Towards a weakerbhfiuihon of solutions

Consider the problem ( net 941 =°

( cp )n ( xio ) = gk )

we have already showed that ucxit ) = g ( × -

q'

( GCSDT ) ( D

is a solution of CCP ) representing a progressive wave with

(2)

velocity q'

Cgcs ) ), being ucx # = g (5) along the

characteristic x= q 'CgC5) )tt 5 coming out of the

point ( 5,0 ).

We can write that (by(D) u is defined

implicitly by the equation G ( × it , a) =o

, beingG Cx it , a) : = a -

g ( × - qku )t ) .

* 3) by (2).

Assuming g and q' to be differentiable

,one can use the

implicit function theorem to find u as a function of Gut )

uwhiu it holds Gwlxitiu )=1 + tq "Cu)g'

( × - gkust ) ¥0

Along the characteristics ×=

q'

C g ( 5) )tt 5 we have

Gucxit , a) = 1 + tq"

Cgt ) ) g' (5)

Note that if q "cg( 3) ) .

g' (5) 20 ( q "og and

g' have the

Same sign ) the solution obtained by the characteristics

method is regular tt > 0 : since d- qllgcs D= g"cg( b) gks)d }

the condition is equivalent to saying that the 9'

CgC 51 ) ,

which is the slope of the characteristic by ( 5,0 ) is always

tn×=q' ogcs .

, )t+s .

°k - singand they do not intersect

×=g' cgcs.yt.is .

91cg ( I ) ) > 9 Egos , ) )Or

)Oz 7

i) a so that

5,

< { ×

1on = arctan ( 5¥ 02 = orctan ( ⇒ , )) .

Theorem : Assume qe CYR ), age

C'

CR ) and

9 "CgC 5) ) .

g'

(5) 70 in R.

Then the equation G ( at ,u)=o

defines u=u(x ,t ) as the unique solution of @P) .

Moreover we C£

( Rx [ °,

to ) ).

Proof : Along the characteristics 5=× - 9ku) 't we bone

ehcxit ,a) =

lttq "cu ) g'

( × - qkult ) z 1 ttzo.

By the implicit function theorem we home that in

is uniquely given by u=uCx,t) .

Moreover the same theorem

gives if = Gtcxit ,u)= g

'( × - qku) . t ) .q÷Guertin ) ittq

" (a) g'

( x . qkust )

u× = Gxk.tl= g'(x-qka)#Gukrtru ) it tq "Cu)g' ( x - fast )

so that let and u× are continuous functions , being

given by the quotients of continuous function , with

denominator 21. D

what happens if q"

C g (5) ) .

g'

B) so ?

Jn this case the characteristics will meet

along a shock line. If we denote by ts the

breaking time e

,the first time when the stock

Starts and by xs the point where the shock

line shorts, we have that at this point we

have a solution of the followingequationi

0=Gµl×, to ), a) = it tC5 ) q

" ( g ( 51 ) g'

(5) .

Assume nowthat we are solving the ¢P) for xe [ a , b)

-1

we have that t (5) =

-.

Observe now

9"

Cgcs ) )g' (5)that ZCJ ) : -q "CgC 3 ) ) .

g'

(5) e C ( [ a ,b] ), and assume

that Z obtains the maximum only at the point SM .

Then tssgmiwabztcsj2¥ '

The starting point of the shook belongs to the ehoract.

× =

q'( g ( Sn ) )t the

,so that g=

9kFBmD.tst5mo-pxs.9k9@t5m.zC Jm )

Geometrically ,the point ( xs ,ts ) is the point of the envelope&the characteristics boring minimal ×

.

)

In the examples considered so for we have raised'

several issues such as ; 1) the characteristics might

not cover the whole 112×6 , a) half - phone,

2) the

characteristicmight collide generating discontinuities

.

1) =D Rarefaction .

waves2) =D Shook - waves

Some questions are in order :

( 1) Zn which sense are they solutions ?

(2) Are they unique solutions ?

(3) If not,

is there any criterion that wwght helpen choosing one oranother solution

°

?

Consider

|net qcu)× =

0 xer,

t > o (e)

hC×c° ) = g ( × ) ×€R( CP )

Lfu E C'

( Rx G , ex ) ) solves CCP ) wesay that u is

a CLASSICAL SOLUTION.

Let ers take now we C'

in Rx Centre ) ,with compact support .

We saythat or

is a TEST FUNCTION. Multiply (E) by N and integrate

on Rx G ,+ a) to get

§P{r[ uetqcuk ] rdxdt - o ( w )

Integrating by parts

§Taut rdxdt =- §o°{ru rtokdt

tfnnvdxbt?and

by or being compactly supported in R×[ at ⇒, ftp.NGto

so that |=

tgoogpgraedt - fucxidrlxp) =

R

= - +§{runfdxdt-

fggcx) rcxio )

.

Again by integration by ports one has that

f+°{rqw× vokdt = - §fasqcu) rxdxdt

Finally ( w ) becomes §°fa@nftqcu )r×) okdtt

,,{ FKWKP )d⇐0

This is an integral equation for w,

that holds to test.

Note that to satisfy ( w ) u does not need to be differentiablesince no derivative of u appear in the equation .

On the other hand, if u is regular one can use the

integrationby parts back word to find that tv test

§°{p[ uttqlw } ] rdxdt -

§[ gcx ) -

ucxiotfvcxedx

=0

Consider now those test functions w i v(xp)=o .

For such tests we have [ { ( ¥+941 ] vokdt = o.

Using the following ( strong ) version of the fundamentaltheorem of the calculus of variations

,

i.e. :

fir - > R,

RCR"

open

becontinuous.

Then

§ frdx =o fv test in R ⇒ F ⇒ 0.

The previous theorem, applied for R= Rx Coit a) gives

µ+tq(u)× = o in Rx Gita ),

and the equation( w ) becomes f [ g ( × ) - ucx , o ) ] ok

,o ) ok = o

R

and by the same theorem for RKR and N=r ( × ,o )

we have that gcx ) = ucxp ).

so that we have proved that CCP ) is equivalentto ( w ) if u is regular .

Definition

=: A function u

,bounded in Rx [9tx)

is a week solution of CCP ) if ( w ) holds for all

test functions r In Rx Gio ) with compact support.