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Does Laplace and Poisson's equation have to be for static condition?

  1. Aug 23, 2013 #1
    I read from the PDE book about Laplace equation in static condition ie ##\frac {\partial U}{\partial t}=0##.

    But is it true that even if U is time varying ie ##U=U(x,y,z,t)##, you can still have Laplace and Poisson's equation at t=k where k is some fixed value. ie:

    [tex]\nabla^2U(x,y,z,t)|_{t=k}=0\;\hbox{ or }\;\nabla^2U(x,y,z,t)|_{t=k}=f(x,y,z)[/tex]

    In English, it's the divergence of the gradient at a point (x,y,z) at time t=k.

    Thanks
     
  2. jcsd
  3. Aug 23, 2013 #2

    jasonRF

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    I'm not sure what you are saying here. Laplace's equation never has a [itex]\frac{\partial}{\partial t}[/itex] term in it. If you could provide an extra sentence or two of context it would be much easier to understand what you are asking and why.

    Of course the answer is yes. You can have Poisson's equation
    [tex]
    \nabla^2 V(x,y,z,t) = g(x,y,z,t),
    [/tex]
    Here [itex]\nabla^2 = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}[/itex] has no [itex]t[/itex] component. I have read some of your other posts so I see that you know electrodynamics. Consider the Coulomb gauge ( [itex] \nabla \cdot \mathbf{A}=0 [/itex]; see Griffith's book which I believe you are familiar with), under which the scalar potential in time varying situations does indeed satisfy such an equation with [itex] g = -\rho / \epsilon [/itex]. (Of course the vector potential in this case satisfies a very messy equation).

    Is this what you are referring to, or are you thinking of a limit of a wave or diffusion equation for a fixed time? In that case, the fact that you only care about one time does not automatically imply that the time derivatives evaluated at that time are zero and hence can be dropped from the fundamental equation you want to solve.

    jason
     
  4. Aug 23, 2013 #3
    Hi Jason, Thanks for taking the time to answer my question. I asked because the PDE books stated of non time varying situation when using Laplace equation. If They can only used in static condition, the use will be limited.

    I understand Laplace and Poisson's equation has no time involved. My main question is whether Laplace and Poisson's equation can be used for time varying case by just set t=k, a constant to eliminate the t. Or else, the use of Laplace and Poisson's equation will be limited....to only static case.

    Also from the wave equation:

    [tex]\frac{\partial ^2U(xyt)}{\partial x^2}+\frac{\partial ^2U(xyt)}{\partial y^2}=\nabla^2U(xyt)=\frac 1{c}\frac{\partial ^2U(xyt)}{\partial t^2}=A,\;\hbox { where } A \;\hbox { can only be a constant.}[/tex]

    This further link Poisson's to time varying.

    Thanks

    Alan
     
    Last edited: Aug 23, 2013
  5. Aug 24, 2013 #4

    SteamKing

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    I think you are trying to work with some type of diffusion equation. The classic example pertains to the conduction of heat in 3-D bodies.

    See: http://en.wikipedia.org/wiki/Heat_equation

    Similar techniques used to solve Laplace or Poisson equations involving only spatial dimensions can be adapted to solve diffusion equations.
     
  6. Aug 24, 2013 #5
    Thanks for your time. I am really asking a general statement. I want to confirm Laplace and Poisson's equation works in time varying situation with restriction. As I stated before, if they only work with static situation, it would be quite useless.

    Like in heat equation, it is related to ##\frac{\partial U}{\partial t}##.

    Thanks
     
  7. Aug 24, 2013 #6

    jasonRF

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    Both this statement and the one you make in your last post do not give any context - I am not sure what you are asking. I thought I gave you a concrete example of a Poisson equation (and hence a Laplace equation) that holds for a time varying physical quantity of interest (scalar electric potential under the Coulomb gauge).


    Why can A only be a constant? Consider a solution of your wave equation:
    [tex]
    U(x,y,t) = \cos \left( \omega \left(\frac{x + y}{c \sqrt{2}} - t \right) \right).
    [/tex]
    You should be able to show that this satisfies the 2D wave equation, and that even for a fixed time it is not true that [itex]\nabla^2 U(x,y,t) [/itex] is a constant.

    I hope that helped.

    jason
     
  8. Aug 24, 2013 #7
    Thanks, I am just confused and making some statement to get clarification. So Both Laplace and Poisson equation work in time varying condition as you indicated that U is also time dependent. Poisson and Laplace just ignore the t and concentrate on the location portion ( x,y,z in this case) and say NOTHING about the t. Then the other part of the equation deal with the t.

