Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Multiplying out differential operators

  1. Apr 16, 2014 #1
    In this video at around 9:00 , Carl Bender demonstrates a method of solving y''+a(x)y'+b(x)y=0.



    He first rewrites it in terms of differential operators

    D2+a(x)D+b(x))y(x)=0,

    then factors it

    (D+A(x))(D+B(x))y=0

    then multiplies it out to determine B(x). I thought we would get

    (D2+DB+AD+AB)y=0

    but at 15:29, he says that D, when it acts on B, either it acts on B or it 'goes past B' and acts on y and because of that, we get two terms, BD and D', so the result is

    (D2+BD+B'+AD+AB)y=0

    Why doesn't the operator just act on B?
    If it only acts on B, then shouldn't BD disappear somehow (and vice-versa)?
    Also, this would mean for D to act on something, it has to be the right? (DA≠AD?)

    Thanks
     
    Last edited by a moderator: Sep 25, 2014
  2. jcsd
  3. Apr 16, 2014 #2
    it's because of the chain rule. Note that there is a difference between
    [itex]\partial(b)y[/itex] and [itex]\partial(by)[/itex]. First write it like this, it makes it more clear:
    [itex](\partial + a)(\partial y +by)=0[/itex]
    In this case, the differential operator acts on both b and y, so you have [itex]\partial(by)=\partial(b)y + \partial(y)b[/itex]
     
  4. Apr 17, 2014 #3
    Thanks, I think I get it - you mean the product rule, yeah?

    (D+A)(D+B)y
    =(D+A)(Dy+By)
    = D2y+D(By)+ADy+ABy
    = D2y+yB'+By'+ADy+ABy
    = D2y+yB'+BDy+ADy+ABy
    = (D2+B'+BD+AD+AB)y
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Multiplying out differential operators
  1. Differential operator (Replies: 7)

Loading...