Fundamental solution of Fokker - Planck equation

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1 Fundamental solution of Fokker - Planck equation Igor A. Tanski 2242, Ivanteevka Moscow region (Russia) Centralny projed, 4/69 tanski@protek.ru ZAO CVProtek ABSTRACT Fundamental solution of Fokker - Planck equation is built by means of the Fourier transform method. The result is checked by direct calculation. Keywords Fokker-Planck equation, fundamental solution, Fourier transform, exact solution We see from recent publications (ref. [-5]) that the Fokker - Planck equation is an object of steady investigations. It is a very useful modelling tool in many fields. Therefore it is desirable to have exact expression for fundamental solution. The object of our considerations is a special case of Fokker - Planck equation, which describes evolution of 3D continuum of non-interacting particles imbedded in a dense medium without outer forces. The interaction between particles and medium causes combined diffusion in physical space and velocities space. The only force, which acts on particles, is damping force proportional to velocity. where For this special case the Fokker - Planck equation is: n t + v n x x + v n y y + v n n = n(t, x, y,, v x, v y, v )-density; t -time variable; x, y, -space coordinates; v x, v y, v -velocities; -coefficient of damping; k -coefficient of diffusion. (v v x n) + x v y (v y n) + (v v n) = k 2 n v 2 x + 2 n v 2 y + 2 n. () We shall search solution of equation () for unlimited space < x < +, < y < +, < < +, (2) < v x < +, < v y < +, < v < +. v 2 Let us denote by n 0 initial density

2 -2- n 0 (x, y,, v x, v y, v ) = n(0, x, y,, v x, v y, v ). (3) We shall use the Fourier transform method. Let us denote by N Fourier transform of density (2π ) 6 and by N 0 -Fourier transform of initial density: where p x, p y, p -space coordinates momentum variables; q x, q y, q -velocities momentum variables. N = N(t, p x, p y, p, q x, q y, q ) = (4) exp( i(xp x + yp y + p + v x q x + v y q y + v q ))ndxdyddv x dv y dv. N 0 ( p x, p y, p, q x, q y, q ) = N(0, p x, p y, p, q x, q y, q ) (5) Multiplying () by exp( i(xp x + yp y + p + v x q x + v y q y + v q )) and integrating over whole space and velocities, we obtain N t + ( q x p x ) N + ( q q y p y ) N + ( q x q p ) N = k y q q2 x + q 2 y + q 2 N. (6) The equation (6) is a linear differential equation of first order, soitcan be solved by the method of characteristics. Relations on the characteristics are dt = dp x 0 = dp y 0 = dp 0 = dq x q x p x = dq y q y p y = dq q p = dn/n k (q 2 x + q 2 y + q 2 ). (7) First three equations (7) have three integrals: p x = const; (8) p y = const; p = const. If we combine (8) with next three equations (7), we get three further integrals ( q x p x ) e t = const; (9) ( q y p y ) e t = const; ( q p ) e t = const. We solve (9) for q x, q y, q and obtain q x = p x + e t q x0 p x ; (0) q y = p y + e t q y0 p y ; q = p + e t q 0 p.

3 -3- On the other hand, if we solve (9) for q x0, q y0, q 0,weobtain q x0 = p x + e t q x p x ; () q y0 = p y + e t q y p y ; q 0 = p + e t q p. To get the last integral of (7), we replace current velocities momentum variables q x, q y, q by initial velocities momentum variables q x0, q y0, q 0 in q 2 x + q 2 y + q 2 q 2 x + q 2 y + q 2 = 2 p2 x + p 2 y + p 2 + (2) + 2 e t p x(q x0 p s ) + p y(q y0 p s ) + p (q 0 p s ) + +e 2 t (q x0 p x )2 + (q y0 p y )2 + (q 0 p )2. Integrating (2) in t, weobtain the last integral ln(n) + k t ( 2 p2 x + p 2 y + p 2 ) + (3) e t p x(q x0 p x ) + p y(q y0 p y ) + p (q 0 p ) + + e2 t 2 (q x0 p x )2 + (q y0 p y )2 + (q 0 p )2 = const. We replace in (3) initial values of velocities momentum variables by their current values ln(n) + k t ( 2 p2 x + p 2 y + p 2 ) + (4) p x(q x p x ) + p y(q y p y ) + p (q p ) (q x p x )2 + (q y p y )2 + (q p )2 = const. To determine the constant term in (4), we write the same expression for initial values and equate both expressions ln(n) + k t ( 2 p2 x + p 2 y + p 2 ) + (5) p x(q x p x ) + p y(q y p y ) + p (q p ) +

