CC BY
24
Chaikovskii Dmitrii Aleksandrovich, East China Normal University, PhD., student, Department of Mathematics E-mail: mitichya@yandex.ru Ming Kang Ni, East China Normal University, Professor of Math, Department of Mathematics
Shanghai, China
INTERNAL LAYERS FOR A SINGULARLY PERTURBED SECOND-ORDER QUASILINEAR DIFFERENTIAL EQUATION WITH DISCONTINUOUS RIGHT-HAND SIDE WITH NEUMANN AND DIRICHLET BOUNDARY CONDITIONS
Abstract: In this work we research a singularly perturbed boundary value problem for a quasilinear ordinary differential equation of the second order with Neumann and Dirichlet boundary conditions. We have considered new class of problems, in which nonlinearities undergo discontinuities, which leads to the appearance of sharp transition layers in the vicinity of the discontinuity points. The existence is proved and an asymptotic expansion of solutions with an internal transition layer is constructed.
Keywords: singular perturbations, asymptotic approximations.
Introduction 1. Formulation of the problem
In this paper we study the boundary value prob- Consider the following boundary-value problem:
lem for a singularly perturbed ordinary differential jsy = A(y,t)y + f (y,t), 0 < t < 1,
equation of the second order, which can be regarded | y (o,£) = y0, y' (l,e) = y \
as a stationary equation of the reaction-advection- where £ > 0 is a small parameter. We assume that the
diffusion type with discontinuous reactive and ad- functions A(y,x) and f (y,x) are defined every-
vective terms. Problems of this type belong to the where in the domain D ={(y,x)l Iy x[0;1]}, where Iy
class of discontinuous dynamical systems, which is the permissible interval for changing the function
are widely used as mathematical models in the field y(x) and sufficiently smooth in D, except for the part
of mechanics,) e^ronics biology, etc. [l, 5-22] of the line y e Iy, x = x0, where x0 is the interior
Boundary problems are of particular interest for ap- point of [0;1], which lies inside the domain D, and
plications whose solutions exhibit large gradients in divides the domain D into two parts:
the neighborhood ofany ofthe interior points ofthe d{-) ={y x )ly e I 0< x < x } D(+) =
domain of definition. In this case one speaks of the ¡, . y 0
— {( y x )|y g I x ^ x ^ 1}
existence of an inner transition layer. Here consider (V' / / y o - - y
i-nr . . . . .,r,r Also suppose that the following conditions are satisa differential equation that contains a term with the rr &
first derivative having the meaning of advection. To fied: t (-) (-)
this type is the Burgers equation known from the a(y,t) = <j ^ ^ (y^) ^ , f (y,t) =
acoustics problems [2, 100-105]. ' [a(+)(y,t), (y,t)e D(+),
(y,t), (y,t) D(-), [f(+)(y,t), (y,t) D(+),
A« (y ,t o ) A (+)(y ,t 0 ), f (-)(y ,to ) f (+)(y ,t o ).
Suppose that on a segment 0 " x " x0 there exists a solution y = ^(x) e Iy of the Cauchy problems AH (y,x)y + f(-)(y,x) = 0, y(0) = y0 and on the segment x0 " x " 1 there exists a solution y = ç2(x) e Iy of the Cauchy problem A(+) (y,x)y' + f{+)(y,x) = 0, y'(1) = y1 and we will assume: Ç (x0 )<Ç (x0 ), A(-) ( (x),x) > 0, A(+) (2 (x),x) < 0, 0 < x <1.
2. Attached system
Consider the auxiliary system
= z, A = A()(y,x0)z, -œ<r<+œ. (l) dr dr
We divide each of the second equations of systems (l) by the first, and we arrive at differential equations of the first order with respect to the functions z ( y ), which determine the phase trajectories of these systems on the plane (y,A), dz/dy = A(T)(y,x0). Points (p(x0),0), (p2(x0),0) on the phase plane (y,z) are the rest points, and because functions A (+)(y, x0) is continuous in <Pi(x0) < y <y2(x 0 ), then there exist the phase trajectory:
5\y)= J A\s,x0)ds.
