Citation for published version (APA):
Brands, J. J. A. M. (1983). On the differential equation yy" + 2xy' = 0. (Eindhoven University of Technology : Dept of Mathematics : memorandum; Vol. 8313). Technische Hogeschool Eindhoven.
Document status and date: Published: 01/01/1983
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Memorandum 1983-13
ON THE DIFFERENTIAL EQUATION yy" + 2xy' = 0
by
J.J.A.M. Brands
August 1983
Eindhoven University of Technology
Department of Mathematics and Computing Science PO Box 513, 5600 MB Eindhoven
by
J.J.A.M. Brands
Department of Mathematics, Eindhoven University of Technology, The Netherlands
Abstract
The asymptotic behaviour for x + 00 of solutions y(x) of the initial value
problem yy" + 2xy' == 0, yeO) = (l > 0, y'(O) = 15, and also the asymptotic
behaviour 15 + +00 and for 15 + -00 of L
:=
lim y(x) is investigated.x-+<»
1. Introduction
In this paper we consider the following initial value problem
(1) yy II + 2xy' = 0 (x > 0)
(2) yeO)
=
(l > 0, y' (0) ==s .
Equation (1) occurs in boundary layer theol"Y, (see [1], p. 22 - 23) and arises
fran the diffusion equation
~
=
l
(D
de)
dt dZ dZ •
In boundary layer theory one is also interested in the boundary value
problem (1), (3) with
(3)
yeo)
=
a. > 0 y(oo)=
L > 0 .A case of special interest is
(2')
yeo)
=
y' (0)=
y ,since all solutions of (1), (2) are expressible in solutions of (1), (2') by means of the transformation (11) (see Section 2). We shall study the
asymptotic behaviour of solutions y(x,y) of (1), (2') for x ~ 00. Moreover,
we shall pay attention to the asymptotic behaviour of the limit
L(y)
:=
lim y(x,y) for y ~ 00 and y ~ -00, since physicists seem interestedX-7<O
2. Results
In Section 3 the following fundamental results are proved. There exists a
unique solution Y(',a,S) of (1), (2), defined on [o,~). This solution
y(x,a,S) tends monotonically to a positive limit L(a,S) if x +~. The
transformation rule
(I 1) y(x,a ,S) = a yea
-i
x,y) L(a,S) = a L(y)enables us to consider only the solution y(.,y) of (1), (2') and its limit
L(y). This limit is a continuous increasing bijection L : ~ + (o,~). Hence
the boundary value problem (1), (3) has a unique solution.
In Section 4 the asymptotic behaviour of y(x,y) for x + ~ is determined.
(20) where co A =
~
1f!
L! Y exp [2f
s (L -1 - (y (s , y ) ) - 1 ) dS] •o
(x +~) ,In Sections 5 and 6 bounds for the limit L(y) are determined.
(36) 2 LO 10g(1 + LO Y ) 2 < L(y) - LO y 2 < 47 La 10g(1 + 3(16) -1 2 y) + 47 La
(y > 1.8)
(56) 1 - y -2 < L(y) exp
U
y2 +D
< I + y-2 (y < -7) ,where LO
=
0.3574 .•• denotes the limit of the special solution y of (1)3. Preliminaries
We shall inves tigate existence and uniqueness properties, and prove the continuous dependence of the limit as functions on the initial values.
(4) Theorem. There exists a unique solution y of the initial value problem
(1), (2) which is defined on [0,00). This solution y is positive and
mono-tone, and y(x) tends to a positive limit L if x + 00.
Proof. According to well-known existence and uniqueness theorems (see for example Sections 1.1 and 1.2 of [2J), there is a positive member, say 0,
such that there is a unique solution, say y, of (I), (2), defined on [0,0).
Let I be the maximal interval of tbe form [O,a), (possibly a may be 00) on
which y is positive and satisfies (1), (2). Dividing both sides of (1) by
yy',
integrating over [O,x] with x E: I we getx
(5) Y t (x)
=
S exp [-2f
s(y(s» -1 ds ] (x E: I) ,°
a result also correct if y'
=
0. It follows that(6)
°
~ Y I (x) sgn B ~I
a
I
(x E: I) •Let a > 0, a e I. Dividing both sides of (1) by xy integrating from a to x,
X E: I, x > a, taking exponentials, we find
x
(7) y(x) = yea) exp [!a-t yt(a) -
~x-l
y'(x)-!
f
S-2 Y' (S)dS] •From (6) and (7) we infer that
(8) y(x) > yea) expUa
s
-1 y I (a) ] (a < x, x ~ I) ,where S holds i f B ;,: 0 and> i f B < O. It follows from (5) and (8) that y
is bounded, and increasing if
B
> 0, and that y is bounded away from zeroand decreasing if B < 0, and that I = [0,00). Clearly lim y(x) exists and
is positive.