    In another word:
    [tex]\frac{1}{c}\frac{\partial^2 U(xyzt)}{\partial t^2}=\frac{\partial^2 U(xyzt)}{\partial x^2}+\frac{\partial^2 U(xyzt)}{\partial y^2}=\nabla^2U(xyzt)[/tex]
    In this equation, ##\nabla^2U(xyzt)## only deal with the physical location, but it is linked back to time by ##\frac{\partial^2 U(xyzt)}{\partial t^2}##

    I should have said ##A## is not a function of (x,y,z,t) instead of a constant. The reason I said that is because the only way to make
    [tex]\frac{1}{c}\frac{\partial^2 U(xyzt)}{\partial t^2}=\frac{\partial^2 U(xyzt)}{\partial x^2}+\frac{\partial^2 U(xyzt)}{\partial y^2}=\nabla^2U(xyzt)=A[/tex]
    Is to have ##A## not a function of (x,y,z,t). Or else, the relation will not hold for all x,y,z and t.

    Again, thanks for your time.
     
  9. Aug 24, 2013 #8

    jasonRF

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    This is not correct. I don't think you really looked at the example I gave you, for which
    [tex]
    \nabla^2 U(x,y,t) = \frac{1}{c^2} \frac{\partial^2 U(x,y,t)}{\partial t^2} = - \frac{\omega^2}{c^2} \cos \left( \omega \left( \frac{x + y}{c \sqrt{2}} - t\right) \right).
    [/tex]
    This holds for all x,y and t, and is NOT a constant. It is a cosine whose argument contains all of the variables x,y and t.

    jason
     
  10. Aug 25, 2013 #9
    Here is a possible example of what you may be driving at. Suppose you are solving the transient heat conduction equation for the region within a slab of material. Initially the slab is at temperature U = 0 throughout, and, at time zero, you apply a constant heat flux to the top and bottom surfaces of the slab for all time. Eventually, at long times, the temperature will approach a solution for which U = kt + f(z), where z is the spatial position measured from the base of the slab. So the partial derivative of U with respect to t will be k (k is determined by the rate at which heat is being added), and the transient heat conduction equation will reduce to a Poisson equation in f(z).

    Chet
     
  11. Aug 26, 2013 #10
    I actually look at your example, I was agreeing with you that U can be a function of t and ##\nabla^2U## is not a constant.
     
  12. Aug 26, 2013 #11
    I think I am getting it. ##\nabla^2U## is not more and no less different from...say..##\frac{\partial U}{\partial x}##. All it say is an operation on ##U##.

    In another words, ##\nabla^2U ## involves operation of ##\frac{\partial ^2}{\partial x^2}## and ##\frac{\partial ^2}{\partial y^2}##. It does not imply anything about whether U is a function of t. Just like ##\frac{\partial ^2U(x,y,t)}{\partial x^2}## only look at the x in U, the others are just being treated as constants.
     
    Last edited: Aug 26, 2013
  13. Aug 26, 2013 #12

    jasonRF

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    Sorry - I should have been more careful in my reply. You made the statement that for
    [tex]
    \nabla^2 U = \frac{1}{c^2} \frac{\partial^2 U}{\partial t^2} = A,
    [/tex]
    A could could not be a function of (x,y,z,t). However, the general case is that A is a function of all the variables. The most common example is the plane wave solution that you must be very familiar with from the EM you have studied. Indeed, the example I gave was just a simple plane wave solution.

    Jason
     
  14. Aug 26, 2013 #13

    jasonRF

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    Yes, that is how partial derivatives work. I am thinking that you may find PDEs pretty rough going if you do not have a solid foundation of basic multivariable calculus. You may want to review it if it has been awhile since you learned it. In any case I'm glad things are starting to make more sense to you.

    best regards,

    jason
     
  15. Aug 26, 2013 #14
    That's the problem where I only use these math on and off. It's not like I use it everyday and just forgot it. It's always rusty!!!
     
  16. Aug 26, 2013 #15

    jasonRF

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    All of us get rusty with stuff we don't use - which is why sometimes a quick review is warranted. Recently I tried to jump back into thermal and statistical physics just for fun and learn some topics I never learned that thoroughly the first time, but found that I had forgotten so many basics that I had to practically start at the beginning all over again. I ended up re-reading much material and solving a bunch of problems on the old stuff I used to know; after that I was able to pick up where I had left of years ago, but it took a lot of time and work!

    I am just "lucky" that I have had to use multivariable calculus so often over the past 20+ years (I am an electrical engineer) that it has always stayed fresh!

    jason
     
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