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Diffusion Problem (Conduction)

  1. Jul 4, 2008 #1
    1. The problem statement, all variables and given/known data

    Since the problem involved formulas I have posted it as an attachment. Please check the attachment first. (I have not scanned this problem, just wrote it in word and posted it)

    3. The attempt at a solution

    My approach on this problem is to start with separation of variables as I don't see any indication that tells me that I am dealing with non homogeneous conditions and can't seperate. I am thinking that in order to get u, I need to separate for sure. Can anyone verify if that seems as a reasonable approach to this problem?
     

    Attached Files:

  2. jcsd
  3. Jul 5, 2008 #2
    I will post my initial approach as an attachment. Do I apply the boundary conditions while I have solved for u?
     

    Attached Files:

  4. Jul 5, 2008 #3

    HallsofIvy

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    You separated variables and determined that the "parts" of the solution are [itex]X(x)= A cos(\lambda x)+ B sin(\lambda x)[/itex] and [/quote]T(t)= Ce^{-\lambda^2 t}.

    Now you can use the boundary conditions, that u(0)= 0 and u(2)= 0, which tell you that X(0)= 0 and X(2)= 0, to determine what values [itex]\lambda[/itex] can have (so that [itex]sin(2\lambda)= 0[/itex]) as well as the fact that the coefficients of [itex]cos(\lamda x)[/itex] are all 0 (so that X(0)= 0).

    Since the exponential will be 1 at t= 0, You can write the initial conditions in terms of a Fourier sine series.
    Since you are told that one of the terms in the expansion is of the form
    [tex]\frac{a}{b\pi^2}sin(\frac{5\pi x}{2}e^(-(c/d)\pi^2t}
     
  5. Jul 5, 2008 #4

    Defennder

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    Woah I'm surprised HallsOfIvy could view the attachment even before it has been approved. Just something available only to mentors I suppose.
     
  6. Jul 5, 2008 #5

    Hootenanny

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    It is the mentors that review and approve (or reject) attachments, so they must be able to view all attachments before they are approved.
     
  7. Jul 5, 2008 #6

    Borek

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    What makes me wonder is that while HOI have already seen the attachement it is still pending approval :smile:

    EOT, we will get banned for OT posts :rolleyes:
     
  8. Jul 5, 2008 #7
    T(t)= Ce^{-\lambda^2 t}.

    Now you can use the boundary conditions, that u(0)= 0 and u(2)= 0, which tell you that X(0)= 0 and X(2)= 0, to determine what values [itex]\lambda[/itex] can have (so that [itex]sin(2\lambda)= 0[/itex]) as well as the fact that the coefficients of [itex]cos(\lamda x)[/itex] are all 0 (so that X(0)= 0).

    Since the exponential will be 1 at t= 0, You can write the initial conditions in terms of a Fourier sine series.
    Since you are told that one of the terms in the expansion is of the form
    [tex]\frac{a}{b\pi^2}sin(\frac{5\pi x}{2}e^(-(c/d)\pi^2t}[/QUOTE]

    Thank you for trying to help me. I have tried a lil bit but I got confused so I guess I wanna make sure that what I am doing is correct.

    X(x)=A*cos(lambda*x)+B*sin(lambda*x)
    X(x=0)=A*cos(0)+B*sin(0)=>A=0
    X(x=2)=0=>B*sin(2*lambda)=0

    Lambda=((n*Pi)/L)^2
    so the eigenfunction EF, EF=B*sin((n*Pi)/L)^2

    How will I move on from here?
     
  9. Jul 5, 2008 #8
    I know that I need to consider temporal and special separately. For the spatial do I have to consider Lambda=0, Lambda<0. Lambda>0 cases?
     
  10. Jul 6, 2008 #9
    Can anyone help me with this problem please? I'm stuck...
     
  11. Jul 9, 2008 #10
    Thank you for trying to help me. I have tried a lil bit but I got confused so I guess I wanna make sure that what I am doing is correct.

    X(x)=A*cos(lambda*x)+B*sin(lambda*x)
    X(x=0)=A*cos(0)+B*sin(0)=>A=0
    X(x=2)=0=>B*sin(2*lambda)=0

    Lambda=((n*Pi)/L)^2
    so the eigenfunction EF, EF=B*sin((n*Pi)/L)^2

    How will I move on from here?[/QUOTE]

    HallsofIvy does what I have written here seem correct to you?
     
  12. Aug 17, 2008 #11
    There seems to be something wrong with my previous account "aeroguy2008" so I have a new account now. How come nobody replies to my questions?
     
  13. Aug 17, 2008 #12

    Astronuc

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    The problem is now one month old, and I suppose people were busy or waiting for some additional work.

    One's solution seems correct.

    One has used the boundary (spatial) conditions to eliminate the cos function and determine the eigenvalue [itex]\lambda[/itex]. One will note that sin (n[itex]\pi[/itex]) = 0, n is an integer = 0, 1, . . . . infty.

    Actually, the spatial function X(x) will be written as an infinite series of sin terms, and each will have a coefficient, Bn. Has one learned how to determine the coefficients?


    From the images:

    [tex]\frac{a}{b\pi^2}sin{\frac{5\pi{x}}{2}}e^{-\frac{c}{d}{\pi^2}t}[/tex]

    [tex]X(x) = A cos {\lambda x} + B sin {\lambda x} = B sin {\lambda x}[/tex]

    [tex]T = C e^{-\lambda^2t}[/tex]

    then

    [tex]X(x) T(t) = C'_n sin\,{\lambda x}\,e^{-\lambda^2{t}}[/tex] and compare to the above expression
     
    Last edited: Aug 17, 2008
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