Invertible Function: Proving Bijection & Finding Inverse

[mr.tea]

Homework Statement

prove, using the definition, that the mapping
$\mathbf{u}=\mathbf{u}(u_1(x_1,x_2),u_2(x_1,x_2))$ where
$u_1=\tan(x_1)+x_2$
$u_2=x_2^3$

is a bijection from the strip $-\frac{pi}{2}<x_1<\frac{pi}{2}$ in the $x_1x_2$-plane onto the entire $u_1u_2$-plane, and find the inverse.

Homework Equations

The Attempt at a Solution

Finding the inverse is not the problem. My problem is that when I am trying to apply the inverse function theorem, I get that at $x_2=0$ the Jacobian is 0. The Jacobian that I have found is:
$\frac{\partial (u_1,u_2)}{\partial (x_1,x_2)}= \begin{vmatrix} \frac{1}{cos^2(x_1)} & 1\\ 0& 3x_2^2\\ \end{vmatrix}$

What am I missing here?

Thank you.

 

[Samy_A]

Homework Statement

prove, using the definition, that the mapping
$\mathbf{u}=\mathbf{u}(u_1(x_1,x_2),u_2(x_1,x_2))$ where
$u_1=\tan(x_1)+x_2$
$u_2=x_2^3$

is a bijection from the strip $-\frac{pi}{2}<x_1<\frac{pi}{2}$ in the $x_1x_2$-plane onto the entire $u_1u_2$-plane, and find the inverse.

Homework Equations

The Attempt at a Solution

Finding the inverse is not the problem. My problem is that when I am trying to apply the inverse function theorem, I get that at $x_2=0$ the Jacobian is 0. The Jacobian that I have found is:
$\frac{\partial (u_1,u_2)}{\partial (x_1,x_2)}= \begin{vmatrix} \frac{1}{cos^2(x_1)} & 1\\ 0& 3x_2^2\\ \end{vmatrix}$

What am I missing here?

Thank you.

Hint:
Let $f: \mathbb R \to \mathbb R \ :\ x \mapsto x³$.
Is $f$ a bijection?
Is the derivative of $f$ different from 0 in every point?

What does this tell us about the inverse function theorem?

 

[mr.tea]

Hint:
Let $f: \mathbb R \to \mathbb R \ :\ x \mapsto x³$.
Is $f$ a bijection?
Is the derivative of $f$ different from 0 in every point?

What does this tell us about the inverse function theorem?

Sorry for the late reply.
the answers are "yes"(but the derivative of f is 0 at 0).
Does this tell us that the inverse function theorem doesn't apply to that point(0 in your example)? Does it mean that we need to prove for that point without using the inverse function theorem?

Thank you.

 

[Samy_A]

Sorry for the late reply.
the answers are "yes"(but the derivative of f is 0 at 0).
Does this tell us that the inverse function theorem doesn't apply to that point(0 in your example)? Does it mean that we need to prove for that point without using the inverse function theorem?

Thank you.

Yes.

Let's look at the precise statement of the inverse function theorem:

Let f : ℝⁿ → ℝⁿ be continuously differentiable on some open set containing a, and suppose that Df(a) is not singular. Then there is some open set V containing a and an open W containing f(a) such that f : V → W has a continuous inverse f⁻¹ : W → V which is differentiable for all y ∈ W.

The Jacobian not being singular is a sufficient condition for a function to have an inverse (locally), not a necessary one.

 

[mr.tea]

Yes.

Let's look at the precise statement of the inverse function theorem:

Let f : ℝⁿ → ℝⁿ be continuously differentiable on some open set containing a, and suppose that Df(a) is not singular. Then there is some open set V containing a and an open W containing f(a) such that f : V → W has a continuous inverse f⁻¹ : W → V which is differentiable for all y ∈ W.

The Jacobian not being singular is a sufficient condition for a function to have an inverse (locally), not a necessary one.

OK, great. I think I have got it.
By the way, do you know a good book for self learning multivariable analysis?(standard course for mathematicians)

Thank you again!

 

[Samy_A]

OK, great. I think I have got it.
By the way, do you know a good book for self learning multivariable analysis?(standard course for mathematicians)

Thank you again!

This insight from @micromass may help.