Using Gronwall Inequality to Prove Uniqueness/Existence

[JMeyer]

Hello,

I need to use the Gronwall inequality to discuss existence/uniqueness of the solution to the initial value problem:

x'(t)=xsin(tx) + t with initial condition x(t₀) = x₀.

I can convert this into integral form

$$x(t) = x_0 + \int\limits_{t_0}^{t} xsin(sx) + s ds$$

Which of course can reduce to

$$x(t) = x_0 + \frac{t^2}{2} - \frac{t_0^2}{2} + \int\limits_{t_0}^{t} xsin(sx) ds$$

But I'm unsure of how to progress from here. Also any links to other good places to study this material would help.

Thank you.

 

[jvicens]

Hello,

Thank you for reaching out to the forum for help with your problem. The Gronwall inequality is a powerful tool in analyzing the existence and uniqueness of solutions to initial value problems. To use it in this case, we need to first rewrite the integral form of the problem in a form that is suitable for the inequality.

We can do this by using the fundamental theorem of calculus to rewrite the integral as:

x(t) = x_0 + \frac{t^2}{2} - \frac{t_0^2}{2} + \int\limits_{t_0}^{t} xsin(sx) ds = x_0 + \frac{t^2}{2} - \frac{t_0^2}{2} + \frac{1}{2} \int\limits_{t_0}^{t} \frac{d}{ds}(s^2xsin(sx)) ds

= x_0 + \frac{t^2}{2} - \frac{t_0^2}{2} + \frac{1}{2} \left(s^2xsin(sx) \bigg|_{t_0}^t - \int\limits_{t_0}^{t} s^2cos(sx) ds \right)

Now, we can use the Gronwall inequality which states that for a continuous function f(t) and constants a and b, if

f(t) \leq a + b \int\limits_{t_0}^{t} f(s) ds

then

f(t) \leq a e^{b(t-t_0)}

In our case, we have f(t) = x(t), a = x_0 + \frac{t_0^2}{2}, and b = \frac{1}{2}.

Therefore, we have:

x(t) \leq x_0 + \frac{t_0^2}{2} + \frac{1}{2} \left(\frac{t^2}{2} + \frac{t_0^2}{2} + \int\limits_{t_0}^{t} s^2cos(sx) ds\right)

Applying the Gronwall inequality, we get:

x(t) \leq \left(x_0 + \frac{t_0^2}{2}\right