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Unformatted text preview: PARTIAL _______
DIEEERENTIAL
EQUATIONS __»___ An Introduction David Colton  University of Delaware The Random House/Birkhéuser Mathematics Series ﬁ Random House
New York 6 / INTRODUCTION and hence we can choose 6¢(x,t)
PthJ) = 90  (1.12)
at
Equation (1.10c) now implies that
p 84>
plow) = [ ,° ]— (m),
f (90) at
and from (1.10a) we have
624) I
E — [f (OMAN) = 0 Setting c2 = f ’(po) and assuming that f ’(p0) > 0 gives the wave equation
A34) = —2 —2 . (1.13) Note that from (1.12) the pressure p,(x,t) also satisfies the wave equation (1.13).
The change of variables t' = ct reduces (1.13) to
6sz A34) atlz , I (1.14)
which is the form of (1.13) we shall consider in the next chapter. ‘ In order to determine ¢(x,t) from (1.13) or (1.14), we must prescribe initial
and boundary data. In particular, at time t = 0 we must be given the initial velocity
and acceleration, i.e. ¢(X,0) and 6¢(x,0)/6t. On the boundary of the region con
taining (or contained in) the ﬂuid there are various possibilities. For example, if the
boundary is rigid and impenetrable, then v1 v = 0 on the boundary where v is the
normal vector, i.e., 6¢/6v = 0. At the other extreme, if the boundary is a fixed
surface which is a site of pressure release so that the pressure vanishes there, then
it suffices to set (I) = O on the boundary. Note that if d>(x,t) is time harmonic, i.e., ¢(x,t) = ¢(X)e_i“",
then ¢(x) satisfies the Helmholtz equation
A34; + k2¢ = 0, where k = w/c. In this case it is no longer necessary to prescribe initial data.
The wave equation also appears in the theory of vibrations and electromagnetic
wave motion (cf. Baldock and Bridgeman). 1.2 FIRST ORDER LINEAR EQUATIONS The simplest type of partial differential equation in two independent variables is the
first order linear equation. Our reason for studying such equations right at the be
ginning is for the DedaEOgical value in introducing characteristic curves and their 3
g
l
:t
’1‘
1;
g
h
i
l
i
t
E 1.2 First Order Linear Equations / 7 importance in solving initial value problems, as well as for the fact that our results
on first order equations will be needed shortly whenwe want to reduce second order
equations in two independent variables to canonical form. Our aim is to show how
the first order linear partial differential equation a(x,y) 6—“ + b(x,y) 6—“ + C(x,y)u = f (XLy) (1.15) 6x 8y
can be solved by reducing it to an ordinary differential equation. We make the as
sumption that the coefficients a(x,y), b(x,y), c(x,y), and f (x, y) are continuously dif
ferentiable functions of x and y in some domain D, and that a(x,y) and b(x,y) never
both vanish at the same point. We further assume that the coefficients a(x,y), b(x,y),
c(x,y), and f (x,y) have real values. We shall show that by a change of variables we can reduce (1.15) to an equation of the form 6
6—: + s(§m)w = t(§,n), (1.16) which is an ordinary differential equation in 1’; depending on the parameter r], where
g and 1] are new independent variables. If a(x,y) or b(x,y) is identically zero, then
(1.15) is already in the form (1.16), and hence we suppose that neither a(x,y) nor b(x,y) is_identically zero.
