Exercise with application of inverse function theorem

Question

Let $F:\mathbb R^2 \to \mathbb R^2$ be a function of class $C^1$ such that $F(3t^3+2,e^{t^2})=(3,6)$ for all $t \in \mathbb R$. Prove that $DF(2,1)$ is not invertible. ($DF(2,1)$ is the jacobian matrix)

My attempt at a solution:

I've tried to solve this problem applying the inverse function theorem, i.e, suppose $DF(2,1)$ is invertible, since $f$ is $C^1$, we are under the hypothesis of the inverse function theorem:

There exist $U$ neighbourhood of $(2,1)$ and $V$ neighbourhood of $f(2,1)$ such that $F:U \to V$ is bijective, the inverse $F^{-1}:V \to U$ is of class $C^1$ and $DF^{-1}(x,y)={DF(x,y)}^{-1}$.

I am trying to get to an absurd and I think that, given the fact that $F(3t^3+2,e^{t^2})=(3,6)$ for all $t \in \mathbb R$, it clearly fails injectivity here.

I am not so sure how to show that $F$ is not injective, I could define $h:\mathbb R \to \mathbb R^2$ as $h(t)=(3t^3+2,e^{t^2})$. If I could find $t_1$ and $t_2$ such that $h(t_1), h(t_2) \in U$, then, I would be done, because $F(h(t_1))=F(h(t_2))$. I don't know how to pick $t_1, t_2$ "appropiately".

I would appreciate if someone could tell me if my idea is correct and give me some help with this last part.

Answer 1

It seems to me that it would be easier just to use the chain rule (considering $F\circ h$) to show directly that $h'(t)\in\ker DF(h(t))$, and $h'(t)\ne 0$ for $t\ne 0$. Thus, if $\det DF(h(t))=0$ for $t\ne 0$, what do you conclude? (Sneaky of the author of the problem to make $h'(0)=0$!)

But, for your approach, since $h(0)\in U$ and $h$ is continuous, there's a whole interval of $t$'s around $0$ with $h(t)\in U$.

Answer 2

It appears to me at first glance that your idea, as a proof strategy, is fundamentally well founded but might be difficult to execute directly. I think it might be simpler to show that $DF$, the Jacobean of $F$, is singular at $(2, 1)$ in a straightforward manner, as follows:

Let us denote the components of $F:\Bbb R^2 \to \Bbb R^2$ by $F_x$ and $F_y$, so that $F(x, y) = (F_x(x, y), F_y(x, y))$ for $(x, y) \in \Bbb R^2$. Consider the curve $\gamma(t) = (3t^3 + 2, e^{t^2})$ in $\Bbb R^2$. Clearly $\gamma(0) = (2, 1)$, so $\gamma(t)$ passes through the point $(2, 1) \in \Bbb R^2$ when $t = 0$. We are given that $F(3t^3 + 2, e^{t^2}) = (3, 6)$ for all $t \in \Bbb R$; writing it out in terms of components yields

$F_x(3t^3 + 2, e^{t^2}) = 3, \tag{1}$

and

$F_y(3t^3 + 2, e^{t^2}) = 6. \tag{2}$

If we now differentiate (1) and (2) with respect to $t$ using the chain rule, we obtain

$(\partial F_x / \partial x)(3t^3 + 2, e^{t^2})(9t^2) + (\partial F_x / \partial y)(3t^3 + 2, e^{t^2})(2te^{t^2}) = 0 \tag{3}$

and

$(\partial F_y / \partial x)(3t^3 + 2, e^{t^2})(9t^2) + (\partial F_y / \partial y)(3t^3 + 2, e^{t^2})(2te^{t^2}) = 0 \tag{4}$

holding for any $t \in \Bbb R$. We can re-write (3), (4) as the matrix-vector equation

$\begin{bmatrix} \partial F_x / \partial x & \partial F_x / \partial y \\ \partial F_y / \partial x & \partial F_y / \partial y \end{bmatrix}_{(3t^3 + 2, \; e^{t^2})} \begin{pmatrix} 9t^2 \\ 2te^{t^2} \end{pmatrix} = 0, \tag{5}$

which indicates the Jacobean of $F$ is singular for any $t \ne 0$, since the vector $(9t^2, 2te^{t^2})^T \ne 0$ for $t \ne 0$. This implies that

$\det(\begin{bmatrix} \partial F_x / \partial x & \partial F_x / \partial y \\ \partial F_y / \partial x & \partial F_y / \partial y \end{bmatrix}_{(3t^3 + 2, \; e^{t^2})}) = 0 \tag{6}$

for any point on the curve $\gamma(t)$ where $t \ne 0$; since the determinant of a square matrix is a continuous function of its entries, letting $t \to 0$ shows

$\det(DF(2, 1)) = \det(\begin{bmatrix} \partial F_x / \partial x & \partial F_x / \partial y \\ \partial F_y / \partial x & \partial F_y / \partial y \end{bmatrix}_{(2, \; 1)}) = 0, \tag{7}$

whence $DF(2, 1)$ is not invertible. QED.

From the other point of view, $F$ is clearly not injective in any neighborhood of $(2, 1)$ since it maps the entire curve $\gamma(t)$ to the single point $(3, 6)$. And as Professor Shifrin points out in his answer, there is an entire interval $I = (-\epsilon, \epsilon)$, $\epsilon > 0$, with $\gamma(I) \subset U$.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!