# Simple question

• Jun 22nd 2010, 10:59 AM
mathman88
Simple question
Why does $f'(x)=0 \; \forall \; x\in\mathbb{R}\implies f(x)=c$?
• Jun 22nd 2010, 11:16 AM
Ackbeet
Integrate both sides. What do you get?
• Jun 22nd 2010, 12:59 PM
mathman88
I understand that, but does that imply that's the unique solution?
• Jun 22nd 2010, 01:01 PM
Ackbeet
Well, you have an infinite family of solutions depending on your initial condition. However, the fact that the RHS of your DE, namely 0, is Lipschitz and continuous, is enough to guarantee the uniqueness of the solution to an initial value problem (an initial value problem is your original DE plus an initial condition to determine the constant of integration). Does that make sense?
• Jun 22nd 2010, 01:23 PM
mathman88
Why does Lipschitz continuity guarantee uniqueness of the solution (up to a constant)?
• Jun 22nd 2010, 01:31 PM
Ackbeet
Because of the Picard-Lindelof theorem. If you're curious, by all means investigate the proof of that theorem.
• Jun 26th 2010, 03:59 PM
HallsofIvy
Quote:

Originally Posted by mathman88
Why does $f'(x)=0 \; \forall \; x\in\mathbb{R}\implies f(x)=c$?

For this specific question, you don't need the full "Picard-Lindelof", just the mean value theorem.

Let $x_0$ and $x_1$ be two distinct values of x: $x_0\ne x_1$.

By the mean value theorem, $\frac{f(x_1)- f(x_0)}{x_1- x_0}= f'(c)$ where c is some number between $x_0$ and $x_1$. If f'(x)= 0 for all x, then f'(x)= 0 so $f(x_1)- f(x_0)= 0$ and $f(x_1)= f(x_0)$.