Practical in Numerical Astronomy, SS 2012 LECTURE 9

Transcription

1 Practical in Numerical Astronomy, SS 01 Elliptic partial differential equations. Poisson solvers. LECTURE 9 1. Gravity force and the equations of hydrodynamics. Poisson equation versus Poisson integral. Numerical solution of the Poisson equation 4. Numerical integration of the Poisson integral Lecturer Eduard Vorobyov. eduard.vorobiev@univie.ac.at 1

2 GRAVITY FORCE and THE EQUATIONS OF HYDRODYNAMICS

3 External gravity vs. self-gravity Although gravity is omnipresent in the Universe, its effect is often simplified or even neglected. However, there are situations when an accurate calculation of gravity is a necessity. In this context, two conceptually distinct descriptions of gravity are often used: external gravity and internal self-gravity. 1 External gravity is an additional external force acting onto the object under consideration. At the same time, the back reaction of the object onto the external perturber is neglected. The typical examples are the motion of gas and stars in the gravitational potential of the dark matter halo or the motion of a planet in the gravitational potential of a star. Internal self-gravity or, simply, self-gravity acts as an intrinsic force, like that of the gas pressure, arising due to gravitational interaction of particles constituting the object under consideration. The typical examples are the gravitational collapse of pre-stellar cores, formation of protostars and planets in protostellar disks or the growth of spiral structures in disk galaxies Myr 0. Myr 0.5 Myr Myr -1 Radial distance (kpc Gyr Gyr Gyr Gyr Radial distance (AU Radial distance (AU Radial distance (kpc Radial distance (kpc Gyr. Gyr.4 Gyr.0 Gyr. Gyr.7 Gyr Gyr 4.1 Gyr

4 If gravity cannot be neglected, the equations of hydrodynamics need to be modified to include the effect of gravitational acceleration g Equations of hydrodynamics in Cartesian coordinates (i,j = x,y,z ( ρu pressure force per unit volume viscous force per unit volume ρ + i = 0 t xi P π ρ ui + ρ ui u j + ρ gi = t x x x ( ij 0 j i j gravity force per unit volume E + E + P u j ui ji uigi = t x j ( π ρ 0 work per unit volume per unit time due to gravity force See Nigel s lecture for more details 4

5 DERIVING THE POISSON EQUATION and POISSON INTEGRAL FROM NEWTON S LAW OF GRAVITY 5

6 Newton s law of gravity for two-body interaction F GMm r = -- scalar form of Newton s law for two-body interaction F GMm = r r vector form of Newton s law for two-body interaction, where r is the position vector of a particle with mass m, and F is the gravity force acting on that object from a particle with mass M. Note the minus sign appearing due to the fact that r and F are pointed in opposite directions. M r F m Applying nd Newton s law, one gets the gravitational acceleration g GMm GM F = mg = r g = r r r In the following slides, boldfaced values are vectors or tensors 6

7 Newton s law of gravity for many interacting bodies j=1 g( r i N = GM ( r - r r - r j= 1 i j i j total acceleration acting on particle i from particles j=(1 N. r i g j=1 i r i -r j rj g j= g j=n j=n j= g( r = G V ρ ( r ( r - r d r - r r acceleration acting at position vector r within a continuous body with density ρ and volume V r = d dx dy dz in Cartesian coordinates r d = dz dr r dϕ in cylindrical coordinates d r = dr r dθ r sin θ dϕ in spherical coordinates r r 7

8 Gravitational potential Gravity is the so-called central force whose magnitude only depends on the distance r between the interacting objects and is directed along the line joining them. Work done by the gravity force has a nice property such that it does not depend on the path but only on the starting and finishing points. This allows us to define the gravitational potential Φ as follows r r r Φ A = Fdr = mgdr = m dr = m Φ( r Φ( r1 r r r r [ ] The minus sign is due to the fact that Φ is negative ρ ( r ( r - r g( r = Φ ( r = G d r, = r - r r V where is the Nabla operator Gravitational potential is a scalar field in the sense that it takes scalar values but, at the same time, is a function of the position vector. It has dimensions of energy per unit mass and is always negative (due to attraction nature of the gravity force. 8

