Let $f_1, f_2:\mathbb{R}^2\to\mathbb{R}$ be the functions $$f_1(x,y)=\frac{3}{2}x-\frac{1}{2}y-\sin{\left(\frac{x+y}{4}\right)}\\ f_2(x,y)=(x-y)(2(x-y)^2+3(x-y)+6)+x+y$$ I want to use the gradient descent to get the solution of $\left(\begin{matrix} f_1(x,y)\\f_2(x,y) \end{matrix}\right)=0.$
I think there should be a way transforming the coordinates by a bijective function $\varphi:\mathbb{R}^2\to\mathbb{R}^2$, s.t. $\tilde f_i(z)=f_i(\varphi^{-1}(z))$ and s.t. there exists a function $F:\mathbb{R}^2\to\mathbb{R}$ where $\nabla F=\left(\begin{matrix} \tilde f_1(x,y)\\\tilde f_2(x,y) \end{matrix}\right)$.
Then we can apply the method of gradient descent to get the zeros of $\nabla F$ and hence the zeros of our system of equations. The method of gradient descent is obvious, but I cannot determine the function $\varphi$.
I am grateful for any hint.
Change variables: $x=s+t,\; y=s-t$. The equations become
$$\eqalign{ s + 2t - \sin(s/2) &= 0\cr 2 t (8 t^2+6 t+6)+2 s &= 0\cr}$$ With $s = -t(8t^2 + 6 t + 6)$ from the second equation, the first becomes $$ g(t) = -8 t^3-6 t^2-4 t+\sin(4 t^3+3 t^2+3 t) = 0$$
For real solutions, $-1 \le \sin(4t^3 + 3t^2 + 3t) \le 1$, so $-1 \le -8 t^3 - 6 t^2 - 4 t\le 1$. Thus $t$ must be between the (unique) real roots of $-8 x^3 - 6 x^2 - 4 x - 1$ and $-8 x^3 - 6 x^2 - 4 x + 1$, approximately $-0.34707$ and $0.18557$. On this interval, it's not hard to show that $g' < 0$ (using the fact that $|\cos| \le 1$). Thus the solution is unique.
And, I repeat, the solution itself is obvious. The first number to try...