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I Non-constant coefficient

  1. Nov 29, 2017 #1
    I was attempting to solve a differential equation using separation of variables' method. After some calculations I got to this:

    $$ \frac 1 {a(r)} \frac k r \frac d {dr} \left(r \frac {da(r)}{dr} \right) = f $$

    and I rearranged to this:

    $$\frac k r \frac {da(r)}{dr} + \frac{d^2a(r)}{dr^2} = fa(r)$$

    ##k## and ##f## are constants. In a more elegant way:

    $$ \ddot a + \frac k r \dot a - fa = 0 $$

    Then I realized I don't know how to solve it because of ## \frac 1/r ##. I never studied how to solve differential equation with non-constant coefficients. Can anybody help me? Or suggest me some reading that can help me?
    Thank you
     
  2. jcsd
  3. Nov 29, 2017 #2

    Orodruin

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  4. Nov 29, 2017 #3
    Thank you. I'll be studying it in the next days... see if I can get it done.
     
  5. Nov 29, 2017 #4
    This looks a lot like a Bessel equation. Are you familiar with Bessel functions?
     
  6. Nov 29, 2017 #5

    Orodruin

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    Not only looks like. It is Bessel's differential equation for ##k = 0## (or modified Bessel's differential equation depending on the signs of ##f## and ##k##), which is the reason I linked to the Wikipedia page on Bessel functions.
     
  7. Nov 29, 2017 #6
    Oops. Missed that. Sorry.
     
  8. Nov 30, 2017 #7
    Hi, thanks for the replies.

    I'm not familiar with Bessel's equation at all (I just happened to see it somewhere some times ago, so I knew there was a solution but I didn't even remember what to search). I checked the wikipedia page and I found out it's used to solve heat/mass transfer problems and it's funny because it is what I'm actually trying to do. So i think I'm on the right spot! :)

    Anyway I think it is a bit too advanced to me right know, but I looked it up on "Mathematical methods for physics and engineering" and I found, like wikipedia says, Bessel's equation is:
    $$ x^2 \ddot y + x \dot y + (x^2-k^2)y = 0 $$

    My function looks instead like:

    $$ kr^2 \ddot a + kr \dot a + fr^2a = 0 → r^2 \ddot a + r \dot a + \frac f k r^2a = 0$$

    Since I started studying it this morning (with no special background in differential equations) I'd like to know if ##\frac f k ## affects the solutions of the equation and how. Thank you

    Ps: since I got here by separating variables ##ƒ## is a constant which I no nothing about, I have to use B.C. later on to find it
     
    Last edited: Nov 30, 2017
  9. Nov 30, 2017 #8

    Orodruin

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    Make a variable substitution ##r = x \sqrt{k/f}##.
     
  10. Nov 30, 2017 #9
    sorry I don't get it. I still have something different from bessel equation
     
  11. Nov 30, 2017 #10

    Orodruin

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    Please quote what you get, otherwise it is impossible to know where you go wrong.
     
  12. Nov 30, 2017 #11
    Sorry, here:

    (In order to make it easier I will put ##h = f/k##)

    $$ \ddot a(r) + \frac 1 r \dot a(r) + ha(r) = 0$$
    $$ r = x \sqrt \frac 1 h $$
    Chain's rule:
    $$ \frac {da(r)} {dr} = \frac d {dx} a \left(x \sqrt {\frac 1 h} \right) \frac {dx} {dx \sqrt {\frac 1 h}} = \sqrt h \frac d {dx} a \left(x \sqrt \frac 1 h \right) $$
    $$ \frac {d^2a(r)} {dr^2} = \frac {d^2} {dx^2} a \left(x \sqrt {\frac 1 h} \right) \left( \frac {dx} {dx \sqrt {\frac 1 h}} \right)^2 + \frac d {dx} a \left(x \sqrt \frac 1 h \right) \frac {d^2x} {d \left( x \sqrt \frac 1 h \right)^2 } = h \frac {d^2} {dx^2} a \left( x sqrt \frac 1 h \right)$$
    finally:
    $$h \frac {d^2} {dx^2} a \left(x \sqrt \frac 1 h\right) + \frac 1 {x \frac 1 h} \sqrt h \frac d {dx} a \left(x \sqrt \frac 1 h \right) + h a \left(x \sqrt \frac 1 h \right) = 0$$
    $$ \frac {d^2} {dx^2} a \left(x \sqrt \frac 1 h\right) + \frac 1 x \frac d {dx} a \left(x \sqrt \frac 1 h \right) + a \left(x \sqrt \frac 1 h \right) = 0 $$

    How should I proceed ?
     
  13. Nov 30, 2017 #12

    Orodruin

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    How is that not Bessel's differential equation?
     
  14. Nov 30, 2017 #13
    But there is ##\sqrt \frac 1 h ## inside the parenthesis... isn't it a problem?

    I mean the solution of Bessel's equation is ##a(x) = c_1J_v(x)+c_2J_{-v}(x) ##, should I replace ##x## with ##x\sqrt \frac 1 h##?
     
    Last edited: Nov 30, 2017
  15. Nov 30, 2017 #14

    Orodruin

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    No, this is not a problem. Just introduce a new function ##h(x) = a(x/\sqrt{h})##. This will lead to ##a(r) = h(r\sqrt{h})##.
     
  16. Nov 30, 2017 #15
    Oh thank you. Just to be sure, the solution is then ##a(r) = cJ_v(r\sqrt h)##?
     
  17. Nov 30, 2017 #16
  18. Nov 30, 2017 #17
    Yes, thank you for mentioning @eys_physics. I'll be studying that too in the future. Btw I didn't understand if you are implying that my solution is wrong because I'm like in the middle of the ocean right now ahahah :frown:
     
  19. Nov 30, 2017 #18
    No, your solutions is not wrong. I said that there are in general more solutions. The ones I mentioned, ##Y_v## are singular at the origin (r=0). So, if you want a regular solution they should be disregarded. But, it depends on the actual problem you want to solve, which wasn't clear from your post.

    By the Frobenius method you can derive the solution of your equation in terms of series expansion. One can then compare with series expansions existing for the Bessel functions. It takes some effort but can be useful to understand better how to solve this kind of differential equations.
     
  20. Nov 30, 2017 #19

    Orodruin

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    This is a common but not completely kosher argument. The question becomes why you want your solution to be regular at r=0. In the end it boils down to the behaviour of the inhomogeneity in your differential equation (typically r=0 is actually just a coordinate singularity in your domain). For example, the Green’s function of the Helmholtz equation in two dimensions involves a Bessel function of the second kind in order to take care of the delta inhomogeneity at r=0. However, as long as your inhomogeneity is sufficiently nice you will be able to assume regularity (and therefore throw away the Bessel functions of the second kind).

    Of course, if r=0 is not part of your domain you cannot apply these arguments and you need to include both linearly independent solutions.
     
  21. Nov 30, 2017 #20
    Thanks, Orodruin for your clarification. I was a bit sloppy with the wording in my post. The word "should" in the quoted sentence wasn't appropriate. I meant that in some cases the irregular solution could be disregarded (depending on the problem statement), but in other cases not as you mention.
     
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