Pl,2 (x0)
3. Asymptotic representation of the solution
The asymptotic approximation of the solution of our problem will be constructed separately from left and right of the point x0 and denote it as y( ^(x,s) on the interval [0,x0] and ,s) on the interval [x0,1]. We continuously associate the functions y( -and y, and also their derivatives [3, 109-122] .(-) _ dy(-) ,+, _ dy<+>
z = ■
dx
, ^ =-
dx
at the point x0 :
y (-)(x 0,^) = y (+)(x 0^) = p (s) (2)
z(-) (x0,e) = z(+) (x0,e) = z (e).
(3)
(4)
The values of p(e) and z(s), respectively, of the functions y(T)(x,s) and z(T)(x,s) at the point x0 are unknown and will be determined in the course of constructing the asymptotic expansion of our problem.
On each of the segments [0, x 0] and [x 0,1] we construct asymptotic approximations of the solutions of the following boundary-value problems (3): sy "(-) = A(-) (y(-), x )y '(-) + f(-) (y(-), x ), 0 < x < x,
y(-) (0,s) = y0, y(-)(x0,s) = p(s),
£y"(+) = A(+) (y(+),x)y,(+) + f(+) (y(+),x), x0 < x < 1,
y(+)(x0,£) = p(e), A'(+)(U) = y1.
Problems (3) for the ordinary differential equations of the second order are equivalent to the following problems for systems of equations of the first order:
y '(-) = z (-), ez '(-) = A (-)(y(-), x )z(-) + f (-)(y(-), x ),
0 < x < x 0, y(-) (0,e) = y0, y(-) (x 0,s) = p (e),
y '(+) = z(+), ez = A m (y m, x )z(+) + f(+) (y(+), xA
x 0 < x < 1, y (+)(x0,e) = p (e), a (+)(i,e) = y1.
Fora detailed description of the behavior of the solution in the vicinity of the transition point layer and boundary parts we introduce an extended variables: t0 = x / £ , T = (x - x0)/ £ , t1 = (1 - x)s .
Asymptotic expansions of the solutions of each of the problems (4) will be constructed in the form of sums of three terms:
y(-) (x,£ ) = y(-) (x,£) + Ly (t0 ,£ ) + Q(-)y (t,£) , z(x ,s) = z(-) (x ,e) + Lz (t0,£) + Q (-)z (t,s) , (5) y ^ (x, £ ) = y ^ (x, £ ) + Q ^ y (t,£)+ Ry (t1 , £ ), z (+)(x ,e) = z (+)(x ,e) + Q (+)z (r,e) + Rz (r1,e), where yW (x ,s
) z (T)( x ,£) the expansion that approximates the solution far from the region of the transition layer, and the functions Q(+)y(t,£) and Q(+)z(t,£ describe the behavior of the solution in the neighborhood (from the left and the right) of the discontinuity point x — x 0 Ly ( Ve) and Lz(T0,£) - the boundary part of the asymptotics near the point x — 0, Ry(T1,e) and
Rz(t1,s) - the boundary part of the asymptotics near the point x = 1.
Each term in (5) will be sought in the form of an expansion in powers of e:
yiT \x ,e) = yf \x ) + sy(\x) + ...,
z(x,s) = z0(T) (x) + sz(T) (x) + ...;
Q {T)y (r,e) = Q^y (T) + sQ^y (t) + ...,
Q (T)z (r,s) = s-'Q^z (r) + Q0T)z (r) + £Q1T)z (r) + ...
Ly (T0 , ^ ) = L0y (T0 ) + ^L1 y (T0 ) +•• •, (-») Lz (t0,s) = s1L_1z (t0 ) + L0z (t0 ) + sLxz (t0 ) + ....
Ry (T1,e) = R0 y (T1 ) + £R1 y (T1) + ■■:
Rz (r1,s) = s-1R-1z (t1 ) + R0z (t1 ) + sR1z (t1 ) + ....
