The symmetries of the Fokker - Planck equation in two dimensions Igor A. Tanski
[email protected] ZAO CV Protek
ABSTRACT We calculate all point symmetries of the Fokker - Planck equation in two-dimensional Euclidean space. General expression of symmetry group action on arbitrary solution of Fokker - Planck equation is presented.
1. The symmetries of the Fokker - Planck equation in two dimensions The object of our considerations is a special case of Fokker - Planck equation, which describes evolution of 2D 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. The 3D variant of this equation was investigated in our work [1]. In this work fundamental solution of 3D equation was obtained by means of Fourier transform. The 1D variant of this equation was investigated in our work [2]. All point symmetries of the Fokker - Planck equation in one-dimensional Euclidean space were calculated. In present work we continue this investigation for more complex 2D equation. The Fokker - Planck equation in two dimensions is 2 2 ∂n ∂n ∂n ∂n ∂n ∂ n ∂ n +u +v − au − av − 2an − k + 2 = 0; 2 ∂u ∂x ∂y ∂u ∂v ∂v ∂t
(1)
where n = n(t, x, y, u, v) - density; t - time variable; x, y - space coordinates; u, v - velocity; a - coefficient of damping; k - coefficient of diffusion. The list of symmetries of the Fokker - Planck equation in one dimension follows. The calculations of symmetries are rather awkward. They are carried out to APPENDIX 1. Instead of classic "ξ − φ " notation we use another ("δ ") notation. This notation was presented in our work [2]. Addition of arbitrary solution v1 = A
∂ ; ∂n
(2)
-2-
where A is arbitrary solution of the (1) equation. Scaling of density v2 = n
∂ ; ∂n
(3)
The reason of symmetries (2-3) existence is linearity of PDE (1). Time shift ∂ ; ∂t
(4)
∂ ∂ ; v5 = . ∂x ∂y
(5)
v3 = Space translations v4 = Space rotation v6 = v
∂ ∂ ∂ ∂ −u +y −x . ∂u ∂v ∂x ∂y
(6)
Transformations (5) and (6) build two-dimensional Euclidean movements group. Extended Galilean transformations, which besides time and space coordinates affect the density v7 =
∂ an ∂ ∂ ∂ an ∂ ∂ +t − (ax + u) ; v8 = +t − (ax + v) . ∂x 2k ∂n ∂v ∂x 2k ∂n ∂u
(7)
Negative exponent transformations - they affect time and space coordinates, contain time-dependent common multiplier (negative exponent). They do not affect density. ∂ ∂ ∂ ∂ v9 = e−at −a + + ; v = e−at −a . ∂u ∂x 10 ∂v ∂y
(8)
Positive exponent transformations - they affect time, space and density, contain time-dependent common multiplier (positive exponent). ∂ a2 ∂ ∂ a2 ∂ ∂ ∂ + − nu + − nv ; v12 = e at a . v11 = e at a ∂v ∂y ∂u ∂x k ∂n k ∂n
(9)
One-parameter groups, generated by vector fields v1 − v12 , are enumerated in the following list. The list contains images of the point (n, t, x, y, u, v) by transformation exp(ε vi ) G 1:
(n + ε A, t, x, y, u, v);
G 2:
(eε n, t, x, y, u, v);
G 3:
(n, t + ε , x, y, u, v);
G 4:
(n, t, x + ε , y, u, v);
G 5:
(n, t, x, y + ε , u, v);
G 6:
(n, t, cos(ε )x + sin(ε )y, − sin(ε )x + cos(ε )y, cos(ε )u + sin(ε )v, − sin(ε )u + cos(ε )v);
G 7:
a 1 (exp − ε (ax + u) + ε 2 (at + 1) n, t, x + ε t, y, u + ε , v); 2 2k
(10)
-3-
G 8:
a 1 ε (ay + v) + ε 2 (at + 1) n, t, x, y + ε t, u, v + ε ); (exp − 2 2k
G 9:
(n, t, x + ε e−at , y, u − ε a e−at , v);
G 10:
(n, t, x, y + ε e−at , u, v − ε a e−at );
G 11:
a2 at 1 e ε u + ε 2 a e at , t, x + ε e at , y, u + ε a e at , v). (n exp − 2 k
G 12:
a2 at 1 (n exp − e ε v + ε 2 a e at , t, x, y + ε e at , u, v + ε a e at ). 2 k For relatively nontrivial integration of G 7 , G 8 , G 11 and G 12 we refer to [2].
The fact, that G i are symmetries of PDE (1) means, that if f (t, x, y, u, v) is arbitrary solution of (1), the functions u(1) :
f (t, x, y, u, v) + ε A(t, x, y, u, v);
u(2) :
eε f (t, x, y, u, v);
u(3) :
f (t − ε , x, y, u, v);
u(4) :
f (t, x − ε , y, u, v);
u(5) :
f (t, x, y − ε , u, v);
u(6) :
f (t, cos(ε )x − sin(ε )y, sin(ε )x + cos(ε )y, cos(ε )u − sin(ε )v, sin(ε )u + cos(ε )v);
u(7) :
a 1 ε (ax + u) − ε 2 (at + 1) f (t, x − ε t, y, u − ε , v); exp − 2 2k
u(8) :
a 1 exp − ε (ay + v) − ε 2 (at + 1) f (t, x, y − ε t, u, v − ε ); 2 2k
u(9) :
f (t, x − ε e−at , y, u + ε a e−at , v);
u(10) :
f (t, x, y − ε e−at , u, v + ε a e−at );
u(11) :
a2 at 1 exp − e ε u − ε 2 a e at f (t, x − ε e at , y, u − ε ae at , v). 2 k
u(12) :
a2 at 1 exp − e ε v − ε 2 a e at f (t, x, y − ε e at , u, v − ε ae at ). 2 k
(11)
where ε - arbitrary real number , also are solutions of (1). Here A is another arbitrary solution of (1). We systematically replaced "old coordinates" by their expressions through "new coordinates". Note, that due to these replacements terms with ε 2 in u(7) , u(8) , u(11 and u(12) change their signs.
