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Laplace equation in a square with mixed boundary conditions

  1. Dec 15, 2012 #1
    The length of the side of the square is a. The boundary conditions are the following:
    (1) the left edge is kept at temperature T=C2
    (2) the bottom edge is kept at temperature T=C1
    (3) the top and right edges are perfectly insulated, that is [tex]\dfrac{\partial T}{\partial x}=0,\dfrac{\partial T}{\partial y}=0[/tex]
    Solve for [tex]T(x,y)[/tex] in steady state.
    The situation is described by the Laplace equation. It would be a bit difficult to directly solve for this problem since this is a mixed BC problem. I want to break it down into several (for example, two) problems that are easier to solve. So far I don't have any idea how to break it down, since the mixed BC makes it difficult to break down the problem. Does anybody have any suggestion?

    Thanks.
     
  2. jcsd
  3. Dec 15, 2012 #2
    What is the value of T(0,0) ( at the left bottom corner) ?
    At the left edge T(0,y)=C2
    At the bottom edge T(x,0)=C1
    Hence T(0,0)=C2=C1 ?
     
  4. Dec 15, 2012 #3
    Not really. IMHO, the value on the corner is not particularly important. And, C1 is not equal to C2.
     
  5. Dec 15, 2012 #4
    Of course, I supposed that C1 and C2 are not equal.
    My remark was about the wording of conditions (1) and (2) : These conditions are NOT for all the edge length : The corner must be excluded.
    It is important to see that the solution will include the Heaviside step function.
    Isn't it ?
     
  6. Dec 15, 2012 #5

    pasmith

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    Certainly
    [tex]
    T(x,y) = \left\{ \begin{array}{r@{\quad}l}
    C_2 & x < y \\
    C_1 & x > y \\
    \end{array}\right.
    [/tex]
    is a weak solution of Laplace's equation which satisfies the boundary conditions.
     
  7. Dec 15, 2012 #6
    Is there a reason why you can't use separation of variables?
     
  8. Dec 15, 2012 #7

    pasmith

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    Actually, no: if [itex]F(x,y)[/itex] is the solution of Laplace's equation in a unit square subject to [itex]F(x,0) = 1[/itex], [itex]F(0,y) = 0[/itex], [itex]\partial F/\partial x(1,y) = 0[/itex] and [itex]\partial F/\partial y(x,1) = 0[/itex] then the solution of the OP's problem is
    [tex]
    T(x,y) = C_1F(x/a,y/a) + C_2F(y/a,x/a)
    [/tex].

    And to find [itex]F(x,y)[/itex], let [itex]k_n = \frac{2n + 1}{2}\pi[/itex] for [itex]n \geq 0[/itex] and use the eigenfunctions
    [tex]
    f_n(x,y) = \sin(k_n x)Y_n(y)
    [/tex]
    where
    [tex]
    Y_n(y) = \cosh(k_ny) - \tanh(k_n)\sinh(k_ny).
    [/tex]
     
  9. Dec 16, 2012 #8
    I think you can write the y-component solution as [tex]\cosh(k_n(a-y))[/tex]
     
  10. Dec 16, 2012 #9
    I think my real problem is:
    when the boundary condition on the bottom edge is T(x,0)=f(x), BC on left edge is T(0,y)=0, BC on right and top edges are perfect insulation (suppose we don't care about the values on the four corners), is it correct to obtain the Fourier coefficients using [tex]b_n\cosh(k_n(a))=\int_0^a\sin(k_nx)f(x)dx[/tex], with [tex]k_n=(n+0.5)\pi[/tex].
    The thing that bothers me is that the integration interval is not a multiple of the period of the basis function [tex]\sin(k_nx)[/tex].
     
  11. Dec 16, 2012 #10

    pasmith

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    But mine has the advantage that [itex]Y(0) = 1[/itex] (although I could write it as [itex]\cosh(k_n(y-a))/\cosh(k_na)[/itex])

    If you're integrating between 0 and a then [itex]k_n = (n+0.5)\pi/a[/itex] and you're missing a factor of [itex]\int_0^a \sin^2(k_nx)\,\mathrm{d}x[/itex] on the left hand side of the first equation, but yes.

    Integrating
    [tex]
    I = \int_0^a \sin(k_nx)\sin(k_mx)\,\mathrm{d}x
    [/tex]
    by parts twice (differentiating the [itex]k_n[/itex] part each time) gives
    [tex]I = \frac{k_n^2}{k_m^2}I[/tex]
    so that [itex]I = 0[/itex] unless [itex]n = m[/itex], when
    [tex]
    \int_0^a \sin^2(k_nx)\,\mathrm{d}x =
    \int_0^a \frac12(1 - \cos(2k_nx))\,\mathrm{d}x =
    \left[\frac12 x - \frac{1}{4k_n} \sin(2k_nx)\right]_0^a = \frac{a}{2}
    [/tex]
    since [itex]2k_n = (2n+1)\pi/a[/itex].
     