The unknown values p(s) and z(s) ofthe functions y(+)(x,s) and z(+)(x,s) at x0 also will be sought in the form of expansions in powers of e: P(£) = P0 +£P1 + •••, z(s) = s~1z_1 + z0 + sz1 + .. (7) We write down the joining conditions (2) with allowance for the expansions (6) - (7):
y0- (x0) + erf- (x0) + ...+ Q0-)y (0) + s(Q[)y (0) + ... = = y0+) (x 0) + (x 0) + ••• + Q0+)y (0) + QV + • • • =
= P0 +^p1 + •••; (8)
zt) (x 0) + sz[-] (x 0) + ... + e~1otiz (0) + + Q0-)z (0) + £Q1-)z (0 ) + ...=
= z(0+)(x 0 ) + szi+\x 0 ) + ^ + s-1Q_\)z (0 ) + •
+ Q0+}z (0 } + £Q1(+)z (0 ) + ...=
= s-z _1 + z 0 + £z1 + .... (9)
4. The regular terms of asymptotic representation
Problems for the terms of the regular part we obtain by substituting the functions y (+)(x and z(x,e) in the form (6) into the system of equations:
^ =y<*>, e^ = A<*>((>,x))*+f(yM,x) . x . x
and by substituting the additional conditions for
x = 0 and x = 1, respectively, of problem (4), and
equating the terms for the same powers of e in both parts of the obtained equalities. As a result, in each i - th order, i = 0,1,..., we will determine functions
yt ^ \x ) as solutions of the Cauchy problem for the differential equations of the first order, and then find
expressions for their derivatives - functions z{ ^ ^(x).
Thus in the 0 - th approximation we obtain the following Cauchy problems:
A-'(y0(-, x + f (70H, x ) = 0,
yH(0) = yi, 0 < x < x0,
A™ ( x ))- + f ((0 x ) = 0. yyl (1) = y1, x0 < x < 1.
We set y0(-) (x) = q>1 (x), -0(+)(x) = p2 (x), and
z0(-) (x) = 9 (x), z0(+) (x) = 92 (x).
The regular terms y{ (x) are defined as solutions of problems:
A^ (( (x),x)) = -W(x)y(T) + F.^ (x),
y1>) = o, .^(1) = 0'
where
W (T) (x) = (x ) x )( (x)+ddyr('2 (x ),x)
and FtiT) (0) are known functions that depend on j (x), j(x), 0 < j < t -1. Function with the superscript " - ", defined for 0 < x < x0, and with the superscript " + " for x0 < x < 1. It is not difficult for them to obtain explicit expressions, for that we will make a replacement
We solve last equation with the help of an integrating factor and the final values y(i~) (x)andy\+) (x) can be represented as:
x
y(-) (x) = expJ - P(-) (5)d5
x x
yj~} (x) = exp{ - P(-) (s) ds ( Jq(-) (s') x
1 1
s'
xexp(-j- P(-) (s )ds)ds '1
s'
Jq(-) (s')exp(-J - P(-) (s)ds)ds'), 0 < x < x0,
x
yl+^(x ) = exp J-P (s )ds x
1
f x s '
x| Jq(s')exp(-[-P(s)ds)ds'-
x 0 < x < 1
and then we can calculate: z(+^ = dyj+^ / dx. 5. Construction of the transition layer
Equations for the coefficients Q{ y (t ), i = 0,1,... and A¿(+)zT),i = -1,0,1... will be obtained by substituting the expansions (6) into systems of equations dQ(t) y
dr
dQ^=A (t )((
dr
= sQ (T)z
(y(+) (x 0 +sr) + Q (+)y, x 0 + 8^
(z \x 0 + st) + Q (t)z ) +
+f (t) (( } (x0 + er) + Q (t V, x0 +ST) --A(t)(((t)(x0 +st),x0 + st~)z)(x0 +sr) -
- f }(r }(x 0 +er), x 0 +st) . (10)
We require that the transition conditions of the standard condition decrease at infinity: Qt(+)y (+œ) = 0,i > 0. Equating the coefficients of £0 in the first equations (10) and in the conditions (8), and also the coefficients of £ in the second equations (10), we obtain the following problems for the principal terms of the expansions (6):
dQTT ^Q^,=()+QVo) №,
dT dT (11)
Q0+)y(O) = Po"^12(xo)> Q0+V(+OT)=Mxo).