We shall denote the solution of (1), (2) by y(x,a,B) and its limit by
L(a,a). In the case of (I), (2') we write y(x,y) and L(y). Simply by
in-spection we can prove
(9) Theorem. For every A >
a
(10) y(x,A 2 a,AB) = A yeA Z - 1 x,a,S) A Z L(a,B) •
Theorem (9) enables us to consider, without loss of generality, only the
initial value problem (I), (2'). We have
(11) y(x,a,S) == a yea
-!
x,y),
L(a,a) .= a L(y) y = a-!
S •(12) Theorem. L(') : lR -I- (0,«» is an increasing continuous bijection.
Proof. Let Y2 > Yl' Then there is a
°
>a
such that y(x,yZ) > y(x'Yl) on [0,0). Suppose that there exists a positive Xo such that y(xO'Y2) == y(xO'Yl)
and y(x'Y2) > y(x'Yl) for
a
< x < xo· Clearly y'(xO'Y2) < y'(xO'YI ) 'Dividing (1) by y and integrating from
a
to x we findx
(13) y'(x,y) - Y + 2xlogy(x,y) == 2
f
s logy(s,y)ds .App lying (13) twi ce wi th y
=
y 2 and y = y 1 and subs trac ting both results we get for x=
xox
y'(xo'Yz) - y'(xO'Y}) - YZ + Y1 = Z
f
slog (y(s,y 2)/(y(s,y1»ds ,a
a contradiction since both sides have different signs. It follows that
In the sequel of the proof we need the formula
(14) log L
=
i
J
a
00
-2
(y - Y f (s) ) s ds ,
where y
=
y(.,y), L=
L(y), and which is obtained by integration of y'y-1 =_~
x-I y", The integral in (14) exists since y"(s) :: -2ys + o(s)(s
+
0) and y - y'(s)=
YSz + 0(s2) (s+
0). Applying (14) to Y2 := y(o'Yz)and Yl := y(o'YI) respectively, and substracting the results we get
00
(15)
a
-I , - 1
I f Y1 > 0 then by (5), Y2 > YZYI YI which implies directly L2 > 1 + YZY1 (LI - 1),
If Y1 "" 0 L2 < (I +
2 then by (5),
Y2
< Y2 exp[-x L2
J
which upon integration givesl
'lr! Y 2) 2 •2 1
I f Y2 "" 0 then by (5)
Yj
> y1exp[-xJ
which leads to L} > 1 + i'lr2y}.-} I f Y
2 < 0 then by (5)
Y2
< Y2Y1Yj
which implies-1 -1
L2 - L 1 < (1 - y 2 Y I ) (1 - L I) < (1 - y 2 Y 1 ) ,
-I -1
and by (15) and (14) logL
2 L} > (I - YlY2 )logL2•
It follows that L(·) is continuous and strictly increasing on lR, and, moreover L(y) + 00 if y + 00, L(y)
+
0 if y + -».4. Asymptotic behaviour
First we investigate the asymptotic behaviour of solutions y := y(.,y) with y ~
o.
Using the inequality y(x) ~ L := L(y) in (5) we get2
y' (x) ~ yexp[-x /L] ,
from which it follows by integration over [x,oo) that
Rence (16) y(x)
=
L + cf(x -1 exp[-x /LJ) 2 (x -)0 00) • It follows that 00 (17) Q := 2f
s«y(s»-1 - L-1)dso
exists and that
00
( 18) 2
f
s«y(s»-1 - L-1)ds=
0'(x-1 exp[-x2/L])x
Using (17) and (18) in (5) we get
(19) Y I (x) = y exp[ -Q - x /L] (1 2 + Cf(x -1 exp[ -x /L]» 2
which upon integration from x to 00 gives
(20)
(x -)0 (0) •
(x -)0 (0) ,
Secondly we consider the case y < O. Then using L < y(x) < 1 in (5) we get y' (x)
~
y exp[ _x2]. By integration over [x,"') it follows that-} 2
y(x) = L + t1(x exp[-x J) (x -+ <XI) •
Repeating the same kind of arguments (after (16» as in the case y ~ 0 we arrive at the same Formula (20) except for the order term which is now d(x-Zexp[_X2(L-1 + 1)]). Hence Formula (16) holds. Again by repeating the arguments after (16) we get (20).
