Bindel, Spring 2014 Applications of Parallel Computers (CS 5220)

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1 Bndel, Sprng 2014 Applcatons of Parallel Computers (CS 5220) 1 Dervng SPH The Naver-Stokes equatons wth gravty are ρa = p + µ 2 v + ρg. The acceleraton s the materal dervatve of velocty, and we usually take an Euleran perspectve and wrte ths as a = Dv Dt = v t + v v. In smoothed partcle hydrodynamcs, though, we take a Lagrangan perspectve, and actually assocate computatonal partcles wth materal ponts. Ths makes t easy to deal wth the left-hand sde of the Naver-Stokes equaton. To compute the spatal dervatves on the rght hand sde of the equaton, we nterpolate pressures and veloctes at the materal partcles to get smoothed felds (hence the name). Then we dfferentate the smoothed felds. For example, suppose we care about some scalar feld A(r). Each partcle j has a mass, a locaton r j, and a value A j = A(r j ). Between partcles, we wrte (1) A S (r) = j A j W (r r j, h), where W s a smoothng kernel wth radus h. The denstes that appear n (1) are themselves are computed usng (1): ρ = ρ S (r ) = j W (r r j, h) = j W (r r j, h). Puttng everythng together, the SPH approxmaton computes feld quanttes at locatons assocated wth computatonal partcles. The governng equatons for the partcles and the assocated quanttes are then (2) (3) ρ a = f pressure f pressure = j + f vscosty + ρ g p + p j 2 W (r r j, h) (4) f vscosty = µ j v j v 2 W (r r j, h), where the pressure and vscous nteracton terms have been symmetrzed to ensure that partcle acts on partcle j n the same way j acts on. To compute the pressure, we use the deal gas equaton of state (5) p = k(ρ ρ 0 ). Of course, ths s not the rght equaton of state for a lqud! Ths equaton s best regarded as a non-physcal approxmaton that s legtmate as long as the artfcal speed of sound s much greater than the veloctes of nterest n the problem (as s the case wth the ncompressble approxmaton that s more commonly used n other settngs).

2 Bndel, Sprng 2014 Applcatons of Parallel Computers (CS 5220) 2 Smoothng kernels One of the key numercal decsons n SPH s the choce of kernels used to nterpolate the felds. We follow the strategy descrbed by Müller, Charypar, and Gross n Partcle-based flud smulaton for nteractve applcatons. In ths paper, the authors use three radally symmetrc dfferent kernels for 3D smulaton, each wth the form { W (r, h) = 1 f(q), 0 q 1 Ch d 0, otherwse where q = r/h = r /h and d = 3 s the dmenson. The kernels are based on the choces f poly6 (q) = (1 q 2 ) 3 for general nterpolaton, f spky (q) = (1 q) 3 for nterpolatng pressures, and f vscosty (q) = q 3 /2 + q 2 + q 1 /2 1 for vscosty computatons. The gradents are gven by and the Laplacans are 2 W (r, h) = W (r, h) = r Ch d+2 { q 1 f (q), 0 q 1 0, otherwse { 1 f (q) + (d 1)q 1 f (q), 0 q 1 Ch d+2 0, otherwse The pressure kernel s desgned wth relatvely steep gradents close to the orgn to prevent the clusterng of computatonal partcles that occurs when pressures are nterpolated wth W poly6. The vscosty kernel s desgned so that the Laplacan wll be postve defnte, ensurng that we don t accdentally get negatve vscous contrbutons that add energy to the system (and compromse stablty). 3 Condensed nteracton force expressons Makng thngs completely explct for the cases we care about most, we have (for 0 q 1) (6) (7) (8) (9) (10) W poly6 (r, h) = πh 3 (1 q2 ) 3 W poly6 (r, h) = πh 5 (1 q2 ) 2 r 2 W poly6 (r, h) = πh 5 (1 q2 )(7q 2 3) W spky (r, h) = 45 (1 q) 2 πh 5 r q 2 W vscosty (r, h) = 45 (1 q) πh5

3 Bndel, Sprng 2014 Applcatons of Parallel Computers (CS 5220) If we substtute (9), (10), and the equaton of state (5) nto (3) and (4), we have f pressure = 45k πh 5 f vscosty = 45µ πh 5 ρ + 2ρ 0 (1 q j ) 2 2 q j j N v v j (1 q j ) j N where N s the set of partcles wthn h of partcle and q j = r j /h, r j = r r j. Puttng these terms together, we have = where f pressure + f vscosty fj nteract = 45 [ k πh 5 (1 q j ) and v j = v v j. We then rewrte (2) as 4 Santy checks a = 1 ρ fj nteract j N r j 2 (ρ + 2ρ 0 ) (1 q j) r j µv j q j j + g. j N f nteract I farly regularly make typographcal and copyng errors when I do algebra and mplement t n code. In order to stay sane when I actually wrte somethng somewhat complcated, I fnd t helpful to put together lttle test scrpts to check my work numercally. For your edfcaton, n ths secton I gve my MATLAB test scrpt correspondng to the dervaton n these notes. The test scrpt s done n MATLAB. I begn by mplementng the functons f(q), the normalzng constants, and the kernel functons for each of the three kernels. fp6 (1-q.^2).^3; fsp (1-q).^3; fv q.^2-0.5*q.^ /q - 1; Cp6 = 64*p/315; Csp = p/15; Cv = 2*p/15; Wp6 1/Cp6/h^3 * fp6( norm(r)/h ); Wsp 1/Csp/h^3 * fsp( norm(r)/h ); Wv 1/Cv/h^3 * fv( norm(r)/h ); ], I computed the normalzaton constants analytcally, but I m prone to algebra mstakes when I compute ntegrals by hand. Let s check aganst MATLAB s quad functon.

