How to describe the region $0<x^3<y<\sqrt{x}<1$ using the coordinates$g(u,v)=(\frac{u}{v},v)$

Question

I need to put the region $0<x^3<y<\sqrt{x}<1$ in terms of u,v using the change of coordinates $g(u,v)=(\frac{u}{v},v)$ but I don't know how to do this. If I just do the change , I get $u\in (0,v)$ and $v\in \left(\frac{u^3}{v^3},\sqrt{\frac{u}{v}}\right)$, and I don't think this is a good answer.

Answer 1

Addendum added to respond to the comment/question of EMS.

---

Unsure if this is what you want. Please advise if you are looking for a different format.

$$v \in (0,1).\tag1 $$

$u^3 < v \implies u < v^{1/3}.$
Also, $v < u^{1/2} \implies v^2 < u.$

Therefore, $v^2 < u < v^{1/3}.$
Therefore,

$$u \in \left(v^2, v^{1/3}\right).\tag2 $$

---

Alternatively,

$$u \in (0,1), ~~v \in \left(u^3, \sqrt{u}\right).$$

---

Another alternative (I am just guessing here).

Let $r$ denote $\dfrac{u}{v}.$

Then you have the change of coordinates:
$g(u,v) = \left(\dfrac{u}{v},v\right) = (r,v).$

Using (2) above, you have that

$v^2 < u < v^{1/3} \implies v < r < v^{-2/3}.$

So, you have coordinates $(r,v)$ where
$v \in (0,1)$ and $r \in \left(v, v^{-2/3}\right).$

---

---

---

$\underline{\textbf{Addendum}}$
Responding to the comment/question of EMS.

could you give me a hint with making an alternative such as the one made in the first alternative?

Let $r$ denote $\dfrac{u}{v}.$

Then you have the change of coordinates:
$g(u,v) = \left(\dfrac{u}{v},v\right) = (r,v).$

At the end of my original answer, the result was

$$v \in (0,1) ~~~~\text{and}~~~~ r \in \left(v, v^{-2/3}\right). \tag3 $$

I am interpreting your request to be that instead, you want to provide a lower and upper bound for $(r)$, and then provide a lower and upper bound for $(v)$, in terms of $(r)$.

Edit
If this is not what you are asking for, then please leave another comment question, and I will add an Addendum-2 to my answer.

---

Using (3) above as a starting point, the goal is to fill in the following blanks:

$$\text{Allowable range for} ~(r)~ \text{is} ~? \tag4 $$

$$\text{Given a specific value for} ~(r),~ \text{the allowable range for} ~(v)~ \text{is} ~? \tag5 $$

First, preliminary results are needed.

---

$\underline{\text{Preliminary results:}}$

Lemma 1
$\displaystyle 0 < a < 1 \implies 1 < a^{-(2/3)}.$
Proof
$\displaystyle 0 < a < 1 \implies 0 < a^{1/3} < 1 \implies 0 < \left[a^{1/3}\right]^2 = a^{2/3} < 1 \implies $
$\displaystyle 1 < \frac{1}{a^{2/3}} = a^{-(2/3)}.$

Lemma 2
$\displaystyle 1 < a \implies 0 < a^{-(3/2)} < 1.$
Proof
$\displaystyle 1 < a \implies 1 < a^{1/2} \implies 1 < \left[a^{1/2}\right]^3 = a^{3/2} \implies $
$\displaystyle 1 > \frac{1}{a^{3/2}} = a^{-(3/2)}.$
Further,
$\displaystyle 1 < a \implies 0 < a \implies 0 < a^{1/2} \implies $
$\displaystyle 0 < \left[a^{1/2}\right]^3 = a^{3/2} \implies 0 < \frac{1}{a^{3/2}} = a^{-(3/2)}.$
Therefore,
$\displaystyle 1 < a \implies 1 > a^{-(3/2)} ~~~~$ and $~~~~\displaystyle 0 < a^{-(3/2)}.$

---

$\underline{\text{Completion of the Addendum}}$
Before attacking (4) and (5) above, the contraints to be satisfied must be listed:

• Constraint C-1 $~: ~0 < v < 1.$
• Constraint C-2 $~: ~v < r.$
• Constraint C-3 $~: \displaystyle ~r < v^{-(2/3)}.$

Now, (4) above will be attacked by the following claim:
Claim
For all $(r)$ such that $0 < r$,
there exists a $(v)$ such that each of C-1, C-2, and C-3 are satisfied.

