_{1}

^{*}

In the investigation, the complex geometric domain is a concave geometrical pattern. Due to the symmetric character, the left side of the geometric pattern,
i.e.
the L-shaped region is calculated in the study. The governing equation is expressed with
Laplace
equations. And the analysis is solved by eigenfunction expansion and point-match method. Besides, visual C^{++} helps obtain the results of numerical calculation. The local values and the mean values of the function are also discussed in this study.

The Laplace equations show an important role in the applied mathematical researches and analysis. Some significant efforts, thus, have been directed towards researches into related fields. For example, Alliney [

Although many researches about Laplace equation under different conditions have been discussed, the Laplace equation with concave domains is also worth discussing. The present paper, thus, will analyze a symmetric domain with complex Laplace equations under two kinds of boundary conditions in order to find local values and the mean values of the function. The analysis of two kinds of boundary conditions, case 1 and case 2, will be specified in the following mathematical formulation. Furthermore, in the present paper, visualization and image processing are obtained from mathematical formulation of the complex Laplace equations on a concave domain. It is hoped that the results can be further applied in engineering and technology, for example, the problem of fluid flow and heat conduction.

The geometric domain in

The analysis of boundary conditions on Case 1:

The boundary condition of the bottom is 1, and the condition of the top is 0. The local function distribution in the solid part of the rectangle is f ( x , y ) . Decompose the concave domain into two parts, and due to the symmetric character, only the left hand of the geometry, i.e. the L-shaped region in

The governing equation for the region is expressed with Laplace equation:

∇ 2 f ( x , y ) = 0 (1)

The L-shape region is composed of two rectangles; the governing equation for the left one is:

∇ 2 f 1 ( x , y ) = 0 (2)

The boundary conditions for Equation (2) are:

f 1 ( x , 0 ) = 1

f 1 ( x , h ) = 0

f 1 ( 0 , y ) = 1 − y h (3)

And the governing equation for the right rectangle is:

∇ 2 f 2 ( x , y ) = 0 (4)

The boundary conditions for Equation (4) are:

f 2 ( x , 0 ) = 1

f 2 ( x , h − b ) = 0

∂ f 2 ∂ x ( w , y ) = 0 (5)

With the boundary conditions Equations (3) and (5), the analytical solution to

Equation (2) is f 1 ( x , y ) , and to Equation (6) is f 2 ( x , y ) . They are as follows:

f 1 ( x , y ) = 1 − y h + ∑ n A n sin ( α n y ) ( e − α n ( x + d ) − e − α n ( x − d ) ) (6)

f 2 ( x , y ) = 1 − y h − b + ∑ m B m sin ( β m y ) ( e β m ( x − 2 w + d ) + e − β m ( x − d ) ) (7)

where the eigenvalues are:

α = n π h

β = m π h − b (8)

Other boundary conditions for governing equations are:

f 1 ( d , y ) = 0 ; h − b ≤ y ≤ h (9)

We choose N points along the boundary at x = d, y i = i h / N and truncate A_{n} to N terms and B_{m} to M terms, where M = int [ ( 1 − b / h ) ] . Substitute the boundary condition into Equation (6), and the following equation can be obtained:

∑ n A n sin ( α n y i ) ( e − 2 α d − 1 ) = y i h − 1 (10)

i = M + 1 to N

Next, the solutions to the two regions of the L-shape domain can be matched along the common boundary conditions [

f 1 ( d , y ) = f 2 ( d , y ) , 0 ≤ y < h − b (11)

∂ f 1 ( d , y ) ∂ x = ∂ f 2 ( d , y ) ∂ x , 0 ≤ y < h − b (12)

Substitute the boundary condition into Equations (6)-(7) and can obtain the following equations:

∑ n A n sin ( α n y i ) ( e − 2 d α − 1 ) − ∑ m B m sin ( β m y i ) ( 1 + e − 2 ( w − d ) β m ) = b y i h ( 1 − h ) (13)

i = 1 to M

∑ n A n α n sin ( α n y i ) ( − 1 − e − 2 d α ) − ∑ m B m β m sin ( β m y i ) ( e − 2 ( w − d ) β m − 1 ) = 0 (14)

i = 1 to M

The mean value for f ( x , y ) is expressed as:

f m e a n = 1 h w − b ( w − d ) [ ∫ 0 d ∫ 0 h f 1 d y d x + ∫ d w ∫ 0 h − b f 2 d y d x ] (15)