-4We have trivial solution n = e2at at our disposal. If we act on this solution by transformations (10), we obtain 4 new solutions: a 1 n = exp 2at − ε (ax + u) − ε 2 (at + 1) ; 2k 2
(12)
a 1 n = exp 2at − ε (ay + v) − ε 2 (at + 1) . 2k 2
(13)
1 a2 at e ε u − ε 2 a e at ; n = exp 2at − k 2
(14)
a2 at 1 n = exp 2at − e ε v − ε 2 a e at . k 2
(15)
a a 1 1 U = eε 2 exp − ε 7 (a x + u) − ε 72 (at + 1) exp − ε 8 (a y + v) − ε 82 (at + 1) × 2 2 2k 2k
(16)
General expression is
a2 at a2 at 1 2 1 2 × exp − e ε 11 (u − ε 7 ) − ε 11 a e at exp − e ε 12 (v − ε 8 ) − ε 12 a e at × k 2 k 2 × f (t − ε 3 , x − ε 4 − ε 7 t − ε 9 e−at − ε 11 e at , y − ε 5 − ε 8 t − ε 10 e−at − ε 12 e at , u − ε 7 + ε 9 a e−at − ε 11 a e at , v − ε 8 + ε 10 a e−at − ε 12 a e at ) + +ε 1 A(t, x, y, u, v); where x = cos(ε 6 )x − sin(ε 6 )y;
(17)
y = sin(ε 6 )x + cos(ε 6 )y;
(18)
u = cos(ε 6 )u − sin(ε 6 )v;
(19)
v = sin(ε 6 )u + cos(ε 6 )v.
(20)
-5-
DISCUSSION Looking at the list of all point symmetries of the Fokker - Planck equation in two-dimensional Euclidean space, we see, that there is no simple way to get, for example, fundamental solution of PDE, using these symmetries. We have not at our disposal such an instrument, as scaling of independent variables t, x, y, u, v. The result (12-15) of action of symmetry group on trivial solution is not very interesting from physical point of view. Indirect way of use of Galilean transformations (7) was demonstrated in [1]. The transformation was used for generalisation of solution, which was obtained in the form of exponent of quadratic form of space coordinates and velocities with time dependent coefficients. There is need of further investigations of Fokker - Planck equation and its set of symmetries, which may lead to another physically interesting results. We can follow the scheme of [7] : to consider invariant solutions for some one-parameter group, thus reduce the independent variables number. To find for obtained in such a way equation all point symmetries - and so long. In the work [7] this scheme was represented for equations of elasticity and plasticity.
ACKNOWLEDGMENTS We wish to thank Jos A. M. Vermaseren from NIKHEF (the Dutch Institute for Nuclear and High-Energy Physics), for he made his symbolic computations program FORM release 3.1 available for download for non-commercial purposes (see [8]). This wonderful program makes difficult task of symmetries search more accessible.
REFERENCES [1]
Igor A. Tanski. Fundamental solution of Fokker - Planck equation. arXiv:nlin.SI/0407007 v1 4 Jul 2004
[2]
Igor A. Tanski. The symmetries of the Fokker - Planck equation in one dimension. arXiv:nlin.CD/0412075 v1 31 Dec 2004
[3]
L. P. Eisenhart, Continuous Groups of Transformations, Princeton University Press, Princeton, 1933.
[4]
L. V. Ovsyannikov, Group analysis of differential equations, Moscow, Nauka, 1978.
[5]
Peter J. Olver, Applications of Lie groups to differential equations. Springer-Verlag, New York, 1986.
[6]
N. H. Ibragimov, CRC Handbook of Lie Group Analysis of Differential Equations, V. 3, CRC Press, New York, 1996.