‘ Let 4) 6 01(0), ll} 6 C‘(D) be such that the Jacobian is not identically zero, and define
«E = 4>(x,y)
n = 4106,11). Then if u(x,y) = w(§,n) we have 6u_6w6¢+6w611 6x 6% ﬁx an 6x
Bu _ 6w ()4) 6w all: ay_ag 6y (may. ' 6 6 6 a a a
<a—¢+b—¢)—W+<a—¢+bi>—w+cw=f. (1.17)
6x 6y 6."; 6x 6y 67]
Hence if we choose tl;(x,y) such that
6 a
a(x,y) i, + b(x,y) —¢ = 0, (1.18) 6x 6y 8 / INTRODUCTION then we arrive at an equation of the form (1.16). The problem now is to construct
a solution 111(x,y) of (1.18) and a function ¢(x,y) such that the Jacobian J(¢,li1) is not
identically zero, g ' ‘ Suppose fer the moment that a solution ii;(x,y) of (1.18) exists such that
amp/6y is not identically zero. Define the curve y = y(x) implicitly by ¢(x,y) = y
where 'y is a constant. Then 6 6
—¢dx+—¢dy=0,
6x 6y which implies that dy _ (NJ/ax a _ _6tir/6y
_ b(x,y)
_ a(x,y) ' Thus ¢(x,y) = y implicitly defines a solution of the ordinary differential equation ﬂ = b(x,y)
dx a(x,y) ' (1.19) Conversely, if ¢(x,y) = 'y implicitly defines a solution of (1.19) such that adj/By
is not identically zero, then ti: satisfies (1.18). Equation (1.19) is called the char
acteristic equation of the partial differential equation (1.15) and defines a one
parameter family of curves called the characteristic curves of the partial differential
equation (1.15). . Having found ¢(x,y) as the solution of (1.19), we can choose the function
¢(x,y) arbitrarily such that the J acobian doesn’t vanish. For example, choose d)(x,y) =
x. The ordinary differential equation (1.17) for w is then 6w C(x,y) _ f (x,y)
— + w —
6% a(x,y) a(x,y) and from the elementary theory of ordinary differential equations the solution of this
equation is E a g .
_ cram f(€,n) (J C(Em) ) ] , — — d d d +d
we 1') exp< I am) g «mew am) 5 g (1‘) where d(n) is an arbitrary function of n and a(§,n) = a(§,y(§m)), c(§,n) = c(§,y(§,n)),
etc., where the function, y(§,n) is determined by solving for y in the equations .5 = x, 1] = 111(x,y). Hence, the general solution of (1.15) is “(16,30 = {W(x,lil(x,y)) + d(li!(x,y))} (120) v(x,lii(x, y)) 1.2 First Order Linear Equations / 9 where ,
«E
_ c(€,n)
Mm) _ “pg arm) dg)
E
wem>= fgm)memda
a(§,n) : EXAMPLE 1
Find the general solution of xux — yuy + u = x. I Solution. The characteristic equation is
it = _2
dx x This is a separable ordinary differential equation, i.e.,
dy dx
7 2 7 '
Integrating both sides gives
log y = —logx + c
where c is an arbitrary constant, or
x)’ = 'Y
where y is (another) arbitrary constant. Hence, setting
a = x '
7i = x)’
in our first order partial differential equation yields
alf+ l w = 1,
6E E
whose solution is w(§,n) = 1E + 1dm),
2 E where d('n) is an arbitrary function of n. The solution of xu, — yuy + u = x is now ‘
given by x 1
“(9600 = E + ;d(x)’) In order to guarantee that u is continuously differentiable we require that the arbitrary
function d be continuously differentiable. I 10 / INTRODUCTION We now turn to the problem of solving initial value problems for first order
linear partial differential equations. The Cauchy problem for the partial differential
equation (1.15) is to find a solution of (1.15) taking on prescribed values d>(x) on
a specified curve C: y = y(x) in the xy plane. The curve C must not be a characteris
tic curve of (1.15) (or tangent to such a curve). On a characteristic curve we have
¢(x,y) = 'y where y is a constant and hence on this curve (1.20) becomes WOW) + 61(1) Wm) Wm) ’ u(x,y) = i.e., u(x,y) does not equal d)(x) on C: y = y(x) unless d)(x) is of the special form , k
W(x 'Y) + Wm) WM) d>(x) = where k is a constant. On the other hand, if d>(x) is of this form, there exist infinitely many solutions of (1.15) given by (1.20) where d(n) is any differentiable function
such that d(y) = k. [I EXAMPLE 2
Find the solution of xux — yuy + u = x such that u(x,y) ‘= x on the curve y = x2. I Solution. As we have shown in the previous example, the general solution of
xux — yuy +. u = x is given by x 1
Hwy) = E + ;d(xy) where d is an arbitrary differentiable function. Hence, on y = x2 we want x+1d(3)
x=—  x,
2 x
i.e., x2 3 = d(x3), 01‘ 1
dx =x2/3.
() 2 Therefore x 1
ux, =—+—x 2/3.
