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Solve partial differential equation

  1. Jun 7, 2014 #1
    1. The problem statement, all variables and given/known data
    Solve PDE ##\frac{\partial u}{\partial t}-k\frac{\partial^2 u}{\partial x^2}=0## for ##0\leq x \leq \pi ## and ##t\geq 0##.

    Also ##\frac{\partial u}{\partial x}(0,t)=\frac{\partial u}{\partial t}(\pi ,t)=0## and ##u(x,0)=sin(x)##.


    2. Relevant equations



    3. The attempt at a solution

    Differential operator ##L## for ##ku_{xx}=u_t## is symmetrical, therfore ##Lu=ku_{xx}=\lambda u##.

    ##ku_{xx}=\lambda u##

    ##ku_{xx}-\lambda u=0## so

    ##u(x)=Ae^{-\sqrt{\frac{\lambda}{k}}x}+Be^{\sqrt{\frac{\lambda}{k}}x}## and

    ##u(x)_x=-A\sqrt{\frac{\lambda}{k}} e^{-\sqrt{\frac{\lambda}{k}}x}+B\sqrt{\frac{\lambda}{k}} e^{\sqrt{\frac{\lambda}{k}}x}##

    Using ##\frac{\partial u}{\partial x}(0,t)=0## gives me that ##A=B## and the second condition gives me:

    ##u(\pi )_x=\sqrt{\frac{\lambda}{k}}Ae^{-\sqrt{\frac{\lambda}{k}}\pi }(e^{2\sqrt{\frac{\lambda}{k}}\pi }-1)##

    therefore

    ##2\sqrt{\frac{\lambda}{k}}\pi =2n\pi i## and finally ##\lambda =-n^2k##

    So eigenfunctions of differential operator ##L## are :

    ##u(x)=Ae^{-\sqrt{\frac{\lambda}{k}}x}+Be^{\sqrt{\frac{\lambda}{k}}x}## where we already know that ##A=B## and inserting ##\lambda =-n^2k## leaves me with:

    ##u_n(x)=A_ncos(nx)##

    IF everything so far is ok, than I shall continue:

    ##ku_{xx}=u_t##

    ##\sum_{n=0}^{\infty }k\lambda u(x)f(t)=\sum_{n=0}^{\infty }u(x){f}'(t)## so

    ##\frac{{f}'(t)}{f(t)}=\lambda k## and therefore

    ##f_n(t)=De^{-n^2k^2t}##


    So ##U(x,t)=\sum_{n=0}^{\infty }u(x)f(t)=\sum_{n=0}^{\infty }Acos(nx)e^{-n^2k^2t}##

    Know I think I should somehow write that ##sin(x)## in ##u(x,0)=sin(x)## in terms of eigenfunctions of ##L## but something has to be wrong, because:

    ##<sin(x),cos(nx)>=<\sum_{n=0}^{\infty }A_ncos(nx),cos(kx)>## where the first scalar product is 0, which means what exactly? Where did I get it wrong?
     
  2. jcsd
  3. Jun 7, 2014 #2

    LCKurtz

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    Science Advisor
    Homework Helper
    Gold Member

    Nothing is wrong. You want $$
    \sin x = A_0 + \sum_{n=1}^\infty A_n \cos(nx)$$You want to extend ##\sin x## to ##(-\pi,0)## as an even function and use the half-range formulas for the ##A_n##.

    [Edit, added] Alternatively if you are thinking in terms of inner products, your orthogonal set on ##(0,\pi)## is the set ##\{\cos(nx)\}##. ##\sin(x)## is not orthogonal to them.
     
    Last edited: Jun 7, 2014
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