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Conditions of the Euler Equation

  1. Jan 31, 2008 #1
    Hi everyone :D

    This is my problem:
    Find conditions on [tex]\alpha[/tex] and [tex]\beta[/tex] in the Euler equation x[tex]^{2}[/tex]y'' + [tex]\alpha[/tex]xy' + [tex]\beta[/tex]y = 0 such that:

    a) All solutions approach zero as x [tex]\rightarrow[/tex] 0
    b) All solutions are bounded as x [tex]\rightarrow[/tex] 0
    c) All solutions approach zero as x [tex]\rightarrow\infty[/tex]

    I don't really know where to start with this, actually, I have no clue where to start.

    Also, what does it really mean for a solution to be bounded? I've been scouring some textbooks for a simple explanation but I can't seem to find it. Does it just mean that the function is constrained to some region?

    Any help would be greatly appreciated :D
     
  2. jcsd
  3. Jan 31, 2008 #2

    HallsofIvy

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    Do you know how to find the general solution to an Euler equation? There are two standard ways. One, the simpler, is to try "y= xr" and get a polynomial equation for r. The other, more complicated, is to make the substitution u= ln(x) which converts the equation into one with constant coefficients.

    It means that the function value has some upper and lower bounds. I would not say "the function is constrained to some region" because that would mean x can only be between two limits. If a function is bounded, then its values do not go to plus or minus infinity. In particular a 'power of x', xr, is bounded as x goes to 0 if and only if r is not negative.

     
    Last edited: Feb 1, 2008
  4. Jan 31, 2008 #3
    Thank you HallsofIvy for your help!

    The polynomial equation for r that you get when you plug in y = x[tex]^{r}[/tex] and try to solve the differential equation is:

    r[tex]_{1}[/tex], r[tex]_{2}[/tex] = [tex]\frac{-(\alpha-1) \pm \sqrt{(\alpha-1)^{2}-4\beta}}{2}[/tex]

    where r1 and r2 can be real and different, real and equal, or complex conjugates.

    Am I supposed to use limits or something to show what is asked in the problem statement?

    Also, would it be fair to say that (a) all solutions approach zero as x[tex]\rightarrow[/tex] 0 and (b) all solutions are bounded as x[tex]\rightarrow[/tex] 0 are roughly the same thing?
     
    Last edited: Jan 31, 2008
  5. Jan 31, 2008 #4
    a implies b, but not the other way around, I believe. So, I suppose if you don't need the weakest possible conditions on alpha and beta, you could just give the same answer, though that seems like a bit of a copout.

    But I completely suck at differential equations, so I should probably stay out of this.
     
  6. Feb 1, 2008 #5

    HallsofIvy

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    Basically, it is a matter of looking at the sign of the powers of x. If all powers of x have positive sign, the solution goes to 0 as x goes to 0 but goes to infinity as x goes to infinity. If all powers of x have negative sign, the solution goes to infinity as as x goes to 0 but goes to 0 as x goes to positive infinity. I'll let you think about what happens if one root is positive and the other negative.

    For powers of x, yes. More generally however, the function cos(x) is bounded for all x and so as x goes to 0 but does not approach 0 there.
     
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