1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Diffusion equation

  1. Apr 6, 2012 #1
    1. The problem statement, all variables and given/known data
    Solve the diffusion equation with the boundary conditions v(0,t)=0 for t > 0 and v(x,0) = c for t=0. The method should be separation of variables.

    2. Relevant equations
    The separation of variables method.

    3. The attempt at a solution
    Attempting a solution of the form XT leads you to an exponential for T and a sinusoidal for X:
    X = Asin(kx) + Bcos(kx)
    where -k^2 was the constant used for solving the two separated differential equations.

    However. My solution manuals writes the constants A and B as a continious function of the parameter k, and I don't understand why. Why do the constants, which are chosen from the boundary conditions have anything to do with k?

    And going further the full solution is then written as a fourier integral from 0 to ∞ of XTB(k)dk

    Where on earth does this come from? Note that A(k)=0 from the boundary conditions.

    Can someone try to explain why you most impose a continuous superposition like the above to get the general solution?
  2. jcsd
  3. Apr 6, 2012 #2


    User Avatar
    Science Advisor

    In general there is NOT a function of the form X= A(t)sin(kx)+ B(t)cos(kx) satisfying the boundary conditions. What you then need to do is to look for a sum of such things (if the boundary conditions require that k be discrete) or an integral of such thing (if the boundary conditions allow k to be continuous- any number on some interval).
  4. Apr 7, 2012 #3
    hmm I see. So the discrete superposition comes from if you get multiple k's satisfying X(0)=0 for instance (as a sinusoidal term).

    But I still don't see how my boundary conditions give a continuous set of parameters k. We wonna solve the diffusion equation for v(0,t)=0 for t>0 and v(x,0)=u. Can you show me exactly how this integral arises?

    I'll do the work I can:

    We get:

    X(x) = Acos(kx) + Bsin(kx)
    T(t) = Cexp(k2t) + Dexp(-k2t)

    v(0,t) implies A=0
    We also want C=0 (I think) since that term wouldn't really make sense, when we are considering thermal conduction (which we are). That leaves us with:

    v(x,t) = B'sin(kx)exp(-k2t) , where B'=BD

    Now my book writes this as:

    v(x,t) = ∫0B(k)sin(kx)exp(-k2t) dk

    I don't see how on earth this arises!!
  5. Apr 7, 2012 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    In general, the solution is a sum (or integral) of terms of the form [itex] B(k) \sin(kx) \exp(-k^2 t), [/itex] where you vary k and the coefficients B(k) are chosen to make v(x,0) have the desired form f(x); that is, you need to be able to say that [itex] f(x) = \sum_{k} B(k) \sin(kx) \text{ or } f(x) = \int B(k) \sin(kx) \, dk [/itex] for all x. If you have a condition on an infinite interval, as in your case of f(x) = c for all x > 0 (or is it for all x ≠ 0 in (-∞,∞)?) then a finite sum (= a Fourier series) will not work; you need an integral (= Fourier integral).

  6. Apr 7, 2012 #5
    hmm.. I still don't really see it. First of all a minor worry: The integral doesn't have the same units as the sum since you are multiplying by dk. So isn't that a problem.
    Secondly, let's try to do it pieceweise:
    The condition v(x,0)=u implies:
    B(k)sin(kx) = u
    How does solving this equation lead you to the integral above?

    I can see how a sum arises, for instance say:

    B(k)sin(kb) = 0

    So a solution would be:

    f = ƩB(k)sin(n∏x/b)

    But how does an integral arise...
  7. Apr 7, 2012 #6

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    I already told you. It is the difference between Fourier series and Fourier integrals. You need to represent a function f(x) in terms of a sum or integral of sin(k*x) and/or cos(k*x), and whether you should use a sum or whether you should use an integral depends on f(x). It is all readily available in books and articles; just look it up, using Google for example.

  8. Apr 7, 2012 #7
    You use the fourier integral when the eigenvalues are continuous, this kind of eigenvalues arises generally when you work with problems extended on an infinite interval. When you have a finite interval, the border conditions gives discrete eigenvalues, so you make a sum, because of the superposition principle, this way a fourier series arises. When you have infinite eigenvalues (for example any positive real is an eigenvalue), you have to make an integral, which is like a continuos sum, if you know the Riemann definition of integral it will be easy to you to get this (the integral can be interpreted as a limit of sums), if not, you can look for it at wikipedia: http://en.wikipedia.org/wiki/Riemann_integral in this case the fourier integral arises.
    Last edited: Apr 7, 2012
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook