# Prove that x(t) is bounded

1. Nov 10, 2013

### motherh

Hi, I want to answer the following question:

x=x(t) is continuous on [0,T) and satisfies

1 ≤ x(t) ≤ C$_{1}$ + C$_{2}$∫$^{t}_{0}$ x(s)(1+logx(s)) ds

for 0 ≤ t < T. Prove x(t) is bounded on [0,T].

Using Gronwall's inequality I get to

x(t) ≤ C$_{1}$exp( C$_{2}$ ∫$^{t}_{0}$ (1+logx(s)) ds )

≤ C$_{1}$exp( C$_{2}$t + C$_{2}$∫$^{t}_{0}$ logx(s) ds )

Can I say that this is less than C$_{1}$exp( C$_{2}$T + ∫$^{T}_{0}$ logx(s) ds ) ?

I'm not too sure where to proceed from here. Would it be helpful to use x(s) > logx(s)?

Any help is appreciated!

Last edited: Nov 10, 2013
2. Nov 15, 2013

### motherh

Can any body help at all? I'm not sure how to bound the integral of logx(s).

3. Nov 15, 2013

### Staff: Mentor

What does logx(s) mean?

4. Nov 15, 2013

### pasmith

You have
$$1 \leq x(t) \leq U(t) = C_1 + C_2 \int_0^t x(s)(1 + \log x(s))\,ds.$$

The worst case scenario is $x(t) = U(t)$, which gives
$$U(t) = C_1 + C_2 \int_0^t U(s)(1 + \log U(s))\,ds.$$

The right hand side is differentiable, so we obtain
$$\frac{dU}{dt} = C_2 U (1 + \log U)$$
This can be solved subject to the initial condition $U(0) = C_1$ to obtain $U(t)$, and one can then check whether
$$\lim_{t \to T^{-}} U(t)$$
is finite.