[7]
B. D. Annin, V. O. Bytev, S. I. Senashov, Group properties of equations of elasticity and plasticity (Gruppovye svojstva uravnenij uprugosti i plastichnosti), Nauka, Novosibirsk, 1985.
[8]
Michael M. Tung. FORM Matters: Fast Symbolic Computation under UNIX. arXiv:cs.SC/0409048 v1 27 Sep 2004
-6-
APPENDIX 1 The infinitesimal invariance criteria for PDE (1) is δ(
−aδ u
∂n ∂n ∂n ∂n ∂n )+δu + uδ ( ) + δ v + vδ ( ) − ∂t ∂x ∂x ∂y ∂y
(A1-1)
∂n ∂n ∂n ∂n ∂2 n ∂2 n − auδ ( ) − aδ v − avδ ( ) − 2aδ n − k(δ ( 2 ) + δ ( 2 )) = 0. ∂u ∂u ∂v ∂v ∂u ∂v
According to [2] (APPENDIX 1, eq. (A1-8) and (A1-18)), we have for variations of derivatives following expression: δ
∂n ∂ ∂ ∂n ∂n ∂ ∂ ∂n = (δ n) + (δ n) − ( (δ x) + (δ x) ) − ∂x ∂x ∂n ∂x ∂x ∂x ∂n ∂x
−
∂ ∂n ∂n ∂ ∂ ∂n ∂n ∂ ( (δ y) + (δ y) ) − ( (δ u) + (δ u) ) − ∂n ∂x ∂u ∂x ∂n ∂x ∂y ∂x
(A1-2)
∂n ∂ ∂ ∂n ∂n ∂ ∂ ∂n ( (δ v) + (δ v) ) − ( (δ t) + (δ t) ); ∂v ∂x ∂n ∂x ∂t ∂x ∂n ∂x
− δ
∂n ∂ ∂ ∂n ∂n ∂ ∂ ∂n = (δ n) + (δ n) − ( (δ x) + (δ x) ) − ∂y ∂y ∂n ∂y ∂x ∂y ∂n ∂y
−
∂n ∂ ∂ ∂n ∂n ∂ ∂ ∂n ( (δ y) + (δ y) ) − ( (δ u) + (δ u) ) − ∂y ∂y ∂n ∂y ∂u ∂y ∂n ∂y
(A1-3)
∂n ∂ ∂ ∂n ∂n ∂ ∂ ∂n ( (δ v) + (δ v) ) − ( (δ t) + (δ t) ); ∂v ∂y ∂n ∂y ∂t ∂y ∂n ∂y
− δ
∂n ∂ ∂ ∂n ∂n ∂ ∂ ∂n = (δ n) + (δ n) − ( (δ x) + (δ x) ) − ∂u ∂u ∂n ∂u ∂x ∂u ∂n ∂u
−
∂ ∂n ∂n ∂ ∂ ∂n ∂n ∂ ( (δ y) + (δ y) ) − ( (δ u) + (δ u) ) − ∂n ∂u ∂u ∂u ∂n ∂u ∂y ∂u
−
∂n ∂ ∂ ∂n ∂n ∂ ∂ ∂n ( (δ v) + (δ v) ) − ( (δ t) + (δ t) ); ∂v ∂u ∂n ∂u ∂t ∂u ∂n ∂u
δ
∂n ∂ ∂ ∂n ∂n ∂ ∂ ∂n = (δ n) + (δ n) − ( (δ x) + (δ x) ) − ∂v ∂v ∂n ∂v ∂x ∂v ∂n ∂v
−
∂n ∂ ∂ ∂n ∂n ∂ ∂ ∂n ( (δ y) + (δ y) ) − ( (δ u) + (δ u) ) − ∂y ∂v ∂n ∂v ∂u ∂v ∂n ∂v
−
(A1-4)
(A1-5)
∂n ∂ ∂ ∂n ∂n ∂ ∂ ∂n ( (δ v) + (δ v) ) − ( (δ t) + (δ t) ); ∂v ∂v ∂n ∂v ∂t ∂v ∂n ∂v
δ
∂n ∂ ∂ ∂n ∂n ∂ ∂ ∂n = (δ n) + (δ n) − ( (δ x) + (δ x) ) − ∂t ∂t ∂n ∂t ∂x ∂t ∂n ∂t
−
∂ ∂n ∂n ∂ ∂ ∂n ∂n ∂ ( (δ y) + (δ y) ) − ( (δ u) + (δ u) ) − ∂n ∂t ∂u ∂t ∂n ∂t ∂y ∂t
(A1-6)
-7-
− δ
∂ ∂n ∂n ∂ ∂ ∂n ∂n ∂ ( (δ v) + (δ v) ) − ( (δ t) + (δ t) ); ∂n ∂t ∂t ∂t ∂n ∂t ∂v ∂t
∂2 n ∂ ∂2 n ∂2 n ∂ ∂ ∂n ∂2 n ∂ ∂ ∂n = ( δ n) − ( ( δ x) + ( δ x) ) − ( (δ y) + (δ y) ) − 2 2 ∂u ∂n ∂u ∂u∂x ∂u ∂n ∂u ∂u∂y ∂u ∂n ∂u
(A1-7)
−
∂ ∂n ∂2 n ∂ ∂ ∂n ∂2 n ∂ ∂ ∂n ∂2 n ∂ ( ( δ u) + ( δ u) ) − ( ( δ v) + ( δ v) ) − ( (δ t) + (δ t) ) − 2 ∂n ∂u ∂u∂v ∂u ∂n ∂u ∂t∂u ∂u ∂n ∂u ∂u ∂u
−
∂2 n ∂n ∂ ∂2 n ∂n ∂ ∂2 n ∂n ∂ ∂2 n ∂n ∂ ∂2 n ∂n ∂ ( δ x) − ( δ y) − ( δ u) − ( δ v) − (δ t) + ∂u2 ∂x ∂n ∂u2 ∂y ∂n ∂u2 ∂u ∂n ∂u2 ∂v ∂n ∂u2 ∂t ∂n +
−
−
∂n ∂2 ∂2 ∂n ∂n ∂2 ∂2 ∂n ( 2 (δ x) + (δ x) ) − ( 2 (δ y) + (δ y) ) − ∂x ∂u ∂n∂u ∂u ∂y ∂u ∂n∂u ∂u
−
∂n ∂2 ∂2 ∂n ∂n ∂2 ∂2 ∂n ( 2 (δ u) + (δ u) ) − ( 2 (δ v) + (δ v) ) − ∂u ∂u ∂n∂u ∂u ∂v ∂u ∂n∂u ∂u
∂n ∂2 ∂2 ∂n ∂n ∂n ∂2 ∂2 ∂n ( 2 (δ t) + (δ t) ) − ( (δ x) + 2 (δ x) ) − ∂t ∂u ∂n∂u ∂u ∂x ∂u ∂n∂u ∂n ∂u
−
∂2 ∂n ∂n ∂n ∂2 ∂2 ∂n ∂n ∂n ∂2 ( (δ y) + 2 (δ y) ) − ( (δ u) + 2 (δ u) ) − ∂n ∂u ∂u ∂u ∂n∂u ∂n ∂u ∂y ∂u ∂n∂u
−
∂2 ∂n ∂n ∂n ∂2 ∂2 ∂n ∂n ∂n ∂2 ( (δ v) + 2 (δ v) ) − ( (δ t) + 2 (δ t) ) − ∂n ∂u ∂t ∂u ∂n∂u ∂n ∂u ∂v ∂u ∂n∂u −
−
∂2 ∂n ∂n ∂2 ∂2 ∂n ∂2 ( δ n) + ( δ n) + ( ( δ n) + (δ n) ) − 2 2 ∂n∂u ∂u ∂u ∂n∂u ∂n ∂u ∂u
