How I got to the problem:

Let $f(x,y)=\frac{1}{\sqrt{(x-a_1)^2+(y-a_2)^2}}+\frac{1}{\sqrt{(x-b_1)^2+(y-b_2)^2}}$, where $a_1,a_2,b_1,b_2 \in \mathbb{R}, a=(a_1,a_2)\neq (b_1,b_2)=b$ are fixed and $x,y \in \mathbb{R}, b \neq (x,y)\neq a$ (You can also see $a$ and $b$ as vectors in $\mathbb{R^2}$. I want to show that the derivative of this function has exactly one point $z=(x_0,y_0)$ where the derivative is zero e.g. $D(f(z))=0$.

One can easily see that

$p_1((x,y))=\frac{\partial}{\partial x}f=\frac{a_1-x}{\left(\left(x-a_1\right){}^2+\left(y-a_2\right){}^2\right){}^{3/2}}+\frac{b_1-x}{\left(\left(x-b_1\right){}^2+\left(y-b_2\right){}^2\right){}^{3/2}}$


$p_2((x,y))=\frac{\partial}{\partial y}f=\frac{a_2-y}{\left(\left(x-a_1\right){}^2+\left(y-a_2\right){}^2\right){}^{3/2}} +\frac{b_2-y}{\left(\left(x-b_1\right){}^2+\left(y-b_2\right){}^2\right){}^{3/2}}$

Before going any deeper, let us look at an example function: I set $a=(0,0)$ and $b=(1,0)$ and get Plot of the function

We can see right easily that the only point where the derivative shoul be zero is exactly between the peaks so we guess that $z=\frac{a+b}{2}$. We set it in and easily verify that $p_1(z)=p_2(z)=0$ (so the total derivative will also be 0 at z). Now we have to proove that this is the only solution which is the hard part for me. We can play a bit around with this and get to two final equations (will be right below this).

Now the real problem:

$(1) 0=\left(a_1-x\right)\left(\left(b_2-y\right){}^2+(\text{b1}-x)^2\right){}^{3/2}+\left(b_1-x\right)\left(\left(a_1-x\right){}^2+\left(a_2-y\right){}^2\right){}^{3/2}$

$(2) 0=\left(b_2-y\right) \left(\left(a_1-x\right){}^2+\left(a_2-y\right){}^2\right){}^{3/2}+\left(a_2-y\right) \left(\left(b_1-x\right){}^2+\left(b_2-y\right){}^2\right){}^{3/2}$

We are done if I can show that $(1)$ and $(2)$ imply that $(x,y)=z=\frac{a+b}{2}$. But even Mathematica fails on that one, I hope you can provide some help on this final step.


Your problem is clearly invariant under translations and rotations, so that you may assume without loss of generality that $a=(0,0)$ and $b=(b_1,0)$ with $b_1>0$. Then equation (2) has only one solution: $y=0$. Equation (1) reduces to $$ -x|b_1-x|^3+(b_1-x)|x|^3=0. $$ There are no solutions with $x<0$ or $x>b_1$. Trivial solutions are $x=0$ and $x=b_1$, but these are not solutions of $\nabla f=0$. The only solution in $(0,b_1)$ is $x=b_1/2$.

  • $\begingroup$ Thank you a lot, I think I was thinking too complicated on this :-). It is really a nice trick to simplify the problem I will surely remember for the future. $\endgroup$ – Listing Apr 24 '11 at 11:37

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