(t)r(t)}{\epsilon + \delta\displaystyle\int_{t_0}^t v(s)r(s)} \leq \delta v(t) \hspace{1cm} \text{for some} \space\ s\in I:=(t_0-\delta,t_0+\delta)$$ Therefore, $$\phi'(t) \leq \delta v(t)$$ Now integrating from $t_0$ to $t$ we have that $$\phi(t) - \phi(t_0) \leq \delta \left| \int_{t_0}^t v(s)ds \right|$$ Note $\phi(t_0)=\ln(\epsilon)$. Therefore, $$\phi(t) \leq \ln(\epsilon) + \delta \left| \int_{t_0}^t v(s)ds \right|$$ Substituting back the definition of $\phi(t)$ and exponentiating we have that: $$\epsilon + \delta \left\|\ \int_{t_0}^t v(s)r(s) \right\|\ \leq \epsilon e^{ \delta \left| \displaystyle\int_{t_0}^t v(s)ds \right|}$$ But by hypothesis, $$0\leq r(t) \leq \epsilon + \delta \left\| \displaystyle\int_{t_0}^t v(s)r(s)ds \right\|.$$ Therefore, we've established that $$r(t)\leq\epsilon \space\ e^{\delta \left| \displaystyle\int_{t_0}^t v(s)ds \right|}.$$

For $t>t_0$ consider $$\phi(t)=\ln\left(ε+δ\int_{t_0}^tv(s)r(s)ds\right).$$ Then $$\phi'(t)=\frac{δv(t)r(t)}{ε+δ\int_{t_0}^tv(s)r(s)ds}\le δv(t)$$ which can now be integrated to obtain the claim.
Your equation for $R'(t)$ does not make any sense, there is no free variable $s$ to bound, and no product to apply any product rule. The next step in the third equation also does not follow neither from the second equation nor is it correct. You should get $$R'(t)=v(t)r(t)\le v(t)(ε+δR(t))$$ where then you get the claim either per separation as above or per the application of the integrating factor $\exp(-\int_{t_0}^tv(s)ds)$.
• I believe I got it with your $\phi(t)$. I edited to include my proof now. Thank you!! – Dragonite Feb 2 '18 at 19:40