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!

LaPlace equation in 1D

  1. Feb 22, 2012 #1
    1. The problem statement, all variables and given/known data

    Solve the Laplace equation in one dimension (x, i.e. (∂^2h)/(∂x^2)= 0)
    Boundary conditions are as follows:
    h= 1m @ x=0m
    h= 13m @ x=10m
    For 0≤x≤5 K1= 6ms^-1
    For 5≤x≤10 K2 = 3ms^-1

    What is the head at x = 3, x = 5, and x = 8?

    What is the Darcy velocity (specific discharge)?


    NOTE: There are multiple steps that will need to be done. Realize that system is heterogeneous. In a multiple layer system with steady-state conditions, Darcy velocity in one layer must equal the Darcy velocity in the other layers

    2. Relevant equations

    h(x) = ho - [(h0 - hD )/D]*x

    3. The attempt at a solution

    I tried to use the equation above subbing in 3, 5, and 8 for the x and using 10m as D, 1m as h0, and 13m for hD

    Then I used the specific specific discharge for the Darcy's velocity (q=K(dh/dL))

    That was apparently all wrong. Apparently this needs to be broken into 2 systems, coupled. Each individual system can be treated as homogenous. So it need two separate LaPlace equations? I really don't know what to do with this problem, please help!
     
  2. jcsd
  3. Feb 24, 2012 #2
    Since the discharge goes through both portions of the domain, you have continuity of the dependent variable h at the interface. In addition you also have continuity of discharge at the interface.

    When you integrate the Laplace equation in 1-D, you have two constants of integration for each section of the domain. The matchings of discharge and h serve to eliminate some of the integration constants.
     
  4. Feb 24, 2012 #3
    So, what I've done is set q(in) = q(out) => -K1(dh/dl)=K2(dh/dl) => 6((h-1)/5) = 3((13-h)/5) => h(@x=5) = 5m

    Then I set the used set of boundary Conditions (BC) to interpret the solution to LaPlace in 1D (h=cx+D):
    For BC @ x=0 and h=1 =>1=c(0)+D => 1=D
    For BC @ x=5 and h=5 => 5=c(5)+1 => c=4/5
    the I used h=cx+D again to solve for the head at x=3
    =>h(@x=3)=(4/5)(3)+1=3.4

    For the second system I used the point slope eq (y1-y)=m(x1-x) to get c
    => (5-13)/(5-10) = m = 8/5 = c
    Then I used h=cx+D to solve for D
    => 5=(8/5)(5) + D => D=-3
    Then to solve for h at x=8
    h=cx+D => h(@x=8) => (8/5)(8)-3 = 9.8


    To get Darcy's velocity I just used q=K(dh/dl) for each system
    System 1 = 6(4/5) = 4.8 = q
    System 2 = 4(8/5) = 4.8 = q

    Yes?
     
  5. Feb 24, 2012 #4
    I get the same answers as you but went about the problem a little differently.

    I solve the ODE over two separate domains:

    h1(x)=c1+c2*x
    h2(x)=c3+c4*x

    I used the Dirichlet boundary conditions along with continuity of head and continuity of Darcy velocity to solve for the 4 constants of integration. Equations worked out to be:

    h(x) = 1 + .8x 0<x<5

    h(x) = -3 + 1.6x 5<x<10

    The < signs should be weak inequalities. I cannot make them on my computer.
     
    Last edited: Feb 24, 2012
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: LaPlace equation in 1D
  1. Laplace's equation (Replies: 6)

  2. Laplace's Equation (Replies: 1)

Loading...