9 Deriving Poisson equation and Poisson integral Φ ( r = G V ρ ( r ( r - r d r - r r Let us use the following two vector identities 1 r r r = ; = 4 πδ ( r r r r r r proof can be done expanding these identities in Cartesian coordinates applying the first identity ρ( r ( r - r d Φ ( r = r = r - r V ρ( r ρ( r = G d r = G d r r r r r V G V applying the second identity ρ( r ( r - r Φ ( r = Φ ( r = G d r = r - r V = = 4 πg ρ( r δ ( r - r d r 4 πgρ( r we arrive at the Poisson integral we arrive at the Poisson equation Φ ( r = G V ρ ( r d r r r where Φ ( r = 4 πgρ( r, = r r is the Laplace operator 9

10 POISSON EQUATION - a prototype elliptic partial differential equation 10

11 Expansion of the Poisson equation in orthogonal coordinates Φ ( r = 4 πgρ( r where the Laplace operator has the following form in three main orthogonal coordinate systems = + + Cartesian (x,y,z x y z 1 1 = + r + z r r r r ϕ Cylindrical (z,r, ϕ = r sin + θ + r r r r sinθ θ θ r sin θ ϕ Spherical (r, θ, ϕ 11

12 Classification of the Poisson equation The general second-order PDE in two independent variables x and y has the form f f f f f x x y y x y A + B + C + D + E + F f + G = 0, (1 where f (x,y is the unknown function and coefficients A, B, C, D, E, F, and G may be constant or may also depend on x and y A B B C > 0 If, equation (1 is said to be a linear elliptic PDE. The Poisson equation is the prototype example. In the example below, the determinant is equal to unity Φ Φ + = 4π Gρ x y 1

13 Elliptical PDEs are different from parabolic PDEs (e.g., diffusion equation or hyperbolic PDEs (wave equation, transport equation in the sense that the latter are the initial value problems while the former is the boundary value problem. = + t x y f f f Φ Φ Φ + + = 4π Gρ x y z t=t 1 t=t 0 x y z Φ is known x y Φ is known at the cube faces but is unknown inside the cube Solving for the diffusion equation means advancing the solution in time starting from the initial layer of known values of f at t=t 0. That is why this sort of problems is called the initial value problem. Boundary values also need to be known but only one layer at a time. Solving for the Poisson equation means finding all the values of the potential Φ inside the cube at once using precalculated boundary values. Usually, one cannot march from the outer boundary towards the center in the same sense as the initial value problem can be integrated forward in time.

14 NUMERICAL SOLUTION of the POISSON EQUATION 14

15 dy dx Discretization of the Poisson equation Suppose we need to find the gravitational potential on a mesh with x grid zones of equal size in each direction. The grid zone size is dx and dy in the x- and y-directions. The first step would be to discretize the D Poisson equation. Φ 1, j= j= Φ Φ Φ = + = 4π Gρ x y Discretizing the first derivative Φ x i+ 1/ = Φ i+ 1, j i, j dx Φ boundary value, Φ 0, i=0 i=1 i= i= boundary layer Φ x 1+ 1/, j=1 Discretizing the second derivative Φ x Φ x Φ i+ 1/, j i 1/, j = x dx For a simple case of D Cartesian equidistant mesh we obtain a five-zone molecule Φi+ 1, j Φ i, j + Φi 1, j Φi, j+ 1 Φ i, j + Φi, j 1 Φ = + = dx dy 4π Gρ i, j 15

16 The discretized Poisson equation on equidistant Cartesian mesh It is convenient to express the discretized Poisson equation as follows a Φ + a Φ + a Φ + a Φ + a Φ = S 1 i+ 1, j i, j+ 1 i, j 4 i 1, j 5 i, j 1 i, j (1 dy dx dx dy dy dx a1 =, a =, a =, a4, a5, Si, j 4π Gρij dx dy dx dy + dy dx = = = dx dy Expanding equation (1 for each grid zone (i =1,, and j =1,,, we obtain ( a Φ + a Φ + a Φ + a Φ + a Φ = S, for i = 1, j = 1 1,1 1, 1,1 4 0,1 5 1,0 11 a1φ, + aφ 1, + aφ 1, + a4φ 0, + a5φ 1,1 = S1, for i = 1, j = a1φ, + aφ 1,4 + aφ 1, + a4φ 0, + a5φ 1, = S1, for i = 1, j =. a1φ 4, + aφ,4 + aφ, + a4φ, + a5φ, = S, for i =, j =

