(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

x/√(x^{2}+y_{1}^{2})-(l-x)/√((l-x)^{2}+y_{2}^{2})=0

How do I prove that the above equation has a solution for x in ℝ and that the solution is unique?

(y_{1}, y_{2}, and l are constants.)

2. Relevant equations

x√((l-x)^{2}+y_{2}^{2})-(l-x)√(x^{2}+y_{1}^{2})=0

x√((l-x)^{2}+y_{2}^{2})+x√(x^{2}+y_{1}^{2})=l√(x^{2}+y_{1}^{2})

x[1+√((l-x)^{2}+y_{2}^{2})/√(x^{2}+y_{1}^{2})]=l=f(x)

I actually don't think it's plausible to isolate x, since it'll result in a quartic equation, which is messy.

3. The attempt at a solution

Existence

Observe that the square root of a real number is always positive. Then lim(x→-∞)f(x)=-∞ and lim(x→∞)f(x)=∞. Since l is a continuous function, there must be at least one value of x that equals the constant l.

Uniqueness

I was told that in general, existence proofs start with the assumption that two different values x_{1}and x_{2}satisfy the equation, and then show that the two values are actually the same, but I really don't feel like solving a quartic equation.

So I'm going to make an observation. Observe that f(x) is a purely increasing function. Hence, df/dx is always positive.

df/dx=[1+√((l-x)^{2}+y_{2}^{2})/√(x^{2}+y_{1}^{2})]+x*(d/dx)[√((l-x)^{2}+y_{2}^{2})/√(x^{2}+y_{1}^{2})]

(d/dx)[√((l-x)^{2}+y_{2}^{2})/√(x^{2}+y_{1}^{2})]=-(l-x)/[√((l-x)^{2}+y_{2}^{2})√(x^{2}+y_{1}^{2})]-x√((l-x)^{2}+y_{2}^{2})/(x^{2}+y_{1}^{2})^{3/2}

df/dx=[1+√((l-x)^{2}+y_{2}^{2})/√(x^{2}+y_{1}^{2})]+x*[-(l-x)/[√((l-x)^{2}+y_{2}^{2})√(x^{2}+y_{1}^{2})]-x√((l-x)^{2}+y_{2}^{2})/(x^{2}+y_{1}^{2})^{3/2}]

So messy...

I'm not even sure if df/dx is always positive. It just appears to be that way when I graph it.

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By the way, I know this "proof" isn't rigorous at all, but I've never taken an analysis course. I also don't know how large the difference is between "sufficient" and "complete", but I think the assumption of continuity and limits is sufficient here.

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# Homework Help: Existence and Uniqueness

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