Question

_{for b part}for b part

Problem 2. The following non-linear system is given: y} = 11y2 + x3 - 1 yż = x2y1 +z1-1 (a) Determine if this system is solvable with respect to y = (y1, y2) locally around the point Z1 = (1,1,1,1) So, you have to determine if there are two functions yi = y111, 12) and y2 = y1 (11, 12) defined locally around x1 = (1, 1) and satisfying the given equations. If yes, determine if the function 0: R2 + R2, 0(21,02) = (y1 (11, 12), y2(21, 12) is invertible around (1,1) and if yes, find its full derivative at (1,1). (b) Repeat (a) for the points z2 = (1,1,0,0) and z3 = (0,0,-1,-1).

Answer 1

Solution :

Given that,

Statement :$Let\ f:R^{n+ m}\rightarrow R^m$ be a continuously differentiable function,

Let have coordinates (x,y)

Rn+m

Fix a point
(a, b) = (a1, ...., An, 61, ...., bn) with fla, b) = 0
,where $0\epsilon\ R^m$ is the zero vector.

If the Jacobian matrix $J_{fy}(a,b)=[(\frac{\partial f_i }{\partial y_j})(a,b)]$ is invertible, thten there exists an open set U of RK containing a such that there exixts a unique continuosly differentiable function $g:U\rightarrow R^m such\ that g(a)=b\ and\ f(x,g(x))=0\ for\ all\ x\epsilon U$

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here

$n=2,m=2,(x,y)\epsilon \mathbb{R}^4,with\ X=(x_1,x_2),y=(y_1,y_2)\epsilon \mathbb{R}^2,z_1=(1,1,1,1)$

with 21 = (a, b), a = (1,1)¢R,b= (1, 1ER?)

$f:R^{2+2}\rightarrow R^2,f(x_1,x_2,y_1,y_2)=(x_1y_2+x_2^2-y_1^3-1,x_2y_1+x_1^2-y_2^3-1)$

= fi = 1192 + - y - 1, f2 = 1241 + x - y - 1

Fix a point(1,1,1,1)

$a=(1,1)\epsilon \mathbb{R}^2,b=(1,1)\epsilon \mathbb{R}^2,f(1,1,1,1)=(0,0)$

$J_{fy}=\begin{bmatrix} \frac{\partial f_1}{\partial y_1}&\frac{\partial f_1}{\partial y_2} \\ \frac{\partial f_2}{\partial y_1} & \frac{\partial f_2}{\partial y_2} \end{bmatrix}$

$\frac{\partial f_1}{\partial y_1}=-3y_1^2,\ \frac{\partial f_1}{\partial y_2}=x_1$

$\frac{\partial f_2}{\partial y_1}=x_2,\ \frac{\partial f_2}{\partial y_2}=-3y_2^2$

$det(J_{fy})=9y_1^2y_2^2-x_1x_2$

$det(J_{fy}(1,1,1,1))=8\neq 0$

$\therefore j_{fy}(1,1,1,1)is\ invertible.$

by implicit function therom,there exixts an open set U of $\mathbb{R}$

containing (1,1) such tha there exixts a unique continuously differentiable function $\varphi :U\rightarrow R^2 such\ that\ \varphi (1,1)=(1,1)$

$\varphi :U\rightarrow R^2\Rightarrow \varphi (x_1,x_2)=(y_1(x_1,x_2),y_2(x_1,x_2)))$ with y(1,1)=1,z(1,1)=1 that is, there exixst a differenable functions $y(x_1,x_2),z(x_1,x_2)$ in the neighbourthood of(1,1) so that y(1,1)=1, z(1,1)=1.

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(b.)

also we have $\frac{\partial \varphi }{\partial x}=-[J_{fy}(x_1,x_2,\varphi (x_1,x_2))]_{2\times 2}^{-1}[\frac{\partial f}{\partial x}(x_1,x_2,\varphi (x_1,x_2))]_{2\times 2}$

$\frac{\partial f}{\partial x}(1,1)=-[J_{fy}(1,1,1,1)]_{2\times 2}^{-1}[\frac{\partial f}{\partial x}(1,1,1,1)]_{2\times 2}$

$J_{fy}=\begin{bmatrix} -3y_1^2 &x_1 \\ x_2 &-3y_2^2 \end{bmatrix}$

$\Rightarrow J_{fy}(1,1,1,1)=\begin{bmatrix} -3 &_1 \\ _1 &-3 \end{bmatrix}$

$[J_{fy}(1,1,1,1)]_{}{2\times 2}^{-1}=\frac{1}{8}\begin{bmatrix} -3 &-1 \\ -1& -3 \end{bmatrix}$

$\frac{\partial f}{\partial x}=\begin{vmatrix} \frac{\partial f_1}{\partial x_1} &\frac{\partial f_1}{\partial x_2} \\ \frac{\partial f_2}{\partial x_1} &\frac{\partial f_2}{\partial x_2} \end{vmatrix}$

$=\bigl(\begin{smallmatrix} y_2 &2x_2 \\ x_1&y_1 \end{smallmatrix}\bigr)$

$\frac{\partial f}{\partial x}=\bigl(\begin{smallmatrix} 1 &2 \\ 2 & 1 \end{smallmatrix}\bigr)$

$\frac{\partial \varphi }{\partial x}(1,1)=-\frac{1}{8}\bigl(\begin{smallmatrix} -3 &-1 \\ -1 & -3 \end{smallmatrix}\bigr)\bigl(\begin{smallmatrix} 1 &2 \\ 2 & 1 \end{smallmatrix}\bigr)$

$=-\frac{1}{8}\bigl(\begin{smallmatrix} -5 &-7 \\ -7 & -5 \end{smallmatrix}\bigr)$