17 ( Rearranging system (, we obtain nine equations for nine unknown Φ i,j a Φ + a Φ + 0Φ + a Φ + 0Φ + 0Φ + 0Φ + 0Φ + 0Φ = S a Φ a Φ 1,1 1, 1, 1,1,,,1,, ,1 5 1,0 a Φ + a Φ + a Φ + 0Φ + a Φ + 0Φ + 0Φ + 0Φ + 0Φ = S a Φ 5 1,1 1, 1,,1 1,,,1,, 1, 4 0, 0Φ + a Φ + a Φ + 0Φ + 0Φ + a Φ + 0Φ + 0Φ + 0Φ = S a Φ a Φ 1,1 5 1, 1,,1, 1,,1,, 1, 1,4 4 0,. 0Φ + 0Φ + 0Φ + 0Φ + 0Φ + a Φ + 0Φ + a Φ + a Φ = S a Φ a Φ 1,1 1, 1,,1, 4,,1 5,,, 1 4,,4 Discretized Poisson equation turns into a system of equations which can be expressed in the matrix form A Φ S a,b a 0 a b 0 0 a 5 a,b a 0 a b 0 Φ 11 Φ 1 S 11,b S 1,b 0 a 5 a,b 0 0 a b a a,b a 0 a a 4 0 a 5 a a 0 a 1 0 Φ 1 Φ 1 Φ S 1,b S 1,b S,b 0 0 a 4 0 a 5 a,b 0 0 a 1 0 b 0 0 a a,b a 0 Φ Φ 1 S,b S 1,b 0 0 b 0 0 a 4 0 a 5 a,b a Φ S,b b 0 0 a 4 0 a 5 a,b Φ S,b 17

18 Boundary conditions The subscript b indicates that these values can be affected by the boundary conditions a,b a 0 a b 0 0 a 5 a,b a 0 a b 0 0 a 5 a,b 0 0 a b a a,b a 0 a a 4 0 a 5 a a 0 a a 4 0 a 5 a,b 0 0 a 1 0 b 0 0 a a,b a b 0 0 a 4 0 a 5 a,b a b 0 0 a 4 0 a 5 a,b Φ 11 Φ 1 Φ 1 Φ 1 Φ Φ Φ 1 Φ Φ S 11,b S 1,b S 1,b S 1,b S,b S,b S 1,b S,b S,b Consider the second equation of system ( a Φ + a Φ + a Φ + 0Φ + a Φ + 0Φ + 0Φ + 0Φ + 0Φ = S a Φ 5 1,1 1, 1,,1 1,,,1,, 1, 4 0, The value Φ 0, needs to be specified a priori because we have 9 equation for 9 unknowns and Φ 0, is already the 10 th unknown value.

19 Periodic boundary conditions Φ 0, = Φ, j= j= Φ 0, Φ, j=1 i=0 i=1 i= i= boundary layer With periodic boundary conditions, you wrap around the computational mesh and need not to invoke any special technique to find Φ 0,. Example: phi-angle in polar and cylindrical coordinates a Φ + a Φ + a Φ + 0Φ + a Φ + 0Φ + 0Φ + 0Φ + 0Φ = S a Φ 5 1,1 1, 1,,1 1,,,1,, 1, 4 0, a Φ + a Φ + a Φ + 0Φ + a Φ + 0Φ + 0Φ + a Φ + 0Φ = S 5 1,1 1, 1,,1 1,,,1 4,, 1, 19

20 Neumann boundary conditions Axis of symmetry j= The gradient of Φ across the boundary is zero Φ 0, Φ 1, j= Φ + r 0 1/, = 0 Φ = Φ 0, 1, j=1 i=0 i=1 i= i= boundary layer With Neumann boundary conditions, you reflect the computational mesh off the boundary and need not to invoke any special technique to find Φ 0,. Example: axis of symmetry, equatorial symmetry. a Φ + a Φ + a Φ + 0Φ + a Φ + 0Φ + 0Φ + 0Φ + 0Φ = S a Φ 5 1,1 1, 1,,1 1,,,1,, 1, 4 0, ( a Φ + a + a Φ + a Φ + 0Φ + a Φ + 0Φ + 0Φ + a Φ + 0Φ = S 5 1,1 4 1, 1,,1 1,,,1 4,, 1, 0

