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Simultaneous Equations (How to test for redundancy)

  1. Sep 6, 2015 #1
    1. The problem statement, all variables and given/known data
    I have three equations:

    ## F = ρw(x_0 y^2_0 - x_1 y^2_1) + \frac{1}{2} γ w (x^2_0 - x^2_1)## ----- 1

    ##y_0 = y_1 \frac{x_1}{x_0}## ----- 2

    ##\frac{y^2_0}{2} + gx_0 = \frac{y^2_1}{2} + gx_1## ------ 3

    2. Relevant equations
    N/A

    3. The attempt at a solution
    My goal is to have ##F## expressed without either ##y^2_0## or ##y^2_1## involved in the equation. My problem is that equations 2 and 3 involve both ##y^2_0## and ##y^2_1## so when it comes to eliminate one of them in equation 1, I re-introduce the other in the equation (and vice-versa). My question is whether or not it's legal to simultaneously solve equations 2 and 3 in two different ways (have ##y^2_0## and ##y^2_1## as subjects), then substitute each equation back in equation 1? Does solving the same simultaneous equation twice to obtain two equations with different subjects make them redundant?
     
  2. jcsd
  3. Sep 6, 2015 #2

    Ray Vickson

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    If you substitute ##y_0## from eq. (2) into eq. (3), that will give you an equation involving ##x_0, x_1, y_1## alone, which you can then solve for ##y_1##, That will give you an expression for ##y_1## in terms of ##x_0,x_1## only; substituting that into eq. (2) will give you ##y_0## in terms of ##x_0,x_1##. Messy, but do-able.
     
  4. Sep 6, 2015 #3
    Hi Ray, thanks for the response.

    I understand I can get equation 2 in terms of ##x_0 , x_1##, however my ultimate goal is to get ##F## in terms of ##x_0, x_1## . I'm worried that using the same equation twice will create redundancy. Would I be able to use the resulting ##y_0## and ##y_1## in equation 1 with no hassle?
     
  5. Sep 6, 2015 #4

    Ray Vickson

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    There is no redundancy in the method I suggested (which, by the way, is 100% standard). Eq (2) gives you ##y_0## in terms of ##x_0,x_1,y_1##. Putting that into eq. (3) allows you to get ##y_1## in terms of ##x_0,x_1##: ##y_1 = Y_1(x_0,x_1)## for some explicit function ##Y_1##. If you have numerical values for the inputs ##g,w,\rho, \gamma##, I can give you arbitrary numerical values for ##x_0,x_1##, and you can use your function ##Y_1## to get a unique, well-defined numerical value for ##y_1##. (Actually, there are two unique values with opposite signs, depending on which square root of ##y_1^2## you choose.) You can then take your now-known numerical values of ##x_0, x_1,y_1## and use eq. (2) to get the numerical value for ##y_0##. Now you can put all those value into your ##F##. Except for the "sign" issue, where is the redundancy in any of that? If you have some reason for choosing, say, the positive square root for ##y_1## then all uncertainty and ambiguity disappears.
     
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