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Is wave and heat equation with zero boundary Poisson Equation?

  1. Dec 3, 2013 #1
    I have two questions:
    (1)As the tittle, if [itex]u(a,\theta,t)=0[/itex], is
    [tex]\frac{\partial{u}}{\partial {t}}=\frac{\partial^2{u}}{\partial {r}^2}+\frac{1}{r}\frac{\partial{u}}{\partial {r}}+\frac{1}{r^2}\frac{\partial^2{u}}{\partial {\theta}^2}[/tex]
    and
    [tex]\frac{\partial^2{u}}{\partial {t}^2}=\frac{\partial^2{u}}{\partial {r}^2}+\frac{1}{r}\frac{\partial{u}}{\partial {r}}+\frac{1}{r^2}\frac{\partial^2{u}}{\partial {\theta}^2}[/tex]
    Just Poisson Equation
    [tex]\nabla^2u=h(r,\theta,t)[/tex]
    Where
    [tex] h(r,\theta,t)=\frac{\partial{u}}{\partial {t}}\;\hbox { or }\;h(r,\theta,t)=\frac{\partial^2{u}}{\partial {t}^2}\;\hbox{ respectively.}[/tex]


    (2)AND if [itex]u(a,\theta,t)=f(r,\theta,t)[/itex], then we have to use superposition of Poisson with zero boundary plus Dirichlet with [itex]u(a,\theta,t)=f(r,\theta,t)[/itex]?

    That is

    [tex] u(r,\theta,t)=u_1+u_2[/tex]
    where
    [tex]\nabla^2u_1=h(r,\theta,t)\;\hbox { with }\;u(a,\theta,t)=0[/tex]
    and
    [tex]\nabla^2u_2=0\;\hbox { with }\;u(a,\theta,t)=f(r,\theta,t)[/tex]

    Thanks
     
    Last edited: Dec 3, 2013
  2. jcsd
  3. Dec 3, 2013 #2

    jasonRF

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    Gold Member

    No. Wave, diffusion, and Poisson's equations are all fundamentally different kinds of PDEs (lookup classification or characteristics in one of your PDE books). However, you can pretend that [itex]h[/itex] is an arbitrary function and formally solve your first two equations. However, your "solution" will be in the form of an integral equation for [itex]u[/itex] (actually, the integral will contain time derivatives of [itex]u[/itex] ...). This kind of trick might be helpful if you have nice approximate/numerical techniques that work well for the integral equation; then again it might just lead to a different mess that you still do not know how to solve.

    I'm not sure I understand question 2.

    jason
     
  4. Dec 3, 2013 #3
    Thanks for you reply. The reason I asked is when I look at the examples and exercises I see a definite relation where if the boundary is not zero, they use superposition of two where

    (1) Assuming [itex] \frac{\partial{u}}{\partial {t}}\neq 0[/itex] and then solve it as if it's a Poisson problem with zero boundary.

    (2) Assuming [itex] \frac{\partial{u}}{\partial {t}}=0[/itex] and then solve it as if it's a Dirichlet problem with non zero boundary.

    The methodology is exactly the same as solving Poisson problem with non zero boundary.

    The second part is just what I said above that solving Heat and Wave problem with non zero boundary is exactly like solving Poisson problem with non zero boundary.

    Thanks
     
  5. Dec 3, 2013 #4
    If this way of superposition technique might not work well with Heat and Wave problem, what is the standard way of solving Heat and Wave problems with non zero boundary?
     
  6. Dec 3, 2013 #5

    jasonRF

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    Gold Member

    Yes, you can use superposition to solve linear problems. We find a particular solution, then add an appropriate homogeneous solution such that the total solution satisfies the equation and all appropriate conditions.

    I just didn't understand how (2) related to (1) ... or are they independent questions altogether?
    jason
     
  7. Dec 3, 2013 #6
    Thanks for the reply

    (2) is just another way to say about non zero boundary [itex]u(a,\theta,t)=f(\theta, t)[/itex]. Which you already confirmed that the method of solving is using superposition just like solving Poisson problem with non zero boundary.

    You answered all my questions:

    1) That Heat and Wave equations are not Poisson equation.

    2) The method of solving Heat and Wave with non zero boundary is by superposition of the homogeneous solution plus particular solution.......Just like solving Poisson with non zero boundary.
     
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