Question

The electrostatic potential U(x,y) in a rectangular region (0≤x≤20m, 0≤ y ≤40m) is governed by the 2 dimensional Laplace partial differential equation: d^2 U(x,y)/dx^2 + d^2 U(x,y)/dy^2 =0.

Determine Potential U(x,y) in the rectangular region 0 ≤ x ≤ 20 and 0 ≤ y ≤ 40 for the following boundary conditions:

U(0, y) = U(20, y) = U(x,0) = 0 Volts and U(x,40) = 110V

Answer 1

Potential U inside the rectangular area (0<= x<=a, 0<= y <= b) can be found as :  
    U(x,y) = U₁(x,y) + U₂(x,y) + U₃(x,y) + U₄ (x,y)   --- (1)

U₂(x,y) is found by setting Potential on boundaries x=0, x=a, y=0 to  zero and finding the solution to the Laplace 2-dimensional Equation.  Others in (1) are found by setting appropriate boundary conditions to 0.  We are given conditions:  

        U₁(0,y)=0,  U₂(x,b) = 110 V,  U₃(a,y)=0 , U₄(x,0) = 0
        a = 20 units,  b = 40 units

Laplace PDE 2-dim of 2nd degree:
           δ²U(x,y)/δx² + δ²U(x,y)/δy² = 0  --- (2)

Simple solutions exist like U₂(x,y) = a (x² - y²)  or, U(x,y) = P(x)+Q(y),
   P(x), Q(y) are 2nd degree polynomials in x, y respectively.
But the boundary conditions cannot be satisfied by these solutions.

So we use separation of variables method.
  U₂(x,y) = X(x) * Y(y)   ---- (3)

So  differentiating and substituting in (2), we get:
    X''(x) Y(y) + Y''(y) X(x) = 0   -- (4)
=>    X''(x) /X(x) = - Y''(y)/ Y(x) = λ²  (say)   constant ≠ 0.
We choose the (+ or -) sign of λ² in such a way that in the direction with both boundaries are zero, we get Sine as solution.  In the direction of non-zero boundary value, we get Sinh as solution.

     X''(x) + λ² X(x) = 0    --- (5)              Y''(y) - λ² Y(y) = 0     --- (6)

Solutions are:   X(x) = A Cos(λx) + B Sin(λx)       ---(7)   
                         Y(y) = C Cosh(λx) + D Sinh(λx)   ---- (8)

Apply boundary conditions now.
U₂(0,y) = 0 => X(0) = 0 =>  A=0.
U₂(x,0)= 0 => Y(0)= 0   =>  C = 0         =>  U₂(x,y)= BD Sin(λx) Sinh(λy)  -- (9)
U₂(a,y) = 0 => X(a) = 0         Sin(λa) = 0    =>   λ = nπ/a ,   n = 1, 2,3, 4....
U₂(x,b) = 110V =>       Sin(nπ x/a) Sinh(2πb/a)  = 110  --- (10)
    This cannot be solved with one value of n.  Fourier sources is to be used.
  
     Let   U₂ⁿ (x,y) = Sin(nπ x/a)  Sinh (2π y/a)
             
[tex]U(x,y)=\Sigma_{n=1}^\infty\ c_n\ Sin(\frac{2\pi}{a}x)\ Sinh(\frac{2\pi}{a}y) ---- (11)\\\\c_n=\frac{2}{a\ Sinh(\frac{2\pi}{a}b)} \int_0^a {U_2(x,b)*Sin(\frac{n\pi}{a}x)} \, dx\\\\=\frac{2*110}{a\ Sinh(\frac{2\pi}{a}b)} \int_0^a {Sin(\frac{n\pi}{a}x)} \, dx\\\\=\frac{220}{a\ Sinh(\frac{2\pi}{a}b)}\frac{a}{n\pi} [-Cos(\frac{n\pi}{a}x)}]_0^a\\\\=\frac{220*2}{n\pi\ Sinh(\frac{2\pi}{a}b)} --- (12)\\\\[/tex]

Finally, the solution :
   [tex]U(x,y)=\Sigma_{n=1}^\infty\ \frac{440}{n\pi\ Sinh(\frac{2\pi}{a}b)}\ Sin(\frac{2\pi}{a}x)\ Sinh(\frac{2\pi}{a}y) ---- (13)\\\\U(x,y)=\Sigma_{n=1}^\infty\ \frac{440}{n \pi\ Sinh(4\pi)}\ Sin(\frac{\pi}{10}x)\ Sinh(\frac{\pi}{10}y) ----(14)\\ [/tex]