5. Bounds fo-r the limi t if y > 0
By (10) we have
(21) L(I,y) =y L(y 2 -2 ,1).
We shall prove that there are positive constants C1,C 2 such that for a > 0 sufficiently small
For simplicity we w-rite y, L, Yo' LO instead of y(-;a,l), L(a, 1), Y(';O,l), L(O,I) -respectively. It is easily seen that
(23) YO(x) = x - x 2 + ~(x 3 ) (x .} 0) •
Clearly, Formula (5) with
a
= 1 holds for yO' It follows by (5) thatinitially y'(x) > Yo (x) , since initially y(x) > YO(x). By standa-rd reasoning we can conclude that
(24) , y(x) > YO(x) + a (x > 0) •
For late-r use we shall prove the following facts:
(25) (26) > x - x 2 + -x 1 3 3 -2x YO(x) < ~(1 - e ) (27) La
=
0.35742210059 ••• 2 > x - x (0 < x:S: I) (x > 0)Proof of (25), (26) and (27). From YO > 0 and yo(O)
=
0 it follows that YO(x) > xyo(x) ex> 0). From (1) we deduce YO > -2 which leads to YO > x - x2• Using this last result in (5) we find YO > (1 - x)2(0 < x :s;; 1) whence YO > x - x2 +
1-
x3 (0 < x:S;; 1). Clearly LO > YO ( 1) >~
•-2x -2x
Using YO(x) < x (x> 0) in (5) we get YO < e and YO <
!(l -
e ). Hence LO < ~. A numerical computation gives (27). We also needLet 0 < as;
~.
The line y ... a - a2 + (1 - 2a)(x - a) is tangent to the curve y=
x - x2 in the point (a,a - a2). It follows by (Z5) that2
YZ(x) := y(x;a ,1 - 2a) equals YO(x) for some x ... b < a since Y2(a) < YO(a). At x
=
b we have YO(b)=
Y2(b) and YO(b) > Yi(b). We show that YO(x) > Y2(x) for all x > b. Immediately to the right of x=
b we have YO(x) > Y2(x). Then in some interval (b ,b + 0)x x
=
YO (b) exp [ -2J
s (yo (s) ) -1 dS] > YO (b) exp [ - 2I
b 0
By standard arguments it follows that YO(x) > Y2(x) (x > b).
Letting x + 00 we find L(a2,1 - 2a) < LO' By (10) we have L(a2,1 - Za)
=
... (I - 2a)2 L(a2(1 - 2a)-2,1). Substituting a'" a,!(I + 2a,!)-1 we get (28).
The function K defined by (29) satisfies (30).
(29) K(x) := y'(x)/YO(x) .
(30) K'(x)/K(x)
=
2x«yO(x» -1 - (y(x» -1 ) , K(O) "" 1 •Using (24) and the inequalities YO(x) < x, YO(x) < 1 for x > 0, we see that
-1
K'/K> 2a.(yo + a) YO'
from which it follows by integration 2a
K > (1 + YO/a.) •
Using (29) and integrating we get
Letting x + ~ we find for all a. > 0
Using that LO < ~ we derive
Now we turn to the proof of the right hand inequality of (22). Let u := y - YO' Then u satisfies
(32) y u" = -2x u' - y" u
1
We want an upper bound for u(4).
From (24) we know that u' > 0, and also y > a + YO' Moreover, we have
1 2 3 1
-yO:S; 2, and, using (30), yO(x) > x(l - x + 3"x ) > 4 x on [0'4J. Hence
(33) un :s; 2(a + 4x) 3 -1 u
We introduce a function v as follows: v(O) == a, v'(O) == 0, v satisfies (32) with equality. Then vex) <:: u(x) on [0
,i
J •
This follows by standard arguments using the differential (in-)equalities for ~ := u'/u, ~ := v'/v. Since v is convex we have v(s) :s; 48 vI + a (] - 4s) (O:S; S :s;!)
where VI := v<!). Hence3 -1
v"(s):S; 2(a +4s) (4sv
I + fl(l - 48». Integrating twice we find
1
from which an upper bound for VI arises, thus also for u(4)' We get
(34) u(t;) 1 < a + a logO + 3(l6a) -1 ) .
