Are there general methods classifying isotopies between two given embeddings?

Question

We can define a different isotopy from i to f by E′(x,y,t)=((1−t)x+tan(tarctan(x/1−y)),(1−t)y) . In this case, we are opening out C left and right of the North pole while simultaneously stretching it and flattening it onto the x -axis, where we arrive at t=1 . However, this time our isotopy does not have fixed points and produces bounded arcs for all t<1 .

In summary, we have two isotopies from i to f . On the one hand E produces all unbounded arcs but one and has fixed points, while E′ produces all bounded arcs but one and has no fixed points. Do features like these may in any way allow us to group/classify the set of isotopies between two given embeddings? Is there a general theory for this in the different categories (TOP, DIFF, PL)?

Following on this, in the related question, the embedding i remains the inclusion, but then an embedding f is defined, which I will call f1 here. The corresponding isotopy from i to f1 is given by E1(x,y,t)=(1−t)i(x,y)+tf1(x,y) . Visually here we are opening up C left and right of the North pole, without much stretching, while simultaneously compressing it until it lies flattened onto the interval (−π/2,π/2)×{0} when we arrive at t=1 . Note how the process produces bounded arcs for all t this time. Moreover, the isotopy has no fixed points.

Answer 1

As an option, you can apply the Cauchy–Schwarz inequality :

$$ \begin{align}\left(2x^2+3y^2+4z^2\right)\left(\frac 12+\frac 13+\frac 14\right)\geqslant (x+y+z)^2\end{align} $$

This implies that:

$$ \begin{align} 2x^2+3y^2+4z^2&\geqslant \frac {13^2}{\frac 12+\frac 13+\frac 14}\\ &=156.\end{align} $$

Equality holds iff:

$$2x=3y=4z\;\wedge\; x+y+z=13.$$

This gives the global minimum:

$$(x,y,z)=(6,4,3)$$

But, observe that the global maximum doesn't exist. Indeed, setting $x=0,\,z=13-y$ with $y\to+\infty,\, z\to-\infty$ yields $2x^2+3y^2+4z^2\to+\infty.$

Answer 2

You can check that the (constant) Hessian matrix of your function $f$ is positive definite, which implies that $f$ is convex in $\mathbb{R}^2$. Since you checked that there is a stationary point, the convexity of $f$ allows you to say that it will be a global minimiser.

The existence of maximisers has already been ruled out by @lonestudent.

Answer 3

You forgot the second derivative test with the discriminant:

$$\Delta=f_{xx}f_{yy}-f_{xy}^2=(4+8)(6+8)-(8)^2=168-64=104>0,$$

and $f_{xx}=12>0$, so that, the point $(6,4,3)$ is a local minumum.

Answer 4

In what follows we use the theory found at this site question

$\quad$Extreme point of quadratic equation

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The plane $x+y+z = 13$ can be sliced into lines, one for each fixed $z =c$.

We can also solve for $y$.

$\quad y = 13 - c -x$

Substituting into the quadratic $2x^2 + 3y^2 + 4z^2$,

$\tag 12x^2+3(13 - c -x)^2 +4c^2= 5 x^2+(6 c- 78) x+7 c^2 - 78 c + 507$

So the maximum on each sliced line is $+\infty$.

Substituting

$$ x = \frac{78-6c}{10}$$

into the rhs of $\text{(1)}$ we get

$$ \frac{26}{5} (c^2 - 6 c + 39)$$

and evaluating this when $c = \frac{6}{2}$ we get the global minimum is $156$.