Chapter3-8
3.8 Laplace’s Equation in Rectangular Coordinates
Figure 3.8-1 A thin rectangular plate with insulated top and bottom surfaces
The two-dimensional steady state heat equation for a thin rectangular plate shown in Figure 3.8-1 is the Laplace’s equation
= 0(3.8-1)
The Laplace’s equation is an elliptic equation for a finite domain that has the following boundary conditions
u(0,y) = u(a,y) = 00 < y < b
u(x,0) = 00 < x < a
andu(x,b) = f2(x)0 < x < a
We assume that u(x,t) can be separated into X(x), a function of x alone,and Y(y), a function of y alone.
u(x,t) = X(x) Y(y)
From the following relations
= T = T
= T = T
Eq. (3.8-1) becomes
= 0
Divide the above equation byXY to obtain
= = 2 = constant
Since the LHS depends on x only, and the RHS depends on y only, they must equal to a constant 2. The constant must be negative for non-trivial solution. The separation constant should be negative for the direction with both homogeneous boundary conditions. For this problem, x-direction has the homogeneous boundary condition at both ends.
The boundary conditions on u(x,t) are changed to the boundary condition on X(x)
u(0,y) = 0 X(0) Y(y) = 0 X(0) = 0
u(a,y) = 0 X(a) Y(y) = 0 X(a) = 0
The ODE with respect to x is
= 2XX = C1cos(x) + C2sin(x)
The constant C1 can be determined from the boundary condition at x = 0
At x = 0, X = 0 = C1(1) + C2(0) C1 = 0
At x = a, X = 0 = C2sin(a)
To avoid the trivial solution, C2 0 and
sin(a) = 0 a = nn =
The solution with respect to x is
X = Xn= C2sin, n = 1, 2, 3, …
The ODE with respect to y is
= 2 Y = An’cosh(ny) + Bn’sinh(ny)
The constant An’ can be determined from the boundary condition at y = 0,
Y(0) = 0 = An’Y = Bn’sinh(ny)
The product solution is then
un(x,t) = Bnsinsinh
The general solution for the steady two dimensional heat equation is
u(x,y) = Bnsinsinh
The constants Bncan be determined from the boundary condition at y = b
u(x,b) = f2(x) = Bnsinsinh
The coefficients Bn sinh are then obtained from the Fourier sine series expansion of f2(x)
Bn =
Non-homogeneous Boundary Conditions
Figure 3.8-2 A thin rectangular plate with all non-homogeneous boundary conditions
The separation of variables method can be applied directly to an elliptic equation with all homogeneous boundary conditions except one. If there are two or more non-homogeneous boundary conditions, the original problem will need to be decomposed into sub problems that will have all homogeneous boundary conditions except one as shown in Figure 3.8-2.
The two-dimensional steady state heat equation for a thin rectangular plate shown in Figure 3.8-2 is the Laplace’s equation
= 0(3.8-1)
The heat equation for this case has the following boundary conditions
u(0,y) = g1(y),u(a,y) = g2(y),0 < y < b
u(x,0) = f1(x),u(x,b) = f2(x),0 < x < a
The original problem with function u is decomposed into four sub-problems with new functions u1, u2, u3, and u4. The boundary conditions for the sub-problems are given in Figure 3.8-2 where only one non-homogeneous boundary condition exits for each sub-problem. For example, the function u1 has u1(x,0) = f1(x) as the only non-homogeneous boundary condition. The original function u is related to the new functions by
u =u1 + u2 + u3 + u4
The function u2 is already evaluated in the previous example where
u2(x,y) = Bnsinsinh
The constants Bncan be determined from the boundary condition at y = b
u(x,b) = f2(x) = Bnsinsinh
The coefficients Bn sinh are then obtained from the Fourier sine series expansion of f2(x)
Bn =
The function u4 is the same as u2 when y is interchanged with x and b is interchanged with a. The boundary condition f2(x) is replaced by the boundary condition g2(y). Therefore
u4(x,y) = Dnsinsinh
The constants Dncan be determined from the boundary condition at x = a
u(a,y) = g2(y) = Bnsinsinh
The coefficients Dn sinh are then obtained from the Fourier sine series expansion of g2(y)
Dn =
u1 is the same as u2 with y replaced by b-y
u1(x,y) = Ansinsinh
The constants Ancan be determined from the boundary condition at y = 0
u1(x,0) = f1(x) = Ansinsinh
The coefficients An sinh are then obtained from the Fourier sine series expansion of f1(x)
An =
u3 is the same as u4 with x replaced by a-x
u3(x,y) = Cnsinsinh
The constants Cncan be determined from the boundary condition at x = 0
u(0,y) = g1(y) = Cnsinsinh
The coefficients Cn sinh are then obtained from the Fourier sine series expansion of g1(y)
Cn =
Thus the original solution can be constructed from the solutions of the sub-problems.
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