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Existence and Uniqueness

  1. Feb 11, 2012 #1
    1. The problem statement, all variables and given/known data

    x/√(x2+y12)-(l-x)/√((l-x)2+y22)=0

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

    (y1, y2, and l are constants.)

    2. Relevant equations

    x√((l-x)2+y22)-(l-x)√(x2+y12)=0

    x√((l-x)2+y22)+x√(x2+y12)=l√(x2+y12)

    x[1+√((l-x)2+y22)/√(x2+y12)]=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 x1 and x2 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+y22)/√(x2+y12)]+x*(d/dx)[√((l-x)2+y22)/√(x2+y12)]

    (d/dx)[√((l-x)2+y22)/√(x2+y12)]=-(l-x)/[√((l-x)2+y22)√(x2+y12)]-x√((l-x)2+y22)/(x2+y12)3/2

    df/dx=[1+√((l-x)2+y22)/√(x2+y12)]+x*[-(l-x)/[√((l-x)2+y22)√(x2+y12)]-x√((l-x)2+y22)/(x2+y12)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.

    -----

    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.
     
    Last edited: Feb 11, 2012
  2. jcsd
  3. Feb 11, 2012 #2

    Dick

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    Science Advisor
    Homework Helper

    You gave up way too easily in trying to isolate x. It's actually really easy. Write the two terms on two different sides of the '=' and square both sides. It falls part pretty easily and the higher degree stuff cancels to give you a quadratic. An EASY quadratic.
     
  4. Feb 11, 2012 #3
    Actually, the problem I was trying to solve was

    x/(c√(x2+y12))-(l-x)/(v√((l-x)2+y22))=0

    I didn't type out the extra coefficients c and v last night because I didn't think it'd make a difference, but apparently it does. Sorry about that. The above definitely does not reduce to an easy quadratic, or I would have done it. :(

    Edit: After taking the derivative, I've found that it equals

    [1/(c√(x2+y12))][1-x2/(x2+y12)] + [1/(v√((l-x)2+y22))][1-(l-x)2/((l-x)2+y22)], which is always positive, since:

    x2/(x2+y12)

    and (l-x)2/((l-x)2+y22)

    can never be greater than 1.
     
    Last edited: Feb 11, 2012
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