Fundamental solution of degenerated Fokker - Planck equation

Fundamental solution of degenerated F okker - Planck equation Igor A. T anski Moscow ,R ussia tanski.igor .arxiv@gmail.com ABSTRA CT Fundamental solution of degenerated F okker - Planck equation is b uilt by means of the Fourier transform method. The result is check ed by direct calculation. Ke ywords Fo kker-Planck equation, fundamental solution, F ourier transform, exact solution The object of our considerations is a special case of Fokk er - Planck equation. The Fokker - Planck equation describes e vo lution of 3D continuum of non-interacting particles imbedded in a dense medium without outer forces. The interaction between particles and medium causes combined dif fusion in ph ysical space and velocities space. The only force, which acts on particles, is damp- ing force proportional to velocity .F undamental solution of Fokk er - Planck equation is b uilt in our work [1]. In this paper we consider the special case of the Fokk er - Planck equation with zero damping force. We c all it degenerated F okker - Planck equation. Fo rt his special case the equation reads: ∂ n ∂ t + v x ∂ n ∂ x + v y ∂ n ∂ y + v z ∂ n ∂ z = k   ∂ 2 n ∂ v 2 x + ∂ 2 n ∂ v 2 y + ∂ 2 n ∂ v 2 z   .( 1) where n = n ( t , x , y , z , v x , v y , v z )-d ensity; t -t ime v ariable; x , y , z -s pace coordinates; v x , v y , v z -v elocities; k -c oefficient of dif fusion. We s hall search solution of equation (1) for unlimited space − ∞ < x < + ∞ , − ∞ < y < + ∞ , − ∞ < z < + ∞ ,( 2) − ∞ < v x < + ∞ , − ∞ < v y < + ∞ , − ∞ < v z < + ∞ . -2- Let us denote by n 0 initial density n 0 ( x , y , z , v x , v y , v z ) = n (0, x , y , z , v x , v y , v z ). (3) We s hall use the Fourier transform method. Let us denote by N Fo urier transform of density N = N ( t , p x , p y , p z , q x , q y , q z ) = (4) 1 (2 π ) 6 ∞ − ∞ ∫ ∞ − ∞ ∫ ∞ − ∞ ∫ ∞ − ∞ ∫ ∞ − ∞ ∫ ∞ − ∞ ∫ ex p( − i ( xp x + yp y + zp z + v x q x + v y q y + v z q z )) ndxdydzdv x dv y dv z . and by N 0 -F ourier transform of initial density: N 0 ( p x , p y , p z , q x , q y , q z ) = N (0, p x , p y , p z , q x , q y , q z )( 5) where p x , p y , p z -s pace coordinates momentum variables; q x , q y , q z -v elocities momentum v ariables. Multiplying (1) by exp( − i ( xp x + yp y + zp z + v x q x + v y q y + v z q z )) and integrating o ve rw hole space and velocities, we obtain ∂ N ∂ t − p x ∂ N ∂ q x − p y ∂ N ∂ q y − p z ∂ N ∂ q z =− k   q 2 x + q 2 y + q 2 z   N .( 6) The equation (6) is a linear dif ferential equation of first order ,s oi tc an be solv ed by the method of characteristics. Relations on the characteristics are dt = dp x 0 = dp y 0 = dp z 0 = − dq x p x = − dq y p y = − dq z p z =− dN / N k ( q 2 x + q 2 y + q 2 z ) .( 7) First three equations (7) ha ve t hree integrals: p x = const ;( 8) p y = const ; p z = const . If we combine (8) with next three equations (7), we get three further inte grals q x + p x t = const ;( 9) q y + p y t = const ; q z + p z t = const . We s olv e( 9) for q x , q y , q z and obtain q x = q x 0 − p x t ;( 10) q y = q y 0 − p y t ; q z = q z 0 − p z t . On the other hand, if we solv e( 9) for q x 0 , q y 0 , q z 0 ,w eo btain -3- q x 0 = q x + p x t ;( 11) q y 0 = q y + p y t ; q z 0 = q z + p z t . To g et the last integral of (7), we replace current v elocities momentum v ariables q x , q y , q z by initial ve loci- ties momentum variables q x 0 , q y 0 , q z 0 in q 2 x + q 2 y + q 2 z q 2 x + q 2 y + q 2 z = q 2 x 0 + q 2 y 0 + q 2 z 0 − 2 t   q x 0 p x + q y 0 p y + q z 0 p z   + t 2   p 2 x + p 2 y + p 2 z   .( 12) Integrating (12) in t ,w eo btain the last integral ln( N ) + k    t   q 2 x 0 + q 2 y 0 + q 2 z 0   − t 2   q x 0 p x + q y 0 p y + q z 0 p z   + 1 3 t 3   p 2 x + p 2 y + p 2 z      = const .