I have been asked to complete the following problem using the Gronwall Inequality. I'm not sure I applied the conditions appropriately because I feel there's too many gaps in my understanding of what I've done to say this proof is correct.

Let $f(t, x)$ be continuous on $\mathbb{R} \times \mathbb{R}^n$ and there exists $k_1,k_2 > 0$ such that $$\|f(t, x)\| \le k_1 + k_2\|x\|$$ for any $x\in \mathbb{R}^n$ and for all $t \in [t_0, \infty)$.

Use Gronwall's Inequality to show that the solution of $$\dot x = f(t, x), \,\,\,\,x(t_0) = x_0$$ satisfies $$\|x(t)\| \le \|x_0\|e^{k_2(t - t_0)} + {k_1 \over k_2}\left(e^{k_2(t - t_0)} - 1\right)$$ for all $t \ge t_0$ for which the solution exists.

We received a hint that says to convert the differential equation to an integral equation which then the condition implies that $${k_1 \over k_2} + \|x(t)\| \le \|x_0\| + {k_1 \over k_2} + k_2 \int_{t_0}^t \left[{k_1 \over k_2} + \|x(s)\|\right]ds.$$

I'm unfortunately stuck here. I can convert the differential equation to an integral equation quite simply: $$\dot x = f(t, x), \,\,\,\, x(t_0) = x_0 \,\,\, \iff \,\,\, x(t) = x_0 + \int_{t_0}^tf(s, x(s))\,ds.$$

But I'm not understanding the motivation behind the hint. Here's what I did initially: $$\begin{align}\|x(t)\| & = \left\|x_0 + \int_{t_0}^tf(s, x(s))\,ds\right\| \\ &\le \|x_0\| + \int_{t_0}^t\|f(s, x(s))\| \, ds \\ &\le \|x_0\| + \int_{t_0}^t\left[k_1 + k_2\|x(s)\|\right]ds \\ \implies {k_1 \over k_2} + \|x(t)\| & \le \|x_0\| + {k_1 \over k_2} + k_2 \int_{t_0}^t\left[{k_1 \over k_2} + \|x(s)\|\right]\,ds,\end{align}$$

as the hint implies. We can see that since $\|x_0\|$ and $k_2$ are constant, by the Gronwall Inequality, $$\|x(t)\| \le \|x_0\|e^{k_2(t - t_0)} + {k_1 \over k_2}\left(e^{k_2(t - t_0)} - 1\right).$$

The last step feels like magic though, as if I just constructed the correct answer (probably because I did). Can someone explain how we get to the last step appropriately?


1 Answer 1


To see what happens, set $\varphi(t) = \frac{k_1}{k_2} + \|x(t)\|$. Then your last inequality $$ \frac{k_1}{k_2} + \|x(t)\| \leq \|x_0\| + \frac{k_1}{k_2} + k_2 \int_{t_0}^{t} \, \left[ \frac{k_1}{k_2} + \|x(s)\| \right] ds$$ becomes

$$\varphi(t) \leq \Big( \|x_0\| + \frac{k_1}{k_2}\Big) + \int_{t_0}^{t} k_2 \, \varphi(s) \, ds$$ which is the assumption in the integral form of Gronwall's inequality. Now according to Gronwall's inequality, $\varphi(t)$ should satisfy the inequality $$\varphi(t) \leq \left(\|x_0\| + \frac{k_1}{k_2} \right) \exp\left({\int_{t_0}^{t} k_2 \, ds}\right) = \left(\|x_0\| + \frac{k_1}{k_2} \right)\, e^{k_2 (t-t_0)}.$$ Now recall what $\varphi(t)$ was and you get $$\frac{k_1}{k_2} + \|x(t)\| \leq \left(\|x_0\| + \frac{k_1}{k_2} \right)\, e^{k_2 (t-t_0)}$$ which turns into $$\|x(t)\| \leq \left(\|x_0\| + \frac{k_1}{k_2} \right)\, e^{k_2 (t-t_0)} - \frac{k_1}{k_2} = \|x_0\| \, e^{k_2 (t-t_0)} + \frac{k_1}{k_2} \big(e^{k_2 (t-t_0)} - 1\big)$$

  • $\begingroup$ I greatly appreciate your answer. I actually figured out the answer all by myself by complete accident, and I was about to close the question. Thanks for your help! $\endgroup$
    – Decaf-Math
    Sep 8, 2016 at 13:27
  • $\begingroup$ It's good that you figured it out. Now you can compare with another solution :). Cheers! $\endgroup$ Sep 8, 2016 at 13:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .