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.
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.
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.