# Laplace equation in a square with mixed boundary conditions

by samuelandjw
Tags: boundary, conditions, equation, laplace, mixed, square
 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 $$\dfrac{\partial T}{\partial x}=0,\dfrac{\partial T}{\partial y}=0$$ Solve for $$T(x,y)$$ 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.
 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 ?
P: 23
 Quote by JJacquelin 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 ?
Not really. IMHO, the value on the corner is not particularly important. And, C1 is not equal to C2.

 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 ?
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P: 1,004
 Quote by JJacquelin It is important to see that the solution will include the Heaviside step function. Isn't it ?
Certainly
$$T(x,y) = \left\{ \begin{array}{r@{\quad}l} C_2 & x < y \\ C_1 & x > y \\ \end{array}\right.$$
is a weak solution of Laplace's equation which satisfies the boundary conditions.
 Sci Advisor HW Helper Thanks PF Gold P: 5,245 Is there a reason why you can't use separation of variables?
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P: 1,004
 Quote by Chestermiller Is there a reason why you can't use separation of variables?
Actually, no: if $F(x,y)$ is the solution of Laplace's equation in a unit square subject to $F(x,0) = 1$, $F(0,y) = 0$, $\partial F/\partial x(1,y) = 0$ and $\partial F/\partial y(x,1) = 0$ then the solution of the OP's problem is
$$T(x,y) = C_1F(x/a,y/a) + C_2F(y/a,x/a)$$.

And to find $F(x,y)$, let $k_n = \frac{2n + 1}{2}\pi$ for $n \geq 0$ and use the eigenfunctions
$$f_n(x,y) = \sin(k_n x)Y_n(y)$$
where
$$Y_n(y) = \cosh(k_ny) - \tanh(k_n)\sinh(k_ny).$$
P: 23
 Quote by pasmith $$Y_n(y) = \cosh(k_ny) - \tanh(k_n)\sinh(k_ny).$$
I think you can write the y-component solution as $$\cosh(k_n(a-y))$$
 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 $$b_n\cosh(k_n(a))=\int_0^a\sin(k_nx)f(x)dx$$, with $$k_n=(n+0.5)\pi$$. The thing that bothers me is that the integration interval is not a multiple of the period of the basis function $$\sin(k_nx)$$.
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P: 1,004
 Quote by samuelandjw I think you can write the y-component solution as $$\cosh(k_n(a-y))$$
But mine has the advantage that $Y(0) = 1$ (although I could write it as $\cosh(k_n(y-a))/\cosh(k_na)$)

 Quote by samuelandjw 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 $$b_n\cosh(k_n(a))=\int_0^a\sin(k_nx)f(x)dx$$ with $$k_n=(n+0.5)\pi$$
If you're integrating between 0 and a then $k_n = (n+0.5)\pi/a$ and you're missing a factor of $\int_0^a \sin^2(k_nx)\,\mathrm{d}x$ on the left hand side of the first equation, but yes.

Integrating
$$I = \int_0^a \sin(k_nx)\sin(k_mx)\,\mathrm{d}x$$
by parts twice (differentiating the $k_n$ part each time) gives
$$I = \frac{k_n^2}{k_m^2}I$$
so that $I = 0$ unless $n = m$, when
$$\int_0^a \sin^2(k_nx)\,\mathrm{d}x = \int_0^a \frac12(1 - \cos(2k_nx))\,\mathrm{d}x = \left[\frac12 x - \frac{1}{4k_n} \sin(2k_nx)\right]_0^a = \frac{a}{2}$$
since $2k_n = (2n+1)\pi/a$.
 HW Helper Thanks P: 1,004 Just a general point: if $X_k'' = -k^2X_k$, $k > 0$, then $$\int_a^b X''_k X_l\,\mathrm{d}x = [X_lX_k']_a^b - \int_a^b X_k'X_l'\,\mathrm{d}x = [X_lX_k' - X_kX_l']_a^b + \int_a^b X_kX_l''\,\mathrm{d}x$$ so that $$(l^2 - k^2) \int_a^b X_kX_l\,\mathrm{d}x = [X_lX_k' - X_kX_l']_a^b$$ so if $k \neq k$ $$\int_a^b X_kX_l\,\mathrm{d}x = \frac{[X_lX_k' - X_kX_l']_a^b}{(l^2 - k^2)}$$ Now if we require that, for all $k$, $AX_k(a) + BX_k'(a) = 0$ with $A^2 + B^2 \neq 0$ and $CX_k(b) + DX_k'(b) = 0$ with $C^2 + D^2 \neq 0$ then $X_lX_k' - X_kX_l'$ vanishes at $x = a$ and $x = b$ and the functions $X_k$ will be orthogonal with respect to the inner product $$\int_a^b f(x)g(x)\,\mathrm{d}x.$$ Calculating the determinant of a 2x2 matrix shows that if $X_k \neq 0$ then $k$ must be chosen so that $$(AC + BDk^2)\sin(k(b-a)) + (AD + BC)k\cos(k(b-a)) = 0$$
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Emeritus
 Quote by samuelandjw 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 $$\dfrac{\partial T}{\partial x}=0,\dfrac{\partial T}{\partial y}=0$$ Solve for $$T(x,y)$$ 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.
The function u(x,y)= C2 for all x and y satisfies U(0)= C2 and $u_x(1)+ u_y(1)= 0+ 0= 0$. Let v= T- u Then $nabla v= \nabla T= 0$ while v(0, y)= C2- C2= 0 and $v_x(1, y)= T_x(1, y)= 0[/tex]. So let [itex]v(x,y)= \sum_{n=0}^\infty A_n(y) sin(n\pi x/a)$. The differential equation reduces to $\sum_{n=0}^\infty A''_n sin(n\pi x/a)- (n\pi/a)\sum_{n= 0}^\infty sin(n\pi x/a)= 0$ which is the same as $\sum_{n=0}^\infty (A''_n- (n\pi/a)Asin(n\pi x/a)$ which is equivalent to $A''_n- (n\pi/a)A= 0$ for all n. Once you have found the Fourier series for v, add u= C2 to get T(x, y).
 Sci Advisor HW Helper Thanks PF Gold P: 5,245 Here are some additional thoughts on this problem: For the specified boundary conditions, $$T(x,y)= \frac{C1+C2}{2}-\frac{(C1-C2)}{2}G(x,y)$$ 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