How do we find the fixed point of the map F:X-->X for F(f)(x)=1+ integration of f(x-y)dy from 0 to x X=C[0,1/2] Thank you
Last edited by jerad; Jan 15th 2010 at 01:17 PM.
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Originally Posted by jerad How do we find the fixed point of the map F:X-->X for F(f)(x)=1+ integration of f(x-y)dy from 0 to x X=C[0,1/2] Thank you You want such that . Put in the integral, then differentiate the equation you get, this gives you a differential equation satisfied by .
thank you I will try now
Hi LAURENT IS THAT GIVING f '(x)= -f(x)? and the unique fixed point is e^(-x) am I right?
Last edited by jerad; Jan 14th 2010 at 11:57 AM.
Originally Posted by jerad Hi LAURENT IS THAT GIVING f '(x)= -f(x)? and the unique fixed point is e^(-x) am I right? No, it is f'(x)=f(x). The integral becomes , there's no minus sign.
why is the limits from x to 0?
Originally Posted by jerad why is the limits from x to 0? It comes from the change of variable: if y=0, then z=x, and vice-versa.
oh yes. I substitute wrong. Thank you very much. And how do we show F is a contraction on X?
Originally Posted by jerad oh yes. I substitute wrong. Thank you very much. And how do we show F is a contraction on X?
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