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PDE questin

  1. Dec 4, 2005 #1
    Slve by separation of variables

    [tex] \frac{\partial u}{\partial t} - k \frac{\partial^2 u}{\partial x^2} = 0 [/tex] for 0 <x < pi, t > 0

    [itex] u(0,t) = u(\pi,t) = 0 [/itex]
    [itex] u(x,0)= \Sin^2 x [/itex]

    let u (x,t) = X(x) T(t)
    [tex] \frac{X''}{X} = \frac{T'}{T} = -\lambda = \mu^2 [/tex]
    also lambda must be positive (imaginary solution)
    [tex] X(x) = C_{1} \cos(\mu x} + C_{2} \sin(\mu x) [/tex]
    using the boundary conditons
    C1 = 0 and let u = n some positive integer
    [tex] X_{n} (x) = \sin(\mu x) [/tex]
    also solution for T is
    [tex] T_{n} (t) = e^{-k \mu^2 t} [/tex]
    now for T(0) = sin^3 x
    sin ^3 x = 1 ?
    SO x must be pi/2? since 0 <x < pi
    i dont understand how to proceed from here
    I know that i have to use some infinite series hereafter...

    Please help on this!

    Thank you for your help!
  2. jcsd
  3. Dec 4, 2005 #2
    Hi, the PDE you are solving is known as the diffusion equation.

    After you seperate, you have two independent ODE's. A 2nd order in x, and 1st order in t. Now, when you solve for X, while applying the BC's, you will need to get a condition, such that the ODE satisfies (known as an eigenvalue problem)

    you get something like this:
    X'' + Lambda X = 0, with X(0)=X(pi)=0

    so try different cases of Lambda: Lambda = 0, Lambda > 0, and Lambda < 0. You will find that only Lambda > 0 has a solution, and the condition is that Lambda = n^2. See if you can get that far. Let me know if you have any questions...

    The series part that you are taking about is Fourier series, but you cannot get that far unless you solve the eigenvalue problem correctly.

    I see that you have Lambda = n^2. Ok, so now you have a general solution in X. Now, since there are infinite solutions, add them up such that Un (x,t) = Sum Xn(x)Tn(t), then apply IC's.
    Sorry, I have to learn Tex.
    - harsh
    Last edited: Dec 4, 2005
  4. Dec 4, 2005 #3
    The solution i got for Xn i got using the different cases for lambda. Use lambda is a positive number, the way i have assuemd that X''/X = - lambda. I know that much

    As you can see in the end of my post that i have the initial condition for T(0) and that yields the following
    [tex] T_{n} (0)= 1 = \sin^3 (x) [/tex]
    so that means x must be pi/2??

    ok so then
    [tex] u(x,t) = e^{-kn^2 t} \sin(nx) [/tex]
    so then [tex] u_{n} (x,t) = \sum_{n=1}^{\infty} b_{n} e^{-kn^2 t} \sin(nx) [/tex]
    but how would i find bn if
    [tex] b_{n} = \frac{2}{\pi} \int_{0}^{\pi} \sin^3 (x) \sin(nx) dx [/tex]
    this doesnt lead the answer in th book which is
    [tex] \frac{3}{8} e^{-kt} \sin(x) - \frac{1}{4} e^{-9kt} \sin(3x) [/tex]
    Last edited: Dec 4, 2005
  5. Dec 4, 2005 #4
    Apply the intial conditions after you have taken a linear combination of all the solutions. This means

    You know
    u(x,t) = (Summation) (a_n * sin(nx))T_n (term))
    Now apply initial conditions,
    u(x,0) = (Summation) (a_n * sin(nx) = x

    so but x as a fourier representation of sin (nx), so use orthogonality, or use the formula given in the book.

    What is the actual initial condition? its not sin^3 x in your first post.

    I would find a trig identity to simplify that, and you will see that only certain values of n will hold.
    Last edited: Dec 4, 2005
  6. Dec 4, 2005 #5
    what do you mena by
    x as a fourier representation of sin nx?
    Do i not integrate the 2/pi thing?
  7. Dec 4, 2005 #6
    ok ok ok i understand waht you mena now.. BUT

    why is the book only finding c1 , and c3? Why not the c5 c7 and so on? Why isnt it represented as an inifnte series?
  8. Dec 4, 2005 #7

    Ok, this is what you are doing. If your initial condition is such that u(x,o) = sin^3 x. then this is what you are trying to solve:

    sin^3 x = summation a_n sin(nx)

    so, i would find a trig identity for sin^3x, and then solve for a_n

    In particular, I would use this: sin 3t = 3 sin t – 4 sin^3 t. As you can now see, there will only be two sin terms in your final answer: when n = 3, and n = 1.
  9. Dec 4, 2005 #8
    Thats because all the other terms will drop out to zero, since sin^3 x is made up of sin x and sin 3x. Only those terms will stay. Yeah?
  10. Dec 4, 2005 #9
    do you mena this identity is getting used here
    sin 3t = 3 sin t – 4 sin^3 t

    isnt this a bit too obscure an identity to remember on sya.. an exam?
  11. Dec 4, 2005 #10
    ok that solves that
    now i have this other problem
    is [tex] \sin(\pi - x) = \sin(n(\pi-x)) [/tex] true?

    The next question is similar setup except for this condition
    [tex] u(x,0) = X(pi - x) [/tex]
  12. Dec 4, 2005 #11
    On an exam, the prof will usually give you something thats already in such form, but if not, you can prolly derive this. Break up sin^3 into sin^2 and sin, then (1 - cos^) sin, which you can break up by using the identity for cos ^2.

    its a start
  13. Dec 4, 2005 #12
    Again, use an identity for it. If you are unsure, you can prolly convert it to exponential and work with them, but knowing some identities is useful.

    Btw, for what class is this? Is this an ODE course or a PDE course?
  14. Dec 4, 2005 #13
    PDE course.. why?

    i have posteed this quesiton in a new thread
  15. Dec 4, 2005 #14
    more PDE fun

    Solve by separation of varaibles
    [tex] \frac{\partial u}{\partial t} - k \frac{\partial^2 u}{\partial x^2} = 0 [/tex] for 0 <x < pi, t> 0
    [itex] u(0,t) = u(\pi,t) = 0 [/itex]
    [itex] u(x,0) = x (\pi - x) [/itex]

    OK i know the boring part of getting u(x,t) = X(x) T(t)
    the infinite series part is hard part
    [tex] T(0) = 1 = \sin(n(\pi)) [/tex]

    the coefficient [tex] c_{n} = \frac{2}{\pi} \int_{0}^{\pi} \sin(n(\pi -x)) \sin(nx) = \frac{-2}{\pi} \left(\frac{-1 + (-1)^n}{n}\right) \sin(nx) [/tex]
    the -1^n is from the Cos n pi term taht would coem from the integration

    thus n must be odd
    [tex] c_{n} = \frac{4}{\pi} \frac{1}{2n-1} \sin(nx) [/tex]

    is this good so far?
  16. Dec 4, 2005 #15
    Just curious, because I am taking a PDE course right now too.
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