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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.

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    The solution in this case is rather obvious. But if you want to use gradient descent, why not use it on the function $f_1(x,y)^2 + f_2(x,y)^2$? I really doubt you'll find such a $\varphi$ in closed form. – 2017-01-23
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    @RobertIsrael: Thanks. $f_1(x,y)^2+f_2(x,y)^2$ would be an option. Which numerical method are you alluding to in your comment? ("is rather obvious") Maybe Newton's method? Is there an other way beside gradient descent which uses the original approach $\nabla F=\left(\begin{matrix} \tilde f_1(x,y)\\\tilde f_2(x,y) \end{matrix}\right)$? – 2017-01-23
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    Actually in my case it suffices to show that there exists an unique solution. – 2017-01-23
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    "Is rather obvious" is not referring to a numerical method (although you'll see a solution staring at you if you plot the curves...). – 2017-01-23

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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...

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    Now I understand what you meant by "really obvious". :-) Thank you for your answer! – 2017-01-24
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    Your transformation gives $F (s,t)=2t (2t^3+ 2t^2 + 3t+s)+ \frac {s^2}{2}+ 2\cos\frac {s}{2} $. Thank you for your help! – 2017-01-24