  12. Dec 16, 2012 #11

    pasmith

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    Just a general point: if [itex]X_k'' = -k^2X_k[/itex], [itex]k > 0[/itex], then
    [tex]
    \int_a^b X''_k X_l\,\mathrm{d}x = [X_lX_k']_a^b - \int_a^b X_k'X_l'\,\mathrm{d}x
    = [X_lX_k' - X_kX_l']_a^b + \int_a^b X_kX_l''\,\mathrm{d}x
    [/tex]
    so that
    [tex]
    (l^2 - k^2) \int_a^b X_kX_l\,\mathrm{d}x = [X_lX_k' - X_kX_l']_a^b
    [/tex]
    so if [itex]k \neq k[/itex]
    [tex]
    \int_a^b X_kX_l\,\mathrm{d}x = \frac{[X_lX_k' - X_kX_l']_a^b}{(l^2 - k^2)}
    [/tex]
    Now if we require that, for all [itex]k[/itex], [itex]AX_k(a) + BX_k'(a) = 0[/itex] with [itex]A^2 + B^2 \neq 0[/itex] and [itex]CX_k(b) + DX_k'(b) = 0[/itex] with [itex]C^2 + D^2 \neq 0[/itex] then [itex]X_lX_k' - X_kX_l'[/itex] vanishes at [itex]x = a[/itex] and [itex]x = b[/itex] and the functions [itex]X_k[/itex] will be orthogonal with respect to the inner product
    [tex]
    \int_a^b f(x)g(x)\,\mathrm{d}x.
    [/tex]

    Calculating the determinant of a 2x2 matrix shows that if [itex]X_k \neq 0[/itex] then [itex]k[/itex] must be chosen so that
    [tex]
    (AC + BDk^2)\sin(k(b-a)) + (AD + BC)k\cos(k(b-a)) = 0
    [/tex]
     
    Last edited: Dec 16, 2012
  13. Dec 16, 2012 #12

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    The function u(x,y)= C2 for all x and y satisfies U(0)= C2 and [itex]u_x(1)+ u_y(1)= 0+ 0= 0[/itex]. Let v= T- u Then [itex]nabla v= \nabla T= 0[/itex] while v(0, y)= C2- C2= 0 and [itex]v_x(1, y)= T_x(1, y)= 0[/tex].

    So let [itex]v(x,y)= \sum_{n=0}^\infty A_n(y) sin(n\pi x/a)[/itex]. The differential equation reduces to [itex]\sum_{n=0}^\infty A''_n sin(n\pi x/a)- (n\pi/a)\sum_{n= 0}^\infty sin(n\pi x/a)= 0[/itex] which is the same as [itex]\sum_{n=0}^\infty (A''_n- (n\pi/a)Asin(n\pi x/a)[/itex] which is equivalent to [itex]A''_n- (n\pi/a)A= 0[/itex] for all n. Once you have found the Fourier series for v, add u= C2 to get T(x, y).
     
  14. Dec 16, 2012 #13
    Here are some additional thoughts on this problem:

    For the specified boundary conditions,

    [tex]T(x,y)= \frac{C1+C2}{2}-\frac{(C1-C2)}{2}G(x,y)[/tex]

    with

    G(x,0) = -1
    G(0,y) = +1
    G(x,2a) = -1
    G(2a,y) = +1

    The second pair of boundary conditions, by symmetry, automatically guarantee that the zero flux boundary conditions at x = a and y = a are satisfied. In addition to these conditions, the value of the function G is zero along all the diagonals:

    G(ζ,ζ) = 0
    G(ζ,-ζ) = 0

    Chet
     
  15. Dec 18, 2012 #14
    This is a continuation of my previous post. The solution for the function G(x,y) is the analog of the solution to a specific fluid mechanics potential flow problem, and can be expressed analytically in terms of a doubly-infinite sum over an array of point sources and point sinks.

    Instead of a single square of side 2a, imagine an infinite array of such squares placed adjacent to one another in the x-y plane, extending infinitely far in all directions. This forms a grid. Place identical point sources or a point sinks alternately at each of the corners of this grid. Fluid flows out of or into each of these point sources or sinks, respectively. The streamlines of the flow will be the lines of constant G in our problem, and the "stream function" values will be directly proportional to G. The stream function for the entire array of sources and sinks is just the sum of the stream functions for the individual sources and sinks. The stream function for an individual source or sink is simple to express analytically.

    Chet
     
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