Is there a Bound for x(t) in Terms of C_1 and C_2?

[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!

 

[motherh]

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

 

[Mark44]

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].

What does logx(s) mean?

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!

 

[pasmith]

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

Gronwall's inequality will not help you here; your bound is not of the correct form.

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
<br /> 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
<br /> \frac{dU}{dt} = C_2 U (1 + \log U)<br />
This can be solved subject to the initial condition U(0) = C_1 to obtain U(t), and one can then check whether
<br /> \lim_{t \to T^{-}} U(t)<br />
is finite.