Given a function $F$, how to evaluate $\max_{t\in (0,1)}\left\vert \frac{\partial^2 F}{\partial x_i \partial x_j}(ty)\right\vert$

calculusderivativesfunctional-analysismultivariable-calculusreal-analysis

As an "extra problem" for my calculus 2 test class we got the following. Since it is an extra problem, I expected it to be more difficult than the other exercises, but actually but not that much. This is the problem:

Let $F:\mathbb{R}^3\setminus\{0\}\to\mathbb{R}$ such that
$$F:(x_1, x_2, x_3)\mapsto -\frac{1}{|x|^2} +\frac{1}{|x|}\in\mathbb{R}.$$
For a fixed $y\in\mathbb{R}^3\setminus\{0\}$, find
$$\max_{t\in (0,1)}\left\vert \frac{\partial^2 F}{\partial x_i \partial x_j}(ty)\right\vert.$$

To start I computed $\nabla F(x)$ which is
$$\nabla F(x) = \left(\frac{2}{|x|^4}-\frac{1}{|x|^3}\right) x,$$
where $x$ denotes the vector $x=(x_1, x_2, x_3)$.

I went ahead by computing the second partial derivatives. I had
$$\frac{\partial^2 F}{\partial x_i^2} = \frac{2}{|x|^4}-\frac{1}{|x|^3} +\frac{3x_i^2}{|x|^5}-\frac{8x_i^2}{|x|^6}, \quad i\in\{1, 2, 3\}$$
and
$$\frac{\partial^2 F}{\partial x_i\partial x_j} = \frac{3x_i x_j}{|x|^5}-\frac{8x_i x_j}{|x|^6}, \quad i\in\{1, 2, 3\}, i\neq j.$$

I got stuck at this point since I have no idea how to establish the maximum. I’ve been thinking about it unsuccessfully for a couple of days, but I am really interested in the solution.

Could someone please help in finding that maximum? Any hint will be very helpful.

Thank you in advance.

Best Answer

Consider the family of functions

$$G_{ij}(t;y):=\frac{\partial^2 F}{\partial x_i\partial x_j}(ty)$$

Write $$F(x)=f(|x|) \\ f(s)=-\frac{1}{s^2}+\frac{1}{s}$$ Using the chain rule, $$\frac{\partial F}{\partial x_i}(x)=f'(|x|)~\frac{x_i}{|x|}$$

And the product rule, $$ \frac{\partial^2 F}{\partial x_i\partial x_j}(x)=f''(|x|)\frac{x_ix_j}{|x|^2}+f'(|x|)\frac{\frac{\partial x_i}{\partial x_j}|x|-\frac{x_j}{|x|}x_i}{|x|^2} \\ =f''(|x|)\frac{x_ix_j}{|x|^2}-f'(|x|)\frac{x_ix_j}{|x|^3}+\delta_{ij}f'(|x|)\frac{1}{|x|}$$

Now, compute $$f'(s)=\frac{2}{s^3}-\frac{1}{s^2} \\ f''(s)=\frac{-6}{s^4}+\frac{2}{s^3}$$

Hence $$\frac{\partial ^2 F}{\partial x_i\partial x_j}(x)=\frac{x_ix_j}{|x|^2}\left(f''(|x|)-\frac{f'(|x|)}{x}+\delta_{ij}f'(|x|)\frac{1}{|x|}\right) \\ =\frac{x_ix_j}{|x|^2}\left(\frac{-6}{|x|^4}+\frac{2}{|x|^3}-\frac{2}{|x|^4}+\frac{1}{|x|^3}+\delta_{ij}\left(\frac{2}{|x|^4}-\frac{1}{|x|^3}\right)\right) \\ =\frac{x_ix_j}{|x|^5}\left(3-\frac{8}{|x|}+\delta_{ij}\left(\frac{2}{|x|}-1\right)\right) \\ =\frac{x_ix_j}{|x|^5}\left((3-\delta_{ij})+\frac{2}{|x|}\left(\delta_{ij}-4\right)\right)$$

Therefore $$G_{ij}(t;y)=\frac{ty_ity_j}{|ty|^5}\left((3-\delta_{ij})+\frac{2}{|ty|}\left(\delta_{ij}-4\right)\right) \\ =|t|^{-3}\frac{y_iy_j}{|y|}\left((3-\delta_{ij})+t^{-1}\frac{2}{|y|}\left(\delta_{ij}-4\right)\right)$$

So for $i\neq j$ and $t\in(0,1)$, $$|G_{ij}(t;y)|=b~t^{-3}|3-8at^{-1}|$$ Where $a=|y_iy_j/|y|^5|$ and $a=1/|y|$. This clearly is unbounded.

On the other hand for $i=j$, $$|G_{ii}(t;y)|=bt^{-3}|2-6at^{-1}|$$

Same thing.

Appendix: Python code

#!/usr/bin/env python
# coding: utf-8

# In[1]: setup
import matplotlib.pyplot as plt
import numpy as np
from scipy.optimize import NonlinearConstraint, minimize
from scipy.spatial import distance_matrix

N=39
M=2
mask = ~np.eye(N, dtype=bool)

def g(X):
    return np.sum(X**2)-1

def f(X):
    X = X.reshape(N,M)
    D = distance_matrix(X,X,p=2)
    S = np.where(mask, D, np.inf)
    return np.sum(S**(-1/2))

# In[2]: generating regular n-gon
r = N**(-1/2)
phi = np.arange(N)*(2*np.pi/N)
X0 = r* np.stack( (np.sin(phi), np.cos(phi)), axis=1 )

# In[3]: calling solver
sol = minimize(f, X0.flatten(), method='trust-constr',
               constraints = NonlinearConstraint(g, 0, 0))

# In[4]: plotting solution
XS = sol.x.reshape(N,M)
print(F"initial config: f={f(X0):.4f} g={g(X0)}")
print(F"  final config: f={f(XS):.4f} g={g(XS)}")
plt.plot(*X0.T, '+k', *XS.T, 'xr')
plt.legend(["$x^{(0)}$", "$x^*$"])
plt.title(F"{N} points")
plt.axis('square')
plt.savefig(F"configs{N}.png", bbox_inches='tight')
plt.show()

Prove that $u(x)=\log\log\left(1+\frac{1}{\vert x\vert}\right)\in H^1(B(0,1))$

In the computation of $\int_{\Omega}|\nabla f|^2 \,dx$ you forgot $r\,dr$.

Hence the integral converges since $\frac{1}{\log(1+1/r)^2 r(1+r)^2}$ is integrable close to $r=0$ and we get $f\in H^1(\Omega)$.

Best Answer

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