∂2 n ∂ ∂ ∂n ∂2 n ∂ ∂ ∂n ( (δ x) + (δ x) ) − ( (δ y) + (δ y) ) − ∂u∂x ∂u ∂n ∂u ∂u∂y ∂u ∂n ∂u
∂2 n ∂ ∂ ∂n ∂2 n ∂ ∂ ∂n ∂2 n ∂ ∂ ∂n ( ( δ u) + ( δ u) ) − ( ( δ v) + ( δ v) ) − ( (δ t) + (δ t) ); 2 ∂u ∂u ∂n ∂u ∂u∂v ∂u ∂n ∂u ∂t∂u ∂u ∂n ∂u δ
∂2 n ∂ ∂2 n ∂2 n ∂ ∂ ∂n ∂2 n ∂ ∂ ∂n = ( δ n) − ( ( δ x) + ( δ x) ) − ( (δ y) + (δ y) ) − ∂v 2 ∂n ∂v 2 ∂v∂x ∂v ∂n ∂v ∂v∂y ∂v ∂n ∂v
(A1-8)
−
∂ ∂n ∂2 n ∂ ∂ ∂n ∂2 n ∂ ∂ ∂n ∂2 n ∂ ( (δ u) + (δ u) ) − 2 ( (δ v) + (δ v) ) − ( (δ t) + (δ t) ) − ∂n ∂v ∂v ∂v ∂n ∂v ∂t∂v ∂v ∂n ∂v ∂u∂v ∂v
−
∂2 n ∂n ∂ ∂2 n ∂n ∂ ∂2 n ∂n ∂ ∂2 n ∂n ∂ ∂2 n ∂n ∂ ( δ x) − ( δ y) − ( δ u) − ( δ v) − (δ t) + ∂v 2 ∂x ∂n ∂v 2 ∂y ∂n ∂v 2 ∂u ∂n ∂v 2 ∂v ∂n ∂v 2 ∂t ∂n +
∂2 ∂2 ∂n ∂n ∂2 ∂2 ∂n ( δ n) + ( δ n) + ( ( δ n) + (δ n) ) − 2 2 ∂v ∂n∂v ∂v ∂v ∂n∂v ∂n ∂v
−
∂n ∂2 ∂2 ∂n ∂n ∂2 ∂2 ∂n ( 2 (δ x) + (δ x) ) − ( 2 (δ y) + (δ y) ) − ∂x ∂v ∂n∂v ∂v ∂y ∂v ∂n∂v ∂v
−
∂n ∂2 ∂2 ∂n ∂n ∂2 ∂2 ∂n ( 2 (δ u) + (δ u) ) − ( 2 (δ v) + (δ v) ) − ∂u ∂v ∂n∂v ∂v ∂v ∂v ∂n∂v ∂v
-8-
−
∂n ∂2 ∂2 ∂n ∂n ∂n ∂2 ∂2 ∂n ( 2 (δ t) + (δ t) ) − ( (δ x) + 2 (δ x) ) − ∂t ∂v ∂n∂v ∂v ∂x ∂v ∂n∂v ∂n ∂v
−
∂n ∂n ∂2 ∂2 ∂n ∂n ∂n ∂2 ∂2 ∂n ( (δ y) + 2 (δ y) ) − ( (δ u) + 2 (δ u) ) − ∂y ∂v ∂n∂v ∂n ∂v ∂u ∂v ∂n∂v ∂n ∂v
−
∂2 ∂n ∂n ∂n ∂2 ∂2 ∂n ∂n ∂n ∂2 ( (δ v) + 2 (δ v) ) − ( (δ t) + 2 (δ t) ) − ∂n ∂v ∂t ∂v ∂n∂v ∂n ∂v ∂v ∂v ∂n∂v −
−
∂2 n ∂ ∂ ∂n ∂2 n ∂ ∂ ∂n ( (δ x) + (δ x) ) − ( (δ y) + (δ y) ) − ∂v∂x ∂v ∂n ∂v ∂v∂y ∂v ∂n ∂v
∂2 n ∂ ∂ ∂n ∂2 n ∂ ∂ ∂n ∂2 n ∂ ∂ ∂n ( (δ u) + (δ u) ) − 2 ( (δ v) + (δ v) ) − ( (δ t) + (δ t) ); ∂u∂v ∂v ∂n ∂v ∂v ∂v ∂n ∂v ∂t∂v ∂v ∂n ∂v
We eliminate
∂n in (A1-1) using original equation ∂t ∂n ∂n ∂n ∂n ∂2 n ∂2 n ∂n =− u +v − au − av − 2an − k( 2 + 2 ) . ∂x ∂t ∂y ∂u ∂v ∂u ∂v
(A1-9)
Collecting similar terms, we obtain following equations: ∂n ∂n ∂x ∂u ∂2 ∂2 (δ t) + 2k (δ x) = 0; ∂n∂u ∂n∂u
(A1-10)
∂2 ∂2 ( δ t) + k (δ x) = 0; ∂n2 ∂n2
(A1-11)
∂2 ∂2 (δ t) + 2k (δ x) = 0; ∂n∂v ∂n∂v
(A1-12)
∂2 ∂2 (δ t) + k 2 (δ x) = 0; 2 ∂n ∂n
(A1-13)
∂ ∂ ∂ ∂ ∂ (δ t) + 2aun (δ t) + au (δ x) − au2 (δ t) + av (δ x) − ∂v ∂n ∂u ∂u ∂v
(A1-14)
−2ku ∂n ∂n ∂x ∂u2
−ku ∂n ∂n ∂x ∂v −2ku ∂n ∂n ∂x ∂v 2 −ku ∂n ∂x −auv
−2an +uv
∂ ∂2 ∂2 ∂2 ∂2 (δ x) − ku 2 (δ t) − ku 2 (δ t) + k 2 (δ x) + k 2 (δ x) + ∂n ∂u ∂v ∂u ∂v
∂ ∂ ∂ ∂ ∂ ∂ (δ t) − u (δ x) + u (δ t) + u2 (δ t) − v (δ x) + δ u − (δ x) = 0; ∂x ∂t ∂x ∂y ∂t ∂y
-9∂n ∂n ∂y ∂u ∂2 ∂2 (δ t) + 2k (δ y) = 0; ∂n∂u ∂n∂u
(A1-15)
∂2 ∂2 (δ t) + k 2 (δ y) = 0; 2 ∂n ∂n
(A1-16)
∂2 ∂2 (δ t) + 2k (δ y) = 0; ∂n∂v ∂n∂v
(A1-17)
∂2 ∂2 ( δ t) + k (δ y) = 0; ∂n2 ∂n2
(A1-18)
∂ ∂ ∂ ∂ ∂ (δ t) + au (δ y) + 2avn (δ t) + av (δ y) − av 2 (δ t) − ∂u ∂u ∂n ∂v ∂v
(A1-19)
−2kv ∂n ∂n ∂y ∂u2
−kv ∂n ∂n ∂y ∂v −2kv ∂n ∂n ∂y ∂v 2 −kv ∂n ∂y −auv
−2an +uv
∂ ∂2 ∂2 ∂2 ∂2 (δ y) − kv 2 (δ t) − kv 2 (δ t) + k 2 (δ y) + k 2 (δ y) + ∂n ∂u ∂v ∂u ∂v
∂ ∂ ∂ ∂ ∂ ∂ (δ t) − u (δ y) − v (δ y)v (δ t) + v 2 (δ t) + δ v − (δ y) = 0; ∂x ∂x ∂y ∂t ∂y ∂t
∂n ∂n ∂u ∂v 2aku
∂2 ∂2 ∂2 ∂2 (δ t) + 2akv (δ t) + 2k (δ v) + 2k (δ u) = 0; ∂n∂v ∂n∂u ∂n∂u ∂n∂v
(A1-20)
∂2 ∂2 ( δ t) + k (δ u) = 0; ∂n2 ∂n2
(A1-21)
∂ ∂2 (δ u) + 2k 2 (δ t) = 0; ∂n ∂n∂u
(A1-22)
∂2 (δ t) = 0; ∂n∂u
(A1-23)
∂n ∂n ∂u ∂v 2 aku ∂n ∂2 n ∂u ∂u2 2k ∂n ∂2 n ∂u ∂v 2 2k 2 ∂n ∂2 n ∂u ∂u∂v
-10-
2k
∂ (δ v) = 0; ∂n
