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Damped Wave equation, PDE.

  1. Apr 30, 2010 #1
    1. The problem statement, all variables and given/known data

    I have the damped wave equation;

    [tex]u_{tt} = 4 u_{xx} -2 u_{t}[/tex]

    which is to be solved on region 0 < x < 2
    with boundary conditions;

    [tex] u(0,t) = 2, u(2,t) = 1. [/tex]

    i must;
    1) find steady state solution [tex] u_{steady}(x) [/tex] and apply boundary conditions.
    2) find [tex]\theta(x,t) = u(x,t) - u_{steady}[/tex] and find PDE and boundary conditions obeyed by theta.
    3) use theta = f(t)g(x) to obtain 2 seperate ODE's with seperation constant
    4) then i must solve the homogeneous problem for theta and then recover the general solution for u
    5) finally i must find the solution for u using intial conditions
    [tex] u(x,0) = 2 - x/2 , u_{t} (x,0) = x^{2}(2-x) [/tex]
    and im given
    (this is between 2 and 0) [tex] \int x^{2} (2-x) sin(\frac{n\pi}{2} x) dx = \frac{32(1+2(-1)^{n}}{n^{3}\pi^{3}} [/tex]

    3. The attempt at a solution
    1) steady state solution [tex] u_{steady} [/tex] implies [tex] u_{t} = 0 [/tex] which reduces the PDE to [tex] 4 u_{xx} = 0 \rightarrow u_{xx} = 0[/tex] after intergration, [tex] u_{steady} = ax + b [/tex] applying boundary conditions leads to [tex] u_{steady} = - \frac{1}{2}x + 2 [/tex]

    2) As
    [tex] \theta (x,t) = u(x,t) - u_{st} [/tex]
    and [tex] u_{st} [/tex] is a function of x only then [tex] \theta _{t} = \theta _{tt} = 0 [/tex] . Using the fact that [tex] u_{st} = - \frac{1}{2}x + 2 [/tex] leads to [tex] \frac{\partial^{2} u_{st}}{\partial x^{2}} = 0 [/tex] and therefore [tex] theta_{xx} = 0 [/tex]
    which doesnt affect the PDE for u, only changes u to theta.
    Oh and both boundary conditions become 0.

    3) Inserting
    [tex] \theta (x,t) = f(t)g(x) [/tex] into the PDE leads to;
    [tex] f''(t)g(x) = 4f(t)g''(x) - 2g(x)f'(t) [/tex]
    reranging leads to;
    [tex] \frac{f''(t) + 2f(t)}{4f(t)} = \frac{g''(x)}{g(x)} [/tex]
    And each side is a function of its arguement only and are equal to each other, so each side equals the same constant, which for the sake of this question must be [tex] -\Lambda [/tex] to create a trig solution for g(x) (need trig solution as boundary conditions require that g must equal 0 in 2 places g(0) = g(2) = 0) .

    4) this is where i start to struggle, the ODE for g(x) is;
    (note; i have used [tex] \Lambda = \lambda ^{2} [/tex]
    [tex] g''(x) + \Lambda g(x) = 0 [/tex] which leads to [tex] g(x) = Acos(\lambda x) + Bsin(\lambda x) [/tex] with boundary conditions g(0) = g(2) = 0.
    Using those boundary conditions, A = 0 and B = 0 OR [tex] sin(2 \lambda) = 0 [/tex]
    B=0 is trivial case, so it is ignored, if [tex] sin(2 \lambda) = 0 [/tex] then; [tex] 2 \lambda = n \pi [/tex] which leads to [tex] \lambda = \frac{n \pi}{2} [/tex]

    this is all fine and dandy, but then when i look at the ODE for f(t)
    [tex] f''(t) + 2f'(t) + 4 \Lambda f(t) = 0 [/tex] and applying the value for [tex] \Lambda = \lambda ^{2} [/tex] leads to

    [tex] f''(t) + 2f'(t) + 4 \frac{n^{2} \pi ^{2}}{4} f(t) = 0 [/tex]

    Once i cancel the 4 im left with an ODE, i try using [tex] A e^{mx} [/tex] as a solution but it leaves me with;
    [tex] m^{2} + 2m = -n^{2} \pi ^{2} [/tex] which has no solutions for m unless m = 0, so im stuck there, i need to find the seperation constant such that i can get solutions to both of these ODE's :(

    Help is really appreciated, i think its the fact that there is an extra partial u by t in there, only examples we have really are when both partials are second order.
     
    Last edited: Apr 30, 2010
  2. jcsd
  3. Apr 30, 2010 #2
    Anyoen have any ideas? I'll take anything atm.. ive come to a stand still on this one
     
  4. Apr 30, 2010 #3
    That equation with the m's in it is just a quadratic. I think you can solve it.
     
  5. Apr 30, 2010 #4
    the lowest point of m^2 + 2m is 1 and n is an integer therefore -n^2 pi^2 > 1 for all n... unless m can take imaginary values... oh hell i havent been stupid enough to look at that and think "no value of m can surely satisfy that equation m^2 is always positive.." if this what it is i want to be euthinised to better preserve mankinds intelligence
     
  6. Apr 30, 2010 #5
    but m cant take imaginary values as the value for 2m would give an imaginary part, and -n^2 pi^2 is a real number
     
  7. Apr 30, 2010 #6
    [tex] m^2 + 2m + n^2\pi^2 = 0 [/tex]

    [tex] m = \frac { -2 \pm \sqrt{4 - 4n^2\pi^2}}{2} [/tex]

    etc...
     
