Consider the function $$f(x):=\frac{x-x_0}{\Vert x-x_0 \Vert^2} + \frac{x-x_1}{\Vert x-x_1 \Vert^2}$$
for two fixed $x_0,x_1 \in \mathbb R^2$ and $x \in \mathbb R^2$ as well.
Does anybody know what the Hessian of the function
$$g(x):=\Vert f(x) \Vert^2$$
is? It is such a difficult composition of functions that I find it very hard to compute.
The bounty is for a person who fully derives the Hessian of $f. $ Please let me know if you have any questions.
On
Well, you can just compute $g(x)$... I'll denote $x_0,x_1$ by $u,v$.
$$ f(x) = \frac{x-u}{\|x-u\|^2}+\frac{x-v}{\|x-v\|^2}=\left(\frac{x_1-u_1}{(x_1-u_1)^2+(x_2-u_2)^2}+\frac{x_1-v_1}{(x_1-v_1)^2+(x_2-v_2)^2}, \right. $$ $$ \left.\frac{x_2-u_2}{(x_1-u_1)^2+(x_2-u_2)^2}+\frac{x_2-v_2}{(x_1-v_1)^2+(x_2-v_2)^2} \right) $$
and so,
$$ g(x)=\left(\frac{x_1-u_1}{(x_1-u_1)^2+(x_2-u_2)^2}+\frac{x_1-v_1}{(x_1-v_1)^2+(x_2-v_2)^2}\right)^2+ $$ $$ \left(\frac{x_2-u_2}{(x_1-u_1)^2+(x_2-u_2)^2}+\frac{x_2-v_2}{(x_1-v_1)^2+(x_2-v_2)^2} \right)^2 $$
Now you can compute the Hessian matrix.
On
Avoid coordinates. Here's a solution that works with all dot products in any dimension:
Define $h_k(x)=\frac{x-x_k}{\|x-x_k\|^2}$ for $k\in\{0,1\}$. Then we have $\|h_k(x)\|^2=1/\|x-x_k\|^2$ and $$d_ph(x)=p\|h_k(x)\|^2-2h_k(x)\langle p,h_k(x)\rangle$$ and$$d_p\|h_k(x)\|^2=-2\|h_k(x)\|^2\langle p,h_k(x)\rangle.$$
After a straightforward calculation I get $$\frac12\nabla g(x)=\bigl(\|h_0(x)\|^2-\|h_1(x)\|^2\bigr)\cdot\bigl(h_1(x)-h_0(x)\bigr)-2\langle h_0(x),h_1(x)\rangle\bigl(h_0(x)+h_1(x)\bigr).$$
From here feel free to calculate the Hessian: Differentiating $\frac12\nabla g(x)$ again at $q$ we get (omitting the argument $x$ for the sake of readability) $$\begin{align} -&q\left(2\langle h_0,h_1\rangle(\|h_0\|^2+\|h_1\|^2)+(\|h_0\|^2-\|h_1\|^2)^2\right)\\ +&h_0\langle q,4h_0\|h_0\|^2-(2h_0+h_1)\|h_1-h_0\|^2\rangle\\ +&h_1\langle q,4h_1\|h_1\|^2-(2h_1+h_0)\|h_0-h_1\|^2\rangle, \end{align} $$ hence the Hessian at $(p,q)$ is $$\begin{align} 2\langle p,-&q\left(2\langle h_0,h_1\rangle\|(h_0\|^2+\|h_1\|^2)+(\|h_0\|^2-\|h_1\|^2)^2\right)\\ +&h_0\langle q,4h_0\|h_0\|^2-(2h_0+h_1)\|h_1-h_0\|^2\rangle\\ +&h_1\langle q,4h_1\|h_1\|^2-(2h_1+h_0)\|h_0-h_1\|^2\rangle\rangle. \end{align} $$
On
(Edit) RIP bounty, but here's the (correct) solution computed via Sympy/Jupyter.