(A1-24)
2k
∂ (δ x) = 0; ∂n
(A1-25)
2k
∂ (δ y) = 0; ∂n
(A1-26)
2k
∂ (δ t) = 0; ∂n
(A1-27)
∂2 ∂2 ∂2 ∂ ( δ t) + aku ( δ t) + 4akn (δ t) − auv (δ t) + 2 2 ∂u ∂v ∂n∂u ∂y
(A1-28)
∂n ∂2 n ∂u ∂u∂x
∂n ∂2 n ∂u ∂u∂y
∂n ∂2 n ∂u ∂t∂u
∂n ∂u aku +au −2a2 un
∂ ∂ ∂ ∂ ∂ ∂ (δ u) − au (δ t) − au2 (δ t) + av (δ u) − 2an (δ u) − aδ u + a2 uv (δ t) − ∂u ∂t ∂x ∂v ∂n ∂v
∂ ∂2 ∂2 ∂2 ∂ ∂ ∂ ∂ (δ t) + a2 u2 (δ t) − 2k (δ n) + k 2 (δ u) + k 2 (δ u) − u (δ u) − v (δ u) − (δ u) = 0; ∂u ∂n∂u ∂u ∂v ∂x ∂y ∂t ∂n
∂n ∂n ∂u2 ∂v akv
∂2 ∂2 ( δ t) + k (δ v) = 0; ∂n2 ∂n2
(A1-29)
k2
∂2 (δ t) = 0; ∂n2
(A1-30)
k2
∂2 (δ t) = 0; ∂n2
(A1-31)
∂n ∂2 n ∂u2 ∂u2
∂n ∂2 n ∂u2 ∂v 2
∂n ∂u2 2aku
∂2 ∂2 ∂2 ∂2 (δ t) + 2akn 2 (δ t) − k 2 (δ n) + 2k (δ u) = 0; ∂n∂u ∂n ∂n ∂n∂u
(A1-32)
∂2 ∂2 ( δ t) + k (δ u) = 0; ∂n2 ∂n2
(A1-33)
∂n ∂u3 aku
-11∂n ∂2 n ∂v ∂u2 k2
∂2 ∂2 (δ t) + k 2 (δ t) = 0; ∂n∂v ∂n∂v
(A1-34)
2k
∂ ∂2 (δ v) + 2k 2 (δ t) = 0; ∂n ∂n∂v
(A1-35)
2k
∂ (δ u) = 0; ∂n
(A1-36)
2k
∂ (δ x) = 0; ∂n
(A1-37)
2k
∂ (δ y) = 0; ∂n
(A1-38)
2k
∂ (δ t) = 0; ∂n
(A1-39)
∂2 ∂2 ∂2 ( δ t) + akv ( δ t) + 4akn (δ t) − ∂u2 ∂v 2 ∂n∂v
(A1-40)
∂n ∂2 n ∂v ∂v 2
∂n ∂2 n ∂v ∂u∂v
∂n ∂2 n ∂v ∂v∂x
∂n ∂2 n ∂v ∂v∂y
∂n ∂2 n ∂v ∂t∂v
∂n ∂v akv −auv −av 2 +a2 v 2
∂ ∂ ∂ ∂ (δ t) + au (δ v) + av (δ v) − av (δ t) − ∂u ∂v ∂t ∂x
∂ ∂ ∂ ∂ (δ t) − 2an (δ v) − aδ v + a2 uv (δ t) − 2a2 vn (δ t) + ∂y ∂n ∂u ∂n
∂2 ∂2 ∂2 ∂ ∂ ∂ ∂ (δ t) + k 2 (δ v) − 2k (δ n) + k 2 (δ v) − u (δ v) − v (δ v) − (δ v) = 0; ∂u ∂n∂v ∂v ∂x ∂y ∂t ∂v
∂n ∂2 n ∂v 2 ∂u2 k2
∂2 (δ t) = 0; ∂n2
(A1-41)
k2
∂2 (δ t) = 0; ∂n2
(A1-42)
∂n ∂2 n ∂v 2 ∂v 2
-12∂n ∂v 2 2akv
∂2 ∂2 ∂2 ∂2 (δ t) + 2akn 2 (δ t) − k 2 (δ n) + 2k (δ v) = 0; ∂n∂v ∂n ∂n ∂n∂v
(A1-43)
∂2 ∂2 (δ t) + k 2 (δ v) = 0; 2 ∂n ∂n
(A1-44)
∂ ∂ ∂ ∂ (δ t) + akv (δ t) − 2akn (δ t) − ku (δ t) − ∂u ∂v ∂n ∂x
(A1-45)
∂n ∂v 3 akv ∂2 n ∂u2 aku −kv
∂ ∂ ∂ ∂2 ∂2 (δ t) + 2k (δ u) − k (δ t) + k 2 2 (δ t) + k 2 2 (δ t) = 0; ∂y ∂u ∂t ∂u ∂v
∂2 n ∂v 2 aku −kv
∂ ∂ ∂ ∂ (δ t) + akv (δ t) − 2akn (δ t) − ku (δ t) − ∂u ∂v ∂n ∂x
(A1-46)
∂ ∂ ∂ ∂2 ∂2 (δ t) + 2k (δ v) − k (δ t) + k 2 2 (δ t) + k 2 2 (δ t) = 0; ∂y ∂v ∂t ∂u ∂v
∂2 n ∂u∂v 2k
∂ ∂ (δ u) + 2k (δ v) = 0; ∂v ∂u
(A1-47)
2k
∂ (δ x) = 0; ∂u
(A1-48)
2k
∂ (δ y) = 0; ∂u
(A1-49)
2k
∂ (δ x) = 0; ∂v
(A1-50)
2k
∂ (δ y) = 0; ∂v
(A1-51)
∂2 n ∂u∂x
∂2 n ∂u∂y
∂2 n ∂v∂x
∂2 n ∂v∂y
∂2 n ∂t∂u
-13-
2k
∂ (δ t) = 0; ∂u
(A1-52)
2k
∂ (δ t) = 0; ∂v
(A1-53)
∂2 n ∂t∂v
1 2akn
∂2 ∂2 ∂ ∂ ( δ t) + 2akn (δ t) − 2aun (δ t) − au (δ n) − 2 2 ∂u ∂v ∂x ∂u
−2avn
∂ ∂ ∂ ∂ (δ t) − av (δ n) + 2an (δ n) − 2an (δ t) − ∂y ∂v ∂n ∂t
−2aδ n + 2a2 un −k
(A1-54)
∂ ∂ ∂ ∂2 (δ t) + 2a2 vn (δ t) − 4a2 n2 (δ t) − k 2 (δ n) − ∂u ∂v ∂n ∂u
∂2 ∂ ∂ ∂ (δ n) + u (δ n) + v (δ n) + (δ n) = 0. 2 ∂v ∂x ∂y ∂t
From (A1-37 - A1-39), (A1-48 - A1-51) we see, that δ x = δ x(x, y, t); δ y = δ y(x, y, t); δ t = δ t(x, y, t). Using these expressions, we simplify the rest of equations (A1-10 - A1-54). uv
∂ ∂ ∂ ∂ ∂ ∂ (δ t) − u (δ x) + u (δ t) + u2 (δ t) − v (δ x) + δ u − (δ x) = 0; ∂x ∂t ∂x ∂y ∂t ∂y
(A1-55)
uv
∂ ∂ ∂ ∂ ∂ ∂ (δ t) − u (δ y) − v (δ y) + v (δ t) + v 2 (δ t) + δ v − (δ y) = 0; ∂x ∂x ∂y ∂t ∂y ∂t
(A1-56)
∂ ∂ ∂ ∂ (δ t) + au (δ u) − au (δ t) − au2 (δ t) + ∂y ∂u ∂t ∂x