1
Finally we shall show that u(~)/u(4) < C, where C is independent of a for
1
a € (0'3")' Defining W := u'/u we get from (32):
-1 -1 2
W' = 2x(yOY) YO - 2xy q> - W =: F(x,q» w(O)
=
0 .1 -} ~ 3/2 -2
Let ~(x) := (YO + (2:xyo) 2) (yO) . Then ~'(x) - F(x,~(x»
=
(2x) (yO) (yO) +~ -} -} -1 -1
+ (2:xyO) (yO) (x(2(y) - (yO) ) + (2:xy0) ). Clearly ~' _ F(x,~) > 0 for
_I _1 - 1 - 1
o
< x < 2 ~. Further ~' - F(x,~) > 0 for x <:: 2 2 provided that 2(y) - (yO) > 0_1
on [2 2,~). This condition is fullfilled, since by (25) and (28), we have
Further, F(x,O) > 0 (x > 0), and ~ >
°
(x > 0). It follows (see [3] p. 177) that 0 < ~ <4.
Therefore, 1 log(u(oo)/u(4'» < 00 +J
1 -1 (2xy')Z(y) dx=
o
°
4' QO-i
r
-3/2 1 - 2 x (y ') 2dx < 1) 0 4' 1 + 2 (2y ,(.!»!
-log(LO/YO(4'»o
4 1log (LO/YO (4'» + 2 (2y
0
(*»
l .
1 I -~
Using YO(4) > 37/192 (from (25» and YO(4') < e (see proof of (26» we find
(35)
Combining (21), (31), (34) and (35) we can summarize the results of this section by
(y > ] .8) •
Remark. Numerical experiments suggest that
2 L - L
6. Bounds for the limit if y < O.
Integrating (1) in the form yll
=
-2xy y' over [1,x] we get -1(37) y(x)
=
x 1 + yx + 2f
o
2 -I (s - xs) (y (s) ) Y I (s) ds • T.T wr~t~ng , , 2 s - xs ( I X ) 2 1 2 d k' , 1 ' = s - 2 - 4x an ta ~ng exponent~a s we arr~ve at (38) where (39)y(x)
=
exp[2x -2 + 2yx -1 - 2x -2 y(x) + l(x)] , xlex) := 4x-2
J
o
2 -1
(s - ~x) (y(s» y' (s)ds •
Since I(x) < 0 we have
(40) y(x) < exp[2x -2 + 2yx ] -1 (x > 0) •
Obviously, we have the inequality L < y(2\yl-l) < exp[- iy2], but we can do
much better. First we will show that
This inequality states that y(2Iyl-l) is a very good approximation of L for
large
\y
I.Secondly, we will obtain sharp inequalities for y(2Iyl-l) which, by (41), will result in sharp inequalities for L.
Utilizing (5) with x == (l :==2hl-1 - 2Y'2hl-2 and x == S := 2Iy\-1
respectively, and (40) we have
2 2 -2 -1
> yexp[-(S - (l ) exp -2(l - 2yex ] >
/';::" -3 2 > y exp[-81'2 h 1 expCh]]·
Using this in (8) with a == 21yl-1 we get (41).
In order to obtain good estimates for y(2Iy\-I) we have, by (38), to find good estimates for I(2\y\-1). Therefore we put
(42) u(t) := yy(lr[-1 t)(y'(hl-1 t»-1 .
Then, with E := 2\y\-2 we get
t
(43) u(t) = -1 + EtU(t) exp
[ f
(u (s» -1 ds ] u(O) = 1 , anda
2 (44) I(2/yl-l)=
J
(t - I)2(u(t»-ldta
The solution u of (42) has the following global behavioo.r. It starts wi th the value 1, decreases till t
=
to' where it attains a minimum Uo := u(tO) >
a
f
t -1and thereafter it increases very rapidly. Since f(t) := u(t) exp [0· (u(s» ds]
-1 is increasing and f(t
O) == (E to) we have li(t) s -I + tlto (0 S t s to)'
2
from which it follows that u(t) s 1 - t + !t
Ito,
whencea
< Uo < I - !to ' Hence to < 2.We need the relations t (45)
=
e -1f
s (u(s» -1 dso
t t (46)J
( f
s(u(s»-1
ds - e-1)
,o
o
the correctness of both is easily checked by differentiation. Substituting to = to in (42), (45) and (46) we obtain
(47) (u (s» -1 ds
=
-log(e to uO) ,
(48) s (u (s) ) -1 ds
=
e -} - to •(49)
We will now prove that to > 1 for e > 0 sufficiently small. For suppose that to S 1. Then on the one hand we have
2 (s - s )ds ==
o
1 1 2 1
which imp lies uo > 1 -
'2
to -'3
to ~6"
On the other hand we have-1
(u(s» ds
=
-log(e to uO) < -log(e uO) )
-1 -1 . 1
implying that u
In the seque 1 we need the inequali ties
(50)
Using (47), (48) and the inequality u(s);':: - s (0 :S' s :S' 1) we can write
< -I _
to
=(0
s (u(s» -I ds <to
(0
(u(s» -I ds =-to
10g«to
uO)
o
0 1 e-1 - to >J
s(u(s»-lds +o
= f
o
1(s - 1) (u(s» -Ids +
(0
(u(s» -Ids> -I + 10g«to "0)
-I •o
from which (50) follows.
