Uniqueness of weak solutions for Lax Oleinik Forumla

Entropy Condition and Uniqueness of WeakSolutions

Gopikrishnan C R

School of MathematicsIndian Institute of Science Education and Research

Thiruvananthapuram

March 7, 2014

Gopikrishnan C R Weak Solutions

PreliminariesWeak Solutions

AppendixReference

NotationsDefinitions

Notation Definition

Lip(f ) Lipschitz constant associated with a Lipschitz function f

(x , t) Element of R [0,)F R Rg R Ru R [0,) R



AppendixReference


Entropy Solution

We say that a function u L (R (0,)) is an entropy solution ofthe initial-value problem

ut + F (u)x = 0 in R (0,) (1)u = g on R{t = 0} (2)

provided 0

(uvt + F (u)vx)dxdt +

gvdx |t=0 = 0 (3)

for all test functions v : R [0,) R with compact support, and

u(x + z , t)u(x , t) C(

1 +1

t

)z (4)

for some constant C 0 and a.e x ,z R, t > 0, with z > 0.Gopikrishnan C R Weak Solutions


AppendixReference


Mollifiers

Define C(Rn) by

:=

{C exp

(1

|x |21), if |x |< 1

0 if |x | 1(5)

for constant C > 0 selected so thatRn dx = 1.

For each > 0, set

(x) :=1

n(

x

)(6)

We call the standard mollifier. The functions k are C andsatisfy

Rndx = 1 spt() B(0,) (7)



AppendixReference


Outline

Step 1: Derive a one sided jump condition/inequality/estimatefor the Lax - Oleinik Formula. We shall call this estimate as the

Entropy Condition.



AppendixReference


Outline

Step 1: Derive a one sided jump condition/inequality/estimatefor the Lax - Oleinik Formula .We shall call this estimate as the

Entropy Condition.

Step 2: Prove that there is at most one entropy solution.



AppendixReference

One Sided Jump EstimateUniqueness of Entropy Solutions

Lemma

Assume that F is smooth and uniformly convex, and g L(R).Then there exists a constant C such that the function u defined buthe Lax - Oleinik formula satisfies the inequality

u(x + z , t)u(x , t) Ct

z (8)

for all t > 0 and x ,z R,z > 0.

Proof:- The mapping x y(x , t) is non decreasing as wellG = (F )1. Therefore we have,

u(x , t) = G

(xy(x , t)

t

) G

(xy(x + z , t)

t

)for z > 0



AppendixReference


Since G is Lipschitz,

G(

x + zy(x + z , t)t

) Lip(G )z

t

= u(x + z , t) Lip(G )zt

which shows that,

u(x + z , t)u(x , t) Czt



AppendixReference


Theorem

Assume F is convex and smooth. Then there exists - up to a set ofmeasure zero - at most one entropy solution of 2.

Proof:-Step 1Assume that u and v are two entropy solutions of 2. Writew = uv . Observe that for any point (x , t)

F (u(x , t))F (v(x , t)) = 1

0

d

drF (ru(x , t) + (1 r)v(x , t))dr

= (u(x , t)v(x , t)) 1

0F (ru(x , t) + (1 r)v(x , t))dr



AppendixReference


Define,

(u(x , t)v(x , t)) 1

0F (ru(x , t) + (1 r)v(x , t))dr = b(x , t)w(x , t)

Let is a test function . Then we will obtain from 2 and abovedefinition,

0 =

0

(uv)t + [F (u)F (v)]x (9)

=

0

w(t + bx)dxdt (10)



AppendixReference


Step 2Take > 0 and define u = u,v = v , where is thestandard mollifier in x and t variables. Then,

||u ||L ||u||L||v ||L ||v ||L

u u,v v a.e as 0From the Entropy condition we shall obtain (link),

ux(x , t),vx (x , t) C

(1 +

1

t

)ux(x , t) (11)

for an appropriate constant C and all > 0, x R, t > 0.



AppendixReference


Step 3Write

b(x , t) := 1

0F (ru(x , t) + (1 r)v (x , t))dr (12)

Then (10) becomes

0 =

0

w [t + bx ]dxdt +

0

w [bb ]dxdt (13)

Step 4Now choose T > 0 and any smooth function : R (0,T ) Rwith compact support. We choose be the solution of thefollowing terminal value problem,

t + bx = in R (0,T ) (14)

= 0 on R{t = T} (15)



AppendixReference


we shall solve this equation by method of characteristics. For thisfix x R, 0 t T , and denote by x(.) the solution of the ODE,

x(s) = b(x(s),s) (s t) (16)x(t) = x (17)

and set

(x , t) := Tt

(x(s),s)ds (x R,0 t T ) (18)

Then is the unique solution of 16. Observe that hascompact support in R [0,T ).



