2 ux u x () () - polytechnique · DISCONTINUITY 0 ( ) 0 0, .,0 prescribed tx ddx dt dt uuu uu ux ux...
Transcript of 2 ux u x () () - polytechnique · DISCONTINUITY 0 ( ) 0 0, .,0 prescribed tx ddx dt dt uuu uu ux ux...
Dispersive Systems
1) Schrödinger equation 2) Cubic Schrödinger 3) KdV 4) Discreterised hyperbolic equation 5) Discrete systems
KdV
( ) ( )
2
0
0,0
t x xxxu uu uu x u x
ε+ + =
=
DISCONTINUITY
( ) ( )0
00, .
,0 prescribed
t x
d dxdt dt
u uuu u
u x u x
+ =
= =
=
t crit
( ) ( )1 0 xx u x>0u x
Collision of characteristic lines
critt t> No continuous solution for
Solutions of equ.-s of co idevelop discontinuities (shocks).
mpress ble flow
Stokes, Airy, Riemann: Conservation laws
( ) ( )0
, , .t x y zu f g h
f f u g g u etc
+ + + =
= =
Integrate over G: Integrate over G: 0,0,d
dt udV F ndS+ ⋅ =∫ ∫G
( ), , flux ( ), , flux F f g h∂
=G
t t
ributionsu 0
0.
xb
db adt
a
f
udx f f
+ =
+ − =∫
WeWeak solutions, dist
a b( ), dyy t s
dt=
[ ]
[ ]
[ ],
yd udx udx⎡∫ ∫
yb b
t t l rdta y a y
y b
x x l ra y
l a b r l r
u dx u dx s u u
f dx f dx s u u
f f f f s u u
⎤+ = + + − =⎢ ⎥
⎢ ⎥⎣ ⎦
− − + − =
− + − + + −
∫ ∫
∫ ∫
So by the conservation law, [ ] [ ]s f u=
Rankine-Hugoniot condition Example:
( )( )[ ] [ ]
2 2
212
212
12 2
e r e r
e ru u
0
0t x
t x
u u u u
u uu
u u
f u u
s f u − +
+ =
+ =
=
= = =
−
Entropy condition: u u> e r
0t x+ =
( )
1t xu u u+ = 0
log 0
RH condition
log log
xt
e r
e r
e r
u uu
a u
u usu u
u u
+ =
−=
−
+≠
2
Multiply equation
( )2 323
3 3 3 2
2 2
2 0
0
R-H condition2 23 3
t x
t x
e r e e r r
e r e r
u u uu
u u
u u u u u usu u u u
+ =
+ =
− + += =
− +
Which conservation law??
Goodman-Lax; in 0t xu uu+ = discretise space:
( ) ( ), ku k t u tΔ Replace byxu symmetric difference quotient:
1 1 02
.
k kk k
u uu u
ddt
+ −−+ =
Δ
• =
Dispersive Conservation form
( )
1 12 2
1 12 2
0,
3,0 cos2
k kk
kk k
f fu
f u u
u x x
+ +
+ +
−+ =
Δ=
= +
Numerical experiments reveal oscillations and weak convergence for critt t>
1. Explain oscillatory behavior 2. Prove weak convergence 3. Determine weak limit 4. What equation does the weak limit
satisfy?
1 12 2
12
1
112
1 1
02
0
Divide by :
log 02
k kk k
k kk
k kk
k
k kk
u uu u
f fu
f u u
uu uu
+
+ −
++
+ −
−+ =
Δ−
+ =Δ
=
−+ =
Δ
1 12 2
12
1
log 0
2
k kk
k kk
g gu
u ug
+ −
++
−+ =
Δ+
=
( )( )
( ) ( )
If would tend strongly to , both
0
andlog 0
would be satisfied.
t x
t x
u u
u f u
u g u
Δ
+ =
+ =
But they are incompatible for discontinuous solutions.
The solution of the difference equation desperately tries to satisfy two incompatible conservation laws. Hence the oscillations.
