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Non linear system of 4 equations, how to solve it?

  1. Jun 29, 2011 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    I'm stuck in a problem where I deduced a set of non linear equations. I must solve for [itex]A_0[/itex], A_1[/itex], [itex]x_0[/itex] and [itex]x_1[/itex]. I just don't know how to tackle this.
    The system is:
    [itex]A_0+A_1=\frac{2}{3}[/itex]
    [itex]A_0x_0+A_1x_1=0[/itex]
    [itex]A_0x_0^2+A_1x_1^2=\frac{2}{5}[/itex]
    [itex]A_0x_0^3+A_1x_1^3=0[/itex].

    2. Relevant equations No clue.



    3. The attempt at a solution
    Dead ends. Too many variables, no restriction on these variables except that they must be real. I can't divide any equation by any variable since they can be worth 0, I'm really stuck here.
    Thanks for any tip on how to solve this system.
     
  2. jcsd
  3. Jun 29, 2011 #2

    I like Serena

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    Hi fluidistic! :smile:

    The method is: substitution, substitution, substitution.

    Use the first equation to express A1 using A0.
    Substitute in the other 3 expressions.

    Use the new second equation to express A0 using x0 and x1.
    Substitute in the new 3rd and 4th expression.

    Repeat once more and you'll have solved x0 (or x1).

    Use the previous expressions to calculate the other variables.
     
  4. Jun 29, 2011 #3

    fluidistic

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    Thanks a lot for this tip. Actually this is very nasty :D
    I reached [itex]\frac{2x_0 ^3}{3(x_0-1)}+ \left [ \frac{2}{3}- \frac{2}{3(x_0-1)} \right ] \left ( \pm \sqrt {\frac{\frac{2}{5}-\frac{2x_0 ^2}{3(x_0-1)}}{\frac{2}{3}-\frac{2}{3(x_0-1)}}} \right ) ^3=0[/itex]. I'll continue with this :/
     
  5. Jun 29, 2011 #4

    I like Serena

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    Yes, I knew it would be nasty. Didn't want to spoil the surprise. ;)
     
  6. Jun 30, 2011 #5

    hunt_mat

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    Newton's method perhaps?
     
  7. Jun 30, 2011 #6

    I like Serena

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  8. Jun 30, 2011 #7

    hunt_mat

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    There is a trick...
    Divide the second equation by [itex]A_{0}x_{0}[/itex] to find that:
    [tex]
    \frac{a_{1}x_{1}}{A_{0}x_{0}}=-1
    [/tex]
    Now divide the last equation through by [itex]A_{0}x_{0}[/itex] to find that:
    [tex]
    x_{0}^{2}+\frac{A_{1}x_{1}^{3}}{A_{0}x_{0}}=0=>x_{0}^{2}+\frac{A_{1}x_{1}}{A_{0}x_{0}}x_{1}^{2}=0
    [/tex]
    From here it is pretty much straight forward.
     
    Last edited: Jun 30, 2011
  9. Jun 30, 2011 #8

    fluidistic

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    Thanks guys. In fact I need(ed) the exact values and using the values given in Wolfram seems to solve my problem so that the set of equation I've fell over seems right.
    By the way I had no time to continue the algebra to solve for the 4 unknowns. At least I know how to do and I know what I did was ok.
    So we can consider the problem as solved :)

    Edit.: I just read your last post hunt_mat. Wow, nice trick. I instantly get [itex]x_0= \pm x_1[/itex] which indeed simplifies things a lot.
     
  10. Jun 30, 2011 #9

    hunt_mat

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    I told you how to solve the system. I don't think the idea was to use wolframalpha to do your homework...
     
  11. Jun 30, 2011 #10

    fluidistic

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    Yes sorry, just read your post. (I posted almost in same time as you, as a result I missed the trick). I edited my previous post.
     
  12. Jun 30, 2011 #11

    hunt_mat

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    So can you say how [itex]x_{0}[/itex] is related to [itex]x_{1}[/itex]?
     
  13. Jun 30, 2011 #12

    hunt_mat

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    What can I say to such praise.
     
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