Nonhomogeneous Boundary Value Problem

  1. I've got a nonhomogeneous BVP I'm trying to solve. Both my book and my professor tend to focus on the really hard cases and completely skipp over the easier ones like this, so I'm not really sure how to solve it. It's the heat equation in a disk (polar coordinates) with no angle dependence.

    [tex]\frac{\partial^2 u}{\partial r^2} + \frac{1}{r} \frac{\partial u}{\partial r} + Q = \frac{\partial u}{\partial t}[/tex]

    Q is a constant.

    0 < r < 1
    t > 0

    Subject To:

    BC1) u(0, t) bounded
    BC2) u(1, t) = 0
    IC) u(r, 0) = 0

    I start with the homogeneous (Q=0) case and separate variables to get ODE's for T(t) and R(r):

    [tex]rR'' + R' + \lambda^2 r R = 0 [/tex]

    T' + \lambda^2 T = 0

    Solving and applying the BC's, I get

    R_{n}(r)=c_{2}J_{0}(j_{n0}r), n=1,2,3,...[/tex]

    T_{n}(t)=c_{4}e^{-j_{n0}^2 t}, n=1,2,3,...[/tex]

    Where [itex]j_{n0}[/itex] represents the nth zero of the [itex]J_{0}[/itex] Bessel function. Putting them together,

    [tex]u_{hom}(r,t) = \sum \limits_{n=1} ^ {\infty} A_{n} e^{-j_{n0}^2 t} J_{0}(j_{n0}r) [/tex]

    This is where I'm stuck at. In Previous problems, our inhomogeneous portion was a function of both space and time {Q(r,t)} and we would model both u(r,t) and Q(r,t) as infinite series of Bessel functions in r with time-dependent coefficients and substitute them back into the PDE, which would result in the sums and Bessel functions falling out, leaving an ODE we could solve to find the coefficients. That's not an option here since Q isn't a function of space OR time, so I really don't know where to go from here to get the partial portion of the solution.

    Thanks in advance for your help,


    EDIT: Corrected homogeneous solution
    Last edited: Dec 3, 2006
  2. jcsd
  3. HallsofIvy

    HallsofIvy 41,264
    Staff Emeritus
    Science Advisor

    Of course it's a "function of space and time", it's just a constant function! Any function of x and t can be written, as you say, as
    [tex]A_{0} + \sum \limits_{n=1} ^ {\infty} A_{n} e^{-j_{n0}^2 t} J_{0}(j_{n0}r) [/tex]
    What must An, for n> 0, be for that to be a constant? What is A0?
  4. Halls,

    The [itex]A_{n}[/itex]'s are zero in that case.

    I should point that I goofed up when I wrote the homogeneous solution above. The n=0 case shouldn't be an eigenvalue (so no constants in the solution.) I must have been working on too many problems at once and gotten them mixed up. I edited my original post to reflect that correctly.

    So where do I go from here? I've tried substituting the homogeneous solution back into the PDE, but nothing, including the sums, drops out so it's messier than the original PDE and doesn't seem to get me closer to a solution.


  5. HallsofIvy

    HallsofIvy 41,264
    Staff Emeritus
    Science Advisor

    An= 0 for n>0 and obviously A0= C. Just include that when writing your equation as a sum of Bessel functions.
  6. Are you saying I should model [itex]u_{hom}[/itex] and [itex]u_{part}[/itex] as:

    [tex]u_{hom} = \sum \limits_{n=1} ^ {\infty} A_{n} \exp(-j_{n0}^2 t) J_{0}(j_{n0}r)[/tex]

    (There is no n=0 solution for the homogeneous case)


    [tex]u_{part} = c + \sum \limits_{n=1} ^ {\infty} A_{n} \exp(-j_{n0}^2 t) J_{0}(j_{n0}r) = c[/tex]

    (there's only a constant for the partial case)

    resulting in

    [tex]u(r,t) = c + \sum \limits_{n=1} ^ {\infty} A_{n} \exp(-j_{n0}^2 t) J_{0}(j_{n0}r) [/tex]

    How do I solve for c though? If I substitute u(r,t) back into the PDE c gets differentiated out. If I substitute just the partial solution [itex]u_{part}=c{/itex] in, I get Q = 0, which obviously can't be true.
  7. I'm working on the exact same problem, any update?
  8. we can write in the solution of a nonhomogenuos equation as

    u(x)=integral(g(x.x1)*f(x)dx........where f(x) is the sourse term...if sourse term is homogenous equation,what will be the solution?zero?what it means?homogenous equation has no solution?

    then in the series form of greens function if lambda is one of the eigenvalue of the differential operator,den greens function become infinte.then we require that for the nonhomogemous eqn to have solution sourse term and eigenfunction shuold be orhogonal...can anyone explain why it should be orthogonal?

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