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Laplace equation in a square with mixed boundary conditions 
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#1
Dec1512, 12:14 AM

P: 23

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
Dec1512, 01:13 AM

P: 759

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 ? 


#3
Dec1512, 01:32 AM

P: 23




#4
Dec1512, 05:29 AM

P: 759

Laplace equation in a square with mixed boundary conditions
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 ? 


#5
Dec1512, 07:41 AM

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[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. 


#6
Dec1512, 11:07 AM

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PF Gold
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Is there a reason why you can't use separation of variables?



#7
Dec1512, 05:51 PM

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[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] 


#8
Dec1612, 03:39 AM

P: 23




#9
Dec1612, 03:47 AM

P: 23

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]. 


#10
Dec1612, 11:08 AM

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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]. 


#11
Dec1612, 11:55 AM

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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(ba)) + (AD + BC)k\cos(k(ba)) = 0 [/tex] 


#12
Dec1612, 03:52 PM

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PF Gold
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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). 


#13
Dec1612, 09:56 PM

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Here are some additional thoughts on this problem:
For the specified boundary conditions, [tex]T(x,y)= \frac{C1+C2}{2}\frac{(C1C2)}{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 


#14
Dec1812, 07:14 AM

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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 doublyinfinite 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 xy 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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