Limits of Semigroups. ( ),Denote by U t ε the solution operator for
the KdV equation 0,t x xxxu uu uε =+ + (1) that is, the operator
( ) ( ) ( ), : ,0; , ;U t u x u x tε ε ε→ that relates initial values of solutions of (1) to their values at t. Clearly, the operators
( ),U t ε form a semigroup (even a group) in the parameter t:
( ) ( ) ( ), , , U s U t U s tε ε ε (2) = +Lax & Levermore have shown that the limit ( )0 of ,U tε ε→ exits in the weak but for not in the strong topology: critt t> ( ) ( )0 0.,U t tu U uε → (3) Note that since equation (1), written in conservation form
( )212 0,t xxxx
u u uε+ + = is nonlinear, a weak limit of its solutions that is not a strong limit is not a solution of the limiting equation ( )21
2 0.t xu u+ = (4) ( )So 0U t u in (2) is not the solution of (4)
with initial value . 0u Question. Do the limit operators ( )U t defined in (3) form a semigroup:
( ) ( ) ( ).U t U s U t s+ (5) L & L have derived an explicit formula for ( )U , which shows that (t )U is not
continuous in the weak topology. That makes it very dubious that the limit as
t
ε tends zero of (2) is (5). The question can be easily decided by a single calculation.
There is an analogue of this question in the theory of turbulence. Here operators to be considered are the solutions operators ( ),U t R for the Navier-Stokes equation. I
surmise that as the Reynolds number R →∞, the operators ( ),U t R tend weakly but not strongly to a limit ( )U t .
Since the limiting Euler equations are nonlinear conservation laws, ( )U is t not the solution operator for the incompressible Euler equation but a description of turbulent flow. In analogy with the zero dispersion limit of the KdV equation I would surmise that the operators ( )U , the limits of
t( ),U t R , do not form a semigroup,
as conjectured by Heinz Kreiss.
Compactness 0, 0,u uu t t x = ≥ +
can be written as conservation law ( )21
2 0.t xu u+ =
(Initial value ) ( )0,0u x u x= , bounded and zero outside a finite interval. Viscosity method
( ), 0, , ; , t x xxu uu u u u x tε ε ε+ = > =
( ) ( ) ( ) ( )0,0; . lim , ; , , 0u x u x u x t u x tε ε ε= = →boundedly, in 1L norm.
uHow does depend on initial ? 0u Define y( ) ( )0 0
xU x u y d
−∞= ∫ ,
and set 0 01
supu U−= .
For any to solutions ,u υ with initial values 0 0, ,u υ
( ) ( ) 0 0 11, ,u t t uυ υ
−−⋅ − ⋅ ≤ −
Proof. Integrate viscous equation: 21
2 ,t x xxU U Uε+ = and 21
2 .t x xxV V Vε+ = Subtract these equations, and denote
as :U V D− ( )1
2t x xD U V D Dε+ = +By maximum principle,
( ) 0, sup ,D x t D≤ as claimed.
( )Let be any sequence of initial date that tend weakly, that is in the sense of the
0nu
01 nom, to u
−; we saw that then ( ) ( ),nu x t
tends weakly to ( ),u x t for every ( )
0.t > u satisfy the basic differential equation
and nu
( ) ( )( )
( )
212
212
0,
0.
n nt
x
t x
u u
u u
+ =
+ =
( )ntu tends to in the sense of distribution.
Therefore (tu
)2n
xu tends to 2
xu in the sense of
distribution. So ( )2nu tends to 2u + const in the sense of distribution. Since both
and nu u tend to zero as ,x →∞ the
constant is zero. But then must tend to strongly, that is in the 1
nu uL norm.
Summary If the initial values of ( )nu tend to the initial values of u weakly, ( ),nu x tends to (t ),u x strongly, for every
t0.t >
The mapping from initial values ( ),0u x to their values ( ), , 0,u x t is a compact mapping.
t >