4 Bndel, Sprng 2014 Applcatons of Parallel Computers (CS 5220) fprntf( Relerr for normalzaton constants:\n ); nerr_p6 = 4*p*q.^2.*fp6(q)/Cp6, 0, 1 ) - 1; nerr_sp = 4*p*q.^2.*fsp(q)/Csp, 0, 1 ) - 1; nerr_v = 4*p*q.^2.*fv(q)/Cv, 1e-12, 1 ) - 1; fprntf( Cp6: %g\n, nerr_p6); fprntf( Csp: %g\n, nerr_sp); fprntf( Cv: %g\n, nerr_v); Now check that I dd the calculus rght for the gradent and Laplacan of the W poly6 kernel, the gradent of the pressure kernel, and the Laplacan of the vscosty kernel h = rand(1); r = rand(3,1)*h/4; q = norm(r)/h; r2 = r *r; h2 = h^2; dr = norm(r)*1e-4; gwp6_fd = fd_grad(@(r) Wp6(r,h), r, dr); gwsp_fd = fd_grad(@(r) Wsp(r,h), r,dr); lwp6_fd = fd_laplace(@(r) Wp6(r,h), r, dr); lwv_fd = fd_laplace(@(r) Wv(r,h), r, dr); gwp6_ex = -(945/32/p)/h^5 *(1-q^2)^2 * r; gwsp_ex = -45/p/h^5*(1-q)^2/q * r; lwp6_ex = (945/32/p)/h^5 * (1-q^2)*(7*q^2-3); lwv_ex = 45/p/h^5 * (1-q); fprntf( Check kernel dervatves:\n ); fprntf( grad Wp6: %g\n, norm(gwp6_fd-gwp6_ex)/norm(gwp6_ex)); fprntf( grad Wsp: %g\n, norm(gwsp_fd-gwsp_ex)/norm(gwsp_ex)); fprntf( lapl Wp6: %g\n, (lwp6_fd-lwp6_ex)/lwp6_ex); fprntf( lapl Wv: %g\n, (lwv_fd-lwv_ex)/lwv_ex); Now check that f vscosty (q) satsfes the boundary condtons f(1) = 0 f (1) = 0 The frst two condtons we check drectly. dq = 1e-4; fprntf( Relerr for vscosty kernel checks:\n );

5 Bndel, Sprng 2014 Applcatons of Parallel Computers (CS 5220) fprntf( fv (1): %g\n, fv(1) ); fprntf( dfv(1): %g\n, fd_derv(fv,1,dq) ); Now, let me check that I dd the algebra rght n gettng the condensed formula for the nteracton forces. % Set up random parameter choces r_j = rand(3,1); v_j = rand(3,1); k = rand(1); rho0 = rand(1); rho = rand(1); rhoj = rand(1); mass = rand(1); mu = rand(1); q = norm(r_j)/h; % Compute pressures va equaton of state P = k*(rho-rho0); Pj = k*(rhoj-rho0); % Dfferentate the kernels Wsp_x = -45/p/h^5*(1-q)^2/q * r_j; LWv = 45/p/h^5*(1-q); % Do the straghtforward computaton fpressure = -mass*(p+pj)/2/rhoj * Wsp_x; fvscous = -mu*mass*v_j/rhoj * LWv; fnteract1 = fpressure + fvscous; % Do the computaton based on my condensed formula fnteract2 = 45*mass/p/h^5/rhoj * (1-q) *... ( k/2*(rho+rhoj-2*rho0)*(1-q)/q * r_j - mu * v_j ); % Compare fprntf( Relerr n nteracton force check:\n ); fprntf( fnt: %g\n, norm(fnteract1-fnteract2)/norm(fnteract1)); Of course, all the above s supported by a number of lttle second-order accurate fnte dfference calculatons. functon fp = fd_derv(f,r,h)

6 Bndel, Sprng 2014 Applcatons of Parallel Computers (CS 5220) fp = ( f(r+h)-f(r-h) )/2/h; functon fpp = fd_derv2(f,r,h) fpp = ( f(r+h)-2*f(r)+f(r-h) )/h/h; functon del2f = fd_laplace_radal(f,r,h) del2f = fd_derv2(f,r,h) + 2*fd_derv(f,r,h)/r; functon del2f = fd_laplace(f,r,h) e1 = [1; 0; 0]; e2 = [0; 1; 0]; e3 = [0; 0; 1]; del2f = (-6*f(r)+... f(r+h*e1)+f(r+h*e2)+f(r+h*e3)+... f(r-h*e1)+f(r-h*e2)+f(r-h*e3) )/h/h; functon gradf = fd_grad(f,r,h) e1 = [1; 0; 0]; e2 = [0; 1; 0]; e3 = [0; 0; 1]; gradf = [f(r+h*e1)-f(r-h*e1); f(r+h*e2)-f(r-h*e2); f(r+h*e3)-f(r-h*e3)] / 2 / h;