−auv
+av +k
∂2 ∂2 ∂ (δ u) − aδ u − 2k (δ n) + k 2 (δ u) + ∂n∂u ∂u ∂v
∂2 ∂ ∂ ∂ (δ u) − u (δ u) − v (δ u) − (δ u) = 0; 2 ∂v ∂x ∂y ∂t −k
−auv
∂2 (δ n) = 0; ∂n2
∂ ∂ ∂ ∂ (δ t) + au (δ v) + av (δ v) − av (δ t) − ∂u ∂v ∂t ∂x
−av 2 +k
(A1-57)
(A1-58) (A1-59)
∂ ∂2 ∂2 (δ t) − aδ v + k 2 (δ v) − 2k (δ n) + ∂y ∂u ∂n∂v
∂2 ∂ ∂ ∂ (δ v) − u (δ v) − v (δ v) − (δ v) = 0; ∂v 2 ∂x ∂y ∂t −k
∂2 (δ n) = 0; ∂n2
(A1-60)
-14-
−ku
∂ ∂ ∂ ∂ (δ t) − kv (δ t) + 2k (δ u) − k (δ t) = 0; ∂y ∂u ∂t ∂x
(A1-61)
−ku
∂ ∂ ∂ ∂ (δ t) − kv (δ t) + 2k (δ v) − k (δ t) = 0; ∂x ∂y ∂v ∂t
(A1-62)
∂ ∂ (δ u) + 2k (δ v) = 0; ∂v ∂u
(A1-63)
2k −2aun
∂ ∂ ∂ ∂ (δ t) − au (δ n) − 2avn (δ t) − av (δ n) + ∂u ∂y ∂v ∂x
+2an −k
(A1-64)
∂ ∂2 ∂ (δ n) − 2an (δ t) − 2aδ n − k 2 (δ n) − ∂t ∂u ∂n
∂2 ∂ ∂ ∂ (δ n) + u (δ n) + v (δ n) + (δ n) = 0. 2 ∂v ∂x ∂y ∂t
From (A1-58) and (A1-60) we conclude, that δ n = A + nB;
(A1-65)
where A = A(x, y, u, v, t), B = B(x, y, u, v, t). Using this expression, we simplify the rest of equations (A1-55 - A1-64). uv
∂ ∂ ∂ ∂ ∂ ∂ (δ t) − u (δ x) + u (δ t) + u2 (δ t) − v (δ x) + δ u − (δ x) = 0; ∂x ∂t ∂x ∂y ∂t ∂y
(A1-66)
uv
∂ ∂ ∂ ∂ ∂ ∂ (δ t) − u (δ y) − v (δ y) + v (δ t) + v 2 (δ t) + δ v − (δ y) = 0; ∂x ∂x ∂y ∂t ∂y ∂t
(A1-67)
∂ ∂ ∂ ∂ (δ t) + au (δ u) − au (δ t) − au2 (δ t) + ∂y ∂u ∂t ∂x
−auv +av
(A1-68)
∂2 ∂2 ∂B ∂ ∂ ∂ ∂ (δ u) − aδ u + k 2 (δ u) + k 2 (δ u) − 2k −u (δ u) − v (δ u) − (δ u) = 0; ∂u ∂v ∂u ∂x ∂y ∂t ∂v −auv
−av 2
∂ ∂ ∂ ∂ (δ t) + au (δ v) + av (δ v) − av (δ t) − ∂x ∂u ∂v ∂t
(A1-69)
∂ ∂2 ∂2 ∂B ∂ ∂ ∂ (δ t) − aδ v + k 2 (δ v) + k 2 (δ v) − 2k −u (δ v) − v (δ v) − (δ v) = 0; ∂y ∂u ∂v ∂v ∂x ∂y ∂t −ku
∂ ∂ ∂ ∂ (δ t) − kv (δ t) + 2k (δ u) − k (δ t) = 0; ∂x ∂y ∂u ∂t
(A1-70)
−ku
∂ ∂ ∂ ∂ (δ t) − kv (δ t) + 2k (δ v) − k (δ t) = 0; ∂x ∂y ∂v ∂t
(A1-71)
∂ ∂ (δ u) + 2k (δ v) = 0; ∂v ∂u
(A1-72)
2k −2au
∂B ∂ ∂B ∂ ∂ (δ t) − au − 2av (δ t) − av − 2a (δ t) − ∂u ∂y ∂v ∂t ∂x −k
∂2 B ∂2 B ∂B ∂B ∂B − k +u +v + = 0; 2 2 ∂u ∂v ∂x ∂y ∂t
(A1-73)
-15-
−au
∂A ∂A ∂2 A ∂2 A ∂A ∂A ∂A − av − 2aA − k 2 − −k 2 + u +v + = 0. ∂u ∂v ∂u ∂v ∂x ∂y ∂t
(A1-74)
Equation (A1-74) is simply Fokker - Planck equation for A. We solve (A1-66 - A1-67) and find δ u, δ v δ u = −(uv
∂ ∂ ∂ ∂ ∂ ∂ (δ t) − u (δ x) + u (δ t) + u2 (δ t) − v (δ x) − (δ x)); ∂y ∂x ∂t ∂x ∂y ∂t
(A1-76)
δ v = −(uv
∂ ∂ ∂ ∂ ∂ ∂ (δ t) − u (δ y) − v (δ y) + v (δ t) + v 2 (δ t) − (δ y)). ∂x ∂x ∂y ∂t ∂y ∂t
(A1-77)
This gives for (A1-67 - A1-73) −2auv −2k +2u2 v
(A1-78)
∂B ∂2 ∂2 ∂2 ∂2 ∂2 + 2uv (δ t) − 2uv (δ x) + uv 2 2 (δ t) + u 2 (δ t) − 2u (δ x) + ∂u ∂t∂y ∂x∂y ∂y ∂t ∂t∂x
∂2 ∂2 ∂2 ∂2 ∂2 ∂2 ∂2 (δ t) + 2u2 (δ t) − u2 2 (δ x) + u3 2 (δ t) − 2v (δ x) − v 2 2 (δ x) − 2 (δ x) = 0; ∂t∂x ∂x ∂x ∂t∂y ∂y ∂t ∂x∂y −2auv −2k
+u2 v
∂ ∂ ∂ ∂ ∂ (δ t) − au (δ t) − 2au2 (δ t) − a (δ x) − 2k (δ t) − ∂y ∂t ∂x ∂t ∂x
∂ ∂ ∂ ∂ ∂ (δ t) − av (δ t) − 2av 2 (δ t) − a (δ y) − 2k (δ t) − ∂x ∂t ∂y ∂t ∂y
(A1-79)
∂2 ∂2 ∂2 ∂2 ∂B + 2uv (δ t) − 2uv (δ y) + 2uv 2 (δ t) − 2u (δ y) + ∂t∂x ∂x∂y ∂x∂y ∂t∂x ∂v
2 2 2 2 ∂2 ∂2 ∂2 ∂2 2 ∂ 2 ∂ 2 ∂ 3 ∂ ( δ t) − u ( δ y) + v ( δ t) − 2v ( δ y) + 2v ( δ t) − v ( δ y) + v ( δ t) − (δ y) = 0; ∂x 2 ∂x 2 ∂t 2 ∂t∂y ∂t∂y ∂y 2 ∂y 2 ∂t 2