Integrating Equation (15) can obtain the following equation:

f m e a n = 1 h w − b ( w − d ) { h w + b ( d − w ) 2 + ∑ n A n α 2 ( 2 e − d α − e − 2 d α − 1 ) + ∑ m B m β 2 [ cos ( h − b ) β − 1 ] [ e − 2 β ( w − d ) − 1 ] } (16)

The analysis of boundary conditions on Case 2:

The geometric domain is also a concave domain. The outer dimension of the domain is 2 w × h , and there is a concave, whose dimension is 2 ( w − d ) × b . The boundary condition of the bottom is 0, and the condition of the top is 1. The L-shape region, in

The governing equation for the left rectangle is:

∇ 2 f 1 ( x , y ) = 0 (17)

The boundary conditions for Equation (17) are:

f 1 ( x , 0 ) = 0

f 1 ( x , h ) = 1

f 1 ( 0 , y ) = y h (18)

And the governing equation for the right rectangle is:

∇ 2 f 2 ( x , y ) = 0 (19)

The boundary conditions for Equation (19) are:

f 2 ( x , 0 ) = 0

f 2 ( x , h − b ) = 1

∂ f 2 ∂ x ( w , y ) = 0 (20)

With the boundary conditions Equations (18) and (20), the analytical solution to Equation (17) is f 1 ( x , y ) , and to Equation (19) is f 2 ( x , y ) . They are as follows:

f 1 ( x , y ) = y h + ∑ n C n sin ( α n y ) ( e − α n ( x + d ) − e − α n ( x − d ) ) (21)

f 2 ( x , y ) = y h − b + ∑ m D m sin ( β m y ) ( e β m ( x − 2 w + d ) + e − β m ( x − d ) ) (22)

where the eigenvalues are:

α = n π h

β = m π h − b (23)

Other boundary conditions for governing equations are:

f 1 ( d , y ) = 1 ; h − b ≤ y ≤ h (24)

Substitute the boundary condition into Equation (21), and the following equation can be obtained:

∑ n C n sin ( α n y i ) ( 1 − e − 2 α d ) = 1 − y h (25)

i = M + 1 to N

Next, the solutions to the two regions of the L-shape domain can be matched along the common boundary conditions. The conditions can be expressed as:

f 1 ( d , y ) = f 2 ( d , y ) , 0 ≤ y < h − b (26)

∂ f 1 ( d , y ) ∂ x = ∂ f 2 ( d , y ) ∂ x , 0 ≤ y < h − b (27)

Substitute the boundary condition into Equations (21)-(22), and can obtain the following equations:

∑ n C n sin ( α n y i ) ( e − 2 d α − 1 ) − ∑ m D m sin ( β m y i ) ( 1 + e − 2 ( w − d ) β m ) = b y i h ( h − b ) (28)

i = 1 to M

∑ n C n α n sin ( α n y i ) ( − 1 − e − 2 d α ) ) − ∑ m D m β m sin ( β m y i ) ( e − 2 ( w − d ) β m − 1 ) = 0 (29)

i = 1 to M

The mean value for f ( x , y ) is expressed as:

Integrating Equation (30) can obtain the following equation:

The following steps of numerical methods are estimated by using Visual C^{++}:

1) Give the constants h, b, w and d.

2) Set N = 29,

3) Equations (10), (13) and (14) are expressed as the linear system of (N + M) equations to solve coefficients

4) Substitute coefficients _{1}, (Equation (6)) and f_{2} (Equation (7)). This process is repeated at all nodes within the range, i.e.

5) Map the f(x,y) on the entire domain.

6) The average values of function can be calculated from Equation (16).

7) Repeat the previous methods can estimate the results of Case 2.

Results and discussion of Case 1:

domain h will influence the coefficient A_{n} and B_{m}, and then the mean values of

_{n} and B_{m}, and then the mean values of

Results and discussions of Case 2:

The present paper can find the influence of the height and the width of the geometric domain on the function mean values. Besides, the present paper uses the analytical solution of point match methods and numerical methods which can easily compute coefficient A_{n}, B_{m}, C_{n}, D_{m} and function

The authors gratefully acknowledge the support provided to this projects by the Ministry of Science and Technology of the Taiwan under Contract Number MOST 106-2221-E-019-062.

Hu, H.-P. (2018) An Analysis of Double Laplace Equations on a Concave Domain. World Journal of Engineering and Technology, 6, 304-314. https://doi.org/10.4236/wjet.2018.62019