21 Dirichlet boundary conditions boundary layer j= j= Φ B Φ 4, j=1 r B θ Β i=1 i= i= i=4 In the case that the boundary is neither periodic or reflective, multipole expansion is often used to find the potential at the boundary. Example: outer boundaries of the computational domain (away from reflection surfaces or axes of symmetry. 1

22 The multipole expansion formulae (Jackson, Classical Electrodynamics Laplace equation in spherical coordinates (r, θ assuming axial symmetry, i.e. = 0 ϕ 1 1 ( r sinθ Φ Φ = Φ + 0 = r r r sinθ θ θ Seeking solution at the boundary in the form U( r Φ B = P ( θ r l ( l 1 B ( rb, θb + Al rb BlrB Pl (cos θb l= 0 Φ Φ = + P (cos θ The general solution is where l B are the Legendre polynomials, and r B and θ B are the spherical coordinates of boundary grid zones The Legendre polynomials can be calculated using the following recurrent formula

23 To find values of A l and B l, we note that A l must go to zero for r B inf (i.e. for outer boundaries and B l must go to zero for r B 0 (i.e. for boundaries near the coordinate origin l ( l 1 B ( rb, θb + Al rb BlrB Pl (cos θb l= 0 Φ Φ = + (4 This requirement splits equation (4 into the following two equations ( l+ 1 l ( rb r, θb Bl rb Pl (cos θb, ( rb r, θb Al rb Pl (cos θb l= 0 l= 0 Φ > = Φ < = where B l and A l are the so-called interior and exterior multipole moments B = ρ( r, θ r P (cos θ d r, A = ρ( r, θ r P (cos θ d l ( l+ 1 l l l l r In the case of B l, the integration (summation is performed over ALL grid zones with r < r B and in the case of A l over all grid zones with r > r B

24 z j= j= j=1 θ θ B r boundary layer r B Φ B For all grid zones INSIDE the circle, the following equations are to be used ( l+ 1 ( rb r, θb Bl rb Pl (cos θb l= 0 Φ > = B = ρ( r, θ r P (cos θ dv, l l l ij i, j For all grid zones OUTSIDE the circle, the following equations are to be used l B θb l B l θb l= 0 Φ ( r < r, = A r P (cos A = ρ r θ r P θ dv ( l+ 1 l = ρ (, θ l (cos θ ij i. j i=1 i= i= i=4 If we do not take into account the input from grid zones with r > r B, the series may diverge in the case when a substantial mass is located at those zones! In practice, one continues adding higher-order terms in the sum over l until the desired relative accuracy (~ is reached. Note that the accuracy with which you calculate the boundary values will directly affect the accuracy of the potential in the rest of your computational mesh due to the boundary-value nature of the Poisson equation. 4

25 General strategy for solving the Poisson equation Once the Poisson equation is properly discretized, the boundary conditions are specified, and the boundary values of the potential are found, one needs to choose the fastest method for solving the following matrix equation (i.e., a large system of linear equations A Φ = S Discretization and boundary conditions Equidistant grid with periodic boundary conditions Fast Fourier Transform Non-equidistant grid with any boundary conditions Alternative direction implicit method (best in D with axial symmetry Successive overrelaxation (best in D, slow in full D Multigrid methods (fast only on Cartesian geometry, slow on cylindrical and spherical geometries For details on each method see: 1. Press, Teukolsky, Vitterling, Flannery: Numerical Recipes in Fortran.. Bodenheimer, Laughlin, Rozyczka, Yorke: Numerical Methods in Astrophysics. 5

26 The Poisson Integral 6

27 Finding the gravitational potential using the Poisson integral Φ ( r = G V ρ ( r d r - r Let us consider a D Cartesian grid. Substituting the integral with a double sum, we obtain r Φ ( x, y l m = G N 1 N 1 l = 0 m = 0 M ( x, y l m ( x x + ( y y l l m m y m=n-1 Where M ( xl, ym is the mass contained in a grid zone ( l, m m= m=1 m=0 Note that no boundary values of Φ are involved in the summation. This is the fundamental advantage of the Poisson Integral over the Poisson equation. You need not to find the boundary values! l=0 l=1 l= l=n-1 x 7

