二元方程组 渐近

hbghlyj

求证：存在常数$C_1,C_2$，对任意实数$u,v,x,y$满足
\begin{cases}\exp (u)+\exp (2 v)+\exp (x)+\exp (2 y)=4\\\exp (u)+\exp (v)+\exp (x)+\exp (y)=4\end{cases}则有
\[\led
{\abs{u+x}\over x^2+y^2}\le C_1\\
{\abs{v+y}\over x^2+y^2}\le C_2
\endled\]
注：这里的系数是AsymptoticSolve[{Exp[u] + Exp[2 v] + Exp[x] + Exp[2 y] == 4, Exp[u] + Exp[v] + Exp[x] + Exp[y] == 4}, {u, v} -> {0, 0}, {x, y} -> {0, 0}]算出的，换成其它数就不存在$C_1,C_2$了。

hbghlyj

由$e^{2 y}<4$得出$y<\frac{\log4}2$

1. Solve[{Exp[u] + Exp[2 v] + Exp[x] + Exp[2 y] == 4, Exp[u] + Exp[v] + Exp[x] + Exp[y] == 4}, {u, v}, Reals]

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解出$v=\log \left(\frac{1}{2} \left(1\pm\sqrt{1-4 e^y \left(e^y-1\right)}\right)\right)$
那么$v\le\log \left(\frac{1}{2} \left(1+\sqrt{1-4 e^y \left(e^y-1\right)}\right)\right)$
$$\frac{\abs{v+y}}{x^2+y^2}\le\frac{\abs{v+y}}{y^2}$$
要证明$\log \left(\frac{1}{2} \left(1+\sqrt{1-4 e^y \left(e^y-1\right)}\right)\right)+y\over y^2$有界。
由$1-4 e^y \left(e^y-1\right)>0$得出$y < \log(\frac12 + \frac1{\sqrt2})$
Plot[(Log[(1 + Sqrt[1 - 4 E^y (E^y - 1)])/2] + y)/y^2, {y,-1,Log[1/2 + 1/Sqrt[2]]}]

它确实是有界的。

hbghlyj

解出$u=\log \left(\frac{1}{2} \left(7-2 \left(e^x+e^y\right)\pm\sqrt{1-4 e^y \left(e^y-1\right)}\right)\right)$
那么$u\le\log \left(\frac{1}{2} \left(7-2 \left(e^x+e^y\right)+\sqrt{1-4 e^y \left(e^y-1\right)}\right)\right)$
\[{\abs{u+x}\over x^2+y^2}\le ???\]

hbghlyj

求证：存在常数$C_1,C_2$，对任意实数$u,v,x,y$满足 \begin{cases}\exp (u)+\exp (2 v)+\exp (x)+\exp (y)=4\\\exp (u)+\exp (v)+\exp (x)+\exp (2y)=4\end{cases}则有 \[\led {\abs{u+x+3y}\over x^2+y^2}\le C_1\\ {\abs{v-y}\over x^2+y^2}\le C_2 \endled\] 注：这里的系数是AsymptoticSolve[{Exp[u] + Exp[2 v] + Exp[x] + Exp[y] == 4, Exp[u] + Exp[v] + Exp[x] + Exp[2 y] == 4}, {u, v} -> {0, 0}, {x, y} -> {0, 0}]算出的，换其它数就不存在$C_1,C_2$了。

hbghlyj

除了解出$u,v$，有更好的办法证明吗
把exp(2u)改成exp(3u)就是三次方程了，但是仍可以找到$u$的一阶渐近

hbghlyj

求证：存在常数$C_1,C_2$，对任意实数$u,v,x,y$满足
\begin{cases}\exp (u)+\exp (2 v)+\exp (x)+\exp (2 y)=4\\\exp (u)+\exp (v)+\exp (x)+\exp (y)=4\end{cases}则有
\[\led
{\abs{u+x+x^2-2y^2}\over(x^2+y^2)^{3/2}}\le C_1\\
{\abs{v+y+3y^2}\over(x^2+y^2)^{3/2}}\le C_2
\endled\]
注：这系数是AsymptoticSolve[{Exp[u] + Exp[2 v] + Exp[x] + Exp[2 y] == 4, Exp[u] + Exp[v] + Exp[x] + Exp[y] == 4}, {u, v} -> {0, 0}, {{x, y}, {0, 0}, 2}]算的，换成其它数就不存在$C_1,C_2$了。

hbghlyj

同样解出$v=\log \left(\frac{1}{2} \left(1\pm\sqrt{1-4 e^y \left(e^y-1\right)}\right)\right)$
较大的$\log \left(\frac{1}{2} \left(1+\sqrt{1-4 e^y \left(e^y-1\right)}\right)\right)$在$y=0$处展开就是$-y-3y^2+O(y^3)$

1. Series[Log[1/2 (1 + Sqrt[1 - 4 E^y (-1 + E^y)])], {y, 0, 2}]

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解出$u=\log \left(\frac{1}{2} \left(7-2 \left(e^x+e^y\right)\pm\sqrt{1-4 e^y \left(e^y-1\right)}\right)\right)$
较小的$\log \left(\frac{1}{2} \left(7-2 \left(e^x+e^y\right)-\sqrt{1-4 e^y \left(e^y-1\right)}\right)\right)$在$(x,y)=(0,0)$处展开是$-x-x^2+2y^2+O((x^2+y^2)^{3/2})$

1. Series[Log[1/2 (7 - 2 (E^x + E^y) - Sqrt[1 - 4 E^y (-1 + E^y)])], {x, 0, 2}, {y, 0, 2}]

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但是较大的$\log \left(\frac{1}{2} \left(7-2 \left(e^x+e^y\right)+\sqrt{1-4 e^y \left(e^y-1\right)}\right)\right)$在$(x,y)=(0,0)$处展开不是那个式子。