4
$\begingroup$

Given $f: \mathbb{R}^5\to \mathbb{R}^2$, of class $C^1$. Let $a= (1,2,-1,3,0)$; suppose that $f(a) = 0$ and $Df(a)=\begin{bmatrix}1 & 3 & 1 & -1 & 2\\ 0 & 0 & 1 & 2 & -4\end{bmatrix}$

(a) Show there is a function $g : B\to \mathbb{R}^2$ of class $C^1$ defined on an open set $B$ of $\mathbb{R}^3$ such that $f(x_1, g_1(x), g_2(x), x_2, x_3)=0$ for $x=(x_1, x_2, x_3)\in B$, and $g(1, 3, 0)=(2, -1)$.

(b) Find $Dg(1, 3, 0)$

(c) Discuss the problem of solving the equation $f(x) = 0$ for an arbitrary pair of the unknowns in terms of the others, near the point $a$.

For (a), I know that $\frac{\partial f}{\partial x}(a)=\begin{bmatrix}1 & -1 & 2\\ 0 & 2 & -4\end{bmatrix}$ and $\frac{\partial f}{\partial y}(a)=\begin{bmatrix}3 & 1\\ 0 & 1\end{bmatrix}$ and as $\det \frac{\partial f}{\partial y}(a)=3\neq 0$, then I can apply the implicit function theorem to find an open $B\subset \mathbb{R}^3$ and a continuous function $g: B\to \mathbb{R}^2$ such that $f(x_1, g_1(x), g_2(x), x_2, x_3)=0$ for $x=(x_1, x_2, x_3)\in B$, and $g(1, 3, 0)=(2, -1)$.

(b) I know that $Dg(x)=-[\frac{\partial f}{\partial y}(x, g(x))]^{-1}\frac{\partial f}{\partial x}(x, g(x))$, with which $Dg(1,3,0)=-[\frac{\partial f}{\partial y}(1,2,-1,3,0)]^{-1}\frac{\partial f}{\partial x}(1,2,-1,3,0)=-\begin{bmatrix}3 & 1\\ 0 & 1\end{bmatrix}^{-1}\begin{bmatrix}1 & -1 & 2\\ 0 & 2 & -4\end{bmatrix}=\begin{bmatrix}-1/3 & 1 & 2\\ 0 & -2 & 4\end{bmatrix}$

(c) What this says is that I can clear $g_1(x)$ and $g_2(x)$ in terms of $x_1, x_1, x_2$ and $x_3$?

Are you all the right arguments? Thank you very much.

$\endgroup$
  • 1
    $\begingroup$ You may stick to the coordinates $x_1, \ldots , x_5$. $x \in B\subset \mathbb{R}^3 \rightarrow x = (x_1,x_4,x_5)$. Near $a$, $f(x_1,\ldots,x_5) = 0$ defines $g: B \subset \mathbb{R}^3\to \mathbb{R}^2 $ with $f(x_1,g_1(x_1,x_4,x_5),g_2(x_1,x_4,x_5),x_4,x_5) = 0$ So, near $a$, the equation $f(x_1,\ldots,x_5) = 0$ has a unique solution and it can - at least numerically - be solved for $x_2$ and $x_3$ in terms of $x_1,x_4,x_5$. For (c): If $x_2,x_3$ are given in terms of $x_1,x_4,x_5$ via $g$, then they represent the unique solution of $f(x_1,\ldots,x_5) = 0$ (near $a$). $\endgroup$ – trancelocation Feb 18 '18 at 7:13
1
$\begingroup$

a) It is right but it should be $\frac{\partial f}{\partial(x_2,x_3)}(a)=3$ instead of $\frac{\partial f}{\partial y}(a)=3$ as you wrote.

b) This is ok but the matrix multiplication gives me $$ \begin{bmatrix}-\frac{1}{3} & 1 & -2 \\ 0 & -2 & 4\end{bmatrix},$$ which differs from your result in the minus sign of the upper right 2.

c) Compute \begin{equation*} \begin{aligned} & \det \frac{\partial f}{\partial (x_1,x_2)}(\textbf{a}) = 0 & & \det \frac{\partial f}{\partial (x_1,x_3)}(\textbf{a}) = 1 & & \det \frac{\partial f}{\partial (x_1,x_4)}(\textbf{a}) = 2 \\ & \det \frac{\partial f}{\partial (x_1,x_5)}(\textbf{a}) = -4 & & \det \frac{\partial f}{\partial (x_2,x_3)}(\textbf{a}) = 3 & & \det \frac{\partial f}{\partial (x_2,x_4)}(\textbf{a}) = 6 \\ & \det \frac{\partial f}{\partial (x_2,x_5)}(\textbf{a}) = -12 & & \det \frac{\partial f}{\partial (x_3,x_4)}(\textbf{a}) = 3 & & \det \frac{\partial f}{\partial (x_3,x_5)}(\textbf{a}) = -6 \\ & \det \frac{\partial f}{\partial (x_4,x_5)}(\textbf{a}) = 0. & & & & \end{aligned} \end{equation*} Therefore, we do not expect to be able to solve for $x_1$ and $x_2$ in terms of the others near $\textbf{a}$, neither to solve for $x_4$ and $x_5$. For all other pairs, however, we can assure we can solve the equation $f(\mathbf{x})=\mathbf{0}$ in terms of the other three.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.