28 Now let s assume that our computational grid is equidistant. ( x x = x ( l l l l then M Φ = = x ( l l + y ( m m N 1 N 1 N 1 N 1 l m l, m G G l l, m m M l m l = 0 m = 0 l = 0 m = 0 Gl l, m m = G g where 1 ( ( x l l + y m m is the gravitational potential in zone (l,m created by unit mass located in zone is often called the Green function. G l l, m m ( l, m Direct evaluation of the sum in equation (1 takes N operations, where N is the total number of grid zones A much faster way for evaluating the double sum is to use the convolution theorem 8

29 The convolution theorem Suppose, we need to calculate the following double sum N 1 N 1 A B C = l, m l, m l l, m m l ' = N m = N where B and C are periodic discrete functions with a period of N This sum can be calculated using the following three steps B k, n N 1 N 1 = l= N m= N B 1. Take direct Fourier transform of B and C lm e πikl πimn N N e. calculate the product of Fourier transforms B and C A ( N A =B C C kn kn kn. take the inverse Fourier Transform of A N 1 N 1 1 l, m = A k, n k= N n= N e N 1 N 1 = k, n k, n l= N m= N πikl πimn N N e C e πikl πimn N N e 9

30 Note the conceptual similarity between the two sums! The convolution sum N 1 N 1 A B C = l, m l, m l l, m m l ' = N m = N Doubling the computational domain N-1 Φ Our gravitational potential N 1 N 1 = M G l, m l m l l, m m l = 0 m = 0 The problem is that M and G are in general non-periodic and we have to make them periodic in order to use the convolution theorem We double the computational domain and fill zones,, and 4 with Ml m = 0 Φ N 1 N 1 = M G l, m l m l l, m m l = N m = N N Now M is periodic with a period of N. Because M = 0 in zones,,4, the gravitational potential in zone 1 is unaffected by the doubling. 0

31 Re-arranging the computational domain to make periodic G l l, m m l=-n l=- l=- l=1 l=0 l=1 l= l=n-1 m=n-1 m= m=1 1 m= m=-1 m=- m=- m=-n 4 We cannot simply set G to zero in zones,,4 because it is the inverse distance. Gl l, m m = G g 1 ( ( x l l + y m m 1

32 Singularity of the Green function At l = l, m = m, Φ is not defined! l, m M ( xl, ym Φ ( xl, ym = Gg = ( x x + ( y y l l m m However, it is possible to calculate the contribution of the material in the (l,m th cell to the potential in the same cell by assuming constant density within the cell and integrating over the cell area x within an individual cell (l,m and vary in the following limits y l m x x xl xl xl + y y ym ym ym + y y ( xl, ym y m x Σ = const x l x

33 defining x x = x and y y = y l l m m Φ ( x, y = G l m g M ( x, y l m ( x x + ( y y l l m m M ( x, y = Σ( x, y dx dy and noticing that, we obstain l m l m Φ ( x, y = G Σ( x, y l m g l m y x y x dx dy x + y after a few pages of algebra y Φ ( xl, ym = Gg Σ( xl, ym x sinh + y sinh x if l = l and m = m 1 1 x y

34 The extension to D Cartesian coordinates is straightforward, though the integral has to be taken numerically (I couldn t find an analytic solution Φ ( x, y, z = G ρ( x, y, z l m k g l m k z y x z y x dx dy dz x + y + z Problem with large memory requirements: the convolution method takes a lot of memory in D due to doubling of the computational grid. Some remedy: see Hockney and Eastwood: Computer simulations using particles and Bodenheimer, Laughlin, Rozyczka, Yorke: Numerical Methods in Astrophysics. The Fourier transform of the Green function has to be taken only once if the grid is not arbitrarily varying during simulations. This leaves us with FFTs each taking N log N operations where N is the total number of grid zones. On the other hand, the direct summation takes N operations and the Fourier transform technique is faster than the direct summation starting from N > 16. 4

35 Practicum Take a sphere with radius R and density profile of the form: where r is the radial distance from the center of the sphere ρ c =1 and r c =0. R Find the gravitational potential at r > R (outside the sphere using the multipole expansion formula. Split the sphere into N x N zones in spherical coordinates (r, θ. Compare the numerical solution with the analytic one: 5