( y) 2 2x( y)
Note that we must not allow (x,y) to be on the lines x = 0 or y = 0 because there u(x,y) is not continuously differentiable. In particular, no solution to our Cauchv A
E;
It. x:
(
:r . Watqr" W 1 mmmmwmmmwmwmwﬂwm 1.3 Classification of Second Order Equations and Canonical Forms / 11 problem exists in a neighborhood of the origin. This is a consequence of the fact
that at the origin the curve y = x2 is tangent to the characteristic curve y = 0. I For a more complete exposition of first order partial differential equations,
including quasilinear equations, the reader is referred to Courant and Hilbert 1961,
Garabedian, and John. ' 1.3 CLASSIFICATION OF SECOND ORDER EQUATIONS
AND CANONICAL FORMS 1.3.1 Types of Second Order Equations We now consider second order quasilinear partial differential equations in Euclidean
n space R" and show how at a given point in R” they can be classified into four
distinct types. As we shall see later, the type of a partial differential equation dictates
to a large extent the behavior of solutions to these equations. In this sense our clas
sification procedure can be compared in spirit to that of a zoologist who classifies
animals into mammals, birds, fishes, and reptiles and then proceeds to study rep
resentative examples from each of these classes.
‘ Consider the second order partial differential equation 62M Bu 1 au
2 an + f x1....,x,,,—... — =0 (1.21) 6x1 ' ’ax, where 0,, : aij(x1,. . .,x,,) are given real valued functions defined in a domain D C
R” and without loss of generality assume that ail : aﬂ. Let (x‘ﬁ, . .,x2) be a fixed
point of D and consider the quadratic form ll ' 2 awed), . .,x2):,.t,. (1.22) i,j=1
Then (1) Equation (1.21) is of elliptic type at (251),. . .,x?,) if at this point the qua
dratic form (1.22) is nonsingular and definite, i.e., it can be reduced by
a real linear transformation to a sum of n squares all of the same sign. (2) Equation (1.21) is of hyperbolic type at (x?,. . .,x2) if at this point (1.22)
is nonsingular, indefinite, and can be reduced by a real linear transfor—
mation to the sum of n squares, n — l of which are the same sign. (3) Equation (1.21) is of ultrahyperbolic type at (x?,. . .,x3) if at this point
(1.22) is nonsingular, indefinite, and can be reduced by a real linear
transformation to the sum of n (n 2 4) squares with more than one coef
ficient of either sign. * 12 / INTRODUCTION (4) Equation (1.21) is of parabolic type at (x?,. . .,x2) if at this point (1.22)
is singular, i.e., it can be reduced by a real linear transformation to the
sum of fewer than n squares, not necessarily all of the same sign. Equation (1.21) is of one of these types in D if it is of that type at each point
in D. There is no similar classification scheme for partial differential equations of
order greater than two and in the remainder of this book we shall consider only
second order equations. (This does not imply that higher order equations are not of
interest from both a physical and mathematical viewpoint!) El EXAMPLES 3—6 The wave equation
n.n,nn
6x2 ay2 622 6t2 is hyperbolic in any domain, the Laplace equation
ﬂiﬂ+ﬁ=0
6x2 ay2 322 is of elliptic type in any domain, and the heat equation
6214 6214 dzu du
Q + $5 + E = E is of parabolic type in any domain. The equation
@.@=n.@
6x2 ay2 622 at2 is of ultra—hyperbolic type in any domain. [:1 1.3.2 Reduction of Second Order Equations with
Constant Coefficients to Canonical Form This section shows how a partial differential equation with constant coefficients can
be reduced to a particularly simple form by a linear change of variables. In particular,
consider the partial differential equation with real valued constant coefficients au dzu
2 an + b.— + cu = foe... . .,x.>. (1.23)
i,j=l 39‘:an 6x, " n i=1 Define the linear change of variables gk : E Ckixi = 1a2a ' '3”):
i=1 where the real constants c," are to be chosen later such that det IcHI does not vanish. Li é
E
i
r
E
E. .v..—_mrmmm ‘2‘ “Maw w ,. W:Vﬂ»uﬂ.ir:lmu~v2:;ﬁcl«3:23.11r'ra..:y...v'i.51:, m m...“ 1.3 Classification of Second Order Equations and Canonical Forms / 13 Then (1.23) becomes ' 62" +256“+ ﬁg 5) (124)
akl i— C”: ls'':n 
“=1 aékaéz i=1. 3 .