  8. Apr 30, 2010 #7
    aye true, but you know that 4*(n^2)*(pi^2) is greater than 4 for all n > 0 (n is an integer) except when n = 0 but if n = 0 then this is the only solution which leads to g(x) = 0 which then leads to theta = 0 and u = -(0.5)x + 2, ie u is only a function of x then and not of t.
     
  9. Apr 30, 2010 #8
    m is complex for n > 0. That's ok, it's going to be exponentiated in the answer, right? So sines and cosines ensue.
     
  10. Apr 30, 2010 #9
    but then it wouldnt satisfy [tex] m^{2} + 2m + n^{2}\pi^{2} = 0 [/tex] as 2m would be a complex number... literally just occoured to me that m^2 will then have a complex part to cancel out the 2m part.. d'oh!
    cheers mate :D
     
  11. Apr 30, 2010 #10
    why would i need sines and cosines btw? would there be more steps before i have the solution to theta?
     
  12. Apr 30, 2010 #11
    Not really, you're almost done.

    If f(t) = Aemt and you know g(t) all you have to do is write the whole thing down as an infinite sum over all the lambdas. You just will want to move the imaginary part of m inside the summation because it depends on n.
     
  13. Apr 30, 2010 #12
    edit: wrong
     
    Last edited: Apr 30, 2010
  14. Apr 30, 2010 #13
    Do end up with
    [tex] f(t) = Ae^{-1+(1-n^{2}\pi^{2})^{1/2}} + Ae^{-1-(1-n^{2}\pi^{2})^{1/2}} [/tex]
    Or is there individual constants for the positive and minus square root bits?
     
  15. Apr 30, 2010 #14
    What you want to do is factor out a [tex] \sqrt{-1} [/tex] of the expression under the square root sign. Then you have the very basic second-order ODE with constant coefficients where the roots are [tex] a \pm bi [/tex]. You should know from some other class or something that the solution to this is

    [tex] f(t) = e^a (c_1 cos bt + c_2 sin bt) [/tex]

    It's no different here except recall that you're not really solving for general f(t) but for fn(t), so the constants should be indexed by n as well. The constants, suppose we call them an (on cos) and bn on sin, will be the Fourier coefficients of u(x,0).
     
  16. May 3, 2010 #15
    can the cosntants c_1 and c_2 be complex here? because i think thats what im struggling with, also do i end up with;
    [tex] f(t) = e^{-t}(Ae^{+(n^{2}\pi^{2}-1)^{1/2}} + Be^{-(n^{2}\pi^{2}-1)^{1/2}} )
    [/tex]
    Leading to;
    [tex] f(t) = e^{-t} (c_1 cos bt + c_2 sin bt) [/tex]
    [tex] c_1 = A+B [/tex] and [tex] c_2 = i(A-B) [/tex]
    [tex] b = (n^{2}\pi^{2}-1)^{1/2} [/tex]
    not sure if this is right
     
  17. May 3, 2010 #16
    haha looks alot like the DEF coursework in for tomorrow :P
     
  18. May 3, 2010 #17
    LOL hello fellow uni dude, you finished this yet? I think i've cracked it now but not too sure of all the final steps.

    EDIT: keep the name of our uni private please, reasons of privacy and this could be, but i highly doubt it viewed as cheating (im of the mind that im being helped with and not told awnsers :) )
     
  19. May 3, 2010 #18
    No, remember that [itex] sin u = \frac {e^u - e^{0u}}{2i} [/itex]. (Note the i in the denominator).

    So [itex] c_2 [/itex] won't be complex.
     
  20. May 3, 2010 #19
    so;
    [tex] sin(t) = \frac{e^{it} - e^{-it}}{2i} [/tex]
    (rewrote it because i wasnt sure of the 0)
    Now;
    [tex] f(t) = A(cos(t\sqrt{n^{2}\pi^{2}-1}) + \frac{e^{it\sqrt{n^{2}\pi^{2}-1}} - e^{-it\sqrt{n^{2}\pi^{2}-1}}}{2})) + B(cos(t\sqrt{n^{2}\pi^{2}-1}) - \frac{e^{it\sqrt{n^{2}\pi^{2}-1}} - e^{-it\sqrt{n^{2}\pi^{2}-1}}}{2})) [/tex]

    which doesnt seem to have got me anywhere :S
    also i think you mean that 0 to be a minus sign, i changed it :)


    edit; lol i dunno why i/2i = 1/i, its probably something to do with how tired i am... corrected now.
     
    Last edited: May 3, 2010
  21. May 3, 2010 #20
    OH unless you also call for cos to be rewritten in terms of exponentials?
     
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