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I am working on the following problem: Suppose $f\in C(U,\mathbb{R}^n)$ satisfies $|f(t,x)-f(t,y)|\leq L(t)|x-y|$. Show that the solution $\varphi(t,x_0)$, to the equation $$x'=f(t,x) \ \ \ x(t_0)=x_0$$ satisfies $$|\varphi(t,x_0)-\varphi(t,y_0)|\leq |x_0-y_0|e^{|\int_{t_0}^tL(s)ds}$$ From the textbook, we know that $$|\varphi(t,x_0)-\varphi(t,y_0)|\leq |x_0-y_0|e^{L|t-t_0|}$$ where $$L=\sup_{(t,x)\neq (t,y)}|\frac{f(t,x)-f(t,y)}{x-y}|$$

How would I use the above result to prove the problem

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  • $\begingroup$ Do you have some conditions on $L$? (Continuity, integrability, etc.) $\endgroup$
    – copper.hat
    Mar 5, 2015 at 20:57

1 Answer 1

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This is known (among other things) as the Bellman Gronwall lemma.

I need to add some assumptions on $L$. I will assume that $L$ is non negative and is integrable on compact intervals. In particular, this means that the function $t \mapsto \int_{t_0}^t L(\tau) d \tau$ is absolutely continuous and has derivative $L(t)$ for ae. $t$. I am also assuming that the solutions $x,y$ are continuous.

Let $\delta(t) = \|x(t)-y(t)\|$. By the usual estimates on $x(t) = x_0 + \int_{t_0}^t f(\tau, x(\tau))d \tau$ we obtain the following inequality: $\delta(t) \le \Delta(t)$, where $\Delta(t) = \|x_0-y_0\| + \int_{t_0}^t L(s) \delta(s) ds$.

Let $\zeta(t)= \Delta(t) e^{-\int_{t_0}^t L(\tau) d \tau}$. Then $\zeta$ is absolutely continuous, and we have $\dot{\zeta}(t) = e^{-\int_{t_0}^t L(\tau) d \tau}L(t)(\delta(t) - \Delta(t)) \le 0$ for ae. $t$, and so $\zeta(t) \le \zeta(t_0)$ for $t \ge t_0$.

Expanding this yields $\|x(t)-y(t)\| = \delta(t) \le \Delta(t) \le \Delta(t_0) e^{\int_{t_0}^t L(\tau) d \tau} = \|x_0-y_0\|e^{\int_{t_0}^t L(\tau) d \tau} $ for $t \ge t_0$.

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  • $\begingroup$ Thanks, I appreciate your answer $\endgroup$
    – nagnag
    Mar 6, 2015 at 20:43

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