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How to show that the following system of equations has a unique solution $(x,y)$? $x,y$ are scalars.

$x+\frac{3}{4}y+\frac{1}{20}\sin x=0$

$-\frac{37}{40}x+y+\frac{1}{10}\sin y=0$

I tried contraction mapping, but it didn't work.

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    I understood I meant actually what bogus suggested as a solution.2012-10-27

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I solved the top equation for $y$, substituted it into the second equation, and got $\frac{1}{10}\sin(\frac{1}{15}\sin x + \frac{4}{3}x) + \frac{271}{120}x + \frac{1}{15}\sin x = 0$. One solution is $x = 0$, as you found. The derivative of the left-hand side is always positive:

$\frac{1}{10}\cos(\frac{1}{15}\sin x + \frac{4}{3}x)(\frac{1}{15}\cos x+\frac{4}{3})+\frac{271}{120} +\frac{1}{15}\cos x \geq -\frac{1}{10}(\frac{1}{15} + \frac{4}{3}) + \frac{271}{120} - \frac{1}{15} > 0$.

So $0$ is the only solution for $x$, and $y$ has to be $0$ too.

It is late and I don't guarantee against errors in the fractions, but I think the procedure works.

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    I also solved the problem using the Contraction Mapping Theorem, using the distance $d(,) = |a-c| +|b-d|$ and rewriting the equations as $f() = <-\frac{3}{4}y - \frac{1}{20}\sin x, \frac{37}{40}x-\frac{1}{10}\sin y>$. $f$ is a contraction with $\lambda = \frac{39}{40}$.2012-10-27