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Laplace's equation

  1. Sep 8, 2005 #1
    Heat equation

    Given the 3-D rectangular solid with sides of length a, b and c in the x, y and z directions, respectively. Find the function T(x,y,z,t) when
    Laplace(T)=1/K(dT/dt) subject to the following conditions:

    1) Initial conditions: T(x,y,z,0)=0
    2) Boundary conditions
    a. dT/dx + h(1)T = 0 for x=0
    b. T = 0 for x=a
    c. dT/dy = 0 for y=0
    d. dT/dy = 0 for y=b
    e. dT/dz = 0 for z=0
    f. dT/dz + h(2)T = 0 for z=c

    Anyone plese suggest me, I don't understand. Thank you.
     
    Last edited: Sep 8, 2005
  2. jcsd
  3. Sep 8, 2005 #2

    HallsofIvy

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    ?? What have you tried? I don't see any reason why T(x,y,z,t) identically equal to 0 isn't the solution. It satisfies all the conditions, doesn't it?
     
  4. Sep 8, 2005 #3

    sic

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    HallsofIvy: it is, but it's the trivial one.. it's a linear equation, the solution can be multiplied by a constant, and knowing that 0 is a good solution doesn't give you anything..
    danai_pa:
    try the following steps:
    1) solve the boundary. you'll see that on the three of the faces of the cube , T is zero and on all the rest, its derivative is zero
    2) write T as
    sum C(n,m,l) sin(pi n x/a) cos(pi m y/b) cos(pi (l+1/2) z/c)
    it is a general way to write the function taking care of the constraints automatically
    3) now, just calculate the time dependence of C(n,m,l) by applying laplacian to the general notation (pay attention that the functions of the series are orthogonal)
     
  5. Sep 8, 2005 #4
    please help

    I can solve this equation for y only and x , z. I can't find it. Please describe to me. Thank you.
     
  6. Sep 9, 2005 #5
    please help

    Anyone please help me. I can not solve it.
    Thank you
     
  7. Sep 9, 2005 #6

    HallsofIvy

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    Clearly the "existance and uniqueness" theorem applies to this. T(x,y,z,t) identically equal to 0 may be trivial but it's the only one you're going to get!
     
  8. Sep 9, 2005 #7

    sic

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    Actually i didn't pay attention to the condition T(x,y,z,0)=0, which really makes the problem trivial (T=0 always), but if it's so, the boundary conditions have absolutely no meaning.. are you sure there's no typo?
     
  9. Sep 9, 2005 #8

    Astronuc

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    Staff: Mentor

  10. Sep 11, 2005 #9
    I can solve variable of x and y but variable z i can not solve it.
    Anyone please suggest me. Thank you
     
  11. Sep 11, 2005 #10

    HallsofIvy

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    Read back through this. You have already been told the solution several times!

    It really doesn't matter whether the variables are x, y, z, or anything else.
     
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