Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

whereas

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

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.

share|improve this question
add comment

1 Answer

up vote 4 down vote accepted

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 $x=b_1/2$.

share|improve this answer
    
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. –  Listing Apr 24 '11 at 11:37
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.