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 >