( 13) We r eplace in (13) initial v alues of velocities momentum v ariables by their current v alues ln( N ) + k    t   q 2 x + q 2 y + q 2 z   + t 2   q x p x + q y p y + q z p z   + 1 3 t 3   p 2 x + p 2 y + p 2 z      = const .( 14) To d etermine the constant term in (14), we write the same e xpression for initial v alues and equate both ex pressions ln( N ) + k    t   q 2 x + q 2 y + q 2 z   + t 2   q x p x + q y p y + q z p z   + 1 3 t 3   p 2 x + p 2 y + p 2 z      = ln( N 0 ). (15) Solv e( 15) for N N = N 0   p x , p y , p z , q x + p x t , q y + p y t , q z + p z t   × (16) × ex p    − k    t   q 2 x + q 2 y + q 2 z   + t 2   q x p x + q y p y + q z p z   + 1 3 t 3   p 2 x + p 2 y + p 2 z         The N 0 (. . . ) in the (16) means, that one ha ve t oc alculate N 0 from initial density according to (5) and then replace v alues of its arguments by e xpressions (11). Let us specify initial density v alue as product of delta functions n 0 = δ ( x − x 0 ) δ ( y − y 0 ) δ ( z − z 0 ) δ ( v x − x 0 ) δ ( v y − x 0 ) δ ( v z − x 0 ). (17) The Fourier transform of initial density (17) is N 0 = 1 (2 π ) 6 ex p    − i   x 0 p x + y 0 p y + z 0 p z + v x 0 q x + v y 0 q y + v z 0 q z      .( 18) -4- Substituting (11) for N 0 arguments in (18) gi ve s ˆ N 0 = 1 (2 π ) 6 ex p    − i   x 0 p x + y 0 p y + z 0 p z + v x 0 ( q x + p x t ) + v y 0 ( q y + p y t ) + v z 0 ( q z + p z t )      (19) or ˆ N 0 = 1 (2 π ) 6 ex p    − i   p x ( x 0 + v x 0 t ) + p y ( y 0 + v y 0 t ) + p z ( z 0 + v z 0 t ) + ( q x v x 0 + q y v y 0 + q z v z 0 )      .( 20) It is clear that (20) is the Fourier transform of (21) ˆ n 0 = δ ( x − ( x 0 + v x 0 t )) δ ( y − ( y 0 + v y 0 t )) δ ( z − ( z 0 + v z 0 t )) δ ( v x − v x 0 ) δ ( v y − v y 0 ) δ ( v z − v z 0 ). (21) To g et in ve rse Fourier transform of n from its Fourier transform (16) we use tw ok nown results ([1]): 1. A product of 2 functions A ( ω ˙ ) B ( ω )t ransforms to con vo lution a ( x )* b ( x ). 2. The Gaussian exponent of quadratic form e − ω t A ω with matrix A transforms to exponent of quadratic form with in ve rse matrix   π det ( A )   1/2 e − 1 4 x t A − 1 x . In our case the matrix is (see (16)) A = k    t 3 /3 t 2 /2 t 2 /2 t    .( 22) The determinant is equal to det ( A ) = 1 12 k 2 t 4 .( 23) Let us denote by D ex pression D = det ( A ) k 2 = t 4 12 .( 24) The in ve rse matrix is A − 1 = 1 kD    t − t 2 /2 − t 2 /2 t 3 /3    .( 25) Combining (21) with (25) we obtain expression for the fundamental solution: G =   π k √  D   3 ˆ n 0 *e xp    − 1 4 kD    t   x 2 + y 2 + z 2   − t 2   xv x + yv y + zv z   + t 3 3   v 2 x + v 2 y + v 2 z         ;( 26) where * means con vo lution of tw of unctions. The con vo lution of arbitrary function with product of delta functions simplifies to substitution delta function arguments for this function ar guments. Finally ,w eo btain -5- G =   π k √  D   3 ex p    − 1 4 kD    t   ˆ x 2 + ˆ y 2 + ˆ z 2   − t 2   ˆ x ˆ v x + ˆ y ˆ v y + ˆ z ˆ v z   + t 3 3   ˆ v 2 x + ˆ v 2 y + ˆ v 2 z         ;( 27) where ˆ x = x − ( x 0 + tv x 0 ); ˆ y = y − ( y 0 + tv y 0 ); ˆ z = z − ( z 0 + tv z 0 ); (28) ˆ v x = v x − v x 0 ;ˆ v x = v y − v y 0 ;ˆ v x = v z − v z 0 . This is the fundamental solution of the degenerated F okker - Planck equation. Let us check v alidity of solution (27-28). Direct dif ferentiation of (27-28) and substitution to (1) leads to cumbersome calculations. Therefore we use "semi-re ve rse" method. (27) has Gaussian form, so we search Gaussian solutions of (1): n = ex p    A ( t )   x 2 + y 2 + z 2   + 2 B ( t )   xv x + yv y + zv z   + C ( t )   v 2 x + v 2 y + v 2 z   + 3 Q ( t )    .