Does this function $f(x,y)=e^{(x-y)^2}(x^4+y^4-\sin(y^3))$ have a global minimum?

Question

Deifne the function $f:\mathbb{R}^2\to\mathbb{R}$ by $f(x,y)=e^{(x-y)^2}(x^4+y^4-\sin(y^3))$ and my question is this: does this function have a global minimum? I am only familiar with standard vector-calculus techniques and tricks, but none works here. A graph-calculator indicates that $f$ does have a global minimum indeed. Can anyone tell me the tool/theorem that is used here to provide the existence of the global minimum?

P.S: I found this on a sheet of "non-linear optimization" exercises (i am in the dark on that part of mathematics), so, I guess that is relevant.

Answer 1

You have $$f(x,y)\geq x^4+y^4-1\qquad\bigl((x,y)\in{\mathbb R}^2\bigr)$$ and $f(0,0)=0$. The square $Q:=[-2,2]^2$ is compact, and on $\partial Q$ and outside of $Q$ we have $f(x,y)\geq15$. Since $f$ is continuous it assumes a global minimum value $\leq0$ on $Q$, and this will take place in an interior point of $Q$. This is then also the global minimum of $f$ on ${\mathbb R}^2$. We therefore have to determine the critical points of $f$ in $Q$ and to compare the values of $f$ in these points.

Answer 2

If we restrict to a closed bounded region $D$ (such as closed rectangle or closed disk), then there is no issue ensuring the existence of global minimum, thanks to Extreme Value Theorem for continuous functions. (Locating such one is a different question, though.)

That being said, the only possible way your function $f$ fails to have a global minimum is that it keeps producing lower values as we progressively enlarge the region $D$. One way to prevent this from happening is to show that $f(\mathrm{x}) \to +\infty$ as $\|\mathrm{x}\|\to\infty$, so that $f$ can attain small values only on a bounded region. (Such functions are often called coercive.)

Keeping this in mind, let us show that this is indeed the case for OP's function $f$. Either by AM-GM inequality or by Cauchy-Schwarz inequality, we have $x^4 + y^4 \geq \frac{1}{2}(x^2 + y^2)^2$. Along with trivial bounds $e^{(x-y)^2} \geq 1$ and $-\sin(y^3) \geq -1$, we get

$$ f(x, y) \geq \frac{1}{2}(x^2 + y^2)^2 - 1. $$

Since this bound diverges to $+\infty$ as $\|(x, y)\| \to \infty$, we also get $f(x, y) \to +\infty$ as $\|(x, y)\| \to \infty$ as well. Together with Extreme Value Theorem, this tells that $f$ achieves a global minimum.

Numerical simulation using Mathematica suggests that $f$ attains the unique global minimum around the point $(-0.349458, 0.755851)$. The following is the density plot of $f$.

Due to the rapid increase in values of $f$, large values are cropped out to emphasize near-minimum behavior of $f$.

Answer 3

Update Note on initial value.

I have some ideas. Note $(x^4+y^4)= (x^2-y^2)^2+2xy$ and $(x-y)$ appears in the exponent. So ignoring the $\sin$ term, there is a symmetry in swapping x and y.

Initial seed cannot be $(0,0)$ it does not converge. Some other value is needed for $(x_0,y_0)$.

$x_{n+1}=x_n-f(x_n,y_n)/\frac{\partial f}{\partial x_n}$

$y_{n+1}=y_n-f(x_n,y_n)/\frac{\partial f}{\partial y_n}$

That should converge pretty quick.