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Solving the differential equation m^2 = 0?

  1. Nov 26, 2013 #1
    I have a problem that starts with the equation:

    [itex]r\frac{d^2u}{dr^2} + \frac{du}{dr} = 0 [/itex]

    The solution I'm looking at says to do a substitution, letting [itex]u = r^m[/itex], which after differentiation and simplification results in:

    [itex]m^2 = 0 [/itex]

    Up until this point, I understand (well, actually I don't understand how to know to do this substitution, but I understand how the differentiation and simplification happened), but then the next step says that because [itex]m^2 = 0[/itex]:

    [itex]u(r) = c_1 + c_2 lnr [/itex]

    Question 1: How do you know to choose [itex]u = r^m[/itex] as the substitution?

    Question 2: Why was that form of a solution chosen?

    Question 3: Is there a name for this type of substitution and solution?

    Thank you!
     
  2. jcsd
  3. Nov 26, 2013 #2

    LCKurtz

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    Actually, that wouldn't have my first suggestion. I would have said let ##y = u'## making the equation ##ry' + y = 0## and notice that ##ry'+y=(ry)' = 0## so ##ry = C## and ##y = u' = \frac C r##. Hence ##u = C\ln r + D##.

    If you multiply the original equation by ##r## so it looks like ##r^2u'' + ru' = 0## it becomes an Euler equation for which a standard method would be to try ##r^m##, if you have studied those.
     
  4. Nov 26, 2013 #3

    Simon Bridge

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    (1) The approach is to guess some sort of solution to the DE. It's not actually a substitution.
    A good guess to try is one that makes every term in the DE contain the same function.

    This particular guess is suggested by ##\sum_n r^{n-1}\frac{d^n u}{dr^n}=0## and works because ##r^{n-1}r^{m-n}=r^{m-1}## ... i.e. the same function for every term in the DE.
    Basically - when you see a DE in that pattern, it is likely that u is a polynomial (but may not be).

    [ref Euler equation]

    (2) $$r\frac{d^2 u}{dr^2}+\frac{du}{dr}=0\\
    u(r)=r^{m}\implies [m(m-1)+m]=0\\

    \therefore m^2 = 0$$... which tells you that u=1 (or any constant) is a solution ... if somewhat trivial.

    So the guess is unhelpful.

    Try to think of something else. i.e. notice that the LHS in the DE is a product-rule expansion:
    $$r\frac{d^2 u}{dr^2}+\frac{du}{dr}=\frac{d}{dr}\left( r\frac{du}{dr}\right)\\
    \frac{d}{dr}\left( r\frac{du}{dr}\right)=0\implies\cdots$$

    (3) it's called "trial and error" ;)

    IIRC: In the case where a guess of the form u=rm for the above form of DE fails, the solution must therefore be of form u=a+b.ln|r|.
     
    Last edited: Nov 26, 2013
  5. Nov 26, 2013 #4
    Thanks for your reply LCKurtz!

    In your solution, [itex]C[/itex] and [itex]D[/itex] are not constants, like they are in [itex]u(r) = c_1 + c_2 lnr[/itex], correct?

    We haven't studied Euler equations as far as I can remember, which is probably why I'm so confused! Is [itex]u(r) = c_1 + c_2 lnr[/itex] a standard solution for a Euler equation?

    I'm really confused how to jump from a characteristic equation of [itex]m^2 = 0[/itex] to [itex]u(r) = c_1 + c_2 lnr[/itex]...
     
  6. Nov 26, 2013 #5

    Simon Bridge

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    C and D are constants just like yours - it's common to use caps for the arbitrary constants in DE solutions.
    The jump is because it is always the case when m=0 so you just remember it like you jump from seeing six rows of seven objects to knowing you have 42 objects.

    note: LaTeX can format many functions better if you put a backslash in front of the function abbreviation ... i.e. \ln|r| gets you ##\ln|r|##. Also works for log, exp, sin, arcsin, etc.
     
  7. Nov 26, 2013 #6

    LCKurtz

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    ##C## and ##D## are constants.

    For constant coefficient DE's you may recall that when the characteristic equation has a double root ##r = \{c,c\}## you have solutions of ##e^{cx}## and ##xe^{cx}##. When you study Euler equations you will learn that when you have a double root of ##r = \{c,c\}## when you try ##x^r##, you get solutions ##x^c## and ##x^c\ln x##. Your text will probably show you why when you get to that section.
     
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