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I am looking for a function $f:[0,1]\rightarrow\mathbb R$ which satifies $\int_{0}^{1}\frac{\sin(f(t)-y)}{2}\,\mathrm dy=f(t)$ for $t\in[0,1]$.

The first thing I do is to define a function $A:M\rightarrow M$ where $M=C([0,1])$ with $A(f)(t)=\int_{0}^{1}\frac{k(f(t),y)}{2}\,\mathrm dy$.

$k$ is Lipschitz continuous to the first variable with Lipschitz constant $1$. $k(x,y)=\sin(x-y)$.

Therefore I am looking for a unique function $f$ which satifies $A(f)=f$. How can it be calculated explicitly?

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Yes I mentioned that. It is Lipschitz with respect to first variable with L=1 –  Voyage Mar 10 '13 at 15:14
    
Oh yes,sorry, I edited it. –  Voyage Mar 10 '13 at 15:16
    
You also wamt $A(f)=f$, and not $K(f)=f$. –  1015 Mar 10 '13 at 15:23

1 Answer 1

up vote 3 down vote accepted

You have $$ |A(f)(t)-A(g)(t)|\leq \frac{1}{2}\int_0^1|k(f(t),y)-k(g(t),y)|dy $$ $$ = \frac{1}{2}\int_0^1|\sin(f(t)-g(t))|dy\leq \frac{1}{2}|f(t)-g(t)| $$ for all $t\in [0,1]$.

So $$ \|A(f)-A(g)\|_\infty\leq \frac{1}{2}\|f-g\|_\infty $$ which shows that $A$ is a contraction on the complete normed vector space $C([0,1])$ equipped with the sup norm.

By Banach fixed point theorem, there exists a unique fixed point $f$. Uniqueness is easy if you plug two fixed points in the estimate. Existence goes by choosing any $f_0$ and then considering the recursive sequence $f_{n+1}=A(f_n)$. It is a Cauchy sequence, so it converges. And it must converge to a fixed point. This is your $f$.

If you start with $f_0(t)=0$, you get a sequence of constant functions. Therefore the limit $f$ is constant equal to $C$. So it satisfies $$ C=\frac{1}{2}\int_0^1\sin(C-y)dy=\frac{1}{2}(\cos(C-1)-\cos C)). $$ I don't think we can find a closed form. But we can cheat and use Wolfram Alpha to find $$ C\simeq -0.364838. $$

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Thank you very much. It helped a lot. –  Voyage Mar 10 '13 at 15:35
    
@Voyage You're welcome. Glad it helped. –  1015 Mar 10 '13 at 15:40

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