Given two functions $f(x_1,x_2,x_3...x_n)$ and $g(x_1,x_2,x_3...x_n)$ in $n$-dimensional space, is there a straightforward way to find the set of points where $f$ and $g$ have the same output and the same input?
In 2D space, the intersection of two curves is “usually” a discrete set of points that can be solved for and manually plugged back into the equation if you're looking for points where the curves intersect at an interesting value. In 3D, the intersection of two surfaces is a curve or discrete set of curves and you can find the intersection of that curve with a plane if you want to know where the two surfaces intersect at a particular value.
For this problem, I have I have two functions of four variables. I want to find the set of coordinates where these functions both have the same input coordinates and both have a particular output value $U$, but I don't know where to begin. Are numerical methods my only hope? Or maybe there's just some standard terminology I'm missing.
For reference, here are the equations:
$P(w,x,y,z)=\frac{1}{\sqrt{wxyz}}$
$Q(w,x,y,z)=\frac{1}{\sqrt{2}}\left(\frac{1}{wy}+\frac{1}{wz}-\frac{A}{xz}\right)$
There is not a general procedure that works for any functions. But there is certainly much that can be done in your particular example.
You set the functions equal to each other: $$\frac 1{\sqrt{wxyz}} = \frac 1{\sqrt 2}\left(\frac 1{wy}+\frac 1{wz}-\frac{A}{xz}\right)$$ A change of variables makes this a little prettier: $$\frac 1{rstu} = \frac 1{\sqrt 2}\left(\frac 1{r^2t^2}+\frac 1{r^2u^2}-\frac{A}{s^2u^2}\right)$$ Multiply through by $r^2s^2t^2u^2\sqrt 2$: $$\sqrt 2 rstu = s^2u^2 + s^2t^2 - Ar^2t^2$$
You can write this as a quadratic equation in each of the variables in terms of the others: $$(At^2)r^2 + (\sqrt 2 stu)r - s^2(u^2 + t^2) = 0\\ (u^2 + t^2)s^2 - (\sqrt 2 rtu)s - Ar^2t^2 = 0\\ (Ar^2 - s^2)t^2 + (\sqrt 2 rsu)t - s^2u^2 = 0\\ (s^2)u^2- (\sqrt 2 rst)u + s^2t^2 - Ar^2t^2 = 0$$ For real solutions, each of the discriminants must be non-negative: $$2s^2t^2u^2 +4(At^2s^2)(u^2 + t^2) \ge 0\\ 2r^2t^2u^2 +4(Ar^2t^2)(u^2 + t^2) \ge 0\\ 2r^2s^2u^2 + 4(Ar^2-s^2)s^2u^2 \ge 0\\ 2r^2s^2t^2 + 4(s^2t^2)(Ar^2-s^2)\ge 0$$ Simplifying and switching back to $x,y,z,w$, the first two conditions turn out to be the same, as are the second two conditions. So it reduces to $$z + 2A(z + y) \ge 0\\w + 2(Aw-x) \ge 0$$
Since $y,z > 0$ already, if $A \ge 0$, the first is automatically satisfied. The second requires $$x \le \left(\frac 12 + A\right)w$$
You can choose the values of any three of $w, x, y, z$ freely, as long as those two inequalities are observed. The fourth variable will have two possible values as determined by the corresponding quadratic equation in $r,s,t,u$. For example, if you've picked positive values for $x, y, z$, and set $s = \sqrt x, t = \sqrt y, u = \sqrt z$, then the value of $r$ is found by $$r = \dfrac{-\sqrt 2stu \pm \sqrt{2s^2t^2u^2 +4(At^2s^2)(u^2 + t^2)}}{2At^2}\\ r = \frac{su}{\sqrt 2At}\left(-1 \pm\sqrt{1 + 2A\left(1 + \frac{t^2}{u^2}\right)}\right)$$ Squaring both sides and substituting for $w,x,y,z$, $$w = \frac {xz}{2A^2y}\left(1 \pm \sqrt{1 + 2A\left(1 + \frac yz\right)}\right)^2$$ Similar equations can be derived to find each of the variables in terms of the other three.