where
dkl : 2 aljcklclj’
i,j=1
and the constants l3, and function f” (51,. . .,§,,) can be expressed in terms of the b, and f (x1,. . .,x,,), respectively. From linear algebra it is known that we can always choose
the CU such that ' ~ {0 when k 75 l
aid =
M when k = l
where M = 0, 1, or —1 and (1.24) becomes
6214  6L! ; xka—gi + ‘1 [LEE + cu = f(g,,. . .,g,). (1.25) The signs and/or vanishing of the M clearly determine the type of equation
(1.25) and hence of equation (1.23), since from the definition of type a real linear ll ll , change of variables cannot change the type of a partial differential equation. Equation (1.25) is called the canonical form of (1.23). In the case of equation (1.23) with
variable'coefficients, it is possible to reduce the equation to canonical form at each
given point (x?,. . .,xg) of D, but one cannot in general find a single transformation
that will work for all points in a neighborhood of (xi), . .,x2). An exception is the
case n = 2, if we allow our transformation to be nonlinear. We shall now show how
this can be done. 1.3.3 Reduction Of Second Order Equations in Two
Independent Variables to Canonical Form Consider the partial differential equation dzu dzu 6214 Bu 614
A——~+ZB—+C—+F x,y,u,—— , = 0 1.26)
6x2 axay 6y2 6x 6y ( where A = A(x,y), B = B(x,y), and C = C(x,y) are realvalued twice continuously
differentiable functions of x and y in some domain D C R2 and A, B, and C do
not all vanish at the same point. The factor of two appearing in the coefficient of
aZu/axay is purely for notational convenience to avoid having a factor of four ap— ‘ pearing repeatedly in our subsequent formulas. (The reader is warned to remember this factor when doing the exercises at the end of the chapter.) The quadratic form
corresponding to (1.26) is At? + 2Br1t2 + Cti ; , (1.27) 14 / INTRODUCTION (1 .26) is of hyperbolic type if B2 — AC > 0, of parabolic type if B2 — AC = 0, and
of elliptic type if B2 —— AC < 0.
Now make the change of variables «E = any) (1.28)
n = n(x,y)
such that §(x,y), n(x,y) E C‘(D) and the Jacobian
6:; 65,
a? 5
an 6n
E 5
does not vanish for (x,y) E D. Then (1.26) becomes  6214  6214  6211 _ au au
A ——2 + 23 + C —2 + F §,n,u,—,— = 0 6% Béan 6n 65, an where , 6 6 6 a 6 6
A(g,.)=.._€_€+23_€_é+c_€é
axax 6x6y ayay  a a a a ’ a a
C(§,n)=A—n—D+ZB—ﬂ—n—+C—nﬂ
axax axay ﬂy ﬂy B(€m)=A§§a_n+B<§a_n+a_§§ﬂ> +Ca_ga_n 6x 6x 6x 6y 6y 6x 6y 8y ’ and it can be verified directly that
3'2 — AC = (32 — AC)JZ; thus the type remains unchanged under the change of variables (1.28)! We shall show
that §(x,y) and n(x,y) can be chosen such that one and only one of the following
conditions is satisfied: (1)A=0, =0
(2)A=0, ~=0
(3)A=C, 3:0. Case 1 (B2  AC > 0). In this case (1.26) is of hyperbolic type in D. We shall
show how to choose €(x,y) and n(x,y) such that A = C‘ = 0. Assume that this is not
already the case in equation (1.26) and without loss of generality assume that A does
not vanish in a neighborhood of (x0,yo) E D. Motivated by the equations for A and E’.
E
i 1"N"'v“‘1v4rr marmnm ~ ‘riLﬁ‘ixb‘F‘rmm’min'mmawmwmwmxm'wtkga 3 mm '2'“??? «m3 1.3 Classification of Second Order Equations and Canonical Forms / 15 6‘, consider the first order partial differential equation a a ' a a a a
A—¢—¢+ZB—¢£+C—9—(B=O. (1.30)
6x 6x 8x 6y 6y 6y
Since 132 — AC > 0 we can write (1.30) as
a a a a
(Aa—¢+ (3+ VBz—AC)39><Aa—¢+ (B— \/B2—AC)a—¢) =0,
x , y X y and solutions of a .