−5ku
∂ ∂ ∂ ∂ (δ t) − 3kv (δ t) + 2k (δ x) − 3k (δ t) = 0; ∂x ∂y ∂x ∂t
(A1-80)
−3ku
∂ ∂ ∂ ∂ (δ t) − 5kv (δ t) + 2k (δ y) − 3k (δ t) = 0; ∂y ∂y ∂t ∂x
(A1-81)
−2ku
∂ ∂ ∂ ∂ (δ t) − 2kv (δ t) + 2k (δ x) + 2k (δ y) = 0; ∂y ∂x ∂y ∂x
(A1-82)
−2au
∂ ∂B ∂ ∂B ∂ (δ t) − au − 2av (δ t) − av − 2a (δ t) − ∂x ∂u ∂y ∂v ∂t
(A1-83)
−k
∂2 B ∂B ∂B ∂B ∂2 B −k 2 +u +v + = 0. 2 ∂v ∂x ∂y ∂t ∂u
Now we can collect similar terms in (A1-80 - A1-82) and so split them into nine equations: −5k
∂ (δ t) = 0; ∂x
(A1-84)
−3k
∂ (δ t) = 0; ∂y
(A1-85)
-16-
2k
2k
2k
∂ ∂ (δ x) − 3k (δ t) = 0; ∂t ∂x
(A1-86)
−3k
∂ (δ t) = 0; ∂x
(A1-87)
−5k
∂ (δ t) = 0; ∂y
(A1-88)
∂ ∂ (δ y) − 3k (δ t) = 0; ∂y ∂t
(A1-89)
−2k
∂ (δ t) = 0; ∂y
(A1-90)
−2k
∂ (δ t) = 0; ∂x
(A1-91)
∂ ∂ (δ x) + 2k (δ y) = 0. ∂y ∂x
(A1-92)
From (A1-84 - A1-85), (A1-87 - A1-88), (A1-90 - A1-91) we see, that δ t = δ t(t), which results in further simplifications −au
∂ ∂ ∂B ∂2 ∂2 ∂2 (δ t) − a (δ x) − 2k − 2uv (δ x) + u 2 (δ t) − 2u (δ x) − ∂t ∂t ∂u ∂x∂y ∂t ∂t∂x −u2 −av −u2
−au
(A1-93)
2 ∂2 ∂2 ∂2 2 ∂ ( δ x) − 2v ( δ x) − v ( δ x) − (δ x) = 0; ∂t∂y ∂y 2 ∂t 2 ∂x 2
∂ ∂ ∂B ∂2 ∂2 (δ t) − a (δ y) − 2k − 2uv (δ y) − 2u (δ y) − ∂t ∂t ∂v ∂x∂y ∂t∂x
(A1-94)
2 ∂2 ∂2 ∂2 ∂2 2 ∂ ( δ y) + v ( δ t) − 2v ( δ y) − v ( δ y) − (δ y) = 0; ∂x 2 ∂t 2 ∂t∂y ∂y 2 ∂t 2
2k
∂ ∂ (δ x) − 3k (δ t) = 0; ∂t ∂x
(A1-95)
2k
∂ ∂ (δ y) − 3k (δ t) = 0; ∂y ∂t
(A1-96)
2k
∂ ∂ (δ x) + 2k (δ y) = 0; ∂x ∂y
(A1-97)
∂B ∂B ∂ ∂2 B ∂2 B ∂B ∂B ∂B − av − 2a (δ t) − k 2 − k 2 + u +v + = 0. ∂u ∂v ∂t ∂u ∂v ∂x ∂y ∂t
(A1-98)
We integrate (A1-95 - A1-96) and find δ x = C + 3/2x
∂ (δ t); ∂t
(A1-99)
δ y = D + 3/2y
∂ (δ t); ∂t
(A1-100)
-17where C = C(y, t), D = D(x, t). We substitute these expressions to (A1-93 - A1-94), (A1-97 - A1-98) and obtain −3/2ax
∂2 ∂ ∂C ∂B ∂3 ( δ t) − au ( δ t) − a − 2k − 3/2x (δ t) − ∂t 2 ∂t ∂t ∂u ∂t 3 −2u
−3/2ay
2 ∂2 ∂2 C ∂2 C 2 ∂ C ( δ t) − 2v − v − = 0; ∂t 2 ∂t∂y ∂y 2 ∂t 2
∂ ∂D ∂B ∂3 ∂2 ( δ t) − av ( δ t) − a − 2k − 3/2y (δ t) − ∂t ∂t ∂v ∂t 3 ∂t 2 −2u
We find
(A1-102)
∂2 D ∂2 D ∂2 ∂2 D − u2 − 2v ( δ t) − = 0; ∂t∂x ∂x 2 ∂t 2 ∂t 2 ∂C ∂D + 2k = 0; ∂y ∂x
(A1-103)
∂B ∂ ∂2 B ∂2 B ∂B ∂B ∂B ∂B − av − 2a (δ t) − k 2 − k 2 + u +v + = 0. ∂v ∂t ∂u ∂v ∂x ∂y ∂t ∂u
(A1-104)
2k −au
(A1-101)
∂B ∂B from (A1-101) and from (A1-102): ∂u ∂v ∂B 1 ∂2 ∂ ∂C ∂3 = (−3/4ax 2 (δ t) − 1/2au (δ t) − 1/2a − 3/4x 3 (δ t) − ∂u k ∂t ∂t ∂t ∂t −u
2 ∂2 C ∂2 C ∂2 2 ∂ C ( δ t) − v − 1/2v − 1/2 ); ∂t∂y ∂y 2 ∂t 2 ∂t 2
∂B 1 ∂2 ∂ ∂D ∂3 = (−3/4ay 2 (δ t) − 1/2av (δ t) − 1/2a − 3/4y 3 (δ t) − ∂v k ∂t ∂t ∂t ∂t −u
(A1-105)
(A1-106)
∂2 D ∂2 D ∂2 ∂2 D − 1/2u2 − v ( δ t) − 1/2 ). ∂t∂x ∂x 2 ∂t 2 ∂t 2
Differentiating (A1-105) by v we have ∂2 B 1 ∂2 C ∂2 C = (−v 2 − ); ∂u∂v k ∂y ∂t∂y
(A1-107)
Differentiating (A1-106) by u we have 1 ∂2 D ∂2 D ∂2 B = (−u − ). ∂x 2 ∂t∂x ∂u∂v k
(A1-108)
We know, that C = C(y, t), D = D(x, t) and so we conclude from (A1-107 - A1-108) C = C 1 y + E;
(A1-109)
D = C 2 x + F;
(A1-110)
1 ∂2 ∂ ∂E ∂3 ∂2 ∂2 E ∂B = (3/2ax 2 (δ t) + au (δ t) + a − 3/2x 3 (δ t) − 2u 2 (δ t) − 2 ); ∂t ∂t ∂t ∂t ∂t ∂t ∂u 2k
(A1-111)
where E = E(t), F = F(t), C 1 + C 2 = 0. We find derivatives of B.