Assuming that the claim is proven, this will resolve (4) above, since it will be established that the allowable range for $(r)$ is any $(r)$ such that $0 < r.$

Proof of Claim
To prove this claim, each of the following cases must be examined separately:

• Case 1: $0 < r \leq 1.$
• Case 2: $1 < r.$

$\underline{\text{Case 1:} ~0 < r \leq 1}$

Choose $~\displaystyle v = \frac{r}{2} \implies 0 < v < r \leq 1.$
Therefore, C-1 and C-2 are both satisfied.
Further, by Lemma 1, since $0 < v < 1, ~ r \leq 1 < v^{-(2/3)}.$
Therefore, C-3 is also satisfied.

$\underline{\text{Case 2:} ~1 < r}$

By Lemma 2, $~0 < r^{-(3/2)} < 1.$
Choose $~\displaystyle v = \frac{r^{-(3/2)}}{2} \implies 0 < v < 1 < r.$
Therefore, Therefore, C-1 and C-2 are both satisfied.
Further,
since $~\displaystyle 0 < v < r^{(-3/2)} < 1,$
$\displaystyle v^{-1} = \frac{1}{v} > \frac{1}{r^{(-3/2)}} = r^{(3/2)} > 1 \implies $
$~\displaystyle v^{-(2/3)} = \left[v^{(-1)}\right]^{(2/3)} > \left[r^{(3/2)}\right]^{(2/3)} = r.$
Therefore, C-3 is also satisfied.

---

Now, with (4) above resolved, (5) will be attacked. That is, a lower and upper bound for $(v)$, in terms of $(r)$ will be established. To do this, two cases must be considered:

• Case 1: $0 < r \leq 1.$
• Case 2: $1 < r.$

$\underline{\text{Case 1:} ~0 < r \leq 1}$
From C-1 and C-2 above, you must have that
$0 < v < r \leq 1.$
Then, by Lemma 1, $~\displaystyle r \leq 1 < v^{-(2/3)} \implies $
C-3 is automatically satisfied.

$\underline{\text{Case 2:} ~1 < r}$
From C-1 and C-2 above, you must have that
$0 < v < 1 < r.$

However, by Lemma 2, $~\displaystyle 0 < r^{-(3/2)} < 1.$
Therefore, if
$~\displaystyle r^{-(3/2)} \leq v < 1$,
then
$\displaystyle 1 < v^{-1} = \frac{1}{v} \leq \frac{1}{r^{-(3/2)}} = r^{(3/2)} \implies $
$\displaystyle v^{-(2/3)} = \left[v^{-1}\right]^{(2/3)} \leq \left[r^{(3/2)}\right]^{(2/3)} = r \implies $
C-3 is violated.

Alternatively, if
$~\displaystyle 0 < v < r^{-(3/2)} < 1$,
then
$\displaystyle 1 < r^{(3/2)} = \frac{1}{r^{-(3/2)}} < \frac{1}{v} = v^{-1} \implies $
$\displaystyle r = \left[r^{(3/2)}\right]^{(2/3)} < \left[v^{-1}\right]^{(2/3)} = v^{-(2/3)} \implies $
C-3 is satisfied.

Therefore, in Case 2, you must have that
$~\displaystyle 0 < v < r^{-(3/2)}.$

---

$\underline{\text{In Summary}}$

$(r)$ can be any value such that $0 < r.$

When $r \leq 1$, you must have that
$0 < v < r.$

When $1 > r$, you must have that
$~\displaystyle 0 < v < r^{-(3/2)}.$