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Transformation/change of variables in this differential equation?

  1. Nov 1, 2009 #1
    Transformation/change of variables in this differential equation!?

    Maybe my background is just weak...I was thinking about this for almost 1.5 hours already, but I still end up totally confused. Perhaps this is because I was never able to understand the ideas of a function and change of variables completely...perhaps I have a serious conceptual flaw.

    1. The problem statement, all variables and given/known data
    Consider the following partial differential equation with initial values(IV) and boundary conditions(BC):
    ut - k uxx = x + 2t, 1<x<7, t>0
    BC: u(1,t) = u(7,t) = 0
    IV: u(x,0) = x+5
    Our goal is to transform the above to the interval [0,6].
    Let w = x-1.
    Transform the whole problem to the interval w E [0,6]. (write in terms of w)

    2. Relevant equations/concepts
    No knowledge of partial differential equation is needed. Anyone with a solid background in mutlivariable calculus and change of variables should be able to answer this question. (I think)

    3. The attempt at a solution
    ux=uw dw/dx = (uw) (1) = uw
    uxx = ...(apply chain rule again) = uww
    [On the left side, think of u as u(x,t). On the right side, think of u as u(w,t)]

    BC:
    x=1 <=> w=0
    x=7 <=> w=6
    So the boundary conditions get transformed to u(0,t)=u(6,t)=0 [here think of u as u(w,t)]

    IV:
    We know u(x,0) = x+5
    => u(w,0) = w+5
    [I believe the logic in this step cannot be wrong, consider e.g. f(4z)=cos(4z), now how do we find f(4z-y)? Of course, f(4z-y)=cos(4z-y). How do we find f(z)? Surely, f(z)=cos(z). Right??]

    So my final answer is: [here think of u as u(w,t)]
    ut - k uww = w+1+2t, 0<w<6, t>0
    BC: u(0,t) = u(6,t) = 0
    IV: u(w,0) = w+5

    However, I really have some bad feeling that the result u(w,0) = w+5 is wrong, but I don't know where the mistake is.
    I tried to calculate it in a different way and the answer is the same.
    u(w,0)
    = u(x-1,0) = (x-1)+5
    => u(w,0) = w+5

    Can someone please kindly explain why and where my mistake is? What is the correct answer?

    Any help is greatly appreciated!
     
  2. jcsd
  3. Nov 1, 2009 #2
    Re: Transformation/change of variables in this differential equation!?

    Can somebody please help?? I know it's suuposed to be simple, but I can't figure it out...

    In short, my problem is the following:
    Suppose u(x,0) = x+5.
    Let w = x-1.
    Does it imply that u(w,0) = w+5?

    Could someone please answer this?
     
  4. Nov 2, 2009 #3

    lanedance

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    Re: Transformation/change of variables in this differential equation!?

    I haven't had a chance to read the top bit but say you have u(x,t) such that
    u(x,0) = x+5

    and you want to subsitute
    w = x-1
    then
    x = w+1

    so I would substitute in
    u(w=x+1,0) = (w+1)+5 = w+6
     
  5. Nov 2, 2009 #4

    lanedance

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    Re: Transformation/change of variables in this differential equation!?

    will try & have a better look tonight
     
  6. Nov 2, 2009 #5
    Re: Transformation/change of variables in this differential equation!?

    Maybe it's u(w,0) = x+6. But can you please tell me where my mistake is? Also, HOW did you get u(w,0) = x+6? You said w=x+1, but I think we're supposed to have w=x-1?

    We know u(x,0) = x+5
    => u(w,0) = w+5 ?
    I believe the logic in this step cannot be wrong, consider e.g. f(4z)=cos(4z), now how do we find f(4z-y)? Of course, f(4z-y)=cos(4z-y). How do we find f(z)? Surely, f(z)=cos(z). Right??


    u(w,0)
    = u(x-1,0) = (x-1)+5
    => u(w,0) = w+5 ?
     
  7. Nov 3, 2009 #6

    lanedance

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    Re: Transformation/change of variables in this differential equation!?

    ok so how about we try and recude the variable confusion, by introducing another function f(w,t) defined by
    [tex]
    f(w,t) = u(x(w),t)[/tex]


    so the variable change is
    [tex]w(x) = x-1[/tex]
    and re-arranging in case we need it later
    [tex]x(w) = 1+w[/tex]
    also differentiating gives
    [tex] \frac{dx}{dw} = 1 [/tex]

    then, chain rule differentiation gives
    [tex] f_{ww} = u_{xx}[/tex]
    [tex] f_{t} = u_{t}[/tex]

    so the differential equation becomes
    [tex] f_t- kf_{ww} = x(w) -2t = 1+w-2t[/tex]

    the boundary & initial conditions becomes
    [tex] f(0,t) = f(6,t) = 0[/tex]
    and
    [tex] f(w,0) = u(x(w),0) = x(w) + 5 = w+6 [/tex]
     
    Last edited: Nov 3, 2009
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