x,y在单位圆盘内,求u/v的范围

hbghlyj

$x^2+y^2\le1$,
$(u,v)$是方程组$\cases{x^2+u (y+1)=y^2+v (x+1)\\u^2+u (x+1)=v^2-v (y+1)}$的解, 且 $u>0,v>0$,
求$u\over v$的范围

hbghlyj

解出$u,v$

1. sol = Assuming[Element[{x, y}, Disk[]],
2.   Solve[{x^2 + u (y + 1) == y^2 + v (x + 1),
3.     u^2 + u (x + 1) == v^2 - v (y + 1), u > 0, v > 0}, {u, v}]]

Copy the Code

发现只有1组解.

1. FullSimplify[u/v/.sol[[1]]]

Copy the Code

\[\small\frac uv=-\frac{-\sqrt{5 x^4-4 x^3 (2 y+1)-2 x^2 y (3 y+2)+4 x (y (y (2 y+3)+2)+2)+(y (y+2)+2) (y (5 y+2)+2)}-x (x+2)+y (y+2)}{2 \left(-x^2+x y+x+y^2+y+1\right)}\]
令$f(x,y)=\frac uv$

1. f = -((-x (2 + x) +
2.        y (2 + y) - \[Sqrt](5 x^4 - 4 x^3 (1 + 2 y) -
3.           2 x^2 y (2 + 3 y) + (2 + y (2 + y)) (2 + y (2 + 5 y)) +
4.           4 x (2 + y (2 + y (3 + 2 y)))))/(2 (1 + x - x^2 + y + x y +
5.          y^2)));
6. ArgMax[{f, x^2 + y^2 <= 1}, {x, y}] // RootReduce
7. ArgMin[{f, x^2 + y^2 <= 1}, {x, y}] // RootReduce

Copy the Code

$f(x,y),x^2+y^2\le1$ 当 $(x,y)=\left(\frac{\sqrt{3}}{2},-\frac{1}{2}\right)$取得最大值$2+\sqrt{6}+\sqrt{5+2 \sqrt{6}}$,
当 $(x,y)=\left(-\frac{\sqrt{3}}{2},-\frac{1}{2}\right)$取得最小值$2-\sqrt{6}-\sqrt{5-2 \sqrt{6}}$.

hbghlyj

1. Limit[f, {x, y} -> {0, 0}]

Copy the Code

输出1. 即: 当$(x,y)\to(0,0)$时$\frac uv\to1$
能否从方程组直接得出?

hbghlyj

hbghlyj 发表于 2023-3-9 10:15
输出1. 即: 当$(x,y)\to(0,0)$时$\frac uv\to1$
能否从方程组直接得出?

把$u$换为$f\cdot v$, 消去$v$得

1. Eliminate[{x^2 + u (y + 1) == y^2 + v (x + 1), u^2 + u (x + 1) == v^2 - v (y + 1)} /. u -> f v, v]

Copy the Code

\[\small0=(x-y) (x+y) \left(1-f^2+x+2 f x-f^2 x-x^2+f x^2+f^2 x^2+y-2 f y-f^2 y+x y-f^2 x y+y^2-f y^2-f^2 y^2\right)\]在最后一个因子令$x=y=0$得$1-f^2=0$(因为$u,v>0$,有$f>0$)即$f=1$

看来2#做麻烦了, 不要解出$u,v$而应消去它们