Fundamental solution of degenerated Fokker - Planck equation

The Fokker -Planck equation 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.F undamental solution of Fokker -Planck equation is built in our work [1].

In this paper we consider the special case of the Fokker -Planck equation with zero damping force. We call it degenerated Fokker -Planck equation. Forthis special case the equation reads:

where n = n(t, x, y, z, v x, v y, v z )-density; t -time variable;

x, y, z -space coordinates; v x, v y, v z -velocities;

k -coefficient of diffusion.

We shall search solution of equation ( 1) for unlimited space

Let us denote by n 0 initial density

We shall use the Fourier transform method. Let us denote by N Fourier transform of density N = N (t, p x, p y, p z, q x, q y, q z) = (4)

∫ exp(-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 -Fourier 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 -space coordinates momentum variables; q x, q y, q z -velocities momentum variables.

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 overw hole space and velocities, we obtain

The equation ( 6) is a linear differential equation of first order,s oi tc an be solved by the method of characteristics. Relations on the characteristics are

First three equations (7) have three integrals:

If we combine (8) with next three equations (7), we get three further integrals q x + p x t = const;( 9)

We solve(9) for q x , q y , q z and obtain q x = q x0p x t;( 10) q y = q y0p y t; q z = q z0p z t.

On the other hand, if we solve(9) for q x0 , q y0 , q z0 ,weobtain q x0 = q x + p x t;( 11) q y0 = q y + p y t; q z0 = q z + p z t.

To get the last integral of (7), we replace current velocities momentum variables q x , q y , q z by initial velocities momentum variables q x0 , q y0 , q z0 in q 2 x + q 2 y + q

Integrating (12) in t,weobtain the last integral

We replace in (13) initial values of velocities momentum variables by their current values

To determine the constant term in (14), we write the same expression for initial values and equate both expressions

Solve(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)

The N 0 (. . . ) in the ( 16) means, that one have tocalculate N 0 from initial density according to (5) and then replace values of its arguments by expressions (11).

The Fourier transform of initial density (17) is

Substituting (11) for N 0 arguments in (18) gives

It is clear that (20) is the Fourier transform of (21

To get inverse Fourier transform of n from its Fourier transform (16) we use twoknown results ([1]):

1. A product of 2 functions A(ω )B(ω )transforms to convolution 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 inverse matrix

x t A -1 x .

In our case the matrix is (see ( 16))

The determinant is equal to

Let us denote by D expression

The inverse matrix is

Combining ( 21) with (25) we obtain expression for the fundamental solution:

where * means convolution of twofunctions.

The convolution of arbitrary function with product of delta functions simplifies to substitution delta function arguments for this function arguments. Finally,weobtain

This is the fundamental solution of the degenerated Fokker -Planck equation.

Let us check validity of solution (27-28). Direct differentiation of (27-28) and substitution to (1) leads to cumbersome calculations. Therefore we use “semi-reverse” method. (27) has Gaussian form, so we search Gaussian solutions of (1):

To get rid of exponents we write

l must satisfy equation

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 = 4kB 2 ;( 32)

It is not easy to solvenonlinear system (32-35), but our task is simpler.W ehav e only to check, that

satisfies both equations (32-35). This result is obvious.

We provedvalidity of (27) for the special case

Forthe common case we prove that differential operators (see [ 43) we build prolongation of differential operator according to Lie prolongation formula (ref. [3])

where u α -aset of dependent variables;

x i -aset of independent variables; D i -full derivation on x i operator; δ u α , δ x k -a ctions of infinitesimal symmetry operator on variables. Weu se this non-standard notation instead of usual ζ α , ξ i to emphasize their nature as small variations of variables. where n t = ∂n ∂t .For second order derivativeswehav e δ (n uu ) = 0, δ (n vv ) = 0, δ (n ww ) = 0(see [3]).

It is easy to calculate, that the action of (45-47) on ( 1) is identical zero.

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