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I am solving a question whose first item is to demonstrate the Banach Fixed Point Theorem, and the second item is as follows:

Show that for any parameter $t \in \mathbb{R}$ the system $$ \begin{cases} x = \frac{1}{2}\sin(x+y) + t - 1 \\ y = \frac{1}{2}\cos(x-y) - t + \frac{1}{2} \end{cases} $$ has a unique solution that depends continuously on the parameter $t$.

Now, this seems clear to me that the question asks one to apply the just proved Banach Fixed Point Theorem to find the existence of fixed points for the map $$ F(x, y) = (\frac{1}{2}\sin(x+y) + t - 1, \frac{1}{2}\cos(x-y) - t + \frac{1}{2}) $$ However, I am not able to show that such a map is a contraction. Any hints wil be the most appreciated.

Thanks in advance.

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    $\begingroup$ To prove that $F$ is a contraction, you could prove that the eigenvalues of its jacobian matrix have absolute values less than $1$. However, this seems messy: WA $\endgroup$ – lhf Sep 12 at 12:02
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By the mean value theorem, for any $a,b \in \mathbb{R}$, $|\sin(a)-\sin(b)| \le |a-b|$. Therefore, $$|(\frac{1}{2}\sin(x+y)+t-1)-(\frac{1}{2}\sin(x'+y')+t-1)| = \frac{1}{2}|\sin(x+y)-\sin(x'+y')|$$ $$\le \frac{1}{2}|(x+y)-(x'+y')| = \frac{1}{2}|(x-x')+(y-y')|.$$ Similarly, $$|(\frac{1}{2}\cos(x-y)-t+\frac{1}{2})-(\frac{1}{2}\cos(x'-y')-t+\frac{1}{2})| = \frac{1}{2}|\cos(x-y)-\cos(x'-y')|$$ $$\le \frac{1}{2}|(x-y)-(x'-y')| = \frac{1}{2}|(x-x')-(y-y')|.$$ Therefore, $$|F(x,y)-F(x',y')| \le \sqrt{\frac{1}{4}|(x-x')+(y-y')|^2+\frac{1}{4}|(x-x')-(y-y')|^2}$$ $$= \frac{1}{2}\sqrt{2|x-x'|^2+2|y-y'|^2} = \frac{\sqrt{2}}{2}|(x,y)-(x',y')|.$$ This shows that $F$ is a contraction with Lipschitz constant at most $\frac{\sqrt{2}}{2}$.

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  • $\begingroup$ I can see flaws in this reasoning. First, the Lipschitz constant which results from your estimation for $\frac{1}{2}\sin(x+y) +t-1$ is equal to 1 (I think it can be corrected so as to get constant equal to $\frac{1}{\sqrt{2}}$). Secondly, it doesn't neccessarily follow that a function is a contraction from the fact that its projections are contractions. $\endgroup$ – Kulisty Sep 11 at 11:32
  • $\begingroup$ @Kulisty Is the first thing an issue? Re second issue: I think it does follow. $|(f_1(x,y),f_2(x,y))-(f_1(x',y'),f_2(x',y'))| = |(f_1(x,y)-f_1(x',y'),f_2(x,y)-f_2(x',y'))| \le |(f_1(x,y)-f_1(x',y'))|+|f_2(x,y)-f_2(x',y')| \le C_1|(x,y)-(x',y')|+C_2|(x,y)-(x',y')| = C|(x,y)-(x',y')|$. Where's the issue? $\endgroup$ – mathworker21 Sep 11 at 11:38
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    $\begingroup$ The issue is that the constant must be strictly smaller than $1$. Here $C_1=C_2=\frac{1}{\sqrt{2}}$. $\endgroup$ – Kulisty Sep 11 at 13:51
  • $\begingroup$ @Kulisty I fixed my answer. Can you remove your downvote (and upvote if you like the answer)? $\endgroup$ – mathworker21 Sep 12 at 11:39
  • $\begingroup$ Yes, it looks correct now. $\endgroup$ – Kulisty Sep 12 at 14:00

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