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Solving Laplace's equation in cylindrical coordinates

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Let's write V(x,y,x) = X(x) + Y(y) + Z(z)

The Laplacian nabla^2 applied to V must be zero:

Nabla^2[X(x) + Y(y) + Z(z)] = 0

X only depends on x, so the Laplacian applied to X will yield the second derivative w.r.t. x, and similarly, the Laplacian applied to Y yields the second derivative of Y w.r.t. y and the Laplacian applied to Z yields the second derivative of Z w.r.t. z. We thus have:

d^2X/dx^2 + d^2Y/dy^2 + d^2Z/dz^2 = 0 (1)

In general the second derivative of a function will depend on the argument of that function. So, the second derivative of a general function X(x) depends on x, and the second derivative of a general function Y(y) depends on Y. In this case, however, we cannot allow the second derivatives to depend on their arguments, because if there is an x-dependence in d^2X/dx^2 then the other two terms cannot cancel that out to yield a sum of zero because the two other terms don't depend on X. This means that Eq. (1) can only be satisfied if each of the second derivatives is a constant. The sum of the three constants must be zero. So we must have:

d^2X/dx^2 = A (2)

d^2Y/dy^2 = B (3)

d^2Z/dz^2 = C (4)


A + B + C = 0 (5)

If we integrate (2), (3) and (4), we obtain:

X(x) = 1/2A x^2 + A_{1} x + A_{2} (6)

Y(y) = 1/2B y^2 + B_{1} y + B_{2} (7)

Z(z) = 1/2C z^2 + C_{1} z + C_{2} (8)

Suppose we want to satisfy boundary conditions on the faces of a cube at z = 0, z= L, x=0, x=L, y =0 and y = L. Let's say that V is constant on the faces of the cubes but takes different values at each of the six faces. If you take the face z = L and demand that V is constant there then you see that this implies that A, B, A_{1} and B_{1} must be zero, because at z=L, V must not depend on x and y. Similarly, if you take the other faces and combine all the constraints, you find that only the coefficients A_{2}, B_{2} and C_{2} can be nonzero. But with only three constants available you cannot give V different values at the six faces.


In the problem about the cylinder you must ...

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