setting $x_1=0$ and relabelling $x_0$ as $(x_0,y_0)$, $x$ as $(x,y)$,
x,y,x0,y0 = symbols('x y x0 y0')
we have from
f1 = (x-x0)/((x-x0)**2 + (y-y0)**2) + x/(x**2+y**2)
f2 = (y-y0)/((x-x0)**2 + (y-y0)**2) + y/(x**2+y**2)
g = f1*f1 + f2*f2
simplify(g)
$$g(x,y) = \displaystyle \frac{\left(x \left(\left(x - x_{0}\right)^{2} + \left(y - y_{0}\right)^{2}\right) + \left(x - x_{0}\right) \left(x^{2} + y^{2}\right)\right)^{2} + \left(y \left(\left(x - x_{0}\right)^{2} + \left(y - y_{0}\right)^{2}\right) + \left(x^{2} + y^{2}\right) \left(y - y_{0}\right)\right)^{2}}{\left(x^{2} + y^{2}\right)^{2} \left(\left(x - x_{0}\right)^{2} + \left(y - y_{0}\right)^{2}\right)^{2}} $$
simplify(diff(g,x,x)) tells us that $\partial_x^2 g = $
$$
\displaystyle \frac{2 \left(2 \left(x \left(\left(x - x_{0}\right)^{2} + \left(y - y_{0}\right)^{2}\right) + \left(x - x_{0}\right) \left(x^{2} + y^{2}\right)\right) \left(4 x^{3} \left(\left(x - x_{0}\right)^{2} + \left(y - y_{0}\right)^{2}\right)^{3} - 3 x \left(x^{2} + y^{2}\right) \left(\left(x - x_{0}\right)^{2} + \left(y - y_{0}\right)^{2}\right)^{3} + 3 \left(- x + x_{0}\right) \left(x^{2} + y^{2}\right)^{3} \left(\left(x - x_{0}\right)^{2} + \left(y - y_{0}\right)^{2}\right) + 4 \left(x - x_{0}\right)^{3} \left(x^{2} + y^{2}\right)^{3}\right) + 2 \left(y \left(\left(x - x_{0}\right)^{2} + \left(y - y_{0}\right)^{2}\right) + \left(x^{2} + y^{2}\right) \left(y - y_{0}\right)\right) \left(4 x^{2} y \left(\left(x - x_{0}\right)^{2} + \left(y - y_{0}\right)^{2}\right)^{3} - y \left(x^{2} + y^{2}\right) \left(\left(x - x_{0}\right)^{2} + \left(y - y_{0}\right)^{2}\right)^{3} + 4 \left(x - x_{0}\right)^{2} \left(x^{2} + y^{2}\right)^{3} \left(y - y_{0}\right) + \left(x^{2} + y^{2}\right)^{3} \left(- y + y_{0}\right) \left(\left(x - x_{0}\right)^{2} + \left(y - y_{0}\right)^{2}\right)\right) + 4 \left(x y \left(\left(x - x_{0}\right)^{2} + \left(y - y_{0}\right)^{2}\right)^{2} + \left(x - x_{0}\right) \left(x^{2} + y^{2}\right)^{2} \left(y - y_{0}\right)\right)^{2} + \left(2 x^{2} \left(\left(x - x_{0}\right)^{2} + \left(y - y_{0}\right)^{2}\right)^{2} + 2 \left(x - x_{0}\right)^{2} \left(x^{2} + y^{2}\right)^{2} - \left(x^{2} + y^{2}\right)^{2} \left(\left(x - x_{0}\right)^{2} + \left(y - y_{0}\right)^{2}\right) - \left(x^{2} + y^{2}\right) \left(\left(x - x_{0}\right)^{2} + \left(y - y_{0}\right)^{2}\right)^{2}\right)^{2}\right)}{\left(x^{2} + y^{2}\right)^{4} \left(\left(x - x_{0}\right)^{2} + \left(y - y_{0}\right)^{2}\right)^{4}}$$
simplify(diff(g,y,y)) tells us that $\partial_y^2 g = $
$$\displaystyle \frac{2 \left(2 \left(x \left(\left(x - x_{0}\right)^{2} + \left(y - y_{0}\right)^{2}\right) + \left(x - x_{0}\right) \left(x^{2} + y^{2}\right)\right) \left(4 x y^{2} \left(\left(x - x_{0}\right)^{2} + \left(y - y_{0}\right)^{2}\right)^{3} - x \left(x^{2} + y^{2}\right) \left(\left(x - x_{0}\right)^{2} + \left(y - y_{0}\right)^{2}\right)^{3} + \left(- x + x_{0}\right) \left(x^{2} + y^{2}\right)^{3} \left(\left(x - x_{0}\right)^{2} + \left(y - y_{0}\right)^{2}\right) + 4 \left(x - x_{0}\right) \left(x^{2} + y^{2}\right)^{3} \left(y - y_{0}\right)^{2}\right) + 2 \left(y \left(\left(x - x_{0}\right)^{2} + \left(y - y_{0}\right)^{2}\right) + \left(x^{2} + y^{2}\right) \left(y - y_{0}\right)\right) \left(4 y^{3} \left(\left(x - x_{0}\right)^{2} + \left(y - y_{0}\right)^{2}\right)^{3} - 3 y \left(x^{2} + y^{2}\right) \left(\left(x - x_{0}\right)^{2} + \left(y - y_{0}\right)^{2}\right)^{3} + 3 \left(x^{2} + y^{2}\right)^{3} \left(- y + y_{0}\right) \left(\left(x - x_{0}\right)^{2} + \left(y - y_{0}\right)^{2}\right) + 4 \left(x^{2} + y^{2}\right)^{3} \left(y - y_{0}\right)^{3}\right) + 4 \left(x y \left(\left(x - x_{0}\right)^{2} + \left(y - y_{0}\right)^{2}\right)^{2} + \left(x - x_{0}\right) \left(x^{2} + y^{2}\right)^{2} \left(y - y_{0}\right)\right)^{2} + \left(2 y^{2} \left(\left(x - x_{0}\right)^{2} + \left(y - y_{0}\right)^{2}\right)^{2} + 2 \left(x^{2} + y^{2}\right)^{2} \left(y - y_{0}\right)^{2} - \left(x^{2} + y^{2}\right)^{2} \left(\left(x - x_{0}\right)^{2} + \left(y - y_{0}\right)^{2}\right) - \left(x^{2} + y^{2}\right) \left(\left(x - x_{0}\right)^{2} + \left(y - y_{0}\right)^{2}\right)^{2}\right)^{2}\right)}{\left(x^{2} + y^{2}\right)^{4} \left(\left(x - x_{0}\right)^{2} + \left(y - y_{0}\right)^{2}\right)^{4}}$$
there is a certain symmetry in the above, which makes the following output 0:
expr1 = diff(g,x,x)
expr2 = diff(g,y,y)
x3,y3 = symbols('x3 y3')
expr1=expr1.subs(x,x3)
expr1=expr1.subs(y,x)
expr1=expr1.subs(x3,y)
expr1=expr1.subs(x0,x3)
expr1=expr1.subs(y0,x0)
expr1=expr1.subs(x3,y0)
simplify(expr2-expr1)
And finally simplify(diff(g,x,y)) gives $\partial_x\partial_y g = \partial_y \partial_x g = $
$$\displaystyle \frac{4 \left(\left(x \left(\left(x - x_{0}\right)^{2} + \left(y - y_{0}\right)^{2}\right) + \left(x - x_{0}\right) \left(x^{2} + y^{2}\right)\right) \left(4 x^{2} y \left(\left(x - x_{0}\right)^{2} + \left(y - y_{0}\right)^{2}\right)^{3} - y \left(x^{2} + y^{2}\right) \left(\left(x - x_{0}\right)^{2} + \left(y - y_{0}\right)^{2}\right)^{3} + 4 \left(x - x_{0}\right)^{2} \left(x^{2} + y^{2}\right)^{3} \left(y - y_{0}\right) + \left(x^{2} + y^{2}\right)^{3} \left(- y + y_{0}\right) \left(\left(x - x_{0}\right)^{2} + \left(y - y_{0}\right)^{2}\right)\right) + \left(y \left(\left(x - x_{0}\right)^{2} + \left(y - y_{0}\right)^{2}\right) + \left(x^{2} + y^{2}\right) \left(y - y_{0}\right)\right) \left(4 x y^{2} \left(\left(x - x_{0}\right)^{2} + \left(y - y_{0}\right)^{2}\right)^{3} - x \left(x^{2} + y^{2}\right) \left(\left(x - x_{0}\right)^{2} + \left(y - y_{0}\right)^{2}\right)^{3} + \left(- x + x_{0}\right) \left(x^{2} + y^{2}\right)^{3} \left(\left(x - x_{0}\right)^{2} + \left(y - y_{0}\right)^{2}\right) + 4 \left(x - x_{0}\right) \left(x^{2} + y^{2}\right)^{3} \left(y - y_{0}\right)^{2}\right) + 2 \left(x y \left(\left(x - x_{0}\right)^{2} + \left(y - y_{0}\right)^{2}\right)^{2} + \left(x - x_{0}\right) \left(x^{2} + y^{2}\right)^{2} \left(y - y_{0}\right)\right) \left(x^{2} \left(\left(x - x_{0}\right)^{2} + \left(y - y_{0}\right)^{2}\right)^{2} + y^{2} \left(\left(x - x_{0}\right)^{2} + \left(y - y_{0}\right)^{2}\right)^{2} + \left(x - x_{0}\right)^{2} \left(x^{2} + y^{2}\right)^{2} + \left(x^{2} + y^{2}\right)^{2} \left(y - y_{0}\right)^{2} - \left(x^{2} + y^{2}\right)^{2} \left(\left(x - x_{0}\right)^{2} + \left(y - y_{0}\right)^{2}\right) - \left(x^{2} + y^{2}\right) \left(\left(x - x_{0}\right)^{2} + \left(y - y_{0}\right)^{2}\right)^{2}\right)\right)}{\left(x^{2} + y^{2}\right)^{4} \left(\left(x - x_{0}\right)^{2} + \left(y - y_{0}\right)^{2}\right)^{4}}$$
If you choose ${\bf x}_0$ and ${\bf x}_1$ at $(\pm a,0)$ of the $(x,y)$-plane you have $$f(x,y)={(x+a,y)\over(x+a)^2+y^2}+{(x-a,y)\over(x-a)^2+y^2}\ .$$ In the following Mathematica notebook ${\tt gxx}$, ${\tt gxy}$, ${\tt gyy}$ are the entries of the Hessian matrix $$\left[\matrix{g_{xx}&g_{xy}\cr g_{xy}&g_{yy}\cr}\right]\ .$$ Here is the result: