The Laplace operator in polar coordinates in several dimensions∗ Attila M´at´e Brooklyn College of the City University of New York January 15, 2015 Contents 1 Polar coordinates and the Laplacian 1 1.1 Polar coordinates in n dimensions............................. 1 1.2 Cartesian and polar differential operators . ............. 2 1.3 Calculating the Laplacian in polar coordinates . .............. 2 2 Changes of variables in harmonic functions 4 2.1 Inversionwithrespecttotheunitsphere . ............ 4 1 Polar coordinates and the Laplacian 1.1 Polar coordinates in n dimensions Let n ≥ 2 be an integer, and consider the n-dimensional Euclidean space Rn. The Laplace operator Rn n 2 2 in is Ln =∆= i=1 ∂ /∂xi . We are interested in solutions of the Laplace equation Lnf = 0 n 2 that are spherically symmetric,P i.e., is such that f depends only on r=1 xi . In order to do this, we need to use polar coordinates in n dimensions. These can be definedP as follows: for k with 1 ≤ k ≤ n 2 k 2 define rk = i=1 xi and put for 2 ≤ k ≤ n write P (1) rk−1 = rk sin φk and xk = rk cos φk. rk−1 = rk sin φk and xk = rk cos φk. The polar coordinates of the point (x1,x2,...,xn) will be (rn,φ2,φ3,...,φn). 2 2 2 In case n = 2, we can write y = x1, x = x2. The polar coordinates (r, θ) are defined by r = x + y , (2) x = r cos θ and y = r sin θ, so we can take r2 = r and φ2 = θ. In case n = 3, the polar coordinates (r,θ,φ) are called spherical 2 2 2 2 coordinates, and we have y = x1, x = x2, z = x3, r = x + y + z , x = r sin φ sin θ, y = r sin φ cos θ, and x = r cos φ, so we can take r3 = r, φ2 = θ, and φ3 = φ. When using spherical coordinates, one often makes the restrictions r ≥ 0, 0 ≤ θ < 2π, and 0 ≤ φ ≤ φ, although it may occasionally be preferable to change these restrictions. ∗Written for the course Mathematics 4211 at Brooklyn College of CUNY. 1 1.2 Cartesian and polar differential operators The transformation T :(x1,x2,...,xn) → (rn,φ2,...,φn) is not one-to-one, since there are several (0) (0) (0) (0) possible choices are for polar coordinates. If P0 = (x1 ,x2 ,...,xn ) is a point for which rk =6 0 for any k with 2 ≤ k ≤ n then there is a neighborhood of the point P0 in which T can be defined in a one-to-one way, and, furthermore, the transformation T so defined is differentiable any number of times. If this condition is not satisfied, then there is no neighborhood of P0 in which T can be defined in a one-to-one way; further T is not be differentiable at the point P0. We would like to find the partial derivatives of the inverse transformation T −1 of T ; in order to do this, it will be easier to first determine the partial derivatives of T . Using equations (1) with k = n, it is easy to calculate the Jacobian matrix ∂(rn− ,xn) ∂r − /∂r ∂x /∂r sin φ cos φ 1 = n 1 n n n = n n . ∂(rn,φn) ∂rn−1/∂φn ∂xn/∂φn rn cos φn −rn sin φn Equations (1) with k = n can be inverted to determine the functions rn = rn(rn−1,xn) and φn = φn(rn−1,xn) in a neighborhood of a point for which rn =6 0, i.e., we have rn ≡ rn(rn sin φn, rn cos φn) and φn ≡ φn(rn sin φn, rn cos φn) in such a neighborhood. Differentiating the latter equations with respect to rn and φn, it follows from the chain rule of differentiation that the Jacobian matrix ∂(rn,φn)/∂(rn−1,xn) is the inverse of the matrix ∂(rn−1,xn)/∂(rn,φn). That is, −1 ∂(rn,φn) ∂rn/∂rn−1 ∂φn/∂rn−1 sin φn rn cos φn = = −1 . ∂(rn−1,xn) ∂rn/∂xn ∂φn/∂xn cos φn −rn sin φn We can illustrate with polar coordinates in case n = 2 why this is a better approach than calculating the 2 2 Jacobian matrix ∂(rn−1,xn)/∂(rn,φn) directly. In this case, equations (2) can be inverted as r = px + y and θ = arctan(y/x). The problem is that arctan(y/x) is not defined if x = 0, even if y =6 0, when equations (2) are invertible. In fact, we can also take π y − x θ = + arctan 4 y + x instead, and there is no trouble with this expression when x = 0 and y =6 0. This problem is avoided by obtaining the desired Jacobian as the inverse of another Jacobian. So, writing Drn = ∂/∂rn, Dxn = ∂/∂xn, Drn−1 = ∂/∂rn−1, and Dφn = ∂/∂φn, according to the chain rule we have ∂rn ∂ ∂φn ∂ cos φn (3) Drn−1 = + = sin φnDrn + Dφn ∂rn−1 ∂rn ∂rn−1 ∂φn rn and ∂rn ∂ ∂φn ∂ sin φn Dxn = + = cos φnDrn − Dφn ∂xn ∂rn ∂xn ∂φn rn 1.3 Calculating the Laplacian in polar coordinates 2 2 We want to evaluate the differential operator Drn−1 + Dxn . In case n = 2 we have r1 = x1 and so this is in fact the Laplacian; in case n > 2 the formula we obtain will give a recursive equation that will help in evaluating the Laplacian. To speed up the evaluation of the differential operator 2 Drn−1 one may observe that the result will consist of a sum second order differential operators, which 2 may be called the principal part, and a sum of first order differential operators. The principal part results from moving all differential operators to the right (as one of the terms in using the product rule of differentiation), and the first order part results when the the differential operator on the left is used up in differentiating a function following it (as the other term in using the product rule). A brief consideration shows that principal part can be calculated by using the binomial theorem; see [1, pp. 3-5] for details All this assumes that these differential operators are applied to functions that the order of derivatives in mixed second derivatives does not make any difference; for this it is sufficient to assume that the functions are twice continuously differentiable, but the conclusion also holds under weaker assumptions; see [2]. We have 2 cos φn cos φn Drn−1 = sin φnDrn + Dφn sin φnDrn + Dφn rn rn 2 2 2 2sin φn cos φn cos φn 2 = sin φnDrn + Drn Dφn + 2 Dφn rn rn cos φ cos2 φ cos φ sin φ − n n − n n 2 Dφn + Drn 2 Dφn . rn rn rn Similarly, 2 sin φn sin φn Dxn = cos φnDrn − Dφn cos φnDrn − Dφn rn rn 2cos φ sin φ sin2 φ 2 2 − n n n 2 = cos φnDrn Drn Dφn + 2 Dφn rn rn 2 cos φn sin φn sin φn cos φn + 2 Dφn + Drn + 2 Dφn . rn rn rn Adding these two equations, we obtain 2 2 2 1 2 1 (4) Drn−1 + Dxn = Drn + 2 Dφn + Drn . rn rn 2 2 For n = 2, taking have rn−1 = y, xn = x, rn = r = x + y , and φn = θ, this gives the Laplace operator in two dimensions: p 1 1 1 1 (5) D2 + D2 = D2 + D + D2 = D rD + D2. x y r r r r2 θ r r r r2 θ 2 2 For n = 3 with y = x1, x = x2, z = x3 we have θ = φ2, φ = φ3, r = r3, and ρ = x + y = r2 = r sin φ. According to (5) we have p 2 2 2 2 1 1 2 ∆=(D + D )+ D = D + Dρ + Dθ + D x y z ρ ρ ρ2 z 1 1 2 2 1 1 2 2 1 1 2 = Dρ + Dθ +(D + D )= Dρ + D + D + Dr + D , ρ ρ2 ρ z r sin φ r2 sin2 φ θ r r r2 φ where the for the last equation we used (4) and the equation ρ = sin φ. Using (3) with n = 3 to 3 express Dρ on the right-hand side, we obtain we obtain 1 cos φ 1 2 2 1 1 2 ∆= sin φDr + Dφ + D + D + Dr + D , r sin φ r r2 sin2 φ θ r r r2 φ 2 2 cos φ 1 2 1 2 = D + Dr + Dφ + D + D r r r2 sin φ r2 φ r2 sin2 φ θ 1 2 1 1 2 = Drr Dr + Dφ sin φDφ + D . r2 r2 sin φ r2 sin2 φ θ As n gets larger, the expression for the Laplacian gets more complicated; in any case, for any n ≥ 2 we have n 2 2 n − 1 1 (6) Dxi = Drn + Drn + 2 Λn, rn r Xi=1 n where Λn is a second order differential operator involving only the angular variables φ2, φ3, ..., φn, (i.e., it does not involve rn). Indeed, assuming that this is true for n − 1 replacing n, where n> 2, and using the equations rn−1 = rn sin φn and equations (3) and (4), we obtain n n−1 2 2 2 1 n − 2 2 2 − Dxi = Dxi + Dxn = 2 Λn 1 + Drn−1 + Drn−1 + Dxn r − rn− Xi=1 Xi=1 n 1 1 1 n − 2 cos φn 2 1 2 1 = Λ − + sin φ D + D + D + D + D 2 2 n 1 n rn φn rn 2 φn rn rn sin φn rn sin φn rn rn rn 2 n − 1 1 1 2 − − = Drn + Drn + 2 2 Λn 1 +(n 2) tan φnDφn + Dφn .