( 29) To g et rid of exponents we write l = ln ( n ) = A ( t )   x 2 + y 2 + z 2   + 2 B ( t )   xv x + yv y + zv z   + C ( t )   v 2 x + v 2 y + v 2 z   + 3 Q ( t ). (30) l must satisfy equation ∂ l ∂ t + v x ∂ l ∂ x + v y ∂ l ∂ y + v z ∂ l ∂ z = k   ∂ 2 l ∂ v 2 x + ∂ 2 l ∂ v 2 y + ∂ 2 l ∂ v 2 z   + k      ∂ l ∂ v x   2 +   ∂ l ∂ v x   2 +   ∂ l ∂ v x   2    .( 31) instead of equation (1) for n . Substituting (29) for A , B , C , Q in (30) and collecting of similar terms leads to following equations dA dt = 4 kB 2 ;( 32) dB dt = 4 kB C − A ;( 33) dC dt = 4 kC 2 − 2 B ;( 34) dQ dt = 2 kC .( 35) It is not easy to solv en onlinear system (32-35), but our task is simpler .W eh av e only to check, that D ( t ) = t 4 12 ;( 36) A ( t ) =   − 1 4 kD   t =− 3 k t − 3 ;( 37) -6- B ( t ) =   1 4 kD   1 2 t 2 = 3 2 k t − 2 ;( 38) C ( t ) =   − 1 4 kD   1 3 t 3 =− 1 k t − 1 ;( 39) Q ( t ) =− 2l n( t ); (40) satisfies both equations (32-35). This result is obvious. We p ro ve dv alidity of (27) for the special case x 0 = 0; y 0 = 0; z 0 = 0; v x 0 = 0; v y 0 = 0; v z 0 = 0. (41) Fo rt he common case we pro ve t hat dif ferential operators (see [2]) ∂ ∂ x ; ∂ ∂ y ; ∂ ∂ z ;( 42) and t ∂ ∂ x + ∂ ∂ v x ; t ∂ ∂ y + ∂ ∂ v y ; t ∂ ∂ z + ∂ ∂ v z ;( 43) are symmetries of PDE (1). This statement is tri vial for (43) because (1) does not contain x ex plicitly .T o pro ve t he statement for (43) we build prolong ation of differential operator according to Lie prolong ation formula (ref. [3]) δ   ∂ u α ∂ x i   = D i ( δ u α ) − ∂ u α ∂ x k D i ( δ x k ); (44) where u α -as et of dependent variables; x i -as et of independent variables; D i -f ull deri va tion on x i operator; δ u α , δ x k -a ctions of infinitesimal symmetry operator on v ariables. W eu se this non-standard notation instead of usual ζ α , ξ i to emphasize their nature as small v ariations of variables. δ   ∂ u α ∂ x i   -i nduced action of infinitesimal symmetry operator on deri va ti ve s. Fo ro perators (43) Lie formula gi ve s t ∂ ∂ x + ∂ ∂ v x − ∂ n ∂ x ∂ ∂ n t ;( 45) t ∂ ∂ y + ∂ ∂ v y − ∂ n ∂ y ∂ ∂ n t ;( 46) t ∂ ∂ z + ∂ ∂ v z − ∂ n ∂ z ∂ ∂ n t ;( 47) where n t = ∂ n ∂ t .F or second order deri va ti ve sw eh av e δ ( n uu ) = 0, δ ( n vv ) = 0, δ ( n ww ) = 0( see [3]). -7- It is easy to calculate, that the action of (45-47) on (1) is identical zero. We p ro ve d, that dif ferential operators (42-43) are symmetries of PDE (1). The action of operators (42-43) on solutions (41) increases variables x 0, y 0 ... from zero to arbitrary va lues. In this way we get from (41) solutions (27-28). We c hecked the solution (27-28). DISCUSSION The main result of these considerations is the closed form e xpression for fundamental solution of the de generated F okker - Planck equation without forces. To o btain this result we could not simply substitute α = 0t op re vious formulae. Some calculations are necessary .T heir result is e xpression for fundamental solution. This expression ha ve t he same Gaussian form, as the fundamental solution of full equation. Its coef ficient one could get by passage to the limit α → 0, using first terms of exponent T aylor expansion. All coef ficients are polynomials on time variable, no e xponential saturation effect is present. Fundamental solution of the full Fokk er - Planck equation contains time t only as product α t with damping coef ficient. Therefore the limit α → 0a lso depicts the limit t → 0b ehavior of the fundamental solution. REFERENCES [1] Igor A. T anski. Fundamental solution of Fokker - Planck equation. arXi v:nlin.SI/0407007 v1 4 Jul 2004 [2] Peter J. Olver ,A pplications of Lie groups to dif ferential equations. Springer-V erlag, Ne wY ork, 1986. [3] Igor A. T anski. The symmetries of the Fokker - Planck equation in three dimensions. arXi v:nlin.CD/0501017 v1 8 Jan 2005