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I have some troubles with the next exercise. It's so hard. I have been trying and I can't found the solution.

In the next system of equations, can we express in a neighborhood of the point $\overline{w_0}=(\overline{x_0},t_0)=\left(\pi,\displaystyle\frac{\pi}{2},1\right)$ the variables $x$ and $y$? $$ \left\{ \begin{array}{cc} \cos(x)+t\sin(y) & = & 0\\ \sin(x)-\cos(ty) & = & 0 \end{array} \right. $$

Then, we consider the functions $f_1=\cos(x)+t\sin(y)=0$ and $f_2=\sin(x)-\cos(ty)=0$. We want to see that the function $f:\mathbb{R}^3\times\mathbb{R}^3\rightarrow\mathbb{R}^2$ defined how $$f(\overline{x},t)=(f_1(\overline{x},t),f_2(\overline{x},t))$$ is Continuously Differentiable, but is clear, because the partial derivatives of $f$ are continuous. Then, $f\in C^{1}$

In this way, the derivative matrix is $$ \left( \begin{array}{ccc} \sin(x) & t\cos(y) & t\sin(y)\\ \cos(x) & t\sin(y) & y\sin(ty) \end{array} \right) $$

We should to see the sub-matrix

$$ \begin{equation} \left( \begin{array}{cc} \sin(x) & t\cos(y)\\ \cos(x) & t\sin(y) \end{array} \right)_{\overline{w_0}} = \left( \begin{array}{cc} 0 & 0\\ -1 & 1 \end{array} \right) \end{equation} $$

The determinant if this matrix, clearly, is zero and thus, the matrix is not invertible. Then, the hypothesis of the Implicit function theorem are not fulfilled. Then, what can I do?

We are looking for a function $g(t)=(x(t),y(t))$ such that $f(g(t),t)=0$ and moreover $\left( \pi,\frac{\pi}{2}\right)=f(1)=(x(1),y(1))$. If the hypothesis of the theorem are not fulfilled, how can we conclude that we can't express the variables $x$ and $y$? Or, can we although the hypothesis are not fulfilled? Really I need help with this. I really appreciate any help you can provide me.

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  • $\begingroup$ Check your derivative matrix! (Unfortunately the problem remains.) $\endgroup$ – Christian Blatter Apr 29 '17 at 13:22
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If you make the change of variables $$ x= \pi +u, \quad y= \pi/2 +v, \quad t =1+s, $$ then your problem becomes $$ \cos u -(1+s) \cos v =0, \quad \sin u - \sin (v+ \pi s/2 +sv)=0, $$ and your problem is now around $(u,v,s)=(0,0,0)$. But if $s=0$, you have the problem $$ \cos u - \cos v =0, \quad \sin u - \sin v=0, $$ which has solutions $u=v$ for all $u$. So, near $s=u=v=0$ the Implicit function Theorem does not work because there is no map $s \mapsto (u(s),v(s))$ which defines a unique function (you have a line of solutions!)

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