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PDE questions

  1. Jul 9, 2011 #1

    i basically dont know how to do pde's, so i'm learning it from scratch today for my test which is tomorrow (which is in 24hrs from now, for those who dont live in australia), and notes/the internet arent nearly as good as explaining things as people are.

    so how would i go by starting these questions? any tips?
    for the second one (Q4), i know that method of separation of variables basically involves putting all the x's on one side and the y's on the other side.
    i dont understand how to find the boundaries, or what is insulated. i think it has something to do with those u(L,y)=0 etc constraints, which i'll look up.

    some guidance on how to start these questions please?

    as with my other threads, i'll post up my working out as i attempt these questions, to see how i'm going.

  2. jcsd
  3. Jul 10, 2011 #2


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    I would start with the tips in the questions
    - first use fourier series
    - 2nd use separation of variables
    the boundary values are given in the question
    u(x,0) = 0 shows the side with y=0 has u=0
    Last edited: Jul 10, 2011
  4. Jul 10, 2011 #3


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    for the first can you find the complimentary solution no worries? do you know what fourier series is?
  5. Jul 10, 2011 #4
    for the first one (c) would i just first do the left hand side.

    turn that into λ2 + 10λ = 0, solve it, so then λ = 0 and -10
    so then
    yh = Ae0x + Be-10x
    = A + Be-10x

    and so now f(x) = A + Be-10x and work this out as a fourier question, but we have A and B and i dont know what to do wtih them, and i dont know what to do next...
  6. Jul 10, 2011 #5


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    i would try substituting that complimentary solution back into the DE as a check... it doesn't look quite right to me, do you get zero?
  7. Jul 10, 2011 #6
    complimentary solution.. do you mean the λ2 + 10λ = 0, λ = 0 and -10 line?
    or yhomogenous= A + Be-10x?
    both look okay, unless theres something really stupid im doing...
  8. Jul 10, 2011 #7


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    homogenous & complimentary mean the same thing, they are the solutions to the homogenous DE
    [tex] \frac{d^2y}{dx^2}+10y = 0[/tex]

    if I take your homogenous solution and substitute it in
    [tex] y_h = A + B e^{-10x}[/tex]
    [tex](0+ (-10)(-10)Be^{-10x}+10(A+B e^{-10x}) \neq 0[/tex]
  9. Jul 10, 2011 #8


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    also it might be worth including the start of the question

    to use tex, just right click
    use square brackets to open & close
    tex = open
    /tex - close
  10. Jul 10, 2011 #9
    well this is awkward. i cant seem to do something so basic.

    the whole question is:
  11. Jul 10, 2011 #10
    ahhhhhhh im such a retard!
    y turns into 1, not lambda, lol!

    so lambda = +- root (-10)?
    i hate imaginary values in ode's -_-
  12. Jul 10, 2011 #11
    ok so after consulting maths notes from 2 years ago, the roots are lambda = +- root (-10)

    so y = (A*cos[itex]\sqrt{10}[/itex]*x + B*sin[itex]\sqrt{10}[/itex]*x)


    supposed to be e ^ alpha*x(Acosblah + Bcosblah) but alpha = 0, so the e at the beginning becomes 1
  13. Jul 10, 2011 #12


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    looks good up to here

    not quite, I think the assumed form is [itex]e^{\lambda x}[/itex], but as you found [itex]\lambda = \pm \sqrt{-10}= \pm i \sqrt{10}[/itex] the exponential becomes complex which can be equivalently expressed in terms of sinusoids
  14. Jul 10, 2011 #13


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    now i'm guessing you can sketch it no worries... what is the period? this will tell you the fourier components to use. If the function were odd or even about the expansion point, it may save you time as you will only need to consider sin or cos terms

    have you managed to find the fourier expansion?
  15. Jul 10, 2011 #14
    oh, that second quote, i meant if the root is in the form λ = a ± bi. but a=0 in this case, so e^0x is just e^0, which is 1.
    working off this: M9oPo.jpg
    no alpha in our roots, only beta.
  16. Jul 10, 2011 #15


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  17. Jul 10, 2011 #16
    yess i was just going to say, it's in the other thread lol.
    how's my solutions from that thread?

    and are you talking about sketching y = (A*cos√10*x + B*sin√10*x)? because i wouldnt know how to do that..
  18. Jul 10, 2011 #17


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    no the problem says sketch f(x)
  19. Jul 10, 2011 #18
    i think the period would be root 10 or something?

    amplitiude has something to do with the A and B i think...

    edit: oh, well im not sure how to graph that. i can do them separately lol (2, 2/pi sinx, 2/pi sin3x) but not together..
  20. Jul 10, 2011 #19
    hang on sketching as in part a? i've done that.
    part b i think ive done it.
  21. Jul 10, 2011 #20
    ok to avoid confusion..

    question: http://i.imgur.com/FfowZ.jpg

    3. a. bAIUq.png


    so now do i say

    (A*cos[itex]\sqrt{10}[/itex]*x + B*sin[itex]\sqrt{10}[/itex]*x) = 2 + [itex]\frac{2}{Pi}[/itex]*sin(x) + [itex]\frac{2}{Pi}[/itex]*sin(3x)


    edit: ugh the LHS is just y, so i'd have to make y" + 10y equal to f(x)
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