Question

The electrostatic potential V(x,y) in a rectangular region is governed by the two- dimensional Laplace equation: d^2V (x,y)/dx^2 + d^2v (x,y)/dy^2 =0 Determine V(x,y) in a rectangular region 0 ≤ x ≤ 20 and 0 ≤ y ≤ 40 for the following boundary conditions: V ,0( y) =V 20( , y) =V(x )0, = 0 and V (x 40, ) =110V

Answer 1

Potential V inside the rectangular area (0<= x<=a,  =<= y <= b) can be found as :  
    V(x,y) = V₁(x,y) + V₂(x,y) + V₃(x,y) + V₄ (x,y)   --- (1)

V₂(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:
        V₁(0,y)=0,  V₂(x,b) = 110 V,  V₃(a,y)=0 , V₄(x,0) = 0
        a = 20 units,  b = 40 units

Laplace PDE 2-dim of 2nd degree:
           δ²V(x,y)/δx² + δ²V(x,y)/δy² = 0  --- (2)

Simple solutions exist like V₂(x,y) = a (x² - y²)  or, V(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.
  V₂(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.
V₂(0,y) = 0 => X(0) = 0 =>  A=0.
V₂(x,0)= 0 => Y(0)= 0   =>  C = 0         =>  V₂(x,y)= BD Sin(λx) Sinh(λy)  -- (9)
V₂(a,y) = 0 => X(a) = 0         Sin(λa) = 0    =>   λ = nπ/a ,   n = 1, 2,3, 4....
V₂(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   V₂ⁿ (x,y) = Sin(nπ x/a)  Sinh (2π y/a)
             
$V(x,y)=\Sigma_{n=1}^\infty\ c_n\ Sin(\frac{2\pi}{a}x)\ Sinh(\frac{2\pi}{a}y) ---- (11)\\\\c_n=\frac{2}{a\ Sinh(\frac{2\pi}{a}b)} \int_0^a {V_2(x,b)*Sin(\frac{n\pi}{a}x)} \, dx\\\\=\frac{2*110}{a\ Sinh(\frac{2\pi}{a}b)} \int_0^a {Sin(\frac{n\pi}{a}x)} \, dx\\\\=\frac{220}{a\ Sinh(\frac{2\pi}{a}b)}\frac{a}{n\pi} [-Cos(\frac{n\pi}{a}x)}]_0^a\\\\=\frac{220*2}{n\pi\ Sinh(\frac{2\pi}{a}b)} --- (12)$

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