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I have the following problem:

Considerate the following system:

$$x=\cos(u)+\sin(v) $$

$$y=\sin(u)+\cos(v)$$

Show that in a neighborhood of $(u_0, v_0, x_0, y_0)=(0,0,1,1), (u,v)$ can be written as differentiable functions of $x$ and $y$. Furthermore,

$$\left(\frac {\partial u}{\partial x}\right)^2 +\left(\frac {\partial u}{\partial y}\right)^2=\left(\frac {\partial v}{\partial x}\right)^2+\left(\frac {\partial v}{\partial y}\right)^2$$

I made some attempts however I did not succeed in the second part. But related to that , I will try again. I would appreciate any hints in the first part, which I believe is an application of implicit function theorem.

Thanks in advance.

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3 Answers 3

PART $1$

We just need to apply the Implicit Function Theorem. Consider the mapping $F : \mathbb{R}^4 \to \mathbb{R}^2$ given by $F(x,y,u,v) = (\cos u + \sin v-x, \cos v + \sin u-y)$. The set of $(x,y,u,v) \in \mathbb{R}^4$ such that $F(x,y,u,v) = (0,0)$ is the surface you're interested in. Calculating the Jacobian Matrix of $F$:

$$J_F = \left[ \begin{array}{cccc} -1 & 0 & -\sin u & \cos v \\ 0 & -1 & \cos u & -\sin v \end{array} \right] . $$

Clearly the rank of $J_F$ is maximal, i.e. equal to two, for all $x,y,u$ and $v.$ It follows that $F$ has no critical values and that $(0,0)$ must be a regular value. This means that the surface is a smooth parametrisable surface in a neighbourhood of each of its points.

Moreover, at $(x,y,u,v) = (0,0,1,1)$ we see that the last two columns form an invertible two-by-two matrix. By the IFT, we can write $u$ and $v$ as smooth functions of $x$ and $y$. (Notice that this hold for all $u$ and $v$ with $\cos(u+v) \neq 0.$)

PART $2$

Assume that $u$ and $v$ are functions of $x$ and $y$, i.e. $u = u(x,y)$ and $v=v(x,y).$ Consider the equation $x = \cos u + \sin v.$ Differentiating with respect to $x$ gives $1 = -u_x\sin u + v_x\cos v.$ Differentiating with respect to $y$ gives $0 = -u_y\sin u + v_y\cos v.$ Applying a similar process to $y = \sin u + \cos v$ gives $0 = u_x\cos u -v_x\sin v$ and $1 = u_y\cos u - v_y\sin v.$ We get:

$$\left[ \begin{array}{cc} -\sin u & \cos v \\ \cos u & -\sin v \end{array} \right]\left[ \begin{array}{cc} u_x & u_y \\ v_x & v_y\end{array}\right] = \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1\end{array} \right] $$

Assuming that $\sin u \sin v - \cos u \cos v \neq 0$ we can rearrange this to give:

$$\left[ \begin{array}{cc} u_x & u_y \\ v_x & v_y\end{array}\right] = \frac{1}{\cos(u+v)}\left[ \begin{array}{cc} \sin v & \cos v \\ \cos u & \sin u \end{array} \right].$$

It follows that:

$$u_x^2+u_y^2 = v_x^2 + v_y^2 \iff \frac{\sin^2v}{\cos^2(u+v)} + \frac{\cos^2v}{\cos^2(u+v)} = \frac{\cos^2u}{\cos^2(u+v)} + \frac{\sin^2u}{\cos^2(u+v)} \, . $$

Using the fact that $\cos^2\theta + \sin^2\theta = 1$ for all $\theta$, we see that this last statement is indeed true. (Assuming $\cos(u+v) \neq 0.$)

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Working off the definition of the implicit function theorem on Wikipedia, let $\textbf{f}(x,y,u,v)=(x-\cos u-\sin v,y-\sin u-\cos v)$. Then there is some function $\textbf{g}(x,y)=(u(x,y),v(x,y))$ whose graph is the level set $\mathbf{f}(x,y,u,v)=\mathbf{0}$ in the vicinity of a point $(x_0,y_0,u_0,v_0)=(1,1,0,0)$ if the matrix of partial derivatives $\frac{\partial\mathbf{f}}{\partial(u,v)}(x_0,y_0,u_0,v_0)$ is nonsingular. But $\frac{\partial\mathbf{f}}{\partial(u,v)}(1,1,0,0)=\left[\begin{matrix} 0&1\\1&0\end{matrix}\right]$, so it is invertible and all is well.

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Let $\phi: \mathbb{R}^2 \times \mathbb{R}^2 \to \mathbb{R}^2$ be given by $\phi(w,z) = \begin{bmatrix} \cos w_1 + \sin w_2 -z_1 \\ \sin w_1 + \cos w_2 - z_2 \end{bmatrix}$. The system above is equivalent to $\phi(w,z) = 0$, and we have $\phi((0,0),(1,1)) = 0$.

A quick computation gives $\frac{\partial \phi(w,z)}{\partial w} = \begin{bmatrix} -\sin w_1 & \cos w_2 \\ \cos w_1 & -\sin w_2\end{bmatrix}$, $\frac{\partial \phi(w,z)}{\partial z} = -I$. Since $\frac{\partial \phi((0,0),(1,1))}{\partial w} = \begin{bmatrix} 0 & 1 \\ 1 & 0\end{bmatrix}$ is invertible and $\frac{\partial \phi}{\partial w}$ is continuous, we see that it is invertible in a neighborhood of $((0,0),(1,1))$.

Then the implicit function says that there are neighborhoods $U$ of $(1,1)$ and $V$ of $(0,0)$, and a unique differentiable function $\omega:U \to V$ such that $\phi(\omega(z),z) = 0$ for all $z \in U$. (In the question's notation, this can be written as $(u(x,y),v(x,y)) = \omega((x,y))$. Also, note that uniqueness implies $\omega((1,1)) = (0,0)$.) This is the solution to the first part.

The second part is just computation, differentiating $\phi(\omega(z),z) = 0$ using the chain rule gives $\frac{\partial \phi(\omega(z),z)}{\partial w} \frac{\partial \omega(z)}{\partial z}+\frac{\partial \phi(\omega(z),z)}{\partial z} = 0$, from which is follows that $\frac{\partial \omega(z)}{\partial z} = - \frac{\partial \phi(\omega(z),z)}{\partial w}^{-1} \frac{\partial \phi(\omega(z),z)}{\partial z}$. Doing the computation shows that $\frac{\partial \omega(z)}{\partial z} = \frac{1}{\sin w_1 \sin w_2 - \cos w_1 \cos w_2} \begin{bmatrix} -\sin w_2 & -\cos w_2 \\ -\cos w_1 & -\sin w_1\end{bmatrix}$. Since we have $\frac{\partial \omega((x,y))}{\partial z} = \begin{bmatrix} \frac{\partial u(x,y)}{\partial x} & \frac{\partial u(x,y)}{\partial y} \\ \frac{\partial v(x,y)}{\partial x} & \frac{\partial v(x,y)}{\partial y} \end{bmatrix}$, the desired result follows from the identity $(\cos x)^2 + (\sin x)^2 = 1$.

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