Let $F:\mathbb{R}^2\to \mathbb{R}^2$ be the diffeomorphism given by $$F(x,y)=(y+\sin x, x) $$ Find $F^{-1}$.
I know that the answer is $F^{-1}(x,y)=(y,x-\sin y$), this can be shown to be true by taking $F\circ F^{-1}(x,y)=F(F^{-1}(x,y))=F(y,x-\sin y)=(x-\sin y+\sin y,y)=(x,y)$
If the function is linear, we can use $F(x,y)=A\cdot (x,y)$ for $A\in\mathbb{R}^{2\times 2}$ and then find the inverse of $A$, giving us the inverse map
This cannot be done here as it it not linear, is there a standard method to solving non-linear inverse function problems like these, instead of just playing around with the equation until you reach a solution.
I don't think I could solve a more complex problem than this as I kind of used trial and error. Is there a standard concrete process that one can use?
Just found the answer:
Replace $f\left(x,y\right)$ by $\left(u,v\right)$ resulting in: $x+y+1=u$ and $x-y-1=v$
Switch $x$ and $u$ and switch $y$ and $v$ resulting in: $u+v+1=x$ and $u-v-1=y$
Solve for $u$ and $v$ resulting in: $u=\frac{1}{2}x+\frac{1}{2}y$ and $v=\frac{1}{2}x-\frac{1}{2}y-1$
Replace $\left(u,v\right)$ with $f^{-1}\left(x,y\right)$ resulting in: $f^{-1}\left(x,y\right)$=$\left(\frac{1}{2}x+\frac{1}{2}y,\frac{1}{2}x-\frac{1}{2}y-1\right)$
So, same procedure. This gives you the inverse of function $f:\mathbb{R}^{2}\rightarrow\mathbb{R}^{2}$ defined by $f\left(x,y\right)=\left(x+y+1,x-y-1\right)$.