Download Laplace's Equation Solutions in Two-Dimensional Coordinate Systems - Prof. Phillip Duxbury and more Study notes Physics in PDF only on Docsity! PHY481 - Lecture 15 Chapter 5.1-5.3 of PS - We will not cover 5.4, 5.5 Solutions to Laplace’s equations with variations in two directions In two dimensional cases where there is no dependence on one of the directions (e.g. z) but there is dependence on the other two directions e.g. x, y. In this kind of problem, we can choose to use either Cartesian co-ordinates or cylindrical co-ordinates. The domain on which the solution is required usually determines our decision of which co-ordinate system to use. If we are asked to solve a problem in a square or rectangular domain, then it is usually better to go with Cartesian, while on a circular domain it better to use cylindrical co-ordinates. In practice the domain is usually irregular so numerical computations are required. It is relatively simple to do simulations in Cartesian co-ordinates - As we will see later. A. Separation of variables in Cartesian co-ordinates in two dimensions Laplace’s equation is then, ∂2V ∂x2 + ∂2V ∂y2 = 0 (1) This can no longer be directly integrated, so we try to reduce it to a simpler form by assuming a solution of the form V (x, y) = X(x)Y (y). Subsituting this expression into Eq. (), we find that, d2X dx2 Y + X d2Y dy2 = 0 or 1 X d2X dx2 = − 1 Y d2Y dy2 = constant = −k2 (2) The requirement that the last expressions have to equal a constant occurs as if we have f(x) = g(y), then both f and g must be constant. It is convenient to call the constant k2 as we shall see. Also the sign difference between the X equation and the Y equation, and whether k is real or imaginary, is very important and determines whether the solutions to X or Y are oscillatory or decaying functions. From this expression we have, X(x) = A′(k)eikx + B′(k)e−ikx; Y (y) = C ′(k)eky + D′(k)e−ky (3) A more convenient form of these solutions is to write them as combinations of odd and even functions, namely, X(x) = A(k)cos(kx) + B(k)sin(kx); Y (y) = C(k)cosh(ky) + D(k)sinh(ky) (4) 1