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Solving D.E using laplace

  1. May 1, 2007 #1
    du/dt=k^2 d^2u/dx^2

    b.c
    du/dx(o,t)=-qo

    i.c
    u(x,0)=0

    can any one please tell me what U(o,s)

    U(o,s)=integral bwteen o to infinity of du/dx(o,t)e^-st dt

    the reason i asked is if

    du/dt=k^2 d^2u/dx^2

    b.c
    u(0,t)=-qo


    i.c
    u(x,0)=0

    U(o,s)=integral bwteen o to infinity of u(o,t)e^-st dt = -qo/s
     
  2. jcsd
  3. May 1, 2007 #2

    Office_Shredder

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    You said du/dx(0,t) = -q0

    (Note this is a function of t only)

    Make up your mind. And please write out what you're trying to say, because it is quite difficult to read your post. If you could use latex, or at least the sub and sup tags, it would be a lot cleaner.

    In general, to solve the heat equation you want to use separation of variables and fourier series
     
  4. May 1, 2007 #3
    hi i have to solve this heat equation using laplace transforms i have nearly finshed it im stuck on finding U(o,s)

    du/dt=k^2 d^2u/dx^2

    b.c
    du/dx(o,t)=-qo

    i.c
    u(x,0)=0

    can any one please tell me what U(o,s)

    U(o,s)=integral between o to infinity of du/dx(o,t)e^-st dt

    iv uploaded the pdf file of the problem i hope this helps im sorry i dont know how to use latex,thanks for the promt reply
    http://files-upload.com/203463/ass3maths361.pdf.html
     
  5. May 1, 2007 #4

    Office_Shredder

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    I've never done laplace transforms, so I'm making this on the fly using google

    [tex]U(0,s) = \int_0^{\infty}u(0,t)e^{-st}dt[/tex]

    Not

    [tex]\int_0^{\infty}du/dx(0,t)e^{-st}dt[/tex]

    But [tex]dU/dx(0,t) = \int_0^{\infty}du/dx(0,t)e^{-st}dt[/tex]

    So you can solve for dU/dx(0,s), and then integrate to get U.

    Take this with a grain of salt, as it may be completely wrong

    EDIT: Upon further consideration, the last step sounds like junk
     
    Last edited: May 1, 2007
  6. May 2, 2007 #5
    cheerz Shredder.thanks for the help
     
  7. May 3, 2007 #6

    Integral

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    The goal of Laplace transforms is to convert a Differential Equation to a algebraic equation. Office Shredder has presented the basic definitions, but this is not how they are generally used.

    Your questions seem to indicate that you are not at all familiar with Laplace Transforms. Perhaps the best thing you could do is get yourself to your profs office and ask for his advice. He may be able to point you to a reference where you can learn what is needed for his class.

    Meanwhile:
    In his book Operational Mathematics Churchill solves the Heat equation with Laplace Transforms in great detail. He does Laplace transforms on t (time) reducing the problem to a ODE in x, then use standard DE methods to complete the problem.

    Look in your text for a table of LP xforms, you do not need to evaluate the integrals O_S has presented, that is what the table are for.

    IIRC L{f'(t)} = sF(s) + f(0)

    Again if you have NO experience with Laplace Transforms we will not be able to help very much, this is not a good place to learn from scratch. You need to find a good reference and do some reading.


    Something else that would help us to help you is for you to go to the tutorial section and read the how to LaTeX thread.
     
  8. May 3, 2007 #7

    tpm

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    HI..integral, i'm looking for a method different from residue integration to obtain the Integral.

    [tex] \oint _{C} ds f(s) exp(st) [/tex]

    for t-->oo i have tried [tex] \oint _{C} ds exp(log f(s)) exp(st) [/tex]

    but saddle-point method asks me to calculate the roots of

    [tex] \frac{g'(s)}{g(s)}+t=0 [/tex]

    but i'm stuck on this...
     
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