Rectangular Pipe Example

A volume V is bounded by |x|= b and y = 0 and y = a. The electrostatic potential on surfaces are given by f(b,y)=f(-b,y)= V and f(x,0) = f(x,a) = a.
The solution is given by (Choosing a = b = 1 and V =1). (Here we add only first 500 terms of the series, and hope for the best. Notice that the graph is very accurate near x = 0, but not so accurate near x = b and x=-b.)

f[x_, y_] = Underoverscript[∑, n = 0, arg3] (4 Cosh[(2n + 1) π x] Sin[(2 n + 1) π y])/((2n + 1) π Cosh[(2n + 1) π]) ;

Plot3D[Evaluate[f[x, y]], {x, -1, 1}, {y, 0, 1}]

[Graphics:HTMLFiles/Pipes_3.gif]

⁃SurfaceGraphics⁃

ContourPlot[Evaluate[f[x, y]], {x, -1, 1}, {y, 0, 1}, {PlotPoints->50, Contours->15}]

[Graphics:HTMLFiles/Pipes_6.gif]

⁃ContourGraphics⁃

Show[%, ContourShadingFalse]

[Graphics:HTMLFiles/Pipes_9.gif]

⁃ContourGraphics⁃

Open Pipe Example

A volume V is bounded by x= 0 and x = a and y = 0. The electrostatic potential on surfaces are given by f(0,y)=f(a,y)= 0 and f(x,0) = V.
The solution is given by (Choosing a = 1 and V =1).

g[x_, y_] = 2/πArcTan[Sin[π x]/Sinh[π y]] ;

RowBox[{Plot3D, [, RowBox[{Evaluate[g[x, y]], ,, {x, 0, 1}, ,, RowBox[{{, RowBox[{y, ,, 0.001, ,, 1}], }}], ,, RowBox[{ViewPoint, , RowBox[{{, RowBox[{3, ,, 2.4, ,, 2}], }}]}]}], ]}]

[Graphics:HTMLFiles/Pipes_13.gif]

⁃SurfaceGraphics⁃

RowBox[{ContourPlot, [, RowBox[{Evaluate[g[x, y]], ,, {x, 0, 1}, ,, RowBox[{{, RowBox[{y, ,, 0.001, ,, 1}], }}], ,, {PlotPoints->50, Contours->15, ContourShadingFalse}}], ]}]

[Graphics:HTMLFiles/Pipes_16.gif]

⁃ContourGraphics⁃


Created by Mathematica  (February 13, 2007)