6
$\begingroup$

Let X be the set of continuous real valued functions defined on $[0,\frac{1}{2}]$ with the metric $d(f,g):=\sup_{x\in[0,\frac{1}{2}]} |f(x)-g(x)|$.

Define the map $\theta:X\rightarrow X$ such that $$\theta (f)(x)=\int_{0}^{x} \frac{1}{1+f(t)^2} dt$$.

I need to show that $\theta$ is a contraction mapping and that the unique fixed point satisfies $f(0)=0$ and the differential equation $\frac{df}{dx}=\frac{1}{1+f(x)^2}$

So I'm pretty lost on this, I'm quite comfortable proving that things like $f(x)=1+\frac{1}{1+x^4}$ are contraction mapping but I'm a bit confused with this, so

$$d(\theta(f),\theta(g))=\sup|\int_{0}^{x} \frac{1}{1+f(t)^2}-\frac{1}{1+g(t)^2} dt|$$

and so this is: $$=\sup|\int_{0}^{x} \frac{(g(t)-f(t))(g(t)+f(t))}{(1+f(t)^2)(1+g(t)^2)} dt|$$

but I am unsure where to go from here, I know that I need to get this to be something like:

$$\leq\alpha\sup|f(x)-g(x)|$$ where $\alpha$ is the contraction constant?

Thanks very much for any help

$\endgroup$
5
$\begingroup$

you can show that the function $g(a,b)={{a+b}\over{(1+a^2)(1+b^2)}}$ has a maximum at $a=b=1/3$ with maximum value of $27/50$. The last integral is less than $ \sup \int_0^x 27/50 (g(t)-f(t)) dt$. Leading to the result.

$\endgroup$
0
$\begingroup$

In fact, Picard-Lindelof Theorem tells you that what you really need is $H(x)=\dfrac{1}{1+x^2}$ being Lipshitz continuous in a neighbourhood of $0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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