2D Laplace potential distribution

Problem. Inside the rectangle 0 < x < a, 0 < y < b the region is charge-free, so the potential satisfies the 2-D Laplace equation ∂²V/∂x² + ∂²V/∂y² = 0, with V = 0 on x = 0, y = 0, y = b and V = V₀ on x = a.

        y
        ↑ V=0 (y=b)
   b    +--------------+
        |              |
  V=0   |   ∇²V = 0    | V=V0
 (x=0)  |              | (x=a)
        +--------------+ → x
        0 V=0 (y=0)    a

Step 1: Separation of variables

Let V(x, y) = X(x)Y(y). Substituting and dividing by XY:

X''/X + Y''/Y = 0
X''/X = λ² ,  Y''/Y = −λ²      (choose sine in y because V = 0 at y = 0 and y = b)

So X'' − λ²X = 0 and Y'' + λ²Y = 0, giving

X = A cosh λx + B sinh λx   (or C e^(λx) + D e^(−λx))
Y = C cos λy + D sin λy

Step 2: Boundary conditions

• V = 0 at y = 0: Y(0) = 0 gives C = 0, so Y = D sin λy.
• V = 0 at y = b: sin λb = 0, so λ = nπ/b, n = 1, 2, 3, …
• V = 0 at x = 0: X(0) = 0 gives A = 0, so X = B sinh(nπx/b).

The general solution is the sum of all modes:

V(x,y) = Σ(n=1→∞) c_n sinh(nπx/b) sin(nπy/b)

Step 3: Fourier coefficients using V = V₀ at x = a

V₀ = Σ c_n sinh(nπa/b) sin(nπy/b)    for 0 < y < b

Multiply by sin(mπy/b) and integrate over 0 to b (orthogonality):

c_n sinh(nπa/b) · (b/2) = V₀ ∫₀ᵇ sin(nπy/b) dy = V₀ (b/nπ)(1 − cos nπ)

c_n sinh(nπa/b) = (2V₀/nπ)(1 − cos nπ)
                = 4V₀/(nπ)   for n odd
                = 0          for n even

Step 4: Final solution

V(x,y) = (4V₀/π) Σ(n = 1,3,5,…) (1/n) · [sinh(nπx/b) / sinh(nπa/b)] · sin(nπy/b)

Check: at x = a the ratio of sinh terms is 1, and the series (4V₀/π)Σ sin(nπy/b)/n for odd n equals V₀ for 0 < y < b. At x = 0 the sinh is zero, and at y = 0, b the sine is zero, so all other boundary conditions hold.