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As the title suggests this is for exam revision so answers and not just hints are okay. I've got the current two problems I'm doing:

$\mathbf{1}.$ Show that there exist a unique continuous function $f:[0,1]\to\mathbb{R}$ such that $$f(x)=\tfrac{1}{3}f(\tfrac{1}{2}(x+1))+4\cos x$$

and a second problem

$\mathbf{2}.$ Show there exists a unique continuously differentiable function $f:[0,4]\to\mathbb{R}$ such that $$f'(t)=|t-1|-5\cos(f(t))$$ and $f(0)=2.$

For 1, I believe the Banach contraction theorem may be used (since the previous question is "state the Banach contraction theorem/fixed-point principle") but I'm really not sure how to go about applying it to this problem. For the second, it looks like this is of the form of something that would be proved using Picard's theorem. I'm not particularly strong with these types of problem, so I'm not sure of how to solve these, help or example solutions would be very appreciated to help me prepare for this exam. Thanks to any helpers.

Edit: Here is my attempt at a solution for $\mathbf{2}.$ If wrong, corrections appreciated.

If we define $F:[0,4]\times\mathbb{R}\to\mathbb{R}$ by $F(t,f(t))=|t-1|-5\cos f(t)$ then we can show that for all $t\in [0,4],\xi,\eta\in\mathbb{R},$ that $\|F(t,\xi)-F(t,\eta)\|\leq 5\|\xi-\eta\|,$ since $\cos x$ is $1$-Lipschitz. So if we take $0\in[0,4]$ and $2\in\mathbb{R}$ then there exists a $C^1$-function $y:[0,4]\to\mathbb{R}$ such that $y'(t)=F(t,y(t))$ for all $t\in [0,4],$ and $y(0)=2$ by Picard's theorem. I think that's correct, but do correct me if not.

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I dont want to give you a complete solution, since I think you can do it on your own (and learn something by doing so). Here are some hints: 1) To use Banach's Fixed Point Theorem you need to define an operator $T:C([0,1])\to C([0,1])$ with the following properties:

a) Each fixed point of $T$ is a solution to your problem, i.e. $Tf=f$ iff $f(x)=1/3 f((x+1)/2)+4\cos x$ for all $x\in[0,1]$.

b) $T$ is a contraction wrt the maximum norm, i.e. there is a constant $L\in[0,1)$ such that $||Tf-Tg||_\infty\leq L||f-g||_\infty$ for all $f,g\in C([0,1])$.

In general it is very easy to define $T$ such that (a) holds. Can you do that? Then you have to verify (b). After that Banach's Theorem yields your desired unique solution. Why can you apply it now?

2) Your ideas are correct. Just some notes on your notation: Its better to define $F$ via $F(t,s):=|t-1|-5\cos(s)$, so there you can see better whats the first variable and whats the second.

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  • $\begingroup$ I'm not quite sure about the definition of $T.$ Do I need to 'guess' a solution here (ie. suppose a solution exists and then define $T$ that way)? I'm not entirely sure how to actually do that for this type of equation where the solution isn't really apparent to me. Thanks for the method help though, I didn't realise it was that simple. $\endgroup$ – user188382 Oct 29 '14 at 21:06
  • $\begingroup$ Well, (a) sais we must have f(x)=(Tf)(x), and your equation sais $f(x)=1/3 f((x+1)/2)+4\cos(x)$. Now look at the right hand sides of these equations. How do we have to choose $T$? $\endgroup$ – sranthrop Oct 29 '14 at 21:37
  • $\begingroup$ I get the feeling I'm being obtuse, but I can't see any way to define $T$ but to choose the right hand side of the equation to give $Tf.$ $\endgroup$ – user188382 Oct 29 '14 at 21:39
  • $\begingroup$ Oh my wow that was dumb. If we can choose $Tf$ to be just the right-hand side I'm pretty sure the result is blatantly obvious, my god that was silly. Thank you for the help. Will give +1 when I can get back in on my main. $\endgroup$ – user188382 Oct 29 '14 at 21:44

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