A i) + (B + V32 — AC) 5933 = 0 ' (1.31)
6x _ 6y
and
A :1) + (B — V32 e AC) :1) = 0 (1.32)
x y will be solutions of (1.30). These equations can be solved by the methods given in
Section 1.2. Let ¢1(x,y) and ¢2(x,y) be non—constant solutions of ( 1.31) and ( 1.32),
respectively. Then ¢1(x,y) = constant and ¢2(x,y) = constant define the character
istic curves of (1.26), and (1.30) is called the characteristic equation of (1.26).
Since 32 ,~— AC > 0 and d>1(x,y) and d>2(x,y) are not identically equal to a constant, ¢1(x,y) is not equal to ¢2(x,y), and so we have two different families of real char
acteristics. Now let
§= ¢i(x,y) ' (1.33)
"‘l = (13205,” such that 6¢1(x0,y0)/6y and 6¢2(x0,y0)/6y are not equal to zero (this is always pos—
sible by imposing appropriate initial conditions on ¢1(x,y) and ¢2(x,y)). Then from
(1.31) and (1.32) we have «an an
ax ay 2V32 — AC 34)] 34,2
J = = —— — —
64» 6% A 6y 6y
6x 6y and hence the Jacobian J does not vanish in a neighborhood of (x0,y0). Thus under
the change of variables (1.33) we have A = C = 0 in (1.29), and since 32 — AC = (B2 — AC)J2 does not vanish, I? does not equal zero in a neighborhood of (x0,y0).
Dividing (1.29) by ZB now gives azu F<g Bu Bu) (1 34)
= 3 ,u’—7 , '
agan 7‘ ag an I 16 / NTFlODUCTON which is the canonical form for hyperbolic equations. If we now set §=a+B
n=a—B
in (1.34) we arrive at anequation of the form
62M 6214 614 an ‘
h—Z — ‘7 = ¢<asﬁaua—5_>a W 60L 68 Ga BB which is the second canonical form for hyperbolic equations. Case 2 (B2 — AC = 0). In this case (1.36) is of parabolic type. Since A, B, and
C do not vanish at the same time and B2 — AC = 0, it follows that A and C cannot both be zero at the same time. Without loss of generality assume that A does not
vanish at (x0,y0) E D. Then (1.31) and (1.32) are identical, 6 a
A—¢+B—¢=O, (1.36)
8x 6y , and since B2 — AC = 0 any solution of (1.36) also satisfies Bﬁ+Ca—¢=O. (1.37)
6x 6y
Then we can find a solution d>(x,y) of (1.36) such that 6¢(x0,y0)/6y does not equal
zero (and hence 6¢(xo,y0)/6x also does not equal zero) and the characteristic curves
of (1.26) are given by d>(x,y) = constant.
Now let E = ¢(x,y) and let 1] = n(x,y) be any twice continuously differentiable function such that the . Jacobian J of g and 1] does not vanish at (x0,y0). We then have A = 0 in a neigh
borhood of (x0,y0) and  6 6 6 6 6 6 '
B: <A£+Bi>ﬂ+ (B—¢+C—¢>—n=0
6x 6y 6x 6x 6y By
in a neighborhood of (x0,y0). Furthermore, 2
1 a a
C=_<yﬂ+3j>
A 6x 6y is not equal to zero in a neighborhood of ,(x0,y0) since otherwise from (1.37) we
would have J = 0. Dividing by C we have that (1.29) can be written in the form 62¢ F<g 614 Bu) 1 38
— : ’ ,u,—,— 3 I
2 n 6% an ( ) which is the canonical 'form for parabolic equations. E:
i
i
i i
E uromcr'rr‘r m mmhmammmwvmxwwvmmﬂumw' 1.3 Classification of Second Order Equations and Canonical Forms / 17 Case 3 (B2 — AC < 0). In this case (1.26) is of elliptic type. Assume that A, B,
and C are analytic functions of x and y in a neighborhood of (x0,y0) E D, i.e., A,
B, and C have Taylor series expansions in x and y which converge in some neigh
borhood of (x0,y0). Then the coefficients of (1.31) and (1.32) are also analytic and
it can be shown that (1.31) (which is a first order system of partial differential equa