AppendixReference


Step 5We now claim that for each s > 0, there exists a constant Cs suchthat

| x | Cs (19)

on R (s,T ). To prove this first note that if 0 < s t T , tehn

b,x(x , t) = 1

0F (ru + (1 r)v )(rux + (1 r)v x )dr

Ct

Cs



AppendixReference


Now differentiate the PDE in 14 with respect to x:

t x + bx x + bxv

x = x (20)

Now set

a(x , t) := e t tx (x , t) (21)

for

=C

s+ 1 (22)

Then

at + bax = a + e t [ xt + bvxx ]

= a + e t [b,x x +x ]= [ b,x ]a + e tx



AppendixReference


Since has compact support, a attains a nonnegative maximumover R [s,T ] at some point (x0, t0). If t0 = T then x = 0. If0 t0 T then,

at(x0, t0) 0,ax(t0,x0) = 0 (23)Consequently the last expression gives,

[ b,x ] + e t0x 0 (24)But since b ,x Cs and is gien by 22, the inequality 24 implies,

a(x0, t0) e t0 eT ||x ||L (25)A similar argument shows that,

a(x1, t1) eT ||x ||L (26)at any point (x1, t1) where a attains a non positive minimum.These two estimates and definition of a imply the required bound.



AppendixReference


Step 6We will need one more inequality, namely,

| x (x , t)|dx D (27)

for all 0 t and some constatn D, provided t is small enough.To prove this choose > 0 so small taht = 0 on R (0,).Then if 0 t , is constant along the characteristic curvex(.) for t s .Select any partiton x0 < x1 < x2 < < xN . Theny0 < y1 < < yN , where yi := xi (s) for,

x(s) = b(x(s),s) (t s ) (28)x(t) = xi (29)



AppendixReference


Step 7 // Now substitute = in 13 and by the PDE 14, 0

wdxdt =

0

w(b b) x dxdt (31)

= T

w(b b) x dxdt +

0

w(b b) x dxdt(32)

:= I + J (33)

Observations

I 0 as 0 for each > 0.I f0 < < T we see,

|J | C max0t

| x |dx C (34)

Thus 0

wdxdt = 0 (35)

Therefore w = 0 identically a.e.Gopikrishnan C R Weak Solutions


AppendixReference

Why G is Lipschitz?Bounded variation?Estimate 1

Inverse Function Theorem

Assume that f C k (U;Rn) and Jf (x0) 6= 0. Then there exists anopen set V U, with x0 V , and with an open set W Rn with

zo W such that the mapping

f : V W (36)

is one one and onto and the inverse function is also C k .

Any C 1 function is smooth in a closed and bounded and interval.Since F is smooth so is G. Therefore restricting our consideration ofG to some bounded interval we can safely assume G is Lipschitz.



AppendixReference


Why is of bounded variation ?

We have for : R (0,T ) R

(x , t) := Tt

(x(s),s)ds (37)

The integrand is smooth and compactly supported. Therefore isa compactly supported smooth function on R [0,T ). Then wehave the result,

Theorem

If f : [a,b] R has finite derivatives at every point x in [a,b] andf is bounded on [a,b] then f is a function of bounded variation.



AppendixReference


For all t > 0, the mapping x u(x , t) Cxt is a non increasingfunction (easily follows the entropy condition). From this we findthat,

(

u Cxt

)= u C x

t(38)

is also non increasing. This is a smooth function and therefore,

x

(

(u Cx

t

))=u

x C

t 0 (39)

which gives,u

x C

t C + C

t= C

(1 +

1

t

)(40)



AppendixReference

Lawrence C Evans, Partial Differential Equations, GraduateStudies in Mathematics - AMS, Shaper 3, Section 3.4.3 p.n149 - 154.

Utpal Chatterjee, Advanced Mathematical Analysis, AcademicPublishers, Chapter 3 and Chapter 4.

Joel Smoller,Shock Waves and Reaction Diffusion Equations,Springer Verlag, Chapter 16, p.n 281 - 290

Walter Rudin, Real Analysis



AppendixReference

Thanks

University Library, Kerala University

Central Library, IISER TVM



AppendixReference

Thanks

University Library, Kerala University

Central Library, IISER TVM

THANK YOU...!


PreliminariesNotationsDefinitions

Weak SolutionsOne Sided Jump EstimateUniqueness of Entropy Solutions

AppendixWhy G is Lipschitz?Bounded variation?Estimate 1

Reference

Date post:	18-Oct-2015
Category:	Documents
Upload:	lori-hayes
View:	26 times
Download:	1 times

Uniqueness of weak solutions for Lax Oleinik Forumla

Documents