关于x,y的二元方程组

hjnk900

对于 $a\inR$, 考虑一个关于 $x,y$ 的系统：
$$x+y+\sin(xy)=a$$$$\sin(x^2+y)=2a$$
问：对于绝对小的 $a$ （$\epsilon>0,|a|<\epsilon$）, 这个系统有解吗？如果有的话，这个解与 $a$ 连续相关吗？

hjnk900

回复 1# hjnk900

做出来了，用隐函数定理..

hbghlyj

For $a\in\Bbb R$, consider a system of $x,y$.
\begin{align*}x+y+\sin(xy)&=a\\\sin(x^2+y)&=2a\end{align*}
Q: For small $a$ ($\epsilon>0,|a|<\epsilon$), does this system have a solution? If so, is this solution continuously dependent on $a$? Use the implicit function theorem.

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We can define the function $F:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ as follows:
$$ F(x,y) = \begin{pmatrix} x+y+\sin(xy)-a \\ \sin(x^2+y)-2a \end{pmatrix} $$
The system can be rewritten as $F(x,y) = 0$.

First, we check that $F(0,0) = (0,0)$.
Next, we compute the Jacobian matrix of $F$:
$$ J_F(x,y)=\begin{pmatrix} \partial F_1\over\partial x&\partial F_1\over\partial y \\ \partial F_2\over\partial x & \partial F_2\over\partial y\end{pmatrix}= \begin{pmatrix} 1+\cos(xy) y &1+ \cos(xy) x  \\ 2x \cos(x^2+y) & \cos(x^2+y) \end{pmatrix} $$
Evaluate the Jacobian at $(0,0)$:
$$ J_F(0,0) = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} $$
The determinant of $J_F(0,0)$ is non-zero, so the implicit function theorem applies, there exists a neighborhood $U$ of $(0,0)$ and a positive real $\epsilon$ such that there exists a continuously differentiable function $g:(-\epsilon,\epsilon) \rightarrow U$ such that $F(g(a)) = 0$ for all $|a|<\epsilon$.$\newcommand{\R}{\mathbb R }$
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hbghlyj

The Implicit Function Theorem I

A harder example.

Consider the system of equations \begin{align} F_1(x,y,u,v) := xye^u + \sin(v-u) &= 0\\ F_2(x,y,u,v) :=(x+1)(y+2)(u+3)(v+4) - 24 &=0 \end{align} Note that $(0,0,0,0)$ is a solution.

We will answer the following questions:

Does the system of equations implicitly determine $(u,v)$ as a function of $(x,y)$, i.e. $(u,v) = \bff(x,y)$ for $(x,y)$ near $(0,0)$, and with $f(0,0) = (0,0)$?

If so, find a formula for $\partial_x \bff(x,y)$ at $(x,y) = (0,0)$.

Solution. First, let $\bfF = \binom{F_1}{F_2}$. Then we consider the matrix $$ \left(\begin{array}{ll} \partial_u F_1&\partial_v F_1\\ \partial_u F_2&\partial_v F_2 \end{array} \right) \ = \ \left(\begin{array}{cc} xye^u - \cos(v-u)&\cos(v-u)\\ (x+1)(y+2)(v+4)&(x+1)(y+2)(u+3) \end{array} \right) $$ At $(x,y,u,v) = (0,0,0,0)$ this becomes \begin{equation}\label{matrix} \left(\begin{array}{rr} -1&1\\8&6 \end{array} \right). \end{equation} This matrix is invertible, so the theorem guarantees that the equations implicitly determine $(u,v)$ as function of $(x,y)$.

Next we find $\partial_x \bff = \binom{\partial_x f_1}{\partial_x f_2}$, where $\binom uv = \bff(x,y) = \binom{f_1(x,y)}{f_2(x,y)}$ is the implicitly defined function.

We start with the equations \begin{align} xye^u + \sin(v-u) &= 0\\ (x+1)(y+2)(u+3)(v+4) - 24 &=0. \end{align} We next implicitly differentiate everything with respect to $x$, taking care to remember that we are considering $x,y$ as independent variables and $u,v$ as dependent variables, so $\frac{\partial y}{\partial x}= 0$. Then after gathering terms we get \begin{align} ye^u + \left( xy e^u - \cos(v-u)\right) \frac{\partial u}{\partial x} + \cos(v-u)\frac{\partial v}{\partial x} &= 0\\ (y+2)(u+3)(v+4) +(x+1)(y+2)(v+4)\frac{\partial u}{\partial x} +(x+1)(y+2)(u+3)\frac{\partial v}{\partial x} &=0. \end{align} At $(x,y,u,v) = (0,0,0,0)$ this reduces to \begin{align} \left(\begin{array}{rr} -1&1\\ 8&6 \end{array} \right)\binom {\frac{\partial u}{\partial x}} {\frac{\partial v}{\partial x}} = \binom 0{-24}. \label{concrete}\end{align} This can be solved to find that $$ \binom{\frac{\partial u}{\partial x}} {\frac{\partial v}{\partial x}} = \frac{1}{-14} \left( \begin{array}{rr} 6&-1 \\ -8&-1 \end{array} \right)\binom0 {-24} = -\binom{12/7}{12/7}. $$ In solving this, we have used the forumla for the inverse of a $2\times 2$ matrix: $$ \left( \begin{array}{cc} a&b \\ c&d \end{array} \right)^{-1} \ = \ \frac {1}{ad-bc} \left(\begin{array}{rr} d&-b \\ -c&a \end{array}\right) , $$