tions written in complex form) has an analytic solution (in the sense defined above) d>(x,y) = d>1(x,y) + i¢2(x,y) defined in a neighborhood of (x0,y0) such that a¢/ax + 6<1>/6y does not vanish in
this neighborhood. Let §= <i>1(x,y) = Re d>(x,y)
n = <i>2(x;y) = Im d>(x,y). Then it can be shown that the Jacobian of ¢1(x,y) and ¢2(x,y) does not vanish in a
neighborhood of (x0,y0), and if we separate the real and imaginary parts in (1.30)
we obtain ‘ ACE) + 2365?? + C(a—E) = A("_“) + 233111“ + C(61) 6x axdy 6y 6x 6x 6y 6y
and
a 19 a a a a a a + + + C : 0_
6x 6x 6x 6y 6y 6x 6y 8y Hence it follows that A = C and B = 0. But since B2 ~ AC < 0, we have that A
and C do not vanish in a neighborhood of (x0,y0), and if we divide (1.29) by A we
have 62u + 0214 F<g Bu 614) (l 39)
—_ — = a ’u3—9—_ ' 2 n 6“; an I
which is the canonical form for elliptic equations. Note that it may turn out that (1.26) is of different type in different parts of
the domain D. Equation (1.26) is then said to be of mixed type. For example, the
Tricomi equation dzu + 6%! _ 0 y 6x2 6312 is elliptic for y > 0 and hyperbolic for y < 0. I: EXAMPLE 7  Reduce the equation 6214 . dzu 2 6214 614
——2—231nx —cos x—Z—cosx——=0 (1.40)
6x axay 8y 6y to canonical form. 18 / INTRODUCTION I Solution. B2 — AC = sin2 x + cos2 x = 1 and hence (1.40) is of hyperbolic
type. The equations for ¢1(x,y) and d>2(x,y) are a 3
—¢— (sinx— mi): 0
6x 6y
. a
£—(smx+1)—¢=0
ax 6y
To solve these equations consider the characteristic equations
dy
— = — sin x — 1
dx ( )
d
——y = —(sm x + 1)
dx
and solve them to obtain y = x + cos x + c and y = —x + cos x + c, respectively, where c is an arbitrary constant. If we now make the change of variables
5: ¢1(x,y) =x— y + cosx
"n = ¢2(x,y) =x + y  cosx in (1.40) we arrive at the equation azu .
.= 0. ' (141)
agan ‘
We can now actually solve this equation since (1.41) implies that
an
‘r = f (n) (142)
6n , where f(n) is an arbitrary function of n, and (1.42) implies that
Winn) = (WI) + $03), where (Mn) and ME) are arbitrary functions of their indicated variables. Hence the
general solution of ( 1.40) is u(x,y) = d>(x + y — cosx) + [11(x — y + cosx) where we now require that the arbitrary functions (Mn) and 41(23) be twice continu
ously differentiable so that u(x,y) also has this property. I 1.4 FOURIER SERIES AND INTEGRALS For future use, we want to know when a function f (x) of period 211' can be expanded
in a Fourier series, which has the form f(x) = 2 anew. (1.43) 1.4 Fourier Series and Integrals / 19  in
If we can integrate termwise, it follows from the orthogonality of the set .{e } that an f(x)e_i""dx (1'44) —2’1T and henceforth we shall assume that the coefficients {an} in (1.43) are given by
(1 44). Our problem is to determine'under what conditions the equality (1.43) is valid. More precisely, if we define the partial sums of (1.43) by N
SW) = E anew, n=—N “11’ (1.45) when does
lim SN(x) =' f(x)? Nam We begin by noting that from (1.44) and (1.45) we have (1.46) The function . 1 
mm =— 2 e'"" is known as Dirichlet’s kernel. It is easily seen that DN(—x) = DN(x) J’ DN(x)dx = 1. (1'48) " 1T
Furthermore, since N N
_ I . .N 1 _ _Nx _ i(N+l)x
(1 — e”) E em" = E (e‘Nx — e“ + )x) — e l e n=—N n=—N 1 e—iNx _ ei(N+1)x 1 e—i(N+<1/2))x _ ei(N+(1/2))x EWC02=E; 1__eu 211' 1) (1.49) e—i(x/2) _ ei(x/2) (1.47) r ...
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