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Could anyone see how the following two lines follow from Gronwall lemma? I use the usual differential form gronwall lemma from in Evans book. I do not know how to deal with the term involves the sobolev norm $|\cdot|_1$. enter image description here

I think that, if the second term on the lhs can ve removed, then the rhs shall follow. But I would get $\int_0^T$ rather than $\int_s^T$ in the integrand of the 2nd term...What is going on here...

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I don't see any problem, or perhaps I'm missing something in the question. For example if $0\le T$ and one has $$\frac{d}{dt}(z(t)) \le u(t) z(t) + v(t)$$ on $I=[0, T]$, then if $u,v$ are continuous $$ z(T)\le z(0)e^{\int_0^T u(s) ds}+\int_0^T e^{\int_s^T u(r)dr} v(s) ds $$ A question is what weaker conditions can we give on $u, v$ such that this holds ? I would suggest $u, v \in L^1(I)$

To explain the bounds in the integral, let's recall the theory of the linear first order equation $$\frac{d}{dt}(z(t)) = u(t) z(t) + v(t)$$ If $v$ is identically $0$, the solution is $z(t) = C z_0(t) = C e^{U(t)}$ where $U$ is an antiderivative of $u$ and $C$ is an arbitrary constant. When $v \ne 0$, the constant $C$ must be replaced by an arbitrary antiderivative of $\frac{v(t)}{z_0(t)} = v(t) e^{-U(t)}$. The solution becomes $$ z(t) = \left(C + \int_0^t v(s)e^{-U(s)}ds\right) e^{U(t)} $$ Now choosing $U(t) = \int_0^t u(r) dr$ yields the above formula because $$ e^{U(t)-U(s)} = e^{\int_0^tu(r)dr - \int_0^s u(r) dr} = e^{\int_s^t u(r) dr} $$

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    OK. Here is the thing i do not understand. I would have write has as $z(T)\le e^{\int_0^T\phi(s)ds}(z(0)+\int_0^T v(s)ds)$. When you push the exponential insider the integral, the limit of integration changes to $\int_s^T$, why? Could you elaborate a few lines? – 2017-01-03
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    OK. I follow your advice and now I think I got it. Correct me again otherwise. I do not understand $\frac{v(t)}{z_0(t)} = v(t) e^{-U(t)}$ you wrote. – 2017-01-03
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    If you write $z(t) = c(t) z_0(t)$, then $z'(t) = c'(t) z_0(t) + c(t) z'_0(t) = c'(t)z_0(t) + c(t) u(t) z_0(t) = c'(t) z_0(t) + c(t) z(t)$. It follows that $c'(t) z_0(t) = v(t)$, hence $c(t)$ is an antiderivative of $v(t)/z_0(t)$ where $z_0(t)$ is a solution of the homogenous equation. This is the method of [variation of parameters](https://en.wikipedia.org/wiki/Variation_of_parameters). – 2017-01-03
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    yes. This http://www.math.vt.edu/people/dlr/m2k_dfq_linfrd.pdf explains what is going on. – 2017-01-04
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Basically, if we start with a DE \begin{align*} \frac{d}{dt}z(t)=u(t)z(t)+v(t) \end{align*} This boils down to the proof of integrating factor technique in DE theory. So, rearrange and multiply with $e^{-\int u(t)ds}$, we have \begin{align*} e^{-\int u(t)ds}\frac{d}{dt}z(t)-e^{-\int u(t)ds}u(t)z(t)=e^{-\int u(t)dt}v(t) \end{align*} \begin{align*} \Longrightarrow\,\frac{d}{dt}(z(t)e^{-\int u(t)dt})=e^{-\int u(t)dt}v(t) \end{align*} Now integrate from 0 to $T$ we have \begin{align*} Z(T)e^{-\int_{0}^Tu(s)ds}=z(0)+\int_0^T e^{-\int_{0}^s u(r)dr}v(s)ds \end{align*}

\begin{align*} \Longrightarrow\, z(T)&=z(0)e^{\int_{0}^Tu(s)ds}+\int_0^Te^{\int_{0}^Tu(r)dr-\int_{0}^su(r)dr}v(s)ds\\ &=z(0)e^{\int_{0}^Tu(s)ds}+\int_0^Te^{\int_{s}^Tu(r)dr}v(s)ds \end{align*}