Emag assignment

The purpose of the group assignment # 3 was to familiarize ourselves with the contents of Chapter 4 (Boundary Value Problems). Moreover, the group assignment brought out the need to use analytical solutions such as the Separation of Variables, Laplace’s Equations, and Fourier analysis. With these tools along with a little bit of intuition we are able to grasp and appreciate the relevance ofanalytical solutions to boundary value problems.

The first problem in the group assignment consisted of a rectangular conducting trough with the given boundary conditions:
* The conducting walls along x=0 ;x=a and y=0 planes are maintained at a potential V=0.
* The conducting wall along the y=b plane is maintained at a potential V=V1Volts

Applying these distinct boundary conditions wemust yield a V(x,y) as the following:

A pictorial of the trough with the given boundary conditions is given here:

Since the object has a rectangular symmetry and there is no dependence of the potential on the third coordinate z, Laplace’s equation in Cartesian coordinates is written as:
∇2V = ∂2V∂y2 + ∂2V∂y2 = 0 (1.0)The philosophy of solving this equation using the method of separation of variables is to assert that the potential V(x, y) is equal to the product of two terms, X(x) and Y(y), that separately are functions of only one of the independent variables x and y. The potential is then given by,

V(x,y) = X(x)Y(y) (1.1)
This isa critical assertion and our solution depends on its accuracy. We may wonder if other functional forms would work at this stage. They may or they may not. The resulting solutions would be obtained using different combinations might physically not make any sense or they may
not satisfy the given boundary conditions. Therefore, Substitute (1.2) into (1.1),
Y(y) ∂2X(x)∂x2 + X(x) ∂2Y(y)∂y2 = 0(1.2)
Note that the terms are to be differentiated only involve one independent variable. Hence, the partial derivatives can be replaced with ordinary derivatives and this will be done in the subsequent development. The next step in this methodical procedure is to divide both sides of this equation by V(x,y) = X(x)Y(y) . We find that,1X(x) ∂2X(x)∂x2 + 1Y(y) ∂2Y(y)∂y2 = 0 (1.3)
The first term on the left side of (1.4) is independent of the variable y. As far as the variable y, it can be considered to be a constant that we will take to be - ky2. Using a similar argument, the second term on the left side of (1.4) is independent of the variable x and it also can bereplaced with another constant written as + kx2 . Therefore, (1.4) can be written as two ordinary differential equations and one algebraic equation:

∂2X(x)∂x2 + kx2 X(x) = 0 (1.4)
∂2Y(y)∂y2 - ky2 Y(y) = 0 (1.5)
kx2 - ky2 = 0(1.6)
The two second-order ordinary differential equations can be easily solved.
The general solutions are,

X(x) = C1 Sin(kxx) + C2 Cos(kxx) (1.7)
Y(y) = C3 ekyy + C4 e-kyy (1.8)

For the firstboundary condition at x = 0, the potential V(0,y)=0 as illustrated in the following Figure.

We use equation (1.8) discussed earlier with X(x) X (0). Applying this condition yields:
X (0) = C1 Sin (kx0) + C2 Cos (kx0) =C10+C2(1)
X(0)=C2
Since V (0, y) must equal zero this draws the conclusion that C2=0 Simplifying equation (1.8) to the following:
Xx=C1sinkxx
Nothing can be stated...