The next stage in our treatment ~s the proof of
(51 ) to - 1 <
!
e + 4e 2110g
e1
(0 < e :S' 0.05) • (52) Therefore to to (53)f
(s 2 - s)(u(s» -1 ds ;.::f
(8 2 - s) (to + u - s) -1 ds'" 0 1=
(t - l)[(u -1 + to - 1» 1 3 - u J > O + to)log(uO (uO + - - - t 0 2 2 0 0> (t - 1) [to log (u-1 - 1» 1 3 - uOJ .
O (to + - - - t
Using (48), (49), (50) and the inequality u(t) ~ 1 - t on [O,]J we find 1 to(l - to - uO)
=
J
o
2 -I (s - s)(u(s» ds + (s - s)(u(s» 2 -I ds >which with (53) leads to
from which (51) follows. Next we show that
2 1 > - -2 + (54)
f
(t - 1) 2 (u (t) ) -1 d t :s; 3£ 2 to
2 -1 (s - s) (u(s» ds, (0 < £ :s; 0.05) •Since 'tt = 3(11 + 1)2(tu)-1 + (11 + 1)2(11 + 2)u-2 > 0 we have li(t)
-1 -1 -1 I -1 2 2
= to + Uo > Uo for t > to whence u(t) > Uo + IUo (t to) .
Now, using this latter inequality and (50), (51) we have
2 1 < -2 > \i(t O)
=
(t > to)'(55) (s - 1)2(u(s»-lds
=
tf
[(to - 1)2 + 2(ta -
1)(s - to) +a
2 -1
+ (s - to) J(u(s» ds:S;
2 _1 2 1 -2 2
:s; (to - 1),2 2'1r + (2(t
a -
I)ua
+ uO)log(l +Iuo) < 3£ (0 < £:S; 0.05) •. to 2 -1
(48), (51) and (55),
o
1 (I - s)2(u(s»-l ds <J
o
2 -1 (I - s) (1 - s) ds + -1 s(u(s» ds:S: (0 < e: :s: 0.05) ,and, using (52) and (5 I) and (50)
to 1
J
2 -1f
2 -1 (1 - s) (u(s» ds > (1 - s) (~ + t - s) ds=
0 0 0 1 (to - 1) + [( U o + 2t - 2)uO + (t -2 + to) (u O + to - 1) -1) = - - 1) ]log(u O - u > 2 0 0 0 --1 --1 2"2 - "2
e: - 4e: ilog e:I
- -e 1 € € (0 < e: :s: 0.05) •Combining the last two inequalities with (38), (41), (44), (54) and (55) we can obtain
(56) - y -2 L(y) expUy2
+n
< 1+
y-2 (y < -7) •Remark. Numerical experiments suggest that
2 -2
Acknowledgement. The author wants to express his gratitude to J.K.M. Jansen (Department of Mathematics and Computing Science, Eindhoven University of Technology), who has done all numerical calculations which are mentioned in
this paper. Particularly in the case of y < 0, Section 6, the numerical results
strongly confirmed the initially very doubted conjecture that the constant
Reference$ .
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Jongen, P .H.M.; De warmte-overdracht van een gas met vibratierelaksatienaar de schokbuisachterwand.
Thesis 1971, Eindhoven University of Technology, Dep. of Physics, Eindhoven, The Netherlands.
[2J Hale, J .K.; Ordinary differential equations.
(2nd ed.) 1980, Robert E. Krieger publ. cy., New York.
[3J Bruijn, N.G. de; Asymptotic Methods in Analysis. 1981, Dover Publications, Inc. New York.