Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Could anyone help on the following problem?

Let R(t) be the solution to the integral equation: $R(t)=1+\int_{0}^{t}\frac{1}{R(s)}ds$, namely $R(t)=\sqrt{2t+1}$. Assume that X is continuous and positive on$[0,\infty)$ and satisfies: $X(t) \leq 1+\int_{0}^{t}\frac{1}{X(s)}ds$ for $t\geq0$. Does $X(t) \leq R(t)$ follow? Either prove it or give a conterexample.

Thank you so much!

share|improve this question
    
Homework? If so, please tag it accordingly. –  Harald Hanche-Olsen Feb 14 '12 at 21:39
    
Thanks you for adding the homework tag. However, you ought to have kept the differential-equations tag too, or perhaps the integral-equations tag would have been better. Multiple tags are good, so long as they are appropriate. –  Harald Hanche-Olsen Feb 14 '12 at 22:18
    
Hint: Try setting $X(t)=1$ for $t \in [0,T]$. Then ask yourself how big the right hand side of the inequality for $X(T)$ will be compared to your exact formula for $R(T)$. Then be creative with $X(t)$ for $t\geq T$. –  Jeff Feb 14 '12 at 22:44
    
@Jeff Thank you! I constructed one counterexample using on your hint. –  user7762 Feb 14 '12 at 23:42

1 Answer 1

up vote 0 down vote accepted

If $X(t)\ge R(t)$ for all $t$, then I think you can easily prove that $X(t)=R(t)$. So if there is a counterexample, you have to look for one which has $X(t)<R(t)$ at least part of the time.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.