4 (q x p x )2 + (q y p y )2 + (q p )2 = ln(n 0 ) + k 2 2 p x(q x0 p x ) + p y(q y0 p y ) + p (q 0 p ) (q x0 p x )2 + (q y0 p y )2 + (q 0 p )2. Solve (5) for N 0 N = N 0 p x, p y, p, p x + e t (q x p x ), p y + e t (q y p y ), p + e t (q p ) exp k t 2 ( p2 x + p 2 y + p 2 ) ( e t ) p x(q x p x ) + p y(q y p y ) + p (q p ) + (6) + 2 ( e 2 t ) (q x p x )2 + (q x p x )2 + (q x p x )2. The N 0 (... ) in the (6) means, that one have to calculate N 0 from initial density according to (5) and then replace values of its arguments by expressions (). Let us specify initial density value as product of delta functions n 0 = δ (x x 0 )δ (y y 0 )δ ( 0 )δ (v x x 0 )δ (v y x 0 )δ (v x 0 ). (7) The Fourier transform of initial density (7) is N 0 = (2π ) 6 exp( i(x 0 p x + y 0 p y + 0 p + v x0 q x + v y0 q y + v 0 q )). (8) Substituting () for N 0 arguments in (8) gives ˆN 0 = (2π ) 6 exp i x 0 p x + y 0 p y + 0 p + (9) p +v x x0 + e t (q x p x ) + v p y y0 + e t (q y p y ) + v p 0 + e t (q p ) or ˆN 0 = (2π ) exp 6 i p x x 0 + v x0 ( e t ) + p y y 0 + v y0 +e t (q x v x0 + q y v y0 + q v 0 ). ( e t ) + p 0 + v 0 ( e t ). (20)

5 -5- It is clear that (20) is the Fourier transform of (2) ˆn 0 = δ x (x 0 + v x0 ( e t )) δ y (y 0 + v y0 ( e t )) δ ( 0 + v 0 ( e t )) (2) δ v x v x0 e t δ v y v y0 e t δ v v 0 e t. [6]): To get inverse Fourier transform of n from its Fourier transform (6) we use two known results (ref.. A product of 2 functions A(ω )B(ω ) tranforms to convolution independent variables). a(x)*b(x) (where n is a number of (2π ) n 2. The Gaussian exponent of quadratic form e ω t A ω with matrix A transforms to exponent of quadratic form with inverse matrix π det(a) e In our case the matrix is 4 xt A x. t 2 A = k 2 ( 3 e t ) + 2 ( 3 e 2 t ) ( 2 e t ) 2 ( 2 e 2 t ) 2 ( e t ) 2 ( 2 e 2 t ). (22) 2 ( e 2 t ) The determinant is equal to Let us denote by D expression det(a) = k 2 t ( e 2 t ) 2( e t ) (23) D = det(a) k 2 = t ( e 2 t ) 2( e t ) (24) The inverse matrix is A = kd 2 ( e 2 t ) 2 ( e t ) ( e 2 t ) ( 2 e t ) + 2 ( 2 e 2 t ) t 2 2 ( 3 e t ) + 2 (. (25) 3 e 2 t ) Combining (2) with (25) we obtain expression for the fundamental solution: π G = (2π ) 6 k D 3 ˆn 0 *exp 4kD 2 ( e 2 t ) x2 + y (26) 2 ( 2 e t ) ( 2 e 2 t ) xv x + yv y + v + + t 2 2 ( 3 e t ) + 2 ( 3 e 2 t ) v2 x + v 2 y + v 2 ;

6 -6- where * means convolution of two functions. The convolution of arbitrary function with product of delta functions simplifies to substitution delta function arguments for this function arguments. Finally, weobtain π G = (2π ) 6 k D 3 exp 4kD 2 ( e 2 t ) ˆx2 + ŷ 2 + ẑ 2 (27) 2 ( 2 e t ) ( 2 e 2 t ) ˆx ˆv x + ŷ ˆv y + ẑ ˆv + + t 2 2 ( 3 e t ) + 2 ( 3 e 2 t ) ˆv2 x + ˆv 2 y + ˆv 2 ; where ˆx = x (x 0 + v x0 ( e t )); ŷ = y (y 0 + v y0 ( e t )); ẑ = ( 0 + v 0 ( e t )); (28) ˆv x = v x v x0 e t ;ˆv x = v y v y0 e t ;ˆv x = v v 0 e t. This is the fundamental solution of Fokker - Planck equation. Let us check validity of solution (27-28). Direct differentiation of (27-28) and substitution to () leads to cumbersome calculations. Therefore we use "semi-reverse" method. (27) has Gaussian form, so we search Gaussian solutions of (): n = exp A(t) x2 + y B(t) xv x + yv y + v + C(t) v2 x + v 2 y + v 2 + 3Q(t). (29) To get rid of exponents we write l = ln(n) = A(t) x2 + y B(t) xv x + yv y + v + C(t) v2 x + v 2 y + v 2 + 3Q(t). (30) l must satisfy equation l t + v x instead of equation () for n. l x + v y l y + v l v x l v x + v y l v y + v = k 2 l + 2 l + 2 l 2 v 2 x v 2 y v 2 + k l + 2 l + l 2 v x v x v x. l 3 = (3) v Substituting (29) for A, B, C, Q in (30) and collecting of similar terms leads to following equtions db dt da dt = 4k B 2 ; (32) = B A + 4k BC; (33)

