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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
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PreliminariesWeak Solutions
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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
Gopikrishnan C R Weak Solutions
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NotationsDefinitions
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
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NotationsDefinitions
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)
Gopikrishnan C R Weak Solutions
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Outline
Step 1: Derive a one sided jump condition/inequality/estimatefor the Lax - Oleinik Formula. We shall call this estimate as the
Entropy Condition.
Gopikrishnan C R Weak Solutions
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NotationsDefinitions
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.
Gopikrishnan C R Weak Solutions
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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
Gopikrishnan C R Weak Solutions
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One Sided Jump EstimateUniqueness of Entropy Solutions
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
Gopikrishnan C R Weak Solutions
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One Sided Jump EstimateUniqueness of Entropy Solutions
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
Gopikrishnan C R Weak Solutions
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PreliminariesWeak Solutions
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One Sided Jump EstimateUniqueness of Entropy Solutions
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)
Gopikrishnan C R Weak Solutions
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PreliminariesWeak Solutions
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One Sided Jump EstimateUniqueness of Entropy Solutions
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.
Gopikrishnan C R Weak Solutions
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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)
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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 ).
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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
Gopikrishnan C R Weak Solutions
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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
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One Sided Jump EstimateUniqueness of Entropy Solutions
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.
Gopikrishnan C R Weak Solutions
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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)
Gopikrishnan C R Weak Solutions
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One Sided Jump EstimateUniqueness of Entropy Solutions
As is constant along each characteristic curve xi (.) we have
N
i=1
| (xi , t) (xi1, t)|=N
i=1
| (yi ,) (yi1,)|
var ( (.,))
Taking supremum over all such partitions, | x (x , t)|dx = var (., t) var (.,) =
| x (x ,)|dx C
(30)30 holds because has constant support and estimate by theestimate 11 applied on s = .
Gopikrishnan C R Weak Solutions
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PreliminariesWeak Solutions
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One Sided Jump EstimateUniqueness of Entropy Solutions
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
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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.
Gopikrishnan C R Weak Solutions
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Why G is Lipschitz?Bounded variation?Estimate 1
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.
Gopikrishnan C R Weak Solutions
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Why G is Lipschitz?Bounded variation?Estimate 1
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)
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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
Gopikrishnan C R Weak Solutions
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PreliminariesWeak Solutions
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Thanks
University Library, Kerala University
Central Library, IISER TVM
Gopikrishnan C R Weak Solutions
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PreliminariesWeak Solutions
AppendixReference
Thanks
University Library, Kerala University
Central Library, IISER TVM
THANK YOU...!
Gopikrishnan C R Weak Solutions
PreliminariesNotationsDefinitions
Weak SolutionsOne Sided Jump EstimateUniqueness of Entropy Solutions
AppendixWhy G is Lipschitz?Bounded variation?Estimate 1
Reference