# An implicit function theorem question

In my course on multivariate calculus we treat the implicit function theorem and I am stuck on the following question:

Find the values of $a$ and $b$ such that, in a neighbourhood of $(x,y,u,v) = (0,1,1,-1)$, $x$ and $y$ are implicitly defined as $C^1$-functions $f$ and $g$ of $u$ and $v$ by the system of equations : $$\begin{cases} e^x +uy + u^4v-a=0 \\ y\cos(x) +bx +b^2u -2v = 4 \end{cases}$$

NB: check whether all conditions of the implicit function theorem are satisfied and state the conclusions as accurately as possible.

My work so far:

Let $G : \mathbb{R}^4 \to \mathbb{R}^2$ be defined by: $$G(x,y,u,v)= \begin{pmatrix} e^x +uy + u^4v -a\\ y\cos(x) +bx +b^2u -2v \end{pmatrix}$$ Obviously, this function is $C^1$.

After this, I presume I have to form a matrix of partial derivatives and check for what values of $a$ and $b$ that matrix is non-singular.

Could anyone give me some guidance on how to continue?

• Just do what you presume you have to do: this is exactly the way it should be done. – Etienne Feb 26 '14 at 14:06

The Jacobian Matrix of the function $G: \mathbb{R}^4 \to \mathbb{R}^2$ defined by: $$G(x,y,u,v)=\begin{pmatrix} e^x + uy + u^4 \\ y\cos(x) + bx + b^2u-2 \end{pmatrix}$$ is $$DG(x,y,u,v)=\begin{pmatrix} e^x & u & y+4u^3v & u^4 \\ -y\sin(x) + b & \cos(x) &b^2 & -2 \end{pmatrix}.$$ Because all entries are continuous functions, $G$ is a $C^1$ function.
Observe that $$G(0,1,1,-1) = \begin{pmatrix} 1 \\ b^2+3 \end{pmatrix} = \begin{pmatrix} a \\ 4 \end{pmatrix} \iff a = 1 \wedge b^2=1 \Longrightarrow a = 1 \wedge \left( b=-1 \vee b = 1\right).$$ The matrix with partial derivatives of $G$ with respect to $x$ and $y$ at $(0,1,1,-1)$, which is $$\begin{pmatrix} \frac{\partial G}{\partial x}(0,1,1,-1) & \frac{\partial G}{\partial y}(0,1,1,-1) \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ b & 1 \end{pmatrix}$$ is singular if $b=1$ and non-singular if $b=-1$. So for $a=1$ and $b=-1$ all conditions of the implicit function theorem are satisfied and thus there exists an open neighbourhood $B$ of $(1,-1)$, an open neighbourhood $U$ of $(0,1,1,-1)$ and $C^1$ functions $f:B \to \mathbb{R}$ and $g : B \to \mathbb{R}$ such that $$\{ (f(u,v), g(u,v), u, v : (u,v) \in B \} = \{(x,y,u,v) \in U: G(,x,y,u,v) = (1.4) \}$$
That's right. Let $z=[u,v]'$. Then the implicit function theorem implies in your case that
$$\frac{\partial G}{\partial z'} + \begin{bmatrix} \frac{\partial G}{\partial x} \; \frac{\partial G}{\partial y}\end{bmatrix} \begin{bmatrix} \frac{\partial f}{\partial z'} \\ \frac{\partial g}{\partial z'} \end{bmatrix}=0.$$
So if the matrix of partial derivatives (with respect to $x,y$) is invertible then life is wonderful.
• What do you mean with $z=[u,v]'$ and $z'$? – Nigel Overmars Feb 26 '14 at 14:35
• $z$ is a (column) vector with elements $u$ and $v$, $z$ is its transpose. – JPi Feb 26 '14 at 19:40