7 -7- dc dt = 2 C 2 B + 4k C 2 ; (34) dq dt = + 2k C. (35) It is not easy to solve nonlinear system (32), but our task is simpler. Wehave only to check, that D(t) = t ( e 2 t ) 2( e t ) ; (36) A(t) = 4kD 2 ( e 2 t ) ; (37) B(t) = 4kD ( 2 e t ) + 2 ( 2 e 2 t ) ; (38) C(t) = t 4kD 2 2 ( 3 e t ) + 2 ( 3 e 2 t ) ; (39) satisfies equations (32-35). Direct substitution proves this. We proved validity of (27) for the special case Q(t) = ln(d(t)). (40) 2 x 0 = 0; y 0 = 0; 0 = 0; v x0 = 0; v y0 = 0; v 0 = 0. (4) For the common case we prove that differential operators x ; y ; ; (42) and ( e t ) x + e t v x ; ( e t ) y + e t ; v y ( e t ) + e t v ; (43) are symmetries of PDE (). This statement is trivial for (42) because () does not contain x explicitly. To prove the statement for (43) we build prolongation of differential operator according to Lie prolongation formula (ref. [7]) where u -aset of dependent variables; x i -aset of independent variables; D i -full derivation on x i operator; δ u x i = D i(δ u ) u x D k i(δ x k ); (44) δ u, δ x k -actions of infinitesimal symmetry operator on variables. We use this non-standard notation instead of usual ζ, ξ i to empasie their nature as small variations of variables. δ u -induced action of infinitesimal symmetry operator on derivatives. x i

8 -8- For operators (43) Lie formula gives ( e t ) x e t ( e t ) y e t ( e t ) e t + e t v x n x + n v x + e t v y n y + n v y n t ; (45) n t ; (46) + e t v n + n ; (47) v n t where n t = n t.for second order derivatives wehave δ (n uu) = 0, δ (n vv ) = 0, δ (n ww ) = 0. It is easy to calculate, that the action of (45-47) on () is identical ero. We proved, that differential operators (42-43) are symmetries of PDE (). The action of operators (42-43) on solutions (4) increases variables x 0, y 0... from ero to arbitrary values. In this way we get from (4) solutions (27-28). We checked the solution (27-28). DISCUSSION The main result of these considerations is the closed form expression for fundamental solution of Fokker - Planck equation without forces. This result can be helpful for modelling of more sofisticated problems. It demonstartes in simple and clear way the result of combined diffusion in physical space and velocities space and inertial motion. REFERENCES [] Stephen Jewson. Weather forecasts, Weather derivatives, Black-Scholes, Feynmann-Kac and Fokker- Planck. arxiv:physics/03225 v 20 Dec 2003 [2] George Krylov. On Solvable Potentials for One Dimensional Schrödinger and Fokker-Planck Equations. arxiv:quant-ph/ v 8 Dec 2002 [3] Alexei Loinski, Cédric Chauvière. A fast solver for Fokker-Planck equation applied to viscoelastic flows calculations: 2D FENE model, J. Comput. Phys., 89(2) (2003) [4] F. Benamira and L. Guechi. Similarity transformations approach for a generalied Fokker-Planck equation. arxiv:physics/02002 v 30 Nov 200 [5] Luís M. A. Bettencourt. Properties of the Langevin and Fokker-Planck equations for scalar fields and their application to the dynamics of second order phase transitions. arxiv:hep-ph/ v 25 May 2000 [6] W. Feller, An Introduction to Probability Theory and Its Applications. Volume II, John Wiley and Sons, 97. [7] Peter J. Olver, Applications of Lie groups to differential equations. Springer-Verlag, New York, 986.