-18∂B 1 ∂2 ∂ ∂F ∂3 ∂2 ∂2 F = (3/2ay 2 (δ t) + av (δ t) + a − 3/2y 3 (δ t) − 2v 2 (δ t) − 2 ); ∂v 2k ∂t ∂t ∂t ∂t ∂t ∂t
(A1-112)
∂2 B ∂2 B = = 0. ∂u∂v ∂v∂u
(A1-113)
Integration of (A1-111 - A1-112) gives B=G+
∂2 ∂ ∂E ∂3 ∂2 ∂2 E 1 (3/2uax 2 (δ t) + 1/2au2 (δ t) + ua − 3/2ux 3 (δ t) − u2 2 (δ t) − u 2 ) + (A1-114) ∂t ∂t ∂t ∂t ∂t ∂t 2k +
1 ∂2 ∂ ∂F ∂3 ∂2 ∂2 F (3/2vay 2 (δ t) + 1/2av 2 (δ t) + va − 3/2vy 3 (δ t) − v 2 2 (δ t) − v 2 ); 2k ∂t ∂t ∂t ∂t ∂t ∂t
where G = G(x, y, t). Substitution of (A1-114) to (A1-98), collecting and equating to zero similar terms by u, v gives 3/4a2 k −1 x
3 ∂4 ∂G ∂2 2 −1 ∂E −1 −1 ∂ E ( δ t) + 1/2a k − 3/4k x ( δ t) − 1/2k + = 0; 4 3 2 ∂t ∂t ∂t ∂x ∂t
1/2a2 k −1 3/4a2 k −1 y
∂ ∂3 (δ t) − 5/4k −1 3 (δ t) = 0; ∂t ∂t
3 ∂2 ∂4 ∂G 2 −1 ∂F −1 −1 ∂ F ( δ t) + 1/2a k − 3/4k y ( δ t) − 1/2k + = 0; 2 4 3 ∂t ∂t ∂t ∂t ∂y
1/2a2 k −1 −a
∂ ∂3 (δ t) − 5/4k −1 3 (δ t) = 0; ∂t ∂t
∂ ∂2 ∂G (δ t) + 2 2 (δ t) + = 0; ∂t ∂t ∂t
(A1-115)
(A1-116)
(A1-117)
(A1-118)
(A1-119)
Integrate (A1-115), (A1-117) and obtain following expression (H = H(t)): G=H− −
1 ∂E ∂4 ∂3 E ∂2 − 3/8x 2 4 (δ t) − 1/2x 3 − 3/8a2 x 2 2 (δ t) + 1/2xa2 k ∂t ∂t ∂t ∂t
(A1-120)
∂F ∂4 ∂3 F ∂2 1 − 3/8y 2 4 (δ t) − 1/2y 3 . 3/8a2 y 2 2 (δ t) + 1/2ya2 ∂t ∂t ∂t ∂t k
Substitution of (A1-120) to (A1-119), collecting and equating to zero terms by x, y gives 4 ∂2 E −1 ∂ E + 1/2k = 0; ∂t 2 ∂t 4
(A1-121)
5 ∂3 −1 ∂ ( δ t) + 3/8k (δ t) = 0; ∂t 3 ∂t 5
(A1-122)
4 ∂2 F −1 ∂ F + 1/2k = 0; ∂t 2 ∂t 4
(A1-123)
5 ∂3 −1 ∂ ( δ t) + 3/8k (δ t) = 0; ∂t 5 ∂t 3
(A1-124)
−1/2a2 k −1 −3/8a2 k −1
−1/2a2 k −1 −3/8a2 k −1 −a
∂ ∂2 ∂H (δ t) + 2 2 (δ t) + = 0. ∂t ∂t ∂t
(A1-125)
-19-
From (A1-116), (A1-118), (A1-122), (A1-124) we conclude, that δ t = const = C 3 .
(A1-126)
E = C 4 + C 5 t + C 6 e−at + C 7 e at ;
(A1-127)
F = C 8 + C 9 t + C 10 e−at + C 11 e at ;
(A1-128)
From (A1-121), (A1-123) we conclude, that
From (A1-125) and (A1-126) we see, that H = C 12 .
(A1-129)
We obtain using (A1-126 - A1-129) and backward substitution final expressions for variations: δ n = −1/2ak −1 unC 5 − 1/2ak −1 vnC 9 − 1/2a2 k −1 xnC 5 −
(A1-130)
−1/2a2 k −1 ynC 9 − a2 k −1 unC 7 e at − a2 k −1 vnC 11 e at + nC 12 + A;
This ends calculations.
δ x = yC 1 + tC 5 + C 4 + C 6 e−at + C 7 e at ;
(A1-131)
δ y = xC 2 + tC 9 + C 8 + C 10 e−at + C 11 e at ;
(A1-132)
δ u = −aC 6 e−at + aC 7 e at + vC 1 + C 5 ;
(A1-133)
δ v = −aC 10 e−at + aC 11 e at + uC 2 + C 9 ;
(A1-134)
δ t = C3.
(A1-135)