Problemsforfunctions with the index "-" will be solved for t <0, and with the index "+" - for t > 0. We introduce the notation
y 0 (t ) = (x 0) + Q0+)y (t ) and rewrite the tasks (11) with their use:
= Q?z, ) = A«)(y0(t),x0)Q?z,
yo (t) = Po, = (*„)• (12)
The systems of equations (12) coincident with the adjoint systems (1), therefore, for ^(xo)< ))<y2 (xo) functions defined:
Q_1(T)z =
J A(T)(s,x0)ds.
(13)
912 (x
2 0 ' S \ —1
From the sewing conditions (9) in the order s , taking into account the conditions for t = 0 of problems (12), we obtain equalities
p0 p0
J A(-)(s,x0)ds = J A(+)(s,x0)ds = z_1. (14)
91 (x0) 9i(x0)
The first of these equations represents the equation for determining the unknown quantity p0 From the second equation (14), we can determine the coefficient z _( .Aftertheexpressions (13) are obtained for the functions Q-'|)z, we can return to the system (12) and put the Cauchy problems for the functions:
y 0" > )=p0,
dy0 (r) = dr -,
91,2 (x0 )
From the inequalities (13) and the expressions (12) it follows that for the functions Q0(+)y we have the following exponential estimates:
Q0(T)y (t)|< c exp (-k |r|), (15)
where c and k are some positive numbers.
The transition layer functions (t), Q(t)z(t) , k > 1 are determined from the following systems of linear equations:
J A(T)(s,x0 )ds.
dQTV
dT
= Qk-1 z,
= A^ (t) • Q^z + A? (t) • ■ Q:]y + G(-1 (t),
dT (16)
QkT)y(0) = Pk-yt](x0), Q(T)y(+«>) = 0, ( )
where Gk+|(r) are known functions that are recurrently expressed in terms of Qjz, —1< j < k — 2 and Qj(+)y, 0 < j < k -1, and introduced designations
aA « (t) = A« ((T),^)) -t) (T) = A* ( (T), x0 )• The tasks for functions with the superscript "-" are solved on the half-line t < 0, and with the super-(16)
script "+" - on the half-line t > 0 . Using equalities
A(A>(y 0* )(t), x 0))V
0%), x 0 )•
^ V =
dy(T}
dT
+ A f >(( 0T\t), x 0). = A ) (t ) • Q$z + An y) (t ) • • QkT V. (17)
Second equations (16) can be reduced to the following form:
dQ;
gz_d (A (t)-q{: V)
dT dT +Gk3(T).
Integrating in the range from to t taking into account the conditions of problems (16) as t we obtain expressions for the functions
Q&(t) :
T
Qgz = AM(t)-Q^y(t)+JcQ(s)ds. (18)
+ro
We substitute these expressions in the first equation (16) and integrate it with allowance for the initial conditions. Then we obtain the functions Qk+^y (T) in an explicit form:
Q?V(T) = {pk -y(}(xo))e
A-+> (s )ds
T Ja (+\q )dq 5
+ Jes ds JGg(q)dq.
0 + œ
For functions G^|(t) exponential estimates of type (17) are valid; therefore, similar estimates are also valid for the functions Q^y (t) and Q^z (t) . From the matching conditions (9) in the order sk-1, we have the equalities:
zS (Xo ) + Qkiz (0) = zk+1 (xo ) + Q^z (0) = Zk> 1.
Substituting the expressions (18) in these equations, we obtain equations for the unknown quantities pk and zk_1, solving which we find [4, 200-205]
Pk = ( a (po ,xo ) - a« (po ,xo ))-1 (zkl (Xo ) - z-1 (x0 )
+A(po,x0 }yt) (x0 ) - AM (po,x0 )yk (xo ) + + } Gkj(T)dT-J Gk:i(T)dT,
Zk-1 = z£1 (xo ) + A(-) (po >x0 )( - yï] (x0 )) +
+ J Gk:i(T)dT.
—œ
6. Construction of left boundary functions
For L y (
t0 ) we have a problem
+
dL0y _
< dT0
Lo y (0 ) =
Making the
y =02 (0)+ L0y (T0
dz
< dT0
change of z = y in (19) we get
/2 (y,0) = z>
dT„
(19)
variables
(20)
y(0) = y0, y(co) = 02(0).
