Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Have $f'(t) = f(t), f(0) = 1$ where $f \in C([0,c])$ where $0 < c < 1$. Using uniform norm, and Banach fixed point theorem with $(Tf)(x) = 1 + \int_0^x f(t) dt$. Show unique solution exists for the ODE.

I began by showing that $T$ is a contraction. As $||Tf_1 - Tf_2|| = ||\int_0^x (f_1 - f_2)(t)dt|| \le \int_0^x||f_1 - f_2||dt \le c \cdot ||f_1 - f_2||$ . Since $C([0,c])$ is complete, we then apply Banach fixed point theorem to get that there exists a unique $f^* \in C([0,c])$ such that $f^*(t) = 1 + \int_0^tf^*(s)ds$. And therefore, we can apply the fundamental theorem of calculus to see that $f^*(t)$ is differentiable on $(0,c)$ and satisfies the ODE on this interval.

My result is different from Picard's existence in that Picard's theorem works in an epsilon-interval around the initial condition and mine works further. Am I doing something wrong here? Or am I just able to get a stronger result because of more specific assumptions?

share|cite|improve this question

You know what exactly the right hand side is, so you have an explicit bound on $\|Tf_1-Tf_2\|$. However, you still need $c<1$ in order to have a contraction. This means in some sense the epsilon-interval had already been adjusted from beginning in the exercise.

share|cite|improve this answer

Your Answer


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.