We write the problem for Lky (t0 ),k >1 lowing form
d2Lky
dTo2
= /2y (T0 )LJ + hk (T0 ),
hy (0 )=-^i- H=0
where f2y (T0) = f2y (02 (0) + L0y,0,0), and hk (T0) -known functions that are recruited through j)(0 )(k > j ), Lky (T0 )(k -1 > j ).
We can express a solution Lky (t0 ) ing form
Lky(r0) = y^)^ z(tO)s(+) (t0), (21)
where
"0 't (To ) = \~Z{-l)(1)d1\~Z(s)hk (s)ds
and z (5) we can find from (20). And from (21) we obtain an exponential estimate for Lky (t0 ).
7. Construction of right boundary functions
The terms of the boundary layer Ry(t^s) are represented in the following form:
" 1 — x
Ry(Ti,s)=Yf'R'^ Ti =— •
i=0
To determine boundary functions, we need boundary conditions.
Substituting the expressions (5) for y(x,s) into the boundary conditions, we obtain equalities dy^ 1 dQ(+) f 1 V 1 dR
dx
-(U) +
dT
Vs J
s dTx
(0,s) = y\ (22)
We substitute the desired expansions for y , R , Qin this equation and equate terms with identical powers of s in both sides of the equalities. Since the boundary condition is given at the point x = 1, the term dQ(+) mt(1/e,e) on the right-hand side of equation (22) can be dropped. Thus we obtain:
dRo x „ dR, . x , dyo,+) _
—(U) = o, —^ (l,e) = y1 -^jM1),
dT, dT, dx
)(l,s) = -d&(l),i = 1,2,3,... (23) dT, dx
We also add conditions for the vanishing of
boundary functions at infinity:
Rt (Ti,e)(oo) = o, ) = 1,2,3,... (24)
Equating terms with equal powers of s in both
parts of equalities, as a result we obtain an equation
for the function Ro (t1 ) : d2
—Ro (T1 ) = Ro f = f (0(1) + Ro,o,o )-f ((1),o,o ),
where f (0(1),o,o ) = o.
The equations (23), (24) follow the boundary
conditions d R
n-(1) = 0, Ro (+») = 0, y (t, ) = 0(1) + Ro (t, ),
dT
y y '=o-
dr,
= 0.
We introduce a change of variable
J y' = v, a (+<») = 0,
[v' = y' = f (y,0,0), v(0) = 0.
We form the characteristic equation 0-X 1
fy(0(1),0,0) 0-X
Hence we obtain A2 - fy (0(1) ,0,0 ) = 0 A = ^ fy ((1),0,0) .As fy ( (x ), x ,0 ) > 0 (stability condition) is a rest point of the saddle type y(t1 ) = 0(1), V(t1 ) = 0, it follows that R0(t1 ) = 0. Thus, in the boundary-value Neumann problem under boundary conditions, the boundary function in the zeroth approximation R0 is zero. For i — 1 we
obtain the equation for determining R1 (t1 ) :
d2R
dR = R1 f = fy (1)R1, t1 > 0.
From (23) the boundary conditions follow
dt (1)=y1 - £ w, r w=0.
From here R1 (t1 ) = h0 exp (-k0t1 ) / k0, t1 > 0, where h0 = dy0+) (1) / dx - y1 ,k0 = fï) > 0. Thus, the function R1 (t1 ) has an exponential estimate |R1 (t1 )| < cexp(-kT1), T1 > 0, where c, kare positive numbers not depending on £. In different assessments, they can be different [5, 2042-2048]. Similarly, the coefficients are found. Then the remaining coefficients of the expansion for the boundary function R, when i > 2 d2R
t4 = Rf = fu (1)Ri + r{ (T1 ), A > 0,
dT1
A (iH-f-1 (1), R H=0,
0
where ri (t ) recurrently expressed in terms of the usin§ the boundary conditions and the results ob-
c T>(r\ -.-u u tained above, as a result of simple transformations,
function Rj (T ) with numbers j<i. r
8. Numeric Example
We consider the boundary-value problem
we obtain:
s y =
(1 - x)y' + 6(1 -x),0 < x <-,
(x - 3)y ' - 2 (x - 3),-< x < 1,
y (0) = 0, y ' (1) = 1,
y (x ) =
-1
V
y
r x2 3 '
--3x
'2 